Principles of Communication 5Ed R. E. Ziemer, William H. Tranter Solutions Manual

Chapter 2Signal and Linear System Theory 2.1 Problem Solutions Problem 2.1 For the single-sided spectra, write the signal in terms of cosines: x(t) = 10 cos(4πt +π/8) + 6 sin(8πt + 3π/4) = 10 cos(4πt +π/8) + 6 cos(8πt + 3π/4 −π/2) = 10 cos(4πt +π/8) + 6 cos(8πt +π/4) For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s theorem: x(t) = 5 exp[(4πt +π/8)] + 5 exp[−j(4πt +π/8)] +3 exp[j(8πt + 3π/4)] + 3 exp[−j(8πt + 3π/4)] The two sets of spectra are plotted in Figures 2.1 and 2.2. Problem 2.2 The result is x(t) = 4e j(8πt+π/2) + 4e −j(8πt+π/2) + 2e j(4πt−π/4) + 2e −j(4πt−π/4) = 8 cos (8πt +π/2) + 4 cos (4πt −π/4) = −8 sin(8πt) + 4 cos (4πt −π/4) 1 Principles of Communication 5Ed R. E. Zeimer, William H Tranter Solutions Manual 2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY f, Hz f, Hz 0 2 4 6 0 2 4 6 10 5 π/4 π/8 Single-sided amplitude Single-sided phase, rad. Figure 2.1: Problem 2.3 (a) Not periodic. (b) Periodic. To find the period, note that 6π 2π = n 1 f 0 and 20π 2π = n 2 f 0 Therefore 10 3 = n 2 n 1 Hence, take n 1 = 3, n 2 = 10, and f 0 = 1 Hz. (c) Periodic. Using a similar procedure as used in (b), we find that n 1 = 2, n 2 = 7, and f 0 = 1 Hz. (d) Periodic. Using a similar procedure as used in (b), we find that n 1 = 2, n 2 = 3, n 3 = 11, and f 0 = 1 Hz. Problem 2.4 (a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6 Hz, and the phase spectrum consists of a single line of height -π/6 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of 6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -π/6 radians at frequency 6 Hz and a line of height π/6 at frequency -6 radians Hz. (b) Write the signal as x b (t) = 3 cos(12πt −π/2) + 4 cos(16πt) From this it is seen that the single-sided amplitude spectrum consists of lines of height 3 and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists 2.1. PROBLEM SOLUTIONS 3 f, Hz 0 2 4 6 π/4 π/8 Double-sided phase, rad. f, Hz -6 -4 -2 0 2 4 6 5 Double-sided amplitude -π/8 -π/4 -6 -4 -2 Figure 2.2: 4 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY of a line of height -π/2 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of height 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrum consists of a line of height -π/2 radians at frequency 6 Hz and a line of height π/2 radians at frequency -6 Hz. Problem 2.5 (a) This function has area Area = ∞ Z −∞ ² −1 · sin(πt/²) (πt/²) ¸ 2 dt = ∞ Z −∞ · sin(πu) (πu) ¸ 2 du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² →0, the central lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. (b) The area for the function is Area = ∞ Z −∞ 1 ² exp(−t/²)u(t) dt = ∞ Z 0 exp(−u)du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² →0, the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. (c) Area = R ² −² 1 ² (1 −|t| /²) dt = R 1 −1 Λ(t) dt = 1. As ² →0, the function becomes narrower and higher, so it approximates a delta function in the limit. Problem 2.6 (a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9. Problem 2.7 (a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively. The waveform of part (c) is a periodic train of impulses extending from -∞ to ∞ spaced by 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). The waveform of part (e) is a doubly-infinite train of square pulses, each of which is one unit high and one unit wide, centered at · · ·, −6, −3, 0, 3, 6, · · ·. Waveform (f) is a raised cosine of minimum and maximum amplitudes 0 and 2, respectively. 2.1. PROBLEM SOLUTIONS 5 Problem 2.8 (a) The result is x(t) = Re ¡ e j6πt ¢ + 6 Re ³ e j(12πt−π/2) ´ = Re h e j6πt + 6e j(12πt−π/2) i (b) The result is x(t) = 1 2 e j6πt + 1 2 e −j6πt + 3e j(12πt−π/2) + 3e −j(12πt−π/2) (c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequencies of 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height −π/2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 3, 1/2, 1/2, and 3 at frequencies of −6, −3, 3, and 6 Hz, respectively. The double-sided phase spectrum consists of lines of height π/2 and −π/2 at frequencies of −6 and 6 Hz, respectively. Problem 2.9 (a) Power. Since it is a periodic signal, we obtain P 1 = 1 T 0 Z T 0 0 4 sin 2 (8πt +π/4) dt = 1 T 0 Z T 0 0 2 [1 −cos (16πt +π/2)] dt = 2 W where T 0 = 1/8 s is the period. (b) Energy. The energy is E 2 = Z ∞ −∞ e −2αt u 2 (t)dt = Z ∞ 0 e −2αt dt = 1 2α (c) Energy. The energy is E 3 = Z ∞ −∞ e 2αt u 2 (−t)dt = Z 0 −∞ e 2αt dt = 1 2α (d) Neither energy or power. E 4 = lim T→∞ Z T −T dt (α 2 +t 2 ) 1/4 = ∞ P 4 = 0 since lim T→∞ 1 T R T −T dt (α 2 +t 2 ) 1/4 = 0.(e) Energy. Since it is the sum of x 1 (t) and x 2 (t), its energy is the sum of the energies of these two signals, or E 5 = 1/α. 6 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we use P 6 = lim T→∞ 1 2T Z T 0 sin 2 (5πt) dt = lim T→∞ 1 2T Z T 0 1 2 [1 −cos (10πt)] dt = lim T→∞ 1 2T · T 2 − 1 2 sin(20πt) 20π ¸ T 0 = 1 4 W Problem 2.10 (a) Power. Since the signal is periodic with period π/ω, we have P = ω π Z π/ω 0 A 2 |sin(ωt +θ)| 2 dt = ω π Z π/ω 0 A 2 2 {1 −cos [2 (ωt +θ)]} dt = A 2 2 (b) Neither. The energy calculation gives E = lim T→∞ Z T −T (Aτ) 2 dt √ τ +jt √ τ −jt = lim T→∞ Z T −T (Aτ) 2 dt √ τ 2 +t 2 →∞ The power calculation gives P = lim T→∞ 1 2T Z T −T (Aτ) 2 dt √ τ 2 +t 2 = lim T→∞ (Aτ) 2 2T ln Ã 1 + p 1 +T 2 /τ 2 −1 + p 1 +T 2 /τ 2 ! = 0 (c) Energy: E = Z ∞ 0 A 2 t 4 exp(−2t/τ) dt = 3 4 A 2 τ 5 (use table of integrals) (d) Energy: E = 2 Ã Z τ/2 0 2 2 dt + Z τ τ/2 1 2 dt ! = 5τ Problem 2.11 (a) This is a periodic train of “boxcars”, each 3 units in width and centered at multiples of 6: P a = 1 6 Z 3 −3 Π 2 µ t 3 ¶ dt = 1 6 Z 1.5 −1.5 dt = 1 2 W 2.1. PROBLEM SOLUTIONS 7 (b) This is a periodic train of unit-high isoceles triangles, each 4 units wide and centered at multiples of 5: P b = 1 5 Z 2.5 −2.5 Λ 2 µ t 2 ¶ dt = 2 5 Z 2 0 µ 1 − t 2 ¶ 2 dt = − 2 5 2 3 µ 1 − t 2 ¶ 3 ¯ ¯ ¯ ¯ ¯ 2 0 = 4 15 W (c) This is a backward train of sawtooths (right triangles with the right angle on the left), each 2 units wide and spaced by 3 units: P c = 1 3 Z 2 0 µ 1 − t 2 ¶ 2 dt = − 1 3 2 3 µ 1 − t 2 ¶ 3 ¯ ¯ ¯ ¯ ¯ 2 0 = 2 9 W (d) This is a full-wave rectified cosine wave of period 1/5 (the width of each cosine pulse): P d = 5 Z 1/10 −1/10 |cos (5πt)| 2 dt = 2 × 5 Z 1/10 0 · 1 2 + 1 2 cos (10πt) ¸ dt = 1 2 W Problem 2.12 (a) E = ∞, P = ∞; (b) E = 5 J, P = 0 W; (c) E = ∞, P = 49 W; (d) E = ∞, P = 2 W. Problem 2.13 (a) The energy is E = Z 6 −6 cos 2 (6πt) dt = 2 Z 6 0 · 1 2 + 1 2 cos (12πt) ¸ dt = 6 J (b) The energy is E = Z ∞ −∞ h e −|t|/3 cos (12πt) i 2 dt = 2 Z ∞ 0 e −2t/3 · 1 2 + 1 2 cos (24πt) ¸ dt where the last integral follows by the eveness of the integrand of the first one. Use a table of definte integrals to obtain E = Z ∞ 0 e −2t/3 dt + Z ∞ 0 e −2t/3 cos (24πt) dt = 3 2 + 2/3 (2/3) 2 + (24π) 2 Since the result is finite, this is an energy signal. (c) The energy is E = Z ∞ −∞ {2 [u(t) −u(t −7)]} 2 dt = Z 7 0 4dt = 28 J 8 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Since the result is finite, this is an energy signal. (d) Note that Z t −∞ u(λ) dλ = r (t) = ½ 0, t < 0 t, t ≥ 0 which is called the unit ramp. The energy is E = Z ∞ −∞ [r (t) −2r (t −10) +r (t −20)] 2 dt = 2 Z 10 0 µ t 10 ¶ 2 dt = 20 3 J where the last integral follows because the integrand is a symmetrical triangle about t = 10. Since the result is finite, this is an energy signal. Problem 2.14 (a) Expand the integrand, integrate term by term, and simplify making use of the orthogonality property of the orthonormal functions. (b) Add and subtract the quantity suggested right above (2.34) and simplify. (c) These are unit-high rectangular pulses of width T/4. They are centered at t = T/8, 3T/8, 5T/8, and 7T/8. Since they are spaced by T/4, they are adjacent to each other and fill the interval [0, T]. (d) Using the expression for the generalized Fourier series coefficients, we find that X 1 = 1/8, X 2 = 3/8, X 3 = 5/8, and X 4 = 7/8. Also, c n = T/4. Thus, the ramp signal is approximated by t T = 1 8 φ 1 (t) + 3 8 φ 2 (t) + 5 8 φ 3 (t) + 7 8 φ 4 (t) , 0 ≤ t ≤ T where the φ n (t)s are given in part (c). (e) These are unit-high rectangular pulses of width T/2 and centered at t = T/4 and 3T/4. We find that X 1 = 1/4 and X 2 = 3/4. (f) To compute the ISE, we use ² N = Z T |x(t)| 2 dt − N X n=1 c n ¯ ¯ X 2 n ¯ ¯ Note that R T |x(t)| 2 dt = R T 0 (t/T) 2 dt = T/3. Hence, for (d), ISE d = T 3 − T 4 ¡ 1 64 + 9 64 + 25 64 + 49 64 ¢ = 5.208 × 10 −3 T. For (e), ISE e = T 3 − T 2 ¡ 1 16 + 9 16 ¢ = 2.083 × 10 −2 T. 2.1. PROBLEM SOLUTIONS 9 Problem 2.15 (a) The Fourier coefficients are (note that the period = 1 2 2π ω 0 ) X −1 = X 1 = 1 4 ; X 0 = 1 2 All other coefficients are zero. (b) The Fourier coefficients for this case are X −1 = X ∗ 1 = 1 2 (1 +j) All other coefficients are zero. (c) The Fourier coefficients for this case are (note that the period is 2π 2ω 0 ) X −2 = X 2 = 1 8 ; X −1 = X 1 = 1 4 ; X 0 = − 1 4 All other coefficients are zero. (d) The Fourier coefficients for this case are X −3 = X 3 = 1 8 ; X −1 = X 1 = 3 8 All other coefficients are zero. Problem 2.16 The expansion interval is T 0 = 4 and the Fourier coefficients are X n = 1 4 Z 2 −2 2t 2 e −jn(π/2)t dt = 2 4 Z 2 0 2t 2 cos µ nπt 2 ¶ dt which follows by the eveness of the integrand. Let u = nπt/2 to obtain the form X n = 2 µ 2 nπ ¶ 3 Z nπ 0 u 2 cos u du = 16 (nπ) 2 (−1) n If n is odd, the Fourier coefficients are zero as is evident from the eveness of the function being represented. If n = 0, the integral for the coefficients is X 0 = 1 4 Z 2 −2 2t 2 dt = 8 3 The Fourier series is therefore x(t) = 8 3 + ∞ X n=−∞, n6=0 (−1) n 16 nπ e jn(π/2)t 10 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.17 Parts (a) through (c) were discussed in the text. For (d), break the integral for x(t) up into a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry contition. Problem 2.18 This is a matter of integration. Only the solution for part (b) will be given here. The integral for the Fourier coefficients is (note that the period really is T 0 /2) X n = A T 0 Z T 0 /2 0 sin(ω 0 t) e −jnω 0 t dt = − Ae −jnω 0 t ω 0 T 0 (1 −n 2 ) [jnsin(ω 0 t) + cos (ω 0 t)] ¯ ¯ ¯ ¯ T 0 /2 0 = A ¡ 1 +e −jnπ ¢ ω 0 T 0 (1 −n 2 ) , n 6= ±1 For n = 1, the integral is X 1 = A T 0 Z T 0 /2 0 sin(ω 0 t) [cos (jnω 0 t) −j sin(jnω 0 t)] dt = − jA 4 = −X ∗ −1 This is the same result as given in Table 2.1. Problem 2.19 (a) Use Parseval’s theorem to get P |nf 0 | ≤ 1/τ = N X n=−N |X n | 2 = N X n=−N µ Aτ T 0 ¶ 2 sinc 2 (nf 0 τ) where N is an appropriately chosen limit on the sum. We are given that only frequences for which |nf 0 | ≤ 1/τ are to be included. This is the same as requiring that |n| ≤ 1/τf 0 = T 0 /τ = 2. Also, for a pulse train, P total = A 2 τ/T 0 and, in this case, P total = A 2 /2. Thus P |nf 0 | ≤ 1/τ P total = 2 A 2 2 X n=−2 µ A 2 ¶ 2 sinc 2 (nf 0 τ) = 1 2 2 X n=−2 sinc 2 (nf 0 τ) = 1 2 £ 1 + 2 ¡ sinc 2 (1/2) + sinc 2 (1) ¢¤ = 1 2 " 1 + 2 µ 2 π ¶ 2 # = 0.91 2.1. PROBLEM SOLUTIONS 11 (b) In this case, |n| ≤ 5, P total = A 2 /5, and P |nf 0 | ≤ 1/τ P total = 1 5 5 X n=−5 sinc 2 (n/5) = 1 5 n 1 + 2 h (0.9355) 2 + (0.7568) 2 + (0.5046) 2 + (0.2339) 2 io = 0.90 Problem 2.20 (a) The integral for Y n is Y n = 1 T 0 Z T 0 y (t) e −jnω 0 t dt = 1 T 0 Z T 0 0 x(t −t 0 ) e −jnω 0 t dt Let t 0 = t −t 0 , which results in Y n = · 1 T 0 Z T 0 −t 0 −t 0 x ¡ t 0 ¢ e −jnω 0 t 0 dt 0 ¸ e −jnω 0 t 0 = X n e −jnω 0 t 0 (b) Note that y (t) = Acos ω 0 t = Asin(ω 0 t +π/2) = Asin[ω 0 (t +π/2ω 0 )] Thus, t 0 in the theorem proved in part (a) here is −π/2ω 0 . By Euler’s theorem, a sine wave can be expressed as sin(ω 0 t) = 1 2j e jω 0 t − 1 2j e −jω 0 t Its Fourier coefficients are therefore X 1 = 1 2j and X −1 = − 1 2j . According to the theorem proved in part (a), we multiply these by the factor e −jnω 0 t 0 = e −jnω 0 (−π/2ω 0 ) = e jnπ/2 For n = 1, we obtain Y 1 = 1 2j e jπ/2 = 1 2 For n = −1, we obtain Y −1 = − 1 2j e −jπ/2 = 1 2 which gives the Fourier series representation of a cosine wave as y (t) = 1 2 e jω 0 t + 1 2 e −jω 0 t = cos ω 0 t 12 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY We could have written down this Fourier representation directly by using Euler’s theorem. Problem 2.21 (a) Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0 to obtain the series 1 = · · · + 4 25π 2 + 4 9π 2 + 4 π 2 + 4 π 2 + 4 9π 2 + 4 25π 2 + · · · Multiply both sides by π 2 8 to get the series in given in the problem. Therefore, its sum is π 2 8 . (b) Use the Fourier series of a square wave (specialize the Fourier series of a pulse train) with A = 1 and t = 0 to obtain the series 1 = 4 π µ 1 − 1 3 + 1 5 −· · · ¶ Multiply both sides by π 4 to get the series in the problem statement. Hence, the sum is π 4 . Problem 2.22 (a) In the expression for the Fourier series of a pulse train (Table 2.1), let t 0 = −T 0 /8 and τ = T 0 /4 to get X n = A 4 sinc ³ n 4 ´ exp µ j πnf 0 4 ¶ (b) The amplitude spectrum is the same as for part (a) except that X 0 = 3A 4 . Note that this can be viewed as having a sinc-function envelope with zeros at multiples of 4 3T 0 . The phase spectrum can be obtained from that of part (a) by adding a phase shift of π for negative frequencies and subtracting π for postitive frequencies (or vice versa). Problem 2.23 (a) There is no line at dc; otherwise it looks like a squarewave spectrum. (b) Note that x A (t) = K dx B (t) dt where K is a suitably chosen constant. The relationship between spectral components is therefore X A n = K(jnω 0 ) X B n where the superscript A refers to x A (t) and B refers to x B (t). 2.1. PROBLEM SOLUTIONS 13 Problem 2.24 (a) This is the right half of a triangle waveform of width τ and height A, or A(1 −t/τ). Therefore, the Fourier transform is X 1 (f) = A Z τ 0 (1 −t/τ) e −j2πft dt = A j2πf · 1 − 1 j2πfτ ³ 1 −e −j2πfτ ´ ¸ where a table of integrals has been used. (b) Since x 2 (t) = x 1 (−t) we have, by the time reversal theorem, that X 2 (f) = X ∗ 1 (f) = X 1 (−f) = A −j2πf · 1 + 1 j2πfτ ³ 1 −e j2πfτ ´ ¸ (c) Since x 3 (t) = x 1 (t) −x 2 (t) we have, after some simplification, that X 3 (f) = X 1 (f) −X 2 (f) = jA πf sinc (2fτ) (d) Since x 4 (t) = x 1 (t) +x 2 (t) we have, after some simplification, that X 4 (f) = X 1 (f) +X 2 (f) = Aτ sin 2 (πfτ) (πfτ) 2 = Aτsinc 2 (fτ) This is the expected result, since x 4 (t) is really a triangle function. Problem 2.25 (a) Using a table of Fourier transforms and the time reversal theorem, the Fourierr transform of the given signal is X(f) = 1 α +j2πf − 1 α −j2πf Note that x(t) → sgn(t) in the limit as α → 0. Taking the limit of the above Fourier transform as α →0, we deduce that F [sgn(t)] = 1 j2πf − 1 −j2πf = 1 jπf 14 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) Using the given relationship between the unit step and the signum function and the linearity property of the Fourier transform, we obtain F [u(t)] = 1 2 F [sgn(t)] + 1 2 F [1] = 1 j2πf + 1 2 δ (f) (c) The same result as obtained in part (b) is obtained. Problem 2.26 (a) Two differentiations give d 2 x 1 (t) dt 2 = dδ (t) dt −δ (t −2) +δ (t −3) Application of the differentiation theorem of Fourierr transforms gives (j2πf) 2 X 1 (f) = (j2πf) (1) −1 · e −j4πf + 1 · e −j6πf where the time delay theorem and the Fourier transform of a unit impulse have been used. Dividing both sides by (j2πf) 2 , we obtain X 1 (f) = 1 j2πf − e −j4πf −e −j6πf (j2πf) 2 = 1 j2πf − e −j5πf j2πf sinc (2f) (b) Two differentiations give d 2 x 2 (t) dt 2 = δ (t) −2δ (t −1) +δ (t −2) Application of the differentiation theorem gives (j2πf) 2 X 2 (f) = 1 −2e −j2πf +e −j4πf Dividing both sides by (j2πf) 2 , we obtain X 2 (f) = 1 −2e −j2πf +e −j4πf (j2πf) 2 = sinc 2 (f) e −j2πf (c) Two differentiations give d 2 x 3 (t) dt 2 = δ (t) −δ (t −1) −δ (t −2) +δ (t −3) 2.1. PROBLEM SOLUTIONS 15 Application of the differentiation theorem gives (j2πf) 2 X 3 (f) = 1 −e −j2πf −e −j4πf +e −j6πf Dividing both sides by (j2πf) 2 , we obtain X 3 (f) = 1 −e −j2πf −e −j4πf +e −j6πf (j2πf) 2 (d) Two differentiations give d 2 x 4 (t) dt 2 = 2Π(t −1/2) −2δ (t −1) −2 dδ (t −2) dt Application of the differentiation theorem gives (j2πf) 2 X 4 (f) = 2sinc (f) e −jπf −2e −j2πf −2 (j2πf) e −j4πf Dividing both sides by (j2πf) 3 , we obtain X 4 (f) = 2e −j2πf + (j2πf) e −j4πf −sinc (f) e −jπf 2 (πf) 2 Problem 2.27 (a) This is an odd signal, so its Fourier transform is odd and purely imaginary. (b) This is an even signal, so its Fourier transform is even and purely real. (c) This is an odd signal, so its Fourier transform is odd and purely imaginary. (d) This signal is neither even nor odd signal, so its Fourier transform is complex. (e) This is an even signal, so its Fourier transform is even and purely real. (f) This signal is even, so its Fourier transform is real and even. Problem 2.28 (a) Using superposition, time delay, and the Fourier transform of an impulse, we obtain X 1 (f) = e j16πt + 2 +e −j16πt = 4 cos 2 (6πt) The Fourier transform is even and real because the signal is even. (b) Using superposition, time delay, and the Fourierr transform of an impulse, we obtain X 2 (f) = e j12πt −e −j12πt = 2j sin(12πf) The Fourier transform is odd and imaginary because the signal is odd. 16 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (c) The Fourier transform is X 3 (f) = 4 X n=0 ¡ n 2 + 1 ¢ e −j4πnf It is complex because the signal is neither even nor odd. Problem 2.29 (a) The Fourier transform of this signal is X 1 (f) = 2 (1/3) 1 + (2πf/3) 2 = 2/3 1 + [f/ (3/2π)] 2 Thus, the energy spectral density is G 1 (f) = ½ 2/3 1 + [f/ (3/2π)] 2 ¾ 2 (b) The Fourier transform of this signal is X 2 (f) = 2 3 Π µ f 30 ¶ Thus, the energy spectral density is X 2 (f) = 4 9 Π 2 µ f 30 ¶ = 4 9 Π µ f 30 ¶ (c) The Fourier transform of this signal is X 3 (f) = 4 5 sinc µ f 5 ¶ so the energy spectral density is G 3 (f) = 16 25 sinc 2 µ f 5 ¶ (d) The Fourier transform of this signal is X 4 (f) = 2 5 · sinc µ f −20 5 ¶ + sinc µ f + 20 5 ¶¸ 2.1. PROBLEM SOLUTIONS 17 so the energy spectral density is G 4 (f) = 4 25 · sinc µ f −20 5 ¶ + sinc µ f + 20 5 ¶¸ 2 Problem 2.30 (a) Use the transform pair x 1 (t) = e −αt u(t) ←→ 1 α +j2πf Using Rayleigh’s energy theorem, we obtain the integral relationship Z ∞ −∞ |X 1 (f)| 2 df = Z ∞ −∞ df α 2 + (2πf) 2 df = Z ∞ −∞ |x 1 (t)| 2 dt = Z ∞ 0 e −2αt dt = 1 2α (b) Use the transform pair x 2 (t) = 1 τ Π µ t τ ¶ ←→sinc (τf) = X 2 (f) Rayleigh’s energy theorem gives Z ∞ −∞ |X 2 (f)| 2 df = Z ∞ −∞ sinc 2 (τf) df = Z ∞ −∞ |x 2 (t)| 2 dt = Z ∞ −∞ 1 τ 2 Π 2 µ t τ ¶ dt = Z τ/2 −τ/2 dt τ 2 = 1 τ (c) Use the transform pair x 3 (t) = e −α|t| ←→ 2α α 2 + (2πf) 2 The desired integral, by Rayleigh’s energy theorem, is I 3 = Z ∞ −∞ |X 3 (f)| 2 df = Z ∞ −∞ · 1 α 2 + (2πf) 2 ¸ 2 df = 1 (2α) 2 Z ∞ −∞ e −2α|t| dt = 1 2α 2 Z ∞ 0 e −2αt dt = 1 4α 3 (d) Use the transform pair 1 τ Λ µ t τ ¶ ←→sinc 2 (τf) 18 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY The desired integral, by Rayleigh’s energy theorem, is I 4 = Z ∞ −∞ |X 4 (f)| 2 df = Z ∞ −∞ sinc 4 (τf) df = 1 τ 2 Z ∞ −∞ Λ 2 (t/τ) dt = 2 τ 2 Z τ 0 [1 −(t/τ)] 2 dt = 2 τ Z 1 0 [1 −u] 2 du = 2 3τ Problem 2.31 (a) The convolution operation gives y 1 (t) =    0, t ≤ τ −1/2 1 α £ 1 −e −α(t−τ+1/2) ¤ , τ −1/2 < t ≤ τ + 1/2 1 α £ e −α(t−τ−1/2) −e −α(t−τ+1/2) ¤ , t > τ + 1/2 (b) The convolution of these two signals gives y 2 (t) = Λ(t) + tr (t) where tr(t) is a trapezoidal function given by tr (t) =        0, t < −3/2 or t > 3/2 1, −1/2 ≤ t ≤ 1/2 3/2 +t, −3/2 ≤ t < −1/2 3/2 −t, 1/2 ≤ t < 3/2 (c) The convolution results in y 3 (t) = Z ∞ −∞ e −α|λ| Π(λ −t) dλ = Z t+1/2 t−1/2 e −α|λ| dλ Sketches of the integrand for various values of t gives the following cases: y 3 (t) =        R t+1/2 t−1/2 e αλ dλ, t ≤ −1/2 R 0 t−1/2 e αλ dλ + R t+1/2 0 e −αλ dλ, −1/2 < t ≤ 1/2 R t+1/2 t−1/2 e −αλ dλ, t > 1/2 Integration of these three cases gives y 3 (t) =    1 α £ e α(t+1/2) −e α(t−1/2) ¤ , t ≤ −1/2 1 α £ e −α(t−1/2) −e −α(t+1/2) ¤ , −1/2 < t ≤ 1/2 1 α £ e −α(t−1/2) −e −α(t+1/2) ¤ , t > 1/2 2.1. PROBLEM SOLUTIONS 19 (d) The convolution gives y 4 (t) = Z t −∞ x(λ) dλ Problem 2.32 (a) Using the convolution and time delay theorems, we obtain Y 1 (f) = F £ e −αt u(t) ∗ Π(t −τ) ¤ = F £ e −αt u(t) ¤ F [Π(t −τ)] = 1 α +j2πf sinc (f) e −j2πfτ (b) The superposition and convolution theorems give Y 2 (f) = F {[Π(t/2) +Π(t)] ∗ Π(t)} = [2sinc (2f) + sinc (f)] sinc (f) (c) By the convolution theorem Y 3 (f) = F h e −α|t| ∗ Π(t) i = 2α α 2 + (2πf) 2 sinc (f) (d) By the convolution theorem (note, also, that the integration theorem can be applied directly) Y 4 (f) = F [x(t) ∗ u(t)] = X (f) · 1 j2πf + 1 2 δ (f) ¸ = X (f) j2πf + 1 2 X (0) δ (f) Problem 2.33 (a) The normalized inband energy is E 1 (|f| ≤ W) E total = 2 π tan −1 µ 2πW α ¶ 20 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) The result is E 1 (|f| ≤ W) E total = 2 Z τW 0 sinc 2 (u) du The integration must be carried out numerically. Problem 2.34 (a) By the modulation theorem X(f) = AT 0 4 ½ sinc · (f −f 0 ) T 0 2 ¸ + sinc · (f +f 0 ) T 0 2 ¸¾ = AT 0 4 ½ sinc · 1 2 µ f f 0 −1 ¶¸ + sinc · 1 2 µ f f 0 + 1 ¶¸¾ (b) Use the superposition and modulation theorems to get X(f) = AT 0 4 ½ sinc · f 2f 0 ¸ + 1 2 · sinc 1 2 µ f f 0 −2 ¶ + sinc 1 2 µ f f 0 + 2 ¶¸¾ Problem 2.35 Combine the exponents of the two factors in the integrand of the Fourier transform integral, complete the square, and use the given definite integral. Problem 2.36 Consider the development below: x(t) ∗ x(−t) = Z ∞ −∞ x(−λ) x(t −λ) dλ = Z ∞ −∞ x(β) x(t +β) dβ where β = −λ has been substituted. Rename variables to obtain R(τ) = lim T→∞ 1 2T Z T −T x(β) x(t +β) dβ Problem 2.37 The result is an even triangular wave with zero average value of period T 0 . It makes no difference whether the original square wave is even or odd or neither. 2.1. PROBLEM SOLUTIONS 21 Problem 2.38 Fourier transform both sides of the differential equation using the differentiation theorem of Fourier transforms to get [j2πf +a] Y (f) = [j2πbf +c] X (f) Therefore, the transfer function is H (f) = Y (f) X(f) = c +j2πbf a +j2πf The amplitude response function is |H (f)| = q c 2 + (2πbf) 2 q a 2 + (2πf) 2 and the phase response is arg [H (f)] = tan −1 µ 2πbf c ¶ −tan −1 µ 2πf a ¶ Amplitude and phase responses for various values of the constants are plotted below. Problem 2.39 (a) The find the unit impulse response, write H (f) as H (f) = 1 − 5 5 +j2πf Inverse Fourier transforming gives h(t) = δ (t) −5e −5t u(t) (b) Use the transform pair Ae −αt u(t) ←→ A α +j2πf and the time delay theorem to find the unit impulse as h(t) = 2 5 e − 8 15 (t−3) u(t −3) 22 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.40 Use the transform pair for a sinc function to find that Y (f) = Π µ f 2B ¶ Π µ f 2W ¶ (a) If W < B, it follows that Y (f) = Π µ f 2W ¶ because Π ³ f 2B ´ = 1 throughout the region where Π ³ f 2W ´ is nonzero. (b) If W > B, it follows that Y (f) = Π µ f 2B ¶ because Π ³ f 2W ´ = 1 throughout the region where Π ³ f 2B ´ is nonzero. Problem 2.41 (a) Replace the capacitors with 1/jωC which is their ac-equivalent impedance. Call the junction of the input resistor, feedback resistor, and capacitors 1. Call the junction at the positive input of the operational amplifier 2. Call the junction at the negative input of the operational amplifier 3. Write down the KCL equations at these three junctions. Use the constraint equation for the operational amplifier, which is V 2 = V 3 , and the definitions for ω 0 , Q, and K to get the given transfer function. (d) Combinations of components giving RC = 2.3 × 10 −4 seconds and R a R b = 2.5757 will work. Problem 2.42 (a) By long division H (f) = 1 − R 1 /L R 1 +R 2 L +j2πf Using the transforms of a delta function and a one-sided exponential, we obtain h(t) = δ (t) − R 1 L exp µ − R 1 +R 2 L t ¶ u(t) 2.1. PROBLEM SOLUTIONS 23 (b) Substituting the ac-equivalent impedance for the inductor and using voltage division, the transfer function is H (f) = R 2 R 1 +R 2 j2πfL R 1 R 2 R 1 +R 2 +j2πfL = R 2 R 1 +R 2 µ 1 − (R 1 k R 2 ) /L (R 1 k R 2 ) /L +j2πf ¶ Therefore, the impulse response is h(t) = R 2 R 1 +R 2 · δ (t) − R 1 R 2 (R 1 +R 2 ) L exp µ − R 1 R 2 (R 1 +R 2 ) L t ¶ u(t) ¸ Problem 2.43 The Payley-Wiener criterion gives the integral I = Z ∞ −∞ βf 2 1 +f 2 df which does not converge. Hence, the given function is not suitable as the transfer function of a causal LTI system. Problem 2.44 (a) The condition for stability is Z ∞ −∞ |h 1 (t)| dt = Z ∞ −∞ |exp(−αt) cos (2πf 0 t) u(t)| dt = Z ∞ 0 exp(−αt) |cos (2πf 0 t)| dt < Z ∞ 0 exp(−αt) dt = 1 α < ∞ which follows because |cos (2πf 0 t)| ≤ 1. (b) The condition for stability is Z ∞ −∞ |h 2 (t)| dt = Z ∞ −∞ |cos (2πf 0 t) u(t)| dt = Z ∞ 0 |cos (2πf 0 t)| dt →∞ which follows by integrating one period of |cos (2πf 0 t)| and noting that the total integral is the limit of one period of area times N as N →∞. (c) The condition for stability is Z ∞ −∞ |h 3 (t)| dt = Z ∞ −∞ 1 t u(t −1) dt = Z ∞ 1 dt t = ln(t)| ∞ 1 →∞ 24 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.45 The energy spectral density of the output is G y (f) = |H (f)| 2 |X (f)| 2 where X (f) = 1 2 +j2πf Hence G y (f) = 100 h 9 + (2πf) 2 i h 4 + (2πf) 2 i Problem 2.46 Using the Fourier coefficients of a half-rectified sine wave from Table 2.1 and noting that those of a half-rectified cosine wave are related by X cn = X sn e −jnπ/2 The fundamental frequency is 10 Hz. The ideal rectangular filter passes all frequencies less than 13 Hz and rejects all frequencies greater than 13 Hz. Therefore y (t) = 3A π − 3A 2 cos (20πt) Problem 2.47 (a) The 90% energy containment bandwidth is given by B 90 = α 2π tan(0.45π) = 1.0055α (b) For this case, using X 2 (f) = Π(f/2W) , we obtain B 90 = 0.9W (c) Numerical integration gives B 90 = 0.85/τ 2.1. PROBLEM SOLUTIONS 25 Problem 2.48 From Example 2.7 x(t) = A 2 + 2A π · cos (ω 0 t) − 1 3 cos (3ω 0 t) + · · · ¸ From the transfer function of a Hilbert transformer, we find its output in response to the above input to be y (t) = A 2 + 2A π · cos ³ ω 0 t − π 2 ´ − 1 3 cos ³ 3ω 0 t − π 2 ´ + · · · ¸ Problem 2.49 (a) Amplitude distortion; no phase distortion. (b) No amplitude distortion; phase distortion. (c) No amplitude distortion; no phase distortion. (d) No amplitude distortion; no phase distortion. Problem 2.50 The transfer function corresponding to this impulse response is H (f) = 2 3 +j2πf = 2 q 9 + (2πf) 2 exp · −j tan µ 2πf 3 ¶¸ The group delay is T g (f) = − 1 2π d df · −tan µ 2πf 3 ¶¸ = 3 9 + (2πf) 2 The phase delay is T p (f) = − θ (f) 2πf = tan ³ 2πf 3 ´ 2πf Problem 2.51 The group and phase delays are, respectively, T g (f) = 0.1 1 + (0.2πf) 2 − 0.333 1 + (0.667πf) 2 T p (f) = 1 2πf [tan(0.2πf) −tan(0.667πf)] 26 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.52 In terms of the input spectrum, the output spectrum is Y (f) = X(f) + 0.2X (f) ∗ X(f) = 4 · Π µ f −20 6 ¶ +Π µ f + 20 6 ¶¸ +3.2 · Π µ f −20 6 ¶ +Π µ f + 20 6 ¶¸ ∗ · Π µ f −20 6 ¶ +Π µ f + 20 6 ¶¸ = 4 · Π µ f −20 6 ¶ +Π µ f + 20 6 ¶¸ +3.2 · 6Λ µ f −40 6 ¶ + 12Λ µ f 6 ¶ + 6Λ µ f + 40 6 ¶¸ where Λ(t) is an isoceles triangle of unit height going from -1 to 1. The student should sketch the output spectrum given the above analytical result.. Problem 2.53 (a) The output of the nonlinear device is y (t) = 1.075 cos (2000πt) + 0.025 cos (6000πt) The steadtstate filter output in response to this input is z (t) = 1.075 |H (1000)| cos (2000πt) + 0.025 |H (3000)| cos (6000πt) so that the THD is THD = (1.075) 2 |H (1000)| 2 (0.025) 2 |H (3000)| 2 = 43 2 1 + 4Q 2 (1000 −3000) 2 = 1849 1 + 16 × 10 6 Q 2 where H (f) is the transfer function of the filter. (b) For THD = 0.005% = 0.00005, the equation for Q becomes 1849 1 + 16 × 10 6 Q 2 = 0.00005 2.1. PROBLEM SOLUTIONS 27 or 6 × 10 6 Q 2 = 1849 × 2 × 10 4 −1 Q = 1.52 Problem 2.54 Frequency components are present in the output at radian frequencies of 0, ω 1 , ω 2 , 2ω 1 , 2ω 2 , ω 1 − ω 2 , ω 1 + ω 2 ,3ω 1 , 3ω 2 , 2ω 2 − ω 1 , ω 1 + 2ω 2 , 2ω 1 − ω 2 , 2ω 1 + ω 2 . To use this as a frequency multiplier, pass the components at 2ω 1 or 2ω 2 to use as a doubler, or at 3ω 1 or 3ω 2 to use as a tripler. Problem 2.55 Write the transfer function as H (f) = H 0 e −j2πft 0 −H 0 Π µ f 2B ¶ e −j2πft 0 Use the inverse Fourier transform of a constant, the delay theorem, and the inverse Fourier transform of a rectangular pulse function to get h(t) = H 0 δ (t −t 0 ) −2BH 0 sinc [2B(t −t 0 )] Problem 2.56 (a) The Fourier transform of this signal is X (f) = A √ 2πb 2 exp ¡ −2π 2 τ 2 f 2 ¢ By definition, using a table of integrals, T = 1 x(0) Z ∞ −∞ |x(t)| dt = √ 2πτ Similarly, W = 1 2X(0) Z ∞ −∞ |X (f)| df = 1 2 √ 2πτ Therefore, 2WT = 2 2 √ 2πτ √ 2πτ = 1 28 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) The Fourier transform of this signal is X (f) = 2A/α 1 + (2πf/α) 2 The pulse duration is T = 1 x(0) Z ∞ −∞ |x(t)| dt = 2 α The bandwidth is W = 1 2X (0) Z ∞ −∞ |X(f)| df = α 4 Thus, 2WT = 2 ³ α 4 ´ µ 2 α ¶ = 1 Problem 2.57 (a) The poles for a second order Butterworth filter are given by s 1 = s ∗ 2 = − ω 3 √ 2 (1 −j) where ω 3 is the 3-dB cutoff frequency of the Butterworth filter. Its s-domain transfer function is H (s) = ω 2 3 h s + ω 3 √ 2 (1 −j) i h s + ω 3 √ 2 (1 +j) i = ω 2 3 s 2 + √ 2ω 3 s +ω 2 3 Letting ω 3 = 2πf 3 and s = jω = j2πf, we obtain H (j2πf) = 4π 2 f 2 3 −4π 2 f 2 + √ 2 (2πf 3 ) (j2πf) + 4π 2 f 2 3 = f 2 3 −f 2 +j √ 2f 3 f +f 2 3 (b) If the phase response function of the filter is θ (f), the group delay is T g (f) = − 1 2π d df [θ (f)] For the second-order Butterworth filter considered here, θ (f) = −tan −1 Ã √ 2f 3 f f 2 3 −f 2 ! 2.1. PROBLEM SOLUTIONS 29 Figure 2.3: Therefore, the group delay is T g (f) = 1 2π d df " tan −1 Ã √ 2f 3 f f 2 3 −f 2 !# = f 3 √ 2π f 2 3 +f 2 f 4 3 +f 4 = 1 √ 2πf 3 1 + (f/f 3 ) 2 1 + (f/f 3 ) 4 (c) Use partial fraction expansion of H (s) and then inverse Laplace transform H (s) /s to get the given step response. Plot it and estimate the 10% and 90% times from the plot. From the MATLAB plot of Fig. 2.3, f 3 t 10% ≈ 0.08 and f 3 t 90% ≈ 0.42 so that the 10-90 % rise time is about 0.34/f 3 seconds. Problem 2.58 (a) 0.5 seconds; (b) and (c) - use sketches to show. Problem 2.59 (a) The leading edges of the flat-top samples follow the waveform at the sampling instants. (b) The spectrum is Y (f) = X δ (f) H (f) 30 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY where X δ (f) = f s ∞ X n=−∞ X(f −nf s ) and H (f) = τsinc (fτ) exp(−jπfτ) The latter represents the frequency response of a filter whose impulse response is a square pulse of width τ and implements flat top sampling. If W is the bandwidth of X (f), very little distortion will result if τ −1 >> W. Problem 2.60 (a) The sampling frequency should be large compared with the bandwidth of the signal. (b) The output spectrum of the zero-order hold circuit is Y (f) = sinc (T s f) ∞ X n=−∞ X(f −nf s ) exp(−jπfT s ) where f s = T −1 s . For small distortion, we want T s << W −1 . Problem 2.61 The lowpass recovery filter can cut off in the range 1.9 + kHz to 2.1 − kHz. Problem 2.62 For bandpass sampling and recovery, all but (b) and (e) will work theoretically, although an ideal filter with bandwidth exactly equal to the unsampled signal bandwidth is necessary. For lowpass sampling and recovery, only (f) will work. Problem 2.63 The Fourier transform is Y (f) = 1 2 X(f −f 0 ) + 1 2 X(f +f 0 ) +[−jsgn(f) X(f)] ∗ · 1 2 δ (f −f 0 ) e −jπ/2 + 1 2 δ (f +f 0 ) e jπ/2 ¸ = 1 2 X(f −f 0 ) [1 −sgn(f −f 0 )] + 1 2 X(f +f 0 ) [1 + sgn(f +f 0 )] 2.1. PROBLEM SOLUTIONS 31 Problem 2.64 (a) b x a (t) = cos (ω 0 t −π/2) = sin(ω 0 t), so lim T→∞ 1 2T Z T −T x(t) b x(t) dt = lim T→∞ 1 2T Z T −T sin(ω 0 t) cos (ω 0 t) dt = lim T→∞ 1 2T Z T −T 1 2 sin(2ω 0 t) dt = lim T→∞ 1 2T cos (2ω 0 t) 4ω 0 ¯ ¯ ¯ ¯ T −T = 0 (b) Use trigonometric identities to express x(t) in terms of sines and cosines. Then find the Hilbert transform of x(t) by phase shifting by −π/2. Multiply x(t) and b x(t) together term by term, use trigonometric identities for the product of sines and cosines, then integrate. The integrand will be a sum of terms similar to that of part (a). The limit as T →∞ will be zero term-by-term. (c) Use the integral definition of b x(t), take the product, integrate over time to get Z ∞ −∞ x(t) b x(t) dt = A 2 Z ∞ −∞ Π(t/τ) ·Z ∞ −∞ Π(λ/τ) π (t −λ) dλ ¸ dt = A 2 Z τ/2 −τ/2 " Z τ/2 −τ/2 1 π (t −λ) dλ # dt = A 2 Z τ/2 −τ/2 1 π ln ¯ ¯ ¯ ¯ t −τ/2 t +τ/2 ¯ ¯ ¯ ¯ dt = 0 where the result is zero by virtue of the integrand of the last integral being odd. Problem 2.65 (a) Note that F [jb x(t)] = j [−jsgn(f)] X(f). Hence x 1 (t) = 3 4 x(t) + 1 4 jb x(t) →X 1 (f) = 3 4 X (f) + 1 4 j [−jsgn(f)] X(f) = · 3 4 + 1 4 sgn(f) ¸ X (f) = ½ 1 2 X (f) , f < 0 X(f) , f > 0 A sketch is shown in Figure 2.4. 32 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) It follows that x 2 (t) = · 3 4 x(t) + 3 4 jb x(t) ¸ exp(j2πf 0 t) → X 2 (f) = 3 4 [1 + sgn(f −f 0 )] X (f −f 0 ) = ½ 0, f < f 0 3 2 X (f −f 0 ) , f > f 0 A sketch is shown in Figure 2.4. (c) This case has the same spectrum as part (a), except that it is shifted right by W Hz. That is, x 3 (t) = · 3 4 x(t) + 1 4 jb x(t) ¸ exp(j2πWt) → X 3 (f) = · 3 4 + 1 4 sgn(f −W) ¸ X(f −W) A sketch is shown in Figure 2.4. (d) For this signal x 4 (t) = · 3 4 x(t) − 1 4 jb x(t) ¸ exp(jπWt) → X 4 (f) = · 3 4 − 1 4 sgn(f −W/2) ¸ X(f −W/2) A sketch is shown in Figure 2.4. Problem 2.66 (a) The spectrum is X p (f) = X(f) +j [−jsgn(f)] X (f) = [1 + sgn(f)] X (f) The Fourier transform of x(t) is X(f) = 1 2 Π µ f −f 0 2W ¶ + 1 2 Π µ f +f 0 2W ¶ Thus, X p (f) = Π µ f −f 0 2W ¶ if f 0 > 2W (b) The complex envelope is defined by x p (t) = ˜ x(t) e j2πf 0 t 2.1. PROBLEM SOLUTIONS 33 f, Hz -W 0 W 2A A X 1 (f) f, Hz 0 f 0 f 0 - W 3A/2 X 2 (f) f, Hz 0 W 2W 2A A X 3 (f) f, Hz - W/2 0 W/2 3W/2 2A A X 4 (f) Figure 2.4: 34 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Therefore ˜ x(t) = x p (t) e −j2πf 0 t Hence F [˜ x(t)] = F [x p (t)]| f→f+f 0 = Π µ f −f 0 2W ¶¯ ¯ ¯ ¯ f→f+f 0 = Π µ f 2W ¶ (c) The complex envelope is ˜ x(t) = F −1 · Π µ f 2W ¶¸ = 2W sinc (2Wt) Problem 2.67 For t < τ/2, the output is zero. For |t| ≤ τ/2, the result is y (t) = α/2 q α 2 + (2π∆f) 2 × n cos [2π (f 0 +∆f) t −θ] −e −α(t+τ/2) cos [2π (f 0 +∆f) t +θ] o For t > τ/2, the result is y (t) = (α/2) e −αt q α 2 + (2π∆f) 2 × n e ατ/2 cos [2π (f 0 +∆f) t −θ] −e −ατ/2 cos [2π (f 0 +∆f) t +θ] o In the above equations, θ is given by θ = −tan −1 µ 2π∆f α ¶ 2.2 Computer Exercises Computer Exercise 2.1 % ce2_1: Computes generalized Fourier series coefficients for exponentially % decaying signal, exp(-t)u(t), or sinewave, sin(2*pi*t) % tau = input(’Enter approximating pulse width: ’); type_wave = input(’Enter waveform type: 1 = decaying exponential; 2 = sinewave ’); if type_wave == 1 2.2. COMPUTER EXERCISES 35 t_max = input(’Enter duration of exponential: ’); elseif type_wave == 2 t_max = 1; end clf a = []; t = 0:.01:t_max; cn = tau; n_max = floor(t_max/tau) for n = 1:n_max LL = (n-1)*tau; UL = n*tau; if type_wave == 1 a(n) = (1/cn)*quad(@integrand_exp, LL, UL); elseif type_wave == 2 a(n) = (1/cn)*quad(@integrand_sine, LL, UL); end end disp(’ ’) disp(’c_n’) disp(cn) disp(’ ’) disp(’Expansion coefficients (approximating pulses not normalized):’) disp(a) disp(’ ’) x_approx = zeros(size(t)); for n = 1:n_max x_approx = x_approx + a(n)*phi(t/tau,n-1); end if type_wave == 1 MSE = quad(@integrand_exp2, 0, t_max) - cn*a*a’; elseif type_wave == 2 MSE = quad(@integrand_sine2, 0, t_max) - cn*a*a’; end disp(’Mean-squared error:’) disp(MSE) disp(’ ’) if type_wave == 1 plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(’t’), ylabel(’x(t); x_a_p_p_r_o_x(t)’) 36 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY hold plot(t, exp(-t)) title([’Approximation of exp(-t) over [0, ’,num2str(t_max),’] with contiguous rectangular pulses of width ’,num2str(tau)]) elseif type_wave == 2 plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(’t’), ylabel(’x(t); x_a_p_p_r_o_x(t)’) hold plot(t, sin(2*pi*t)) title([’Approximation of sin(2*pi*t) over [0, ’,num2str(t_max),’] with contiguous rectangular pulses of width ’,num2str(tau)]) end % Decaying exponential function for ce_1 % function z = integrand_exp(t) z = exp(-t); % Sine function for ce_1 % function z = integrand_sine(t) z = sin(2*pi*t); % Decaying exponential squared function for ce_1 % function z = integrand_exp2(t) z = exp(-2*t); % Sin^2 function for ce_1 % function z = integrand_sine2(t) z = (sin(2*pi*t)).^2; A typical run follows: >> ce2_1 Enter approximating pulse width: 0.125 Enter waveform type: 1 = decaying exponential; 2 = sinewave: 2 n_max = 8 c_n 0.1250 Expansion coefficients (approximating pulses not normalized): 2.2. COMPUTER EXERCISES 37 Figure 2.5: Columns 1 through 7 0.3729 0.9003 0.9003 0.3729 -0.3729 -0.9003 -0.9003 Column 8 -0.3729 Mean-squared error: 0.0252 Computer Exercise 2.2 % ce2_1: Computes generalized Fourier series coefficients for exponentially % decaying signal, exp(-t)u(t), or sinewave, sin(2*pi*t) % tau = input(’Enter approximating pulse width: ’); type_wave = input(’Enter waveform type: 1 = decaying exponential; 2 = sinewave ’); if type_wave == 1 t_max = input(’Enter duration of exponential: ’); elseif type_wave == 2 t_max = 1; end clf a = []; t = 0:.01:t_max; 38 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY cn = tau; n_max = floor(t_max/tau) for n = 1:n_max LL = (n-1)*tau; UL = n*tau; if type_wave == 1 a(n) = (1/cn)*quad(@integrand_exp, LL, UL); elseif type_wave == 2 a(n) = (1/cn)*quad(@integrand_sine, LL, UL); end end disp(’ ’) disp(’c_n’) disp(cn) disp(’ ’) disp(’Expansion coefficients (approximating pulses not normalized):’) disp(a) disp(’ ’) x_approx = zeros(size(t)); for n = 1:n_max x_approx = x_approx + a(n)*phi(t/tau,n-1); end if type_wave == 1 MSE = quad(@integrand_exp2, 0, t_max) - cn*a*a’; elseif type_wave == 2 MSE = quad(@integrand_sine2, 0, t_max) - cn*a*a’; end disp(’Mean-squared error:’) disp(MSE) disp(’ ’) if type_wave == 1 plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(’t’), ylabel(’x(t); x_a_p_p_r_o_x(t)’) hold plot(t, exp(-t)) title([’Approximation of exp(-t) over [0, ’,num2str(t_max),’] with contiguous rectangular pulses of width ’,num2str(tau)]) elseif type_wave == 2 plot(t, x_approx), axis([0 t_max -inf inf]), xlabel(’t’), ylabel(’x(t); x_a_p_p_r_o_x(t)’) 2.2. COMPUTER EXERCISES 39 hold plot(t, sin(2*pi*t)) title([’Approximation of sin(2*pi*t) over [0, ’,num2str(t_max),’] with contiguous rectangular pulses of width ’,num2str(tau)]) end % Decaying exponential function for ce_1 % function z = integrand_exp(t) z = exp(-t); % Sine function for ce_1 % function z = integrand_sine(t) z = sin(2*pi*t); % Decaying exponential squared function for ce_1 % function z = integrand_exp2(t) z = exp(-2*t); % Sin^2 function for ce_1 % function z = integrand_sine2(t) z = (sin(2*pi*t)).^2; A typical run follows: >> ce2_2 Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: 1 Computer Exercise 2.3 % ce2_4.m: FFT plotting of line spectra for half-rectified, full-rectified sinewave, square wave, % and triangular waveforms % clf I_wave = input(’Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine ’); T = 2; del_t = 0.001; t = 0:del_t:T; 40 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Figure 2.6: L = length(t); fs = (L-1)/T; del_f = 1/T; n = [0 1 2 3 4 5 6 7 8 9]; if I_wave == 1 x = pls_fn(2*(t-T/4)/T); X_th = abs(0.5*sinc(n/2)); disp(’ ’) disp(’ 0 - 1 level squarewave’) elseif I_wave == 2 x = 2*trgl_fn(2*(t-T/2)/T)-1; X_th = 4./(pi^2*n.^2); X_th(1) = 0; % Set n = even coeficients to zero (odd indexed because of MATLAB) X_th(3) = 0; X_th(5) = 0; X_th(7) = 0; X_th(9) = 0; disp(’ ’) disp(’ 0-dc level triangular wave’) elseif I_wave == 3 x = sin(2*pi*t/T).*pls_fn(2*(t-T/4)/T); X_th = abs(1./(pi*(1-n.^2))); 2.2. COMPUTER EXERCISES 41 X_th(2) = 0.25; X_th(4) = 0; % Set n = odd coeficients to zero (even indexed because of MATLAB) X_th(6) = 0; X_th(8) = 0; X_th(10) = 0; disp(’ ’) disp(’ Half-rectified sinewave’) elseif I_wave == 4 x = abs(sin(2*pi*t/T)); X_th = abs(2./(pi*(1-n.^2))); X_th(2) = 0; % Set n = odd coeficients to zero (even indexed because of MATLAB) X_th(4) = 0; X_th(6) = 0; X_th(8) = 0; X_th(10) = 0; disp(’ ’) disp(’ Full-rectified sinewave’) end X = 0.5*fft(x)*del_t; % Multiply by 0.5 because of 1/T_0 with T_0 = 2 f = 0:del_f:fs; Y = abs(X(1:10)); Z = [n’ Y’ X_th’]; disp(’Magnetude of the Fourier coefficients’); disp(’ ’) disp(’ n FFT Theory’); disp(’ __________________________’) disp(’ ’) disp(Z); subplot(2,1,1),plot(t, x), xlabel(’t’), ylabel(’x(t)’) subplot(2,1,2),plot(f, abs(X),’o’),axis([0 10 0 1]),... xlabel(’n’), ylabel(’|X_n|’) A typical run follows: >> ce2_3 Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine 3 Half-rectified sinewave Magnetude of the Fourier coefficients n FFT Theory __________________________ 42 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Figure 2.7: 0 0.3183 0.3183 1.0000 0.2501 0.2500 2.0000 0.1062 0.1061 3.0000 0.0001 0 4.0000 0.0212 0.0212 5.0000 0.0001 0 6.0000 0.0091 0.0091 7.0000 0.0000 0 8.0000 0.0051 0.0051 9.0000 0.0000 0 Computer Exercise 2.4 Make the time window long compared with the pulse width. Computer Exercise 2.5 % ce2_5.m: Finding the energy ratio in a preset bandwidth % I_wave = input(’Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine. ’); tau = input(’Enter pulse width ’); per_cent = input(’Enter percent total desired energy: ’); 2.2. COMPUTER EXERCISES 43 clf T = 20; f = []; G = []; del_t = 0.01; t = 0:del_t:T; L = length(t); fs = (L-1)/T; del_f = 1/T; if I_wave == 1 x = pls_fn((t-tau/2)/tau); elseif I_wave == 2 x = trgl_fn(2*(t-tau/2)/tau); elseif I_wave == 3 x = sin(pi*t/tau).*pls_fn((t-tau/2)/tau); elseif I_wave == 4 x = abs(sin(pi*t/tau).^2.*pls_fn((t-tau/2)/tau)); end X = fft(x)*del_t; f1 = 0:del_f*tau:fs*tau; G1 = X.*conj(X); NN = floor(length(G1)/2); G = G1(1:NN); ff = f1(1:NN); f = f1(1:NN+1); E_tot = sum(G); E_f = cumsum(G); E_W = [0 E_f]/E_tot; test = E_W - per_cent/100; L_test = length(test); k = 1; while test(k) <= 0 k = k+1; end B = k*del_f; if I_wave == 2 tau1 = tau/2; else tau1 = tau; end 44 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY subplot(3,1,1),plot(t/tau, x), xlabel(’t/\tau’), ylabel(’x(t)’), axis([0 2 0 1.2]) if I_wave == 1 title([’Energy containment bandwidth for square pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 2 title([’Energy containment bandwidth for triangular pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 3 title([’Energy containment bandwidth for half-sine pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 4 title([’Energy containment bandwidth for raised cosine pulse of width ’, num2str(tau),’ seconds’]) end subplot(3,1,2),semilogy(ff*tau1, abs(G./max(G))), xlabel(’f\tau’), ylabel(’G’), ... axis([0 10 -inf 1]) subplot(3,1,3),plot(f*tau1, E_W,’—’), xlabel(’f\tau’), ylabel(’E_W’), axis([0 4 0 1.2]) legend([num2str(per_cent), ’ bandwidth % X (pulse width) = ’, num2str(B*tau)],4) A typical run follows: >> ce2_5 Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine. 3 Enter pulse width 2 Enter percent total desired energy: 95 Computer Exercise 2.6 The program for this exercise is similar to that for Computer Exercise 2.5, except that the waveform is used in the energy calculation. Computer Exercise 2.7 Use Computer Example 2.2 as a pattern for the solution (note that in earlier printings of the book “Computer Excercise 2.2” should be “Computer Example 2.2”). 2.2. COMPUTER EXERCISES 45 Figure 2.8: Chapt er 3 Basi c M odul at i on Techni ques 3.1 Pr obl ems Pr obl em 3.1 The demodulated output, in general, is y D (t) = Lp{x c (t) 2 cos[ω c t + θ (t)]} where Lp{•} denotes the lowpass portion of the argument. With x c (t) = A c m(t) cos [ω c t + φ 0 ] the demodulated output becomes y D (t) = Lp{2A c m(t) cos [ω c t + φ 0 ] cos [ω c t + θ (t)]} Performing the indicated multiplication and taking the lowpass portion yields y D (t) = A c m(t) cos [θ (t) −φ 0 ] If θ(t) = θ 0 (a constant), the demodulated output becomes y D (t) = A c m(t) cos [θ 0 −φ 0 ] Letting A c = 1 gives the error ε (t) = m(t) [1 −cos (θ 0 −φ 0 )] The mean-square error is ε 2 (t) ® = D m 2 (t) [1 −cos (θ 0 −φ 0 )] 2 E 1 2 CHAPTER 3. BASIC MODULATION TECHNIQUES where h·i denotes the time-average value. Since the term [1 −cos (θ 0 −φ 0 )] is a constant, we have ε 2 (t) ® = m 2 (t) ® [1 −cos (θ 0 −φ 0 )] 2 Note that for θ 0 = φ 0 , the demodulation carrier is phase coherent with the original modulation carrier, and the error is zero. For θ (t) = ω 0 t we have the demodulated output y D (t) = A c m(t) cos (ω 0 t −φ 0 ) Letting A c = 1, for convenience, gives the error ε (t) = m(t) [1 −cos (ω 0 t −φ 0 )] giving the mean-square error ε 2 (t) ® = D m 2 (t) [1 −cos (ω 0 t −φ 0 )] 2 E In many cases, the average of a product is the product of the averages. (We will say more about this in Chapters 4 and 5). For this case ε 2 (t) ® = m 2 (t) ® D [1 −cos (ω 0 t −φ 0 )] 2 E Note that 1 − cos (ω 0 t −φ 0 ) is periodic. Taking the average over an integer number of periods yields D [1 −cos (ω 0 t −φ 0 )] 2 E = 1 −2 cos (ω 0 t −φ 0 ) + cos 2 (ω 0 t −φ 0 ) ® = 1 + 1 2 = 3 2 Thus ε 2 (t) ® = 3 2 m 2 (t) ® Pr obl em 3.2 Multiplying the AM signal x c (t) = A c [1 +am n (t)] cos ω c t by x c (t) = A c [1 +am n (t)] cos ω c t and lowpass Þltering to remove the double frequency (2ω c ) term yields y D (t) = A c [1 +am n (t)] cos θ (t) 3.1. PROBLEMS 3 ( ) C x t ( ) D y t C R Figure 3.1: For negligible demodulation phase error, θ(t) ≈ 0, this becomes y D (t) = A c +A c am n (t) The dc component can be removed resulting in A c am n (t), which is a signal proportional to the message, m(t). This process is not generally used in AM since the reason for using AM is to avoid the necessity for coherent demodulation. Pr obl em 3.3 A full-wave rectiÞer takes the form shown in Figure 3.1. The waveforms are shown in Figure 3.2, with the half-wave rectiÞer on top and the full-wave rectiÞer on the bottom. The message signal is the envelopes. Decreasing exponentials can be drawn from the peaks of the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-wave rectiÞed x c (t) deÞnes the message better than the half-wave rectiÞed x c (t) since the carrier frequency is effectively doubled. Pr obl em 3.4 Part m 2 n (t) ® a = 0.4 a = 0.6 a = 1 a 1/3 E ff = 5, 1% E ff = 10.7% E ff = 25% b 1/3 E ff = 5.1% E ff = 10.7% E ff = 25% c 1 E ff = 13.8% E ff = 26.5% E ff = 50% Pr obl em 3.5 By inspection, the normalized message signal is as shown in Figure 3.3. Thus m n (t) = 2 T t, 0 ≤ t ≤ T 2 4 CHAPTER 3. BASIC MODULATION TECHNIQUES 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 Figure 3.2: t -1 ( ) n m t T 1 0 Figure 3.3: 3.1. PROBLEMS 5 and m 2 n (t) ® = 2 T Z T/2 0 µ 2 T t ¶ 2 dt = 2 T µ 2 T ¶ 2 1 3 µ T 2 ¶ 3 = 1 3 Also A c [1 +a] = 40 A c [1 −a] = 10 This yields 1 +a 1 −a = 40 10 = 4 or 1 +a = 4 −4a 5a = 3 Thus a = 0.6 Since the index is 0.6, we can write A c [1 + 0.6] = 40 This gives A c = 40 1.6 = 25 This carrier power is P c = 1 2 A 2 c = 1 2 (25) 2 = 312.5 Watts The efficiency is E ff = (0.6) 2 ¡ 1 3 ¢ 1 + (0.6) 2 ¡ 1 3 ¢ = 0.36 3.36 = 0.107 = 10.7% Thus P sb P c +P sb = 0.107 where P sb represents the power in the sidebands and P c represents the power in the carrier. The above expression can be written P sb = 0.107 + 0.107P sb This gives P sb = 0.107 1.0 −0.107 P c = 97.48 Watts 6 CHAPTER 3. BASIC MODULATION TECHNIQUES Pr obl em 3.6 For the Þrst signal τ = T 6 , m n (t) = m(t) and E ff = 81.25% and for the second signal τ = 5T 6 , m n (t) = 1 5 m(t) and E ff = 14.77% Pr obl em 3.7 (a) The Þrst step in the solution to this problem is to plot m(t), or use a root-Þnding algorithm in order to determine the minimum value of m(t). We Þnd that the minimum of m(t) = −11.9523, and the minimum falls at t = 0.0352 and t = 0.0648. Thus the normalized message signal is m n (t) = 1 11.9523 [9 cos 20πt −7 cos 60πt] With the given value of c (t) and the index a, we have x c (t) = 100 [1 + 0.5m n (t)] cos 200πt This yields x c (t) = −14.6415 cos 140πt + 18.8428 cos 180πt +100 cos 200πt +18.8248 cos 220πt −14.6415 cos 260πt We will need this later to plot the spectrum. (b) The value of m 2 n (t) ® is m 2 n (t) ® = µ 1 11.9523 ¶ 2 µ 1 2 ¶ h (9) 2 + (7) 2 i = 0.455 (c)This gives the efficiency E = (0.5) 2 (0.455) 1 + (0.5) 2 (0.455) = 10.213% 3.1. PROBLEMS 7 70 90 100 110 130 -130 –110 –100 –90 -70 A B 50 50 f B A B A B A Figure 3.4: (d) The two-sided amplitude spectrum is shown in Figure 3.4. where A = 14.4615 2 = 7.2307 and B = 18.8248 2 = 9.4124 The phase spectrum results by noting that A is negative and all other terms are positive. (e) By inspection the signal is of the form x c (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap. Thus the Hilbert transform of c(t) is b x c (t) = a(t)bc(t) in which c(t) = 100 cos 200πt. This the envelope is e(t) = 100 q a 2 (t) cos 2 200πt +a 2 (t) sin 2 200πt = 100a(t) where a(t) = [1 + 0.5m n (t)] Pr obl em 3.8 8 CHAPTER 3. BASIC MODULATION TECHNIQUES (a) m n (t) = 1 16 [9 cos 20πt + 7 cos 60πt] (b) m 2 n (t) ® = ¡ 1 16 ¢ 2 ¡ 1 2 ¢ h (9) 2 + (7) 2 i = 0.2539 (c) E ff = 0.25(0.2539) 1+ 0.25(0.2539) = 0.05969 = 5.969% (d) The expression for x c (t) is x c (t) = 100 · 1 + 1 32 (9 cos 20πt + cos 60πt) ¸ cos 200πt = 10.9375 cos 140πt + 14.0625 cos 180πt +100 cos 200πt +14.0625 cos 220πt + 10.9375 cos 260πt Note that all terms are positive so the phase spectrum is everywhere zero. The amplitude spectrum is identical to that shown in the previous problem except that A = 1 2 (10.9375) = 5.46875 B = 1 2 (14.0625) = 7.03125 (e) As in the previous problem, the signal is of the form x c (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass. The spectra of a(t) and c(t) are do not overlap. Thus the Hilbert transform of c(t) is b x c (t) = a(t)bc(t) where c(t) = 100 cos 200πt. This the envelope is e(t) = 100 q a 2 (t) cos 2 200πt +a 2 (t) sin 2 200πt = 100a(t) = 100 · 1 + 1 32 (9 cos 20πt + cos 60πt) ¸ Pr obl em 3.9 The modulator output x c (t) = 40 cos 2π (200) t + 4 cos 2π (180) t + 4 cos 2π (220) t can be written x c (t) = [40 + 8 cos 2π (20) t] cos 2π (200) t 3.1. PROBLEMS 9 or x c (t) = 40 · 1 + 8 40 cos 2π (20) t ¸ cos 2π (200) t By inspection, the modulation index is a = 8 40 = 0.2 Since the component at 200 Hertz represents the carrier, the carrier power is P c = 1 2 (40) 2 = 800 Watts The components at 180 and 220 Hertz are sideband terms. Thus the sideband power is P sb = 1 2 (4) 2 + 1 2 (4) 2 = 16 Watts Thus, the efficiency is E ff = P sb P c +P sb = 16 800 + 16 = 0.0196 = 1.96% Pr obl em 3.10 A = 14.14 B = 8.16 a = 1.1547 Pr obl em 3.11 The modulator output x c (t) = 20 cos 2π (150) t + 6 cos 2π (160) t + 6 cos 2π (140) t is x c (t) = 20 · 1 + 12 20 cos 2π (10) t ¸ cos 2π (150) t Thus, the modulation index, a, is a = 12 20 = 0.6 The carrier power is P c = 1 2 (20) 2 = 200 Watts 10 CHAPTER 3. BASIC MODULATION TECHNIQUES and the sideband power is P sb = 1 2 (6) 2 + 1 2 (6) 2 = 36 Watts Thus, the efficiency is E ff = 36 200 + 36 = 0.1525 Pr obl em 3.12 (a) By plotting m(t) or by using a root-Þnding algorithm we see that the minimum value of m(t) is M = −3.432. Thus m n (t) = 0.5828 cos (2πf m t) + 0.2914 cos (4πf m t) + 0.5828 cos (10πf m t) The AM signal is x c (t) = A c [1 + 0.7m n (t)] cos 2πf c t = 0.2040A c cos 2π (f c −5f m ) t +0.1020A c cos 2π (f c −2f m ) t +0.2040A c cos 2π (f c −f m ) t +A c cos 2πf c t +0.2040A c cos 2π (f c +f m ) t +0.1020A c cos 2π (f c + 2f m ) t +0.2040A c cos 2π (f c + 5f m ) t The spectrum is drawn from the expression for x c (t). It contains 14 discrete components as shown Comp Freq Amp Comp Freq Amp 1 −f c −5f m 0.102A c 8 f c −5f m 0.102A c 2 −f c −2f m 0.051A c 9 f c −2f m 0.051A c 3 −f c −f m 0.102A c 10 f c −f m 0.102A c 4 −f c 0.5A c 11 f c 0.5A c 5 −f c +f m 0.102A c 12 f c +f m 0.102A c 6 −f c + 2f m 0.051A c 13 f c + 2f m 0.051A c 7 −f c + 5f m 0.102A c 14 f c + 5f m 0.102A c (b) The efficiency is 15.8%. Pr obl em 3.13 3.1. PROBLEMS 11 f Filter Characteristic ( ) { } 2 m t ℑ ( ) { } m t ℑ 2 f c f c f W c + f W c − 0 W 2W Figure 3.5: (a) From Figure 3.75 x(t) = m(t) + cos ω c t With the given relationship between x(t) and y (t) we can write y (t) = 4 {m(t) + cos ω c t} + 10 {m(t) + cos ω c t} 2 which can be written y (t) = 4m(t) + 4 cos ω c t + 10m 2 (t) + 20m(t) cos ω c t + 5 + 5 cos 2ω c t The equation for y (t) is more conveniently expressed y (t) = 5 + 4m(t) + 10m 2 (t) + 4 [1 + 5m(t)] cos ω c t + 5 cos 2ω c t (b) The spectrum illustrating the terms involved in y (t) is shown in Figure 3.5. The center frequency of the Þlter is f c and the bandwidth must be greater than or equal to 2W. In addition, f c − W > 2W or f c > 3W, and f c + W < 2f c . The last inequality states that f c > W, which is redundant since we know that f c > 3W. (c) From the deÞnition of m(t) we have m(t) = Mm n (t) so that g (t) = 4 [1 + 5Mm n (t)] cos ω c t It follows that a = 0.8 = 5M 12 CHAPTER 3. BASIC MODULATION TECHNIQUES Thus M = 0.8 5 = 0.16 (d) This method of forming a DSB signal avoids the need for a multiplier. Pr obl em 3.14 H U (f) = 1 − 1 2 sgn(f +f c ) x USB (t) = 1 2 A c m(t) cos ω c t − 1 2 A c b m(t) sinω c t Pr obl em 3.15 For the USB SSB case, the modulator output is a sinusoid of frequency f c +f m , while for the LSB SSB case, the modulator output is a sinusoid of frequency of f c −f m . Pr obl em 3.16 Using Figure 3.13 and the phases given in the problem statement, the modulator output becomes x c (t) = Aε 2 cos [(ω c −ω 1 ) t + φ] + A(1 −ε) 2 cos [(ω c + ω 1 ) t + θ 1 ] + B 2 cos [(ω c + ω 2 ) t + θ 2 ] Multiplying x c (t) by 4 cos ω c t and lowpass Þltering yields the demodulator output y D (t) = Aε cos (ω 1 −φ) +A(1 −ε) cos (ω 1 t + θ 1 ) +Bcos (ω 2 t + θ 2 ) For the sum of the Þrst two terms to equal the desired output with perhaps a time delay, θ 1 must equal−φ. This gives y D (t) = Acos (ω 1 t + φ) +Bcos (ω 2 t + θ 2 ) which we can write y D (t) = Acos ω 1 µ t + θ 1 ω 1 ¶ +Bcos ω 2 µ t + θ 2 ω 2 ¶ For no distortion, y D (t) must be of the form m(t −τ). Thus θ 1 ω 1 = θ 2 ω 2 3.1. PROBLEMS 13 so that θ 2 = ω 2 ω 1 θ 1 Thus, the phase must be linear. The fact that φ = −θ 1 tells us that the Þlter phase response must have odd symmetry about the carrier frequency. Pr obl em 3.17 We assume that the VSB waveform is given by x c (t) = 1 2 Aε cos (ω c −ω 1 ) t + 1 2 A(1 −ε) cos (ω c + ω 1 ) t + 1 2 Bcos (ω c + ω 2 ) t We let y (t) be x c (t) plus a carrier. Thus y (t) = x c (t) +Kcos ω c t It can be shown that y (t) can be written y (t) = y 1 (t) cos ω c (t) +y 2 (t) sinω c t where y 1 (t) = A 2 cos ω 1 t + B 2 cos ω 2 t +K y 2 (t) = µ Aε + A 2 ¶ sinω 1 t − B 2 sinω 2 t Also y (t) = R(t) cos (ω c t + θ) where R(t) is the envelope and is therefore the output of an envelope detector. It follows that R(t) = q y 2 1 (t) +y 2 2 (t) For K large, R(t) = |y 1 (t)|, which is 1 2 m(t) +K, where K is a dc bias. Thus if the detector is ac coupled, K is removed and the output y (t) is m(t) scaled by 1 2 . Pr obl em 3.18 The required Þgure appears in Figure 3.6. Pr obl em 3.19 14 CHAPTER 3. BASIC MODULATION TECHNIQUES ω 1 ω Desired Signal 1 2 ω ω − ω Local Oscillator 2 IF ω ω − 1 2 2ω ω − ω Signal at Mixer Output 1 2 2ω ω − ω Image Signal 2 1 3 2 ω ω − 2 ω ω Image Signal at Mixer Output Figure 3.6: 3.1. PROBLEMS 15 Since high-side tuning is used, the local oscillator frequency is f LO = f i +f IF where fi , the carrier frequency of the input signal, varies between 5 and 25 MHz. The ratio is R = f IF + 25 f IF + 5 where f IF is the IF frequency expressed in MHz. We make the following table f IF ,MHz R 0.4 4.70 0.5 4.63 0.7 4.51 1.0 4.33 1.5 4.08 2.0 3.86 A plot of R as a function of f IF is the required plot. Pr obl em 3.20 For high-side tuning we have f LO = f i +f IF = 1120 + 455 = 1575 kHz f IMAGE = f i + 2f IF = 1120 + 910 = 2030 kHz For low-side tuning we have f LO = f i −f IF = 1120 −455 = 665 kHz f IMAGE = f i −2f IF = 1120 −910 = 210 kHz Pr obl em 3.21 For high-side tuning f LO = f i +f IF = 1120 + 2500 = 3620 kHz f IMAGE = f i + 2f IF = 1120 + 5000 = 6120 kHz For low-side tuning f LO = f i −f IF = 1120 −2500 = −1380 kHz f LO = 1380 kHz f IMAGE = f i −2f IF = 1120 −5000 = −3880 kHz f LO = 3880 kHz 16 CHAPTER 3. BASIC MODULATION TECHNIQUES 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 0 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 0 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1 0 1 Figure 3.7: In the preceding development for low-side tuning, recall that the spectra are symmetrical about f = 0. Pr obl em 3.22 By deÞnition x c (t) = A c cos [ω c t +k p m(t)] = A c cos [ω c t +k p u(t −t 0 )] The waveforms for the three values of k p are shown in Figure 3.7. The top pane is for k p = π, the middle pane is for k p = −π/2, and the bottom pane is for k p = π/4. Pr obl em 3.23 Let φ(t) = β cos ω m t where ω m = 2πf m . This gives x c2 (t) = A c Re n e jωct e jβ cosωmt o Expanding a Fourier series gives e jβ cosωmt = ∞ X n= −∞ C n e jnωmt 3.1. PROBLEMS 17 where C n = ω m 2π Z π/ω m −π/ω m e jβ cosω m t e −jnω m t dt With x = ω m t, the Fourier coefficients become C n = 1 2π Z π −π e jβ cosx e −jnx dx Since cos x = sin ¡ x + π 2 ¢ C n = 1 2π Z π −π e j[β si n(x+ π 2 )−nx] dx With x = x + π 2 , the preceding becomes C n = 1 2π Z 3π/2 −π/2 e j[β sin y−ny+ n π 2 ] dy This gives C n = e j nπ 2 ½ 1 2π Z π −π e j[β si n y−ny] dy ¾ where the limits have been adjusted by recognizing that the integrand is periodic with period 2π. Thus C n = e j nπ 2 J n (β) and x c2 (t) = A c Re ( e jωct ∞ X n= −∞ J n (β) e j(nωmt+ nπ 2 ) ) Taking the real part yields x c2 (t) = A c ∞ X n= −∞ J n (β) cos h (ω c +nω m ) t + nπ 2 i The amplitude spectrum is therefore the same as in the preceding problem. The phase spectrum is the same as in the preceding problem except that nπ 2 is added to each term. Pr obl em 3.24 Since sin(x) = cos(x − π 2 ) we can write x c3 (t) = A c sin(ω c t + β sinω m t) = A c cos ³ ω c t − π 2 + β sinω m t ´ which is x c3 (t) = A c Re n e j(ωct−π/2) e j si n ωmt o 18 CHAPTER 3. BASIC MODULATION TECHNIQUES Since e j sin ω m t = ∞ X n= −∞ J n (β)e jnω m t we have x c3 (t) = A c Re ( e j(ωct−π/2) ∞ X n= −∞ J n (β)e jnωmt ) Thus x c3 (t) = A c ∞ X n= −∞ J n (β) Re n e j(ω c t+ nω m t−π/2) o Thus x c3 (t) = A c ∞ X n= −∞ J n (β) cos h (ω c +nω m ) t − π 2 i Note that the amplitude spectrum of x c3 (t) is identical to the amplitude spectrum for both x c1 (t) and x c2 (t). The phase spectrum of x c3 (t) is formed from the spectrum of x c1 (t) by adding −π/2 to each term. For x c4 (t) we write x c4 (t) = A c sin(ω c t + β cos ω m t) = A c cos ³ ω c t − π 2 + β cos ω m t ´ Using the result of the preceding problem we write x c4 (t) = A c Re ( e j(ωct−π/2) ∞ X n= −∞ J n (β) e j(nωmt+ nπ 2 ) ) This gives x c4 (t) = A c ∞ X n= −∞ J n (β) Re n e j(ωct+ nωmt− π 2 + nπ 2 ) o Thus x c4 (t) = A c ∞ X n= −∞ J n (β) cos h (ω c +nω m ) t + π 2 (n −1) i Compared to x c1 (t), x c2 (t) and x c3 (t), we see that the only difference is in the phase spectrum. Pr obl em 3.25 From the problem statement x c (t) = A c cos [2π (40) t + 10 sin(2π (5) t)] 3.1. PROBLEMS 19 The spectrum is determined using (3.101). Since f c = 40 and f m = 5 (it is important to note that f c is an integer multiple of f m ) there is considerable overlap of the portion of the spectrum centered about f = −40 with the portion of the spectrum centered about f = 40 for large β (such as 10). The terms in the spectral plot, normalized with respect to A c , are given in Table 3.1 for f ≤ 0 (positive frequency terms and the term at dc). The terms resulting from the portion of the spectrum centered at f = 40 are denoted S 40 and the terms resulting from the portion of the spectrum centered at f = −40 are denoted S −40 . Given the percision of the Table of Bessel coefficients (Page 131 in the textbook), we have overlap for |f| ≤ 45. Where terms overlap, they must be summed. The sum is denoted S T in Table 3.1. In developing the Table 3.1 be sure to remember that J −n (β) = −J n (β) for odd n. The power must now be determined in order to Þnd A c . With the exception of the dc term, the power at each frequency is S 2 T A 2 c /2 (the power for negative frequencies is equal to the power at positive frequencies and so the positive frequency power can simply be doubled to give the total power with the dc term excepted). The power at dc is S 2 T A 2 c = (0.636) 2 A 2 c . Carring out these operations with the aid of Table 3.1 gives the total power 0.8036A 2 c = 40 which is A c = r 40 0.8036 = 7.0552 The waveform x c (t) is illustrated in the top pane of Figure 3.8. The spectrum, normalized with respect to A c , is illustrated in the bottom frame. The fact that x c (t) has a nonzero dc value is obvious. Pr obl em 3.26 We are given J 0 (3) = −0.2601 and J 1 (3) = −0.3391. Using J n+ 1 (β) = 2n β J n (β) −J n−1 (β) with β = 3 we have j n+ 1 (3) = 2 3 nJ n (3) −J n−1 (3) With n = 1, J 2 (3) = 2 3 J 1 (3) −J 0 (3) = 2 3 (0.3391) + 0.2601 = 0.4862 20 CHAPTER 3. BASIC MODULATION TECHNIQUES Table 3.1: Data for Problem 3.25 f S 40 S −40 S T 125 0.001 0 0.001 120 0.002 0 0.002 115 0.005 0 0.005 110 0.012 0 0.012 105 0.029 0 0.029 100 0.063 0 0.063 95 0.123 0 0.123 90 0.207 0 0.207 85 0.292 0 0.292 80 0.318 0 0.318 75 0.219 0 0.219 70 −0.014 0 −0.014 65 −0.234 0 −0.234 60 −0.220 0 −0.220 55 0.058 0 0.058 50 0.255 0 0.255 45 0.043 0.001 0.044 40 −0.246 0.002 −0.244 35 −0.043 0.005 −0.038 30 0.255 0.012 0.067 25 −0.058 0.029 −0.029 20 −0.220 0.063 −0.157 15 0.234 0.123 0.357 10 −0.014 0.207 −0.193 5 −0.217 0.292 0.075 0 0.318 0.318 0.636 3.1. PROBLEMS 21 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -1 -0.5 0 0.5 1 Time x c ( t ) -150 -100 -50 0 50 100 150 0 0.2 0.4 0.6 0.8 Frequency M a g n i t u d e Figure 3.8: With n = 2, J 3 (3) = 4 3 J 2 (3) −J 1 (3) = 4 3 (0.04862) + 0.3391 = 0.3091 Finally, with n = 3 we have J 4 (3) = 2J 3 (3) −J 2 (3) = 2 (0.3091) + 0.4862 = 0.1320 Pr obl em 3.27 The amplitude and phase spectra follow directly from the table of Fourier-Bessel coefficients. The single-sided magnitude and phase spectra are shown in Figure 3.9. The magnitude spectrum is plotted assuming A c = 1. 22 CHAPTER 3. BASIC MODULATION TECHNIQUES 600 700 800 900 1000 1100 1200 1300 1400 0 0.1 0.2 0.3 0.4 Frequency M a g n i t u d e 600 700 800 900 1000 1100 1200 1300 1400 -4 -3 -2 -1 0 Frequency P h a s e Figure 3.9: 3.1. PROBLEMS 23 Pr obl em 3.28 The modulated signal can be written x c (t) = Re h 10 + 3e j2π(20)t + 5e −j2π(20)t i e j2π(100)t We will concentrate on the term in brackets, which is the complex envelope described in Chapter 2. Denoting the complex envelope by e x c (t), we can write e x c (t) = [10 + 3 cos 2π (20t) + 5 cos 2π (20) t] +j [3 sin2π (20t) −5 sin2π (20) t] = [10 + 8 cos 2π (20t)] −j [2 sin2π (20) t] It follows from the deÞnition of x c (t) that e x c (t) = R(t) e jφ(t) Thus R 2 (t) = [10 + 8 cos 2π (20t)] 2 + 4 sin 2 2π (20) t = 134 + 160 cos 2π (20) t + 30 cos 2π (40) t This gives R(t) = p 134 + 160 cos 2π (20) t + 30 cos 2π (40) t Also φ(t) = tan −1 −2 sin2π (20) t 10 + 8 cos 2π (20) t Pr obl em 3.29 Since sinx = cos ¡ x − π 2 ¢ , we can write x c (t) = Re n e j200πt h 5e −j40πt + 10 + 3e j(40πt− π 2 ) io or x c (t) = Re © a(t) e j200πt ª where a(t) = (5 cos 40πt + 10 + 3 sin40πt) +j (5 sin40πt −3 cos 40πt) 24 CHAPTER 3. BASIC MODULATION TECHNIQUES Thus if we write x c (t) as x c (t) = R(t) cos (200πt + φ(t)) the envelope, R(t), is given by R(t) = h (5 cos 40πt + 10 + 3 sin40πt) 2 + (5 sin40πt −3 cos 40πt) 2 i1 2 and the phase deviation, φ(t), is given by φ(t) = tan −1 5 sin40πt −3 cos 40πt 10 + 5 cos 40πt + 3 sin40πt The envelope, R(t), can be simpliÞed and expressed in several different forms. Pr obl em 3.30 (a) Since the carrier frequency is 1000 Hertz, the general form of x c (t) is x c (t) = A c cos [2π (1000) t + φ(t)] The phase deviation, φ(t), is therefore given by φ(t) = 20t 2 rad The frequency deviation is dφ dt = 40t rad/sec or 1 2π dφ dt = 20 π t Hertz (b) The phase deviation is φ(t) = 2π (500) t 2 −2π (1000) t rad and the frequency deviation is dφ dt = 4π (500) t −2π (1000) rad/sec = 2000π (t −1) rad/sec or 1 2π dφ dt = 1000 (t −1) Hertz 3.1. PROBLEMS 25 (c) The phase deviation is φ(t) = 2π (100) t = 200πt rad and the frequency deviation is dφ dt = 200π rad/sec or 1 2π dφ dt = 100 Hertz which should be obvious from the expression for x c (t). (d) The phase deviation is φ(t) = 200πt + 10 √ t rad and the phase deviation is dφ dt = 200π + 1 2 (10) t − 1 2 = 200π + 5 √ t rad/sec or 1 2π dφ dt = 100 + 5 2π √ t Hertz Pr obl em 3.31 (a) The phase deviation is φ(t) = 2π(30) Z t 0 (8)dt = 480πt, t ≤ 4 The maximum phase deviation is φ(4) = 480π (4) = 1920π. The required plot is simply φ(t) =    0, t < 0 480πt, 0 ≤ t < 4 1920π, t ≥ 4 (b) The frequency deviation, in Hz, is 1 2π dφ dt = 30m(t) = 0, t < 0 240, 0 ≤ t < 4 0 t ≥ 4 26 CHAPTER 3. BASIC MODULATION TECHNIQUES The required sketch follows simply. (c) The peak frequency deviation is 8f d = 8 (30) = 240 Hertz (d) The peak phase deviation is 2π (30) Z 4 0 (8) dt = 2π (30) (8) (4) = 1920π rad The modulator output power is P = 1 2 A 2 c = 1 2 (100) 2 = 5000 Watts Pr obl em 3.32 (a) The message signal is m(t) =        0, t < 4 t −4, 4 ≤ t ≤ 6 8 −t 6 ≤ t ≤ 8 0, t > 8 The phase deviation is φ(t) = 2πf d Z t 4 (t −4) dt = 60π (−4) (t −4) + 30π ¡ t 2 −16 ¢ = 30π ¡ t 2 −8t + 16 ¢ , 4 ≤ t ≤ 6 φ(t) = 2πf d Z t 6 (8 −t) dt + φ(6) = 120π + 60π (8) (t −6) −30π ¡ t 2 −36 ¢ = 120π −30π ¡ t 2 −16t + 60 ¢ , 6 ≤ t ≤ 8 Also φ(t) = 240π, t > 8 φ(t) = 0, t < 4 The sketches follow immediately from the equations. (b) The frequency deviation in Hertz is 30m(t). (c) The peak phase deviation = 240π rad. (d) The peak frequency deviation = 120 Hertz. 3.1. PROBLEMS 27 (e) The carrier power is, assuming a sufficiently high carrier frequency, P = 1 2 A 2 c = 1 2 (100) 2 = 5000 Watts Pr obl em 3.33 The frequency deviation in Hertz is the plot shown in Fig. 3.76 with the ordinate values multiplied by 25.The phase deviation in radians is given φ(t) = 2πf d Z t m(α) dα = 50π Z t m(α) dα For 0 ≤ t ≤ 1, we have φ(t) = 50π Z t 0 2αdα = 50πt 2 For 1 ≤ t ≤ 2 φ(t) = φ(1) + 50π Z t 1 (5 −α) dα = 50π + 250π (t −1) −25π ¡ t 2 −1 ¢ = −175π + 250πt −25πt 2 For 2 ≤ t ≤ 3 φ(t) = φ(2) + 50π Z t 2 3dα = 225π + 150π (t −2) For 3 ≤ t ≤ 4 φ(t) = φ(3) + 50π Z t 3 2dα = 375π + 100π(t −3) Finally, for t > 4 we recognize that φ(t) = φ(4) = 475π. The required Þgure results by plotting these curves. Pr obl em 3.34 The frequency deviation in Hertz is the plot shown in Fig. 3.77 with the ordinate values multiplied by 10. The phase deviation is given by φ(t) = 2πf d Z t m(α) dα = 20π Z t m(α) dα For 0 ≤ t ≤ 1, we have φ(t) = 20π Z t 0 αdα = 10πt 2 28 CHAPTER 3. BASIC MODULATION TECHNIQUES For 1 ≤ t ≤ 2 φ(t) = φ(1) + 20π Z t 1 (α −2) dα = 10π + 10π ¡ t 2 −1 ¢ −40π(t −1) = 10π ¡ t 2 −4t + 4 ¢ = 10π (t −2) 2 For 2 ≤ t ≤ 4 φ(t) = φ(2) + 20π Z t 2 (6 −2α)dα = 0 + 20π (6) (t −2) −20π ¡ t 2 −4 ¢ = −20π(t 2 −6t + 8) Finally, for t > 4 we recognize that φ(t) = φ(4) = 0. The required Þgure follows by plotting these expressions. Pr obl em 3.35 The frequency deviation in Hertz is the plot shown in Fig. 3.78 with the ordinate values multiplied by 5. The phase deviation in radians is given by φ(t) = 2πf d Z t m(α) dα = 10π Z t m(α) dα For 0 ≤ t ≤ 1, we have φ(t) = 10π Z t 0 (−2α)dα = −10πt 2 For 1 ≤ t ≤ 2 φ(t) = φ(1) + 10π Z t 1 2dα = −10π + 20π (t −1) = 10π (2t −3) For 2 ≤ t ≤ 2.5 φ(t) = φ(2) + 10π Z t 2 (10 −4α)dα = 10π + 10π (10) (t −2) −10π(2) ¡ t 2 −4 ¢ = 10π(−2t 2 + 10t −11) For 2.5 ≤ t ≤ 3 φ(t) = φ(2.5) −10π Z t 2.5 2dα = 15π −20π(t −2.5) = 10π(−2t + 5.5) 3.1. PROBLEMS 29 For 3 ≤ t ≤ 4 φ(t) = φ(3) + 10π Z t 3 (2α −8) dα = 5π + 10π(t 2 −9) −10π(8)(t −3) = 10π(t 2 −8t + 15.5) Finally, for t > 4 we recognize that φ(t) = φ(4) = −5π. The required Þgure follows by plotting these expressions. Pr obl em 3.36 (a) The peak deviation is (12.5)(4) = 50 and f m = 10. Thus, the modulation index is 50 10 = 5. (b) The magnitude spectrum is a Fourier-Bessel spectrum with β = 5. The n = 0 term falls at 1000 Hz and the spacing between components is 10 Hz. The sketch is that of Figure 3.24 in the text. (c) Since β is not ¿ 1, this is not narrowband FM. The bandwidth exceeds 2f m . (d) For phase modulation, k p (4) = 5 or k p = 1.25. Pr obl em 3.37 The results are given in the following table: Par t f d D = 5f d /W B = 2 (D + 1) W a 20 0.004 50.2 kHz b 200 0.04 52 kHz c 2000 0.4 70 kHz d 20000 4 250 kHz Pr obl em 3.38 From x c (t) = A c ∞ X n= −∞ J n (β) cos (ω c + ω m ) t we obtain x 2 c (t) ® = 1 2 A 2 c ∞ X n= −∞ J 2 n (β) We also know that (assuming that x c (t) does not have a signiÞcant dc component - see Problem 3.24) x 2 c (t) ® = A 2 c cos 2 [ω c t + φ(t)] ® 30 CHAPTER 3. BASIC MODULATION TECHNIQUES which, assuming that ω c À 1 so that x c (t) has no dc component, is x 2 c (t) ® = 1 2 A 2 c This gives 1 2 A 2 c = 1 2 A 2 c ∞ X n= −∞ J 2 n (β) from which ∞ X n= −∞ J 2 n (β) = 1 Pr obl em 3.39 Since J n (β) = 1 2π Z π −π e −j(nx−β si n x) dx = 1 2π Z π −π e j(β si n x−nx) dx we can write J n (β) = 1 2π Z π −π cos (β sinx −nx) dx +j 1 2π Z π −π sin(β sinx −nx) dx The imaginary part of J n (β) is zero, since the integrand is an odd function of x and the limits (−π, π) are even. Thus J n (β) = 1 2π Z π −π cos (β sinx −nx) dx Since the integrand is even J n (β) = 1 π Z π 0 cos (β sinx −nx) dx which is the Þrst required result. With the change of variables λ = π −x, we have J n (β) = 1 π Z π 0 cos [β sin(π −λ) −n(π −λ)] (−1) dλ = 1 π Z π 0 cos [β sin(π −λ) −nπ +nλ] dλ Since sin(π −λ) = sinλ, we can write J n (β) = 1 π Z π 0 cos [β sinλ +nλ −nπ] dλ 3.1. PROBLEMS 31 Using the identity cos (u −ν) = cos ucos ν + sinusinν with u = β sinλ +nλ and ν = nπ yields J n (β) = 1 π Z π 0 cos [β sinλ +nλ] cos (nπ) dλ + 1 π Z π 0 sin[β sinλ +nλ] sin(nπ) dλ Since sin(nπ) = 0 for all n, the second integral in the preceding expression is zero. Also cos (nπ) = (−1) n Thus J n (β) = (−1) n 1 π Z π 0 cos [β sinλ +nλ] dλ However J −n (β) = 1 π Z π 0 cos [β sinλ +nλ] dλ Thus J n (β) = (−1) n J −n (β) or equivalently J −n (β) = (−1) n J n (β) Pr obl em 3.40 (a) Peak frequency deviation = 80 Hz (b) φ(t) = 8 sin(20πt) (c) β = 8 (d) P i = 50 Watts, P 0 = 16.76 Watts (e) The spectrum of the input signal is a Fourier_Bessel spectrum with β = 8. The n = 0 term is at the carrier frequency of 500 Hz and the spacing between components is 10 Hz. The output spectrum consistes of the n = 0 term and three terms each side of the n = 0 term. Thus the output spectrum has terms at 470, 480, 490, 500, 510, 520 and 530 Hz. 32 CHAPTER 3. BASIC MODULATION TECHNIQUES Figure 3.10: 3.1. PROBLEMS 33 Pr obl em 3.41 The required spectra are given in Figure 3.10. The modulation indices are, from top to bottom, β = 0.5, β = 1, β = 2, β = 5, and β = 10. Pr obl em 3.42 We wish to Þnd k such that P r = J 2 0 (10) + 2 k X n= 1 J 2 0 (10) ≥ 0.80 This gives k = 9, yielding a power ratio of P r = 0.8747. The bandwidth is therefore B = 2kf m = 2 (9) (150) = 2700 Hz For P r ≥ 0.9, we have k = 10 for a power ratio of 0.9603. This gives B = 2kf m = 2 (10) (150) = 3000 Hz Pr obl em 3.43 From the given data, we have f c 1 = 110 kHz f d 1 = 0.05 f d 2 = n(0.05) = 20 This gives n = 20 0.05 = 400 and f c 1 = n(100) kHz = 44 MHz The two permissible local oscillator frequencies are f ` 0.1 = 100 −44 = 56 MHz f ` 0.2 = 100 + 44 = 144 MHz The center frequency of the bandpass Þlter must be f c = 100 MHz and the bandwidth is B = 2 (D + 1) W = 2 (20 + 1) (10) ¡ 10 3 ¢ or B = 420 kHz 34 CHAPTER 3. BASIC MODULATION TECHNIQUES Pr obl em 3.44 For the circuit shown H (f) = E(f) X (f) = R R+j2πfL + 1 j2πfC or H (f) = 1 1 +j ³ 2πfτ L − 1 2πfτ C ´ where τ L = L R = 10 −3 10 3 = 10 −6 , τ C = RC = ¡ 10 3 ¢ ¡ 10 −9 ¢ = 10 −6 A plot of the amplitude response shows that the linear region extends from approximately 54 kHz to118 kHz. Thus an appropriate carrier frequency is f c = 118 + 54 2 = 86 kHz The slope of the operating characteristic at the operating point is measured from the amplitude response. The result is K D ∼ = 8 ¡ 10 −6 ¢ Pr obl em 3.45 We can solve this problem by determining the peak of the amplitude response characteristic. This peak falls at f p = 1 2π √ LC It is clear that f p > 100 MHz. Let f p = 150 MHz and let C = 0.001 ¡ 10 −12 ¢ . This gives L = 1 (2π) 2 f 2 p C = 1.126 ¡ 10 −3 ¢ We Þnd the value of R by trial and error using plots of the amplitude response. An appropriate value for R is found to be 1 MΩ. With these values, the discriminator constant is approximately K D ≈ 8.5 ¡ 10 −9 ¢ Pr obl em 3.46 3.1. PROBLEMS 35 For A i = A c we can write, from (3.176), x r (t) = A c [cos ω c t + cos (ω c + ω i ) t] which is x r (t) = A c [(1 + cos ω i t) cos ω c t −sinω i t sinω c t] This yields x r (t) = R(t) cos [ω c t + ψ(t)] where ψ(t) = tan −1 · sinω i t 1 + cos ω i t ¸ = tan −1 · tan ω i t 2 ¸ = ω i t 2 This gives y D (t) = 1 2π d dt µ 2πf i t 2 ¶ = 1 2 f i For A i = −A c , we get ψ(t) = tan −1 · −sinω i t 1 −cos ω i t ¸ = −tan −1 · sinω i t 1 + cos ω i t ¸ Since sinx 1 −cos x = cot x 2 = tan ³ π 2 − x 2 ´ we have ψ(t) = tan −1 h −tan ³ π 2 − x 2 í = x 2 − π 2 Thus y D (t) = 1 2π d dt µ 2πf i t 2 − π 2 ¶ = 1 2 f i Finally, for A i À A c we see from the phasor diagram that ψ(t) ≈ θ (t) = ω i t and y D (t) = K D 2π d dt (2πf i t) = f i Pr obl em 3.47 From Example 3.5, m(t) = Au(t). Thus φ(t) = k f Z t Au(α) dα = Ak f t, t ≥ 0 36 CHAPTER 3. BASIC MODULATION TECHNIQUES ( ) t θ ( ) t φ ( ) t ψ f T Ak K 0 t Figure 3.11: Also Θ(s) = AK T k f s 2 (s +K T ) Taking the inverse transform gives θ (t) = Ak f µ t + 1 K T e −K T t − 1 K T ¶ u(t) The phase error is ψ (t) = φ(t) −θ (t). This gives ψ (t) = Ak f K T ¡ 1 −e −K T t ¢ u(t) The maximum phase error occurs as t →∞ and is clearly Ak f /K T . Thus we require 0.2 = Ak f K T and K T = 5Ak f The VCO constant is contained in K T . The required sketch follows. Pr obl em 3.48 For m(t) = Acos ω m t and φ(t) = Ak f Z t cos ω m αdα = Ak f ω m sinω m t 3.1. PROBLEMS 37 and Φ(s) = Ak f s 2 + ω 2 m The VCO output phase is Θ(s) = Φ(s) K T s +K T = Ak f K T (s +K T ) (s 2 + ω 2 m ) Using partial fraction expansion,Θ(s) can be expressed as Θ(s) = Ak f K T K T 2 + ω 2 m · 1 s +K T − s s 2 + ω 2 m + K T s 2 + ω 2 m ¸ This gives, for t ≥ 0 θ (t) = Ak f K T K T 2 + ω 2 m · e −K T t −cos ω m t + K T ω m sinω m t ¸ The Þrst term is the transient response. For large K T , only the third term is signiÞcant. Thus, e ν (t) = 1 K ν dθ dt = Ak f K 2 T K ν ¡ K 2 T + ω 2 m ¢ cos ω m t Also, since K T is large, K 2 T + ω 2 m ≈ K 2 T . This gives e ν (t) ≈ Ak f K ν cos ω m t and we see that e ν (t) is proportional to m(t). If k f = K ν , we have e ν (t) ≈ m(t) Pr obl em 3.49 The Costas PLL is shown in Figure 3.57. The output of the top multiplier is m(t) cos ω c t [2 cos (ω c t + θ)] = m(t) cos θ +m(t) cos (2ω c t + θ) which, after lowpass Þltering, is m(t) cos θ. The quadrature multiplier output is m(t) cos ω c t [2 sin(ω c t + θ)] = m(t) sinθ +m(t) sin(2ω c t + θ) which, after lowpass Þltering, is m(t) sinθ. The multiplication of the lowpass Þlter outputs is m(t) cos θm(t) sinθ = m 2 (t) sin2θ 38 CHAPTER 3. BASIC MODULATION TECHNIQUES A ψ d dt ψ / Figure 3.12: ( ) 0 e t ( ) v e t Phase Detector Loop Filter and Amplifier VCO Bandpass Filter Figure 3.13: as indicated. Note that with the assumed input m(t) cos ω c t and VCO output 2 cos (ω c t + θ), the phase error is θ. Thus the VCO is deÞned by dθ dt = dψ dt = −K ν e ν (t) This is shown below. Since the dψ dt intersection is on a portion of the curve with negative slope, the point A at the origin is a stable operating point. Thus the loop locks with zero phase error and zero frequency error. Pr obl em 3.50 With x(t) = A2πf 0 t, we desire e 0 (t) = Acos 2π ¡ 7 3 ¢ f 0 t. Assume that the VCO output is a pulse train with frequency 1 3 f o . The pulse should be narrow so that the seventh harmonic is relatively large. The spectrum of the VCO output consists of components separated by 3.1. PROBLEMS 39 0 7 3 f 0 1 3 f Bandpass filter passband This component (the fundamental) tracks the input signal f 0 Figure 3.14: 1 3 f 0 with an envelope of sinc(τf), where τ is the pulse width. The center frequency of the bandpass Þlter is 7 3 f 0 and the bandwidth is on the order of 1 3 f 0 as shown. Pr obl em 3.51 The phase plane is deÞned by ψ = ∆ω −K t sinψ (t) at ψ = 0, ψ = ψ ss , the steady-state phase error. Thus ψ ss = sin −1 µ ∆ω K t ¶ = sin −1 µ ∆ω 2π (100) ¶ For ∆ω = 2π (30) ψ ss = sin −1 µ 30 100 ¶ = 17.46 degrees For ∆ω = 2π (50) ψ ss = sin −1 µ 50 100 ¶ = 30 degrees For ∆ω = 2π (80) ψ ss = sin −1 µ 80 100 ¶ = 53.13 defrees 40 CHAPTER 3. BASIC MODULATION TECHNIQUES For ∆ω = −2π (80) ψ ss = sin −1 µ −80 100 ¶ = −53.13 degrees For ∆ω = 2π (120), there is no stable operating point and the frequency error and the phase error oscillate (PLL slips cycles continually). Pr obl em 3.52 From (3.228) Θ(s) Φ(s) = K t F (s) s +K t F (s) = K t ³ s+ a s+ ε ´ s +K t ³ s+ a s+ ε ´ which is Θ(s) Φ(s) = K t (s +a) s (s + ε) +K t (s +a) = K t (s +a) s 2 + (K t + ε) s +K t a Therefore s 2 + 2ζω n s + ω 2 n = s 2 + (K t + ε) s +K t a This gives ω n = p K t a and ζ = K t + ε 2 √ K t a Pr obl em 3.53 Since ω n = 2π (100) we have p K t a = 2π (100) or K t a = 4π 2 ¡ 10 4 ¢ Since ς = 1 √ 2 = K t + ε 2 √ K t a = 1.1K t 2 (2π) (100) we have K t = √ 2 (2π) (100) 1.1 = 807.8 Thus ε = 0.1K t = 80.78 3.1. PROBLEMS 41 ( ) ⋅ z dt ( ) PAM x t x t PWM ( ) x t 2 ( ) Pulse train x t 1 ( ) Figure 3.15: and a = 4π 2 ¡ 10 4 ¢ 807.8 = 488.7 Pr obl em 3.54 From (3.247), we see that the phase deviation is reduced by 1+ 1 2π K D K ν . The VCO constant is 25 Hz /volt. Thus K ν = 2π (25) rad/s/volt we may then write D 1 D 2 = 5 0.4 = 12.5 = 1 + 1 2π K D (2π) (25) which gives 1 + 25K D = 12.5 Thus K D = 0.46 Pr obl em 3.55 A system converting PWM to PAM can be realized as illustrated in Figure 3.15. The operation should be clear from the waveforms shown in Figure 3.16. The integrator can be realized by using a capacitor since, for a capacitor, ν (t) = 1 C Z i (t) dt Multiplication by the pulse train can be realized by sampling the capacitor voltage. Thus, the simple circuit is as illustrated in Figure 3.17. 42 CHAPTER 3. BASIC MODULATION TECHNIQUES T 2T 3T 4T 0 x t PWM ( ) t T 2T 3T 4T 0 x t 1 ( ) t T 2T 3T 4T 0 x t 2 ( ) t T 2T 3T 4T 0 x t PAM ( ) t Figure 3.16: 3.1. PROBLEMS 43 switch closes at t nT = to restart integration sampling switch is closed for nT t nT τ − < < ( ) PAM x t ( ) ( ) PWM x t i t = C Figure 3.17: Pr obl em 3.56 Let A be the peak-to-peak value of the data signal. The peak error is 0.5% and the peak- to-peak error is 0.01 A. The required number of quantizating levels is A 0.01A = 100 ≤ 2 n = q so we choose q = 128 and n = 7. The bandwidth is B = 2Wk log 2 q = 2Wk(7) The value of k is estimated by assuming that the speech is sampled at the Nyquist rate. Then the sampling frequency is f s = 2W = 8 kHz. Each sample is encoded into n = 7 pulses. Let each pulse be τ with corresponding bandwidth 1 τ . For our case τ = 1 nf s = 1 2Wn Thus the bandwidth is 1 τ = 2Wn = 2W log 2 q and so k = 1. For k = 1 B = 2 (8, 000) (7) = 112 kHz Pr obl em 3.57 44 CHAPTER 3. BASIC MODULATION TECHNIQUES Let the maximum error be λA where A is the peak-to-peak message signal. The peak-topeak error is 2λA and the minimum number of quantizing levels q min = A 2λA = 1 2λ The wordlength is given by n = · log 2 1 2λ ¸ where [x] is the smallest integer greater than or equal to x. With k = 1 (as in the previous problem), this gives a normalized bandwidth of B N = B 2W = · log 2 1 2λ ¸ We make the following table λ log ¡ 1 2λ ¢ B N 0.001 8.966 9 0.005 6.644 7 0.01 5.644 6 0.05 3.322 4 0.1 2.322 3 0.2 1.322 2 0.4 0.322 1 0.5 0 0 A plot of B N as a function of λ gives the required plot. Pr obl em 3.58 The message signal is m(t) = 4 sin2π(10)t + 5 sin2π(20)t The derivative of the message signal is dm(t) dt = 80π cos 2π (10) t + 200π cos 2π (20t) The maximum value of dm(t) /dt is obviously 280π and the maximum occurs at t = 0. Thus δ 0 T s ≥ 280π or f s ≥ 280π δ 0 = 280π 0.05π = 5600 3.1. PROBLEMS 45 Output 1 12 9 10 4 3 2 11 5 8 6 7 x W 2 (BW = ) x W 1 (BW = ) x W 3 2 (BW = ) x W 5 4 (BW = ) x W 4 4 (BW = ) Figure 3.18: Thus, the minimum sampling frequency is 5600 Hz. Pr obl em 3.59 One possible commutator conÞguration is illustrated in Figure 3.18. The signal at the point labeled “output” is the baseband signal. The minimum commutator speed is 2W revolutions per second. For simplicity the commutator is laid out in a straight line. Thus, the illustration should be viewed as it would appear wrapped around a cylinder. After taking the sample at point 12 the commutator moves to point 1. On each revolution, the commutator collects 4 samples of x 4 and x 5 , 2 samples of x 3 , and one sample of x 1 and x 2 . The minimum transmission bandwidth is B = X i W i = W +W + 2W + 4W + 4W = 12W Pr obl em 3.60 The single-sided spectrum for x(t) is shown in Figure 3.19. From the deÞnition of y (t) we have Y (s) = a 1 X (f) +a 2 X (f) ∗ X(f) 46 CHAPTER 3. BASIC MODULATION TECHNIQUES 2 f 1 f f W W Figure 3.19: The spectrum for Y (f) is given in Figure 3.20. Demodulation can be a problem since it may be difficult to Þlter the desired signals from the harmonic and intermodulation distortion caused by the nonlinearity. As more signals are included in x(t), the problem becomes more difficult. The difficulty with harmonically related carriers is that portions of the spectrum of Y (f) are sure to overlap. For example, assume that f 2 = 2f 1 . For this case, the harmonic distortion arising from the spectrum centered about f 1 falls exactly on top of the spectrum centered about f 2 . 3.1. PROBLEMS 47 2 2 f 1 2 f f + 2 2 f 2 1 f f − 2 2a W 2 a W 1 a ( ) Y f 2 f 1 f f Figure 3.20: Chapter 4 Probability and Random Variables 4.1 Problem Solutions Problem 4.1 S = sample space is the collection of the 25 parts. Let A i = event that the pointer stops on the ith part, i = 1, 2, ..., 25. These events are exhaustive and mutually exclusive. Thus P (A i ) = 1/25 (a) P(even number) = P (2 or 4 or · · · or 24) = 12/25; (b) P (A 25 ) = 1/25; (c) P (4 or 5 or 9) = 3/25; (d) P(number > 10) = P (11, 12, · · · , or 25) = 15/25 = 3/5. Problem 4.2 (a) Use a tree diagram similar to Fig. 4.2 to show that P (3K, 2A) = 4 52 ½ 4 51 · 3 3 50 3 49 2 48 ¸ + 3 51 · 3 4 50 3 49 2 48 ¸ + 3 51 · 4 50 3 49 2 48 ¸ + 4 51 · 3 3 50 3 49 2 48 ¸¾ = 9.2345 × 10 −6 (b) Use the tree diagram of Fig. 4.2 except that we now look at outcomes that give 4 of a kind. Since the tree diagram is for a particular denomination, we multiply by 13 and get P (4 of a kind) = 13 ½ 4 52 · 3 51 µ 3 2 50 1 49 48 48 ¶ + 48 51 3 50 2 49 1 48 ¸ + 48 52 4 51 3 50 2 49 1 48 ¾ = 0.0002401 1 2 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES (c) The first card can be anything. Given a particular suit on the first card, the probability of suit match on the second card is 12/51; on the third card it is 11/50; on the fourth card it is 10/49; on the fifth and last card it is 9/48. Therefore, P (all same suit) = 1 12 51 11 50 10 49 9 48 = 0.001981 (d) For a royal flush P (A, K, Q, J, 10 of same suit) = 4 52 1 51 1 50 1 49 1 48 = 1.283 × 10 −8 (e) The desired probability is P (Q|A, K, J, 10 not all of same suit) = 4 48 = 0.833 Problem 4.3 P (A, B) = P (A) P (B) P (A, C) = P (A) P (C) P (B, C) = P (B) P (C) P (A, B, C) = P (A) P (B) P (C) Problem 4.4 A and B being mutually exclusive implies that P (A|B) = P (B|A) = 0. A and B being statistically independent implies that P (A|B) = P (A) and P (B|A) = P (B). The only way that both of these conditions can be true is for P (A) = P (B) = 0. Problem 4.5 (a) The result is P (AB) = 1 −P (1 or more links broken) = 1 − ¡ 1 −q 2 ¢ 2 (1 −q) (b) If link 4 is removed, the result is P (AB| link 4 removed) = 1 − ¡ 1 −q 2 ¢ (1 −q) 4.1. PROBLEM SOLUTIONS 3 (c) If link 2 is removed, the result is P (AB| link 2 removed) = 1 − ¡ 1 −q 2 ¢ 2 (d) Removal of link 4 is more severe than removal of link 2. Problem 4.6 Using Bayes’ rule P (A|B) = P (B|A) P (A) P (B) where, by total probability P (B) = P (B|A) P (A) +P ¡ B|A ¢ P ¡ A ¢ = P (B|A) P (A) + £ 1 −P ¡ B|A ¢¤ P ¡ A ¢ = (0.9) (0.4) + (0.4) (0.6) = 0.6 Therefore P (A|B) = (0.9) (0.4) 0.6 = 0.6 Similarly, P ¡ A|B ¢ = P ¡ B|A ¢ P (A) P ¡ B ¢ with P ¡ B ¢ = P ¡ B|A ¢ P (A) +P ¡ B|A ¢ P ¡ A ¢ = (0.1) (0.4) + (0.6) (0.6) = 0.4 Thus P ¡ A|B ¢ = (0.1) (0.4) 0.4 = 0.1 Problem 4.7 (a) P (A 2 ) = 0.3; P (A 2 , B 1 ) = 0.05; P (A 1 , B 2 ) = 0.05; P (A 3 , B 3 ) = 0.05; P (B 1 ) = 0.15; P (B 2 ) = 0.25; P (B 3 ) = 0.6. (b) P(A 3 |B 3 ) = 0.083; P (B 2 |A 1 ) = 0.091; P (B 3 |A 2 ) = 0.333. Problem 4.8 See the tables below for the results. 4 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES (a) P spots up X 1 P (X 1 = x i ) 2 4 1/36 3 9 2/36 4 16 3/36 5 25 4/36 6 36 5/36 7 49 6/36 8 64 5/36 9 81 4/36 10 100 3/36 11 121 2/36 12 144 1/36 (b) P spots up X 2 P (X 2 = x i ) 2 1 3 1 4 1 15/36 5 1 6 1 7 0 8 0 9 0 10 0 21/36 11 0 12 0 Problem 4.9 The cdf is zero for x < 0, jumps by 1/8 at x = 0, jumps another 3/8 at x = 1, jumps another 3/8 at x = 2, and jumps another 1/8 at x = 3. The pdf consists of an impulse of weight 1/8 at x = 0, impulses of weights 3/8 at x = 1 and 2, and an impulse of weight 1/8 at x = 3. Problem 4.10 (a) As x → ∞, the cdf approaches 1. Therefore B = 1. Assuming continuity of the cdf, F x (10) = 1 also, which says A× 10 3 = 1 or A = 10 −3 . (b) The pdf is given by f X (x) = dF X (x) dx = 3 × 10 −3 x 2 u(x) u(10 −x) The graph of this pdf is 0 for t < 0, the quadratic 3 × 10 −3 x 2 for 0 ≤ x ≤ 10, and 0 for t > 10. (c) The desired probability is P (X > 7) = 1 −F X (7) = 1 − ¡ 10 −3 ¢ (7) 3 = 1 −0.343 = 0.657 (d) The desired probability is P (3 ≤ X < 7) = F X (7) −F X (3) = 0.316 Problem 4.11 (a) A = α; (b) B = β; (c) C = γ/2; (d) D = τ/π 4.1. PROBLEM SOLUTIONS 5 Problem 4.12 (a) Factor the joint pdf as f XY (x, y) = √ Aexp[−|x|] √ Aexp[−|y|] With proper choice of A, the two separate factors are marginal pdfs. Thus X and Y are statistically independent. (b) Note that the pdf is the volume between a plane which intersects the x and y coordinate axes one unit out and the f XY coordinate axis at C. First find C from the integral Z ∞ −∞ Z ∞ −∞ f XY (x, y) dxdy = 1 or Z 1 0 Z 1−y 0 C (1 −x −y) dxdy = 1 This gives C = 6. Next find f X (x) and f Y (y) as f X (x) = Z 1−x 0 6 (1 −x −y) dy = ( 3 (1 −x) 2 , 0 ≤ x ≤ 1 0, otherwise and f Y (y) = Z 1−y 0 6 (1 −x −y) dx = ( 3 (1 −y) 2 , 0 ≤ y ≤ 1 0, otherwise Since the joint pdf is not equal to the product of the two marginal pdfs, X and Y are not statistically independent. Problem 4.13 (a) R R f XY (x, y) dxdy = 1 gives C = 1/255; (b) f XY (0.1, 1.5) = 1+1.5 255 = 0.0098 (c) f XY (x, 3) = n (1+3x)/255, 0≤x≤6 0, otherwise (d) By integration, f Y (y) = 6(1+3y) 255 , 0 ≤ y ≤ 5, so the desired conditional pdf is f X|Y (x|y) = f XY (x, y) f Y (y) = 1 +xy 6 (1 + 3y) , 0 ≤ x ≤ 6, 0 ≤ y ≤ 5 Substitute y = 3 to get the asked for result. Problem 4.14 (a) To find A, evaluate the double integral Z ∞ −∞ Z ∞ −∞ f XY (x, y) dxdy = 1 = Z ∞ 0 Z ∞ 0 Axye −(x+y) dxdy = A Z ∞ 0 xe −x dx Z ∞ 0 ye −y dy = A 6 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Thus A = 1. (b) Clearly, f X (x) = xe −x u(x) and f Y (y) = ye −y u(y) (c) Yes, the joint pdf factors into the product of the marginal pdfs. Problem 4.15 (a) Use normalization of the pdf to unity: Z ∞ −∞ αx −2 u(x −α) dx = Z ∞ α αx −2 dx = 2x −1 Z ∞ α = 1 Hence, this is the pdf for any value of α. (b) The cdf is F X (x) = Z x −∞ αz −2 u(z −α) dz = ½ 0, x < α (1 −α/x) , x > α (c) The desired probability is P (X ≥ 10) = 1 −P (X < 10) = 1 −F X (10) = ½ 1, α > 10 α/10, α < 10 Problem 4.16 The result is f Y (y) =    exp(−y/2σ 2 ) √ 2πσ 2 y , y ≥ 0 0, y < 0 Problem 4.17 First note that P (Y = 0) = P (X ≤ 0) = 1/2 For y > 0, transformation of variables gives f Y (y) = f X (x) ¯ ¯ ¯ ¯ dg −1 (y) dy ¯ ¯ ¯ ¯ x=g −1 (y) Since y = g (x) = ax, we have g −1 (y) = y/a. Therefore f Y (y) = exp ³ − y 2 2a 2 σ 2 ´ √ 2πa 2 σ 2 , y > 0 4.1. PROBLEM SOLUTIONS 7 For y = 0, we need to add 0.5δ (y) to reflect the fact that Y takes on the value 0 with probability 0.5. Hence, for all y, the result is f Y (y) = 1 2 δ (y) + exp ³ − y 2 2a 2 σ 2 ´ √ 2πa 2 σ 2 u(y) where u(y) is the unit step function. Problem 4.18 (a) The normalization of the pdf to unity provides the relationship A Z ∞ −∞ e −b|x| dx = 2A Z ∞ 0 e −bx dx = 2A/b = 1 where the second integral follows because of evenness of the integrand. Thus A = b/2. (b) E[X] = 0 because the pdf is an even function of x. (c) Since the expectation of X is zero, σ 2 X = E © X 2 ª = Z ∞ −∞ b 2 x 2 e −b|x| dx = b Z ∞ 0 x 2 e −bx dx = 2/b 2 where evenness of the integrand has again been used, and the last integral can be found in an integral table. Problem 4.19 Use the fact that E n [X −E(X)] 2 o ≥ 0 (0 only if X = 0 with probability one). Expanding, we have E £ X 2 ¤ −{E[X]} 2 > 0 if X 6= 0. Problem 4.20 For the uniform distribution, see Example 4.21. The mean for the Gaussian pdf follows by a change of variables and the normalization property of the Gaussian pdf. Its variance follows by defining the random variable Y = X − m, which has zero mean, and using a tabulated definite integral. The mean and variance of the Rayleigh random variable can be obtained by using tabulated definite integrals. The mean of a Laplacian random variable is zero by the evenness of the pdf. The variance was obtained in Problem 4.18. The mean and variance of the single-sided exponential pdf are easily found in terms of tabulated integrals. The mean of the hyperbolic pdf is zero by virtue of the evenness of the pdf, and its variance is worked out below. The mean and variance of the Nakagami-m pdf can be put in terms of tabulated definite integrals. The mean and variance of the binomial random variable is worked out beginning with (4.162). The mean and variance for the Poisson and geometric distributions follow similarly. For the variance of the hyperbolic pdf, consider E £ X 2 ¤ = Z ∞ −∞ x 2 (m−1) h m−1 dx 2 (|x| +h) m dx = Z ∞ 0 x 2 (m−1) h m−1 dx 2 (x +h) m dx 8 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES where the second integral follows by evenness of the integrand in the first integral, and the absolute value on x is unnecessary because the integration is for positive x. Integrate by parts twice to obtain E £ X 2 ¤ = (m−1) h m−1 (" x 2 (h +h) −m−1 −m+ 1 # ∞ 0 + Z ∞ 0 2x(x +h) −m+1 m−1 dx ) The first term is zero if m ≥ 3. Therefore E £ X 2 ¤ = 2h m−1 Z ∞ 0 x(x +h) −m+1 dx = 2h 2 (m−2) (m−3) , m ≥ 4 Problem 4.21 (a) A = b £ 1 −e −bB ¤ −1 ; (b) The cdf is 0 for x < 0 and 1 for x > B. For 0 ≤ x ≤ B, the result is (A/b) ¡ 1 −e −bx ¢ ; (c) The mean is E[X] = 1 b · 1 −bB e −bB 1 −e −bB ¸ (d) The mean-square is E £ X 2 ¤ = 2A b · 1 b 2 − e −bB b 2 (1 +bB) ¸ − AB 2 b e −bB where A must be substituted from part (a.). (e) For the variance, subtract the result of part (c) squared from the result of part (d). Problem 4.22 (a) The integral evaluates as follows: E © X 2n ª = Z ∞ −∞ x 2n e −x2/2σ 2 √ 2πσ 2 dx = 2 Z ∞ 0 ³ 2 √ 2σy ´ 2n e −y 2 √ π dy = 2 n+1 σ 2n √ π Z ∞ 0 y 2n e −y dy where y = x/ ¡ 2 1/2 σ ¢ . Using a table of definite integrals, the given result is obtained. (b) The integral is zero by virtue of the oddness of the integrand. Problem 4.23 E[X] = Z ∞ −∞ x ½ 1 2 f (x −4) + 1 8 [u(x −3) −u(x −7)] ¾ dx = 1 2 Z ∞ −∞ xf (x −4) dx + 1 8 Z 7 3 xdx = 9 2 4.1. PROBLEM SOLUTIONS 9 E £ X 2 ¤ = Z ∞ −∞ x 2 ½ 1 2 f (x −4) + 1 8 [u(x −3) −u(x −7)] ¾ dx = 127 6 α 2 X = E £ X 2 ¤ −E 2 [X] = 127 6 − 81 4 = 11 12 Problem 4.24 Regardless of the correlation coefficient, the mean of Z is E{Z} = E{3X −4Y } = 3E{X} −4E{Y } = 3 (2) −4 (1) = 2 The variance of Z is var (Z) = E n [Z −E(Z)] 2 o = E n [3X −4Y −3E(X) + 4E(Y )] 2 o = E n £ 3 ¡ X −X ¢ −4(Y −Y ) ¤ 2 o = E © 9 ¡ X −X ¢ −24 ¡ X −X ¢ (Y −Y ) + 16(Y −Y ) 2 ª = 9σ 2 X + 16σ 2 Y −24σ X σ Y σ XY Putting in numbers, the results for the variance of Z are: (a) 107; (b) 83.76; (c) 51.23; (d) 14.05. Problem 4.25 Consider independent Gaussian random variables U and V and define a transformation g 1 (u, v) = x = ρu + p 1 −ρ 2 u and g 2 (u, v) = y = u The inverse transformation is u = y and v = (x −ρy) / ¡ 1 −ρ 2 ¢ 1/2 . Since the transformation is linear, the new random variables are Gaussian. The Jacobian is ¡ 1 −ρ 2 ¢ −1/2 . Thus the joint pdf of the new random variables X and Y is f XY (x, y) = 1 p 1 −ρ 2 e − u 2 +v 2 2σ 2 2πσ 2 ¯ ¯ ¯ ¯ ¯ ¯ u=g −1 1 (x,y) v=g −1 2 (x,y) = exp h − x 2 −2ρxy+y 2 2σ 2 (1−ρ 2 ) i 2πσ 2 (1 −ρ 2 ) Thus X and Y are equivalent to the random variables in the problem statement and which proves the desired result. To see this, note that E{XY } = E n³ ρU + p 1 −ρ 2 V ´ U o = ρσ 2 10 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES which proves the desired result. Problem 4.26 Divide the joint Gaussian pdf of two random variables by the Gaussisn pdf of Y , collect all exponential terms in a common exponent, complete the square of this exponent which will be quadratic in x and y, and the desired result is a Gaussian pdf with E{X|Y } = m x + ρσ x σ Y (Y −m Y ) and var (X|Y ) = σ 2 X ¡ 1 −ρ 2 ¢ Problem 4.27 Convolve the two component pdfs. A sketch of the two component pdfs making up the integrand show that there is no overlap until z > −1/2, and the overlap ends when z > 3.5. The result, either obtained graphically or analytically, is given by f Z (z) =            0, z < −0.5 (z + 0.5) /3, −0.5 ≤ z ≤ 0.5 1/3, 0.5 ≤ z < 2.5 −(z −3.5) /3, 2.5 ≤ z ≤ 3.5 0, z > 3.5 Problem 4.28 (a) E[X] = 0, E £ X 2 ¤ = 1/32 =var[X]; (b) The pdf of Y is f Y (y) = 4 3 e −8|y−3|/3 (c) E[Y ] = E[2 + 3X] = 2, E[Y ] = E h (2 + 3X) 2 i = E £ 4 + 12X + 9X 2 ¤ = 4 + 12E[X] + 9E £ X 2 ¤ = 4 + 12 + 9/32 = 4.28125, σ 2 = 4.28125 −2 2 = 0.28125 Problem 4.29 (a) The characteristic function is M X (jv) = a a −jv 4.1. PROBLEM SOLUTIONS 11 (b) The mean and mean-square values are E[X] = 1/a and E £ X 2 ¤ = 2/a 2 . (c) var[X] = 1/a 2 . Problem 4.30 The required distributions are P (k) = µ n k ¶ p k (1 −p) n−k (Binomial) P (k) = e −(k−np) 2 /[2np(1−p)] p 2πnp (1 −p) (Laplace) P (k) = (np) k k! e −np (Poisson) A comparison is given below: (a) n, p k Binomial Laplace Poisson 3,1/5 0 0.512 0.396 0.549 1 0.384 0.487 0.329 2 0.096 0.075 0.099 3 0.008 0.0014 0.020 (b) n, p k Binomial Laplace Poisson 3,1/10 0 0.729 0.650 0.741 1 0.243 0.310 0.222 2 0.027 0.004 0.033 3 0.001 1×10 −6 0.003 (c) n, p k Binomial Laplace Poisson 10,1/5 0 0.107 0.090 0.135 1 0.268 0.231 0.271 2 0.302 0.315 0.271 3 0.201 0.231 0.180 4 0.088 0.090 0.090 5 0.026 0.019 0.036 6 0.005 0.002 0.012 7 7.9×10 −4 1.3×10 −4 3.4×10 −3 8 7.0×10 −5 4.1×10 −6 8.6×10 −4 9 4.1×10 −6 7.1×10 −8 1.9×10 −4 12 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES (d) n, p k Binomial Laplace Poisson 10,1/10 0 0.348 0.241 0.368 1 0.387 0.420 0.368 2 0.194 0.241 0.183 3 0.057 0.046 0.061 4 0.011 0.003 0.015 5 1.5×10 −3 6×10 −5 3.1×10 −3 6 1.4×10 −4 3.9×10 −7 5.1×10 −4 7 3.6×10 −7 6.3×10 −13 3.4×10 −3 8 9.0×10 −9 1.5×10 −16 1×10 −6 9 1×10 −10 1.2×10 −20 1×10 −7 Problem 4.31 (a) P (5 or 6 heads in 20 trials) = P k=5,6 [20!/k! (20 −k)!] 2 −20 = 0.0518 (b) P (first head at trial 15) = (1/2)(1/2) 14 = 3.052 × 10 −5 (c) P (50 to 60 heads in 100 trials) = P 60 k=50 100! k!(100−k)! 2 −100 Problem 4.32 (a) N p = 26 (25) (24) (21) = 358, 800; (b) N p = (26) 4 = 456, 976; (c) The answers are the reciprocals of the numbers found in (a) and (b). Problem 4.33 (a) The desired probability is P (fewer than 3 heads in 10 tosses) = 2 X k=0 µ 10 k ¶µ 1 2 ¶ 10 = (1 + 10 + 45) 2 −10 = 0.0547 (b) npq = 10 (1/2) (1/2) = 2.25; np = 5. The Laplace-approximated probability is P(< 3 heads in 10 tosses) = 2 X k=0 e −(k−5) 2 /5 √ 5π = 1 √ 5π ³ e −5 +e −16/5 +e −9/5 ´ = 0.0537 (c) The percent error is % error = 0, 0547 −0.0537 0.0547 × 10 2 = 1.84% 4.1. PROBLEM SOLUTIONS 13 Problem 4.34 P (3 or more errors in 10 6 ) = 0.08 Problem 4.35 First note that Z ∞ −∞ e − 1 2 au 2 du = Z ∞ −∞ e − 1 2 a(u−b) 2 du = r 2π a (a) In the joint Gaussian pdf, let u = x −m X σ X ; v = y −m Y σ Y ; k = ³ 2πσ X σ Y p 1 −ρ 2 ´ −1 Then the joint Gaussian pdf becomes f XY (x, y) = k exp · − u 2 −2ρuv +v 2 2 (1 −ρ 2 ) ¸ Complete the square in v to obtain f XY (x, y) = k exp " − (v −2ρu) 2 2 (1 −ρ 2 ) # exp ¡ −u 2 /2 ¢ Use the definite integral above to integrate over v with the result f X (x) = q 2π (1 −ρ 2 ) σ 2 Y exp ¡ −u 2 /2 ¢ = 1 q 2πσ 2 X exp " − (x −m X ) 2 2σ 2 X # The pdf of Y can be found in a similar manner. (b) First find the characteristic function: M X (jv) = E © e jvX ª = Z ∞ −∞ e jvx e − (x−m X ) 2 2σ X q 2πσ 2 X dx Put the exponents together and complete the square in the exponent to get M X (jv) = Z ∞ −∞ 1 q 2πσ 2 X exp · − 1 2 σ 2 X v 2 +jm X v ¸ exp · − 1 2σ 2 X ¡ x −m X −jvσ 2 X ¢ 2 ¸ dx 14 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Use the definite integral given above to get M X (jv) = exp · jvm X − 1 2 σ 2 X v 2 ¸ It is then easy to differentiate and get E[X] = −jM p X (jv) | v=0 = m X and E £ X 2 ¤ = (−j) 2 M q X (jv) | v=0 = σ 2 X +m 2 X from which it follows that the variance is σ 2 X . Problem 4.36 (a) K = 1/π; (b) E[X] is not defined, but one could argue that it is zero from the oddness of the integrand for this expectation. If one writes down the integral for the second moment, it clearly does not converge (the integrand approaches a constant as |x| →∞). (c) The answer is given; (d) Compare the form of the characteristic function with the answer given in (c) and do a suitable redefinition of variables. Problem 4.37 (a) The characteristic function is found from M Y (jv) = E © e jvY ª = E n e jv P n i=1 X 2 i o = E ( n Y i=1 e jvX 2 i ) But E n e jvX 2 i o = Z ∞ −∞ e jvx 2 e −x 2 /2σ 2 √ 2πσ 2 dx = Z ∞ −∞ e −(1/2σ 2 −jv)x 2 √ 2πσ 2 dx = ¡ 1 −j2vσ 2 ¢ −1/2 which follows by using the definite integral defined at the beginning of Problem 4.35. Thus M Y (jv) = ¡ 1 −j2vσ 2 ¢ −N/2 (b) The given pdf follows easily by letting α = 2σ 2 in the given Fourier transform pair. 4.1. PROBLEM SOLUTIONS 15 Figure 4.1: (c) The mean of Y is E[Y ] = Nσ 2 by taking the sum of the expectations of the separate terms of the sum defining Y . The mean-square value of X 2 i is E h ¡ X 2 i ¢ 2 i = (−j) 2 d 2 h ¡ 1 −j2vσ 2 ¢ −1/2 i dv 2 ¯ ¯ ¯ ¯ v=0 = 3σ 4 Thus, var h ¡ X 2 i ¢ 2 i = 3σ 4 −σ 4 = 2σ 4 Since the terms in the sum defining Y are independent, var[Y ] = 2Nσ 4 . The mean and variance of Y can be put into the definition of a Gaussian pdf to get the desired approximation. (d) A comparison of the cases N = 4 and N = 16 is shown in Figure 4.1. Note the extreme difference between exact and approximation due to the central limit theorem not being valid for the former case. (e) Evident by direct substitution. 16 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Problem 4.38 This is a matter of plotting Q(x) = R ∞ x exp(−t 2 /2) √ 2π dx and Q a (x) = exp(−x 2 /2) √ 2πx on the same set of axes. Problem 4.39 By definition, the cdf of a Gaussian random variable is F X (x) = Z x −∞ e − (u−m) 2 2σ 2 √ 2πσ 2 du = 1 − Z ∞ x e − (u−m) 2 2σ 2 √ 2πσ 2 du Change variables to v = u −m σ with the result that F X (x) = 1 − Z x x−m σ e − v 2 2 √ 2π du = 1 −Q µ x −m σ ¶ A plot may easily be obtained by a MATLAB program. It is suggested that you use the MATLAB function erfc(x) = 2 √ π R ∞ x exp ¡ −t 2 ¢ dt. Problem 4.40 The exact probability is P (|X| ≤ kσ X ) = P (−kσ X ≤ X ≤ kσ X ) = ½ kσ x , kσ x ≤ 1 1, kσ x > 1 Chebyshev’s bound is P (|X| ≤ kσ X ) ≥ 1 − 1 k 2 , k > 0 A plot is left to the student. Problem 4.41 (a) The probability is P (X ≤ 11) = Z 11 −∞ e −(x−12) 2/30 √ 30π dx = Q ³ 1/ √ 15 ´ = 1 2 erfc ³ 1/ √ 30 ´ = 0.398 4.1. PROBLEM SOLUTIONS 17 (b) This probability is P (10 < X ≤ 12) = Z 12 10 e −(x−12) 2/30 √ 30π dx = 1 2 −Q ³ 2/ √ 15 ´ = 1 2 erf ³ 2/ √ 30 ´ = 0.197 (c) The desired probability is P (11 < X ≤ 13) = Z 13 11 e −(x−12) 2/30 √ 30π dx = 1 −2Q ³ 1/ √ 15 ´ = erf ³ 1/ √ 30 ´ = 0.204 (d) The result is P (9 < X ≤ 12) = Z 12 9 e −(x−12) 2/30 √ 30π dx = 1 2 −Q ³ 3/ √ 15 ´ = 1 2 erf ³ 3/ √ 30 ´ = 0.281 Problem 4.42 Observe that σ 2 = Z ∞ −∞ (x −m) 2 f X (x) dx ≥ Z |x−m|>kσ (x −m) 2 f X (x) dx ≥ Z |x−m|>kσ k 2 σ 2 f X (x) dx = k 2 σ 2 P {|X −m| > kσ} = k 2 σ 2 {1 −P [|X −m| > kσ]} or P [|X −m| < kσ] ≥ 1 − 1 k 2 18 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Problem 4.43 The mean is 0 by even symmetry of the pdf. Thus, the variance is var [X] = E £ X 2 ¤ = Z ∞ −∞ x 2 ³ a 2 ´ exp(−a1 × 1) dx = 2 ³ a 2 ´ Z ∞ 0 x 2 exp(−ax) dx = 2a Z ∞ 0 ³ y a ´ 2 exp(−y) dy a = 2a µ 1 a 3 ¶µ 1 2 ¶ = 1/a 2 4.2 Computer Exercises Computer Exercise 4.1 % ce4_1.m: Generate an exponentially distributed random variable % with parameter a from a uniform random variable in [0, 1) % clf a = input(‘Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x): ’); N = input(‘Enter number of exponential random variables to be generated: ’); U = rand(1,N); V = -(1/a)*log(U); [M, X] = hist(V); disp(‘ ’) disp(‘No. of random variables generated’) disp(N) disp(‘ ’) disp(‘Bin centers’) disp(X) disp(‘ ’) disp(‘No. of random variable counts in each bin’) disp(M) disp(‘ ’) norm_hist = M/(N*(X(2)-X(1))); plot(X, norm_hist, ’o’) hold plot(X, a*exp(-a*X), ‘—’), xlabel(‘x’), ylabel(‘f_X(x)’),... 4.2. COMPUTER EXERCISES 19 title([‘Theoretical pdf and histogram for ’,num2str(N), ‘ computer generated exponential random variables’]) legend([‘Histogram points’],[‘Exponential pdf; a = ’, num2str(a)]) A typical run follows: >> ce4_1 Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x): 2 Enter number of exponential random variables to be generated: 5000 No. of random variables generated 5000 Bin centers Columns 1 through 7 0.1876 0.5626 0.9375 1.3125 1.6875 2.0625 2.4375 Columns 8 through 10 2.8125 3.1875 3.5624 No. of random variable counts in each bin Columns 1 through 6 2677 1212 590 263 129 71 Columns 7 through 10 34 9 11 4 Current plot held Computer Exercise 4.2 % ce4_2.m: Generate pairs of Gauss. random variables from Rayleigh & uniform RVs % clf m = input(‘Enter the mean of the Gaussian random variables: ’); sigma = input(‘Enter the standard deviation of the Gaussian random variables: ’); N = input(‘Enter number of Gaussian random variable pairs to be generated: ’); U = rand(1,N); V = rand(1,N); R = sqrt(-2*log(V)); X = sigma*R.*cos(2*pi*U)+m; Y = sigma*R.*sin(2*pi*U)+m; disp(‘ ’) disp(‘Covarance matrix of X and Y vectors:’) disp(cov(X,Y)) disp(‘ ’) [MX, X_bin] = hist(X, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); [MY, Y_bin] = hist(Y, 20); 20 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Figure 4.2: norm_MY = MY/(N*(Y_bin(2)-Y_bin(1))); gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); gauss_pdf_Y = exp(-(Y_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); subplot(2,1,1), plot(X_bin, norm_MX, ’o’) hold subplot(2,1,1), plot(X_bin, gauss_pdf_X, ‘—’), xlabel(‘x’), ylabel(‘f_X(x)’),... title([‘Theoretical pdfs and histograms for ’,num2str(N), ‘ computer generated independent Gaussian RV pairs’]) legend([‘Histogram points’],[‘Gauss pdf; \sigma, \mu = ’, num2str(sigma),‘, ’, num2str(m)]) subplot(2,1,2), plot(Y_bin, norm_MY, ’o’) hold subplot(2,1,2), plot(Y_bin, gauss_pdf_Y, ‘—’), xlabel(‘y’), ylabel(‘f_Y(y)’) A typical run follows: >> ce4_2 Enter the mean of the Gaussian random variables: 1 Enter the standard deviation of the Gaussian random variables: 3 Enter number of Gaussian random variable pairs to be generated: 2000 Covarance matrix of X and Y vectors: 8.5184 0.0828 4.2. COMPUTER EXERCISES 21 Figure 4.3: 0.0828 8.9416 Current plot held Current plot held Computer Exercise 4.3 % ce4_3.m: Generate a sequence of Gaussian random variables with specified correlation % between adjacent RVs % clf m = input(‘Enter the mean of the Gaussian random variables: ’); sigma = input(‘Enter the standard deviation of the Gaussian random variables: ’); rho = input(‘Enter correlation coefficient between adjacent Gauss. RVs: ’); N = input(‘Enter number of Gaussian random variables to be generated: ’); var12 = sigma^2*(1 - rho^2); X = []; X(1) = sigma*randn(1)+m; for k = 2:N m12 = m + rho*(X(k-1) - m); X(k) = sqrt(var12)*randn(1) + m12; end 22 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES [MX, X_bin] = hist(X, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); subplot(2,1,1), plot(X_bin, norm_MX, ‘o’) hold subplot(2,1,1), plot(X_bin, gauss_pdf_X, ‘—’), xlabel(‘x’), ylabel(‘f_X(x)’),... title([‘Theoretical pdf and histogram for ’,num2str(N), ‘ computer generated Gaussian RVs’]) legend([‘Histogram points’],[‘Gauss pdf; \sigma, \mu = ’, num2str(sigma),‘, ’, num2str(m)],2) Z = X(1:50); subplot(2,1,2), plot(Z, ‘x’), xlabel(‘k’), ylabel(‘X(k)’) title([‘Gaussian sample values with correlation ’, num2str(rho), ‘ between samples’]) A typical run follows: >> ce4_3 Enter the mean of the Gaussian random variables: 0 Enter the standard deviation of the Gaussian random variables: 2 Enter correlation coefficient between adjacent Gaussian random variables: .7 Enter number of Gaussian random variables to be generated: 2000 Current plot held Computer Exercise 4.4 % ce4_4.m: Testing the validity of the Central Limit Theorem with sums of uniform % random numbers % clf N = input(‘Enter number of Gaussian random variables to be generated: ’); M = input(‘Enter number of uniform RVs to be summed to generate one GRV: ’); % Mean of uniform = 0; variance = 1/12 Y = rand(M,N)-0.5; X_gauss = sum(Y)/(sqrt(M/12)); disp(‘ ’) disp(‘Estimated variance of the Gaussian RVs:’) disp(cov(X_gauss)) disp(‘ ’) [MX, X_bin] = hist(X_gauss, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); gauss_pdf_X = exp(-X_bin.^2/2)/sqrt(2*pi); plot(X_bin, norm_MX, ‘o’) hold 4.2. COMPUTER EXERCISES 23 Figure 4.4: plot(X_bin, gauss_pdf_X, ‘—’), xlabel(‘x’), ylabel(‘f_X(x)’),... title([‘Theoretical pdf & histogram for ’,num2str(N), ‘ Gaussian RVs each generated as the sum of ’, num2str(M), ‘ uniform RVs’]) legend([‘Histogram points’], [‘Gauss pdf; \sigma = 1, \mu = 0’],2) A typical run follows: >> ce4_4 Enter number of Gaussian random variables to be generated: 10000 Enter number of uniform random variables to be summed to generate one GRV: 30 Estimated variance of the Gaussian RVs: 1.0075 Current plot held 24 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Figure 4.5: Chapter 5 Random Signals and Noise 5.1 Problem Solutions Problem 5.1 The various sample functions are as follows. Sample functions for case (a) are horizontal lines at levels A, 0, −A, each case of which occurs equally often (with probability 1/3). Sample functions for case (b) are horizontal lines at levels 5A, 3A, A, −A, −3A, −5A, each case of which occurs equally often (with probability 1/6). Sample functions for case (c) are horizontal lines at levels 4A, 2A, −2A, −4A, or oblique straight lines of slope A or −A, each case of which occurs equally often (with probability 1/6). Problem 5.2 a. For case (a) of problem 5.1, since the sample functions are constant with time and are less than or equal 2A, the probability is one. For case (b) of problem 5.1, since the sample functions are constant with time and 4 out of 6 are less than or equal to 2A, the probability is 2/3. For case (c) of problem 5.1, the probability is again 2/3 because 4 out of 6 of the sample functions will be less than 2A at t = 4. b. The probabilities are now 2/3, 1/2, and 1/2, respectively. c. The probabilities are 1, 2/3, and 5/6, respectively. Problem 5.3 a. The sketches would consist of squarewaves of random delay with respect to t = 0. 1 2 CHAPTER 5. RANDOM SIGNALS AND NOISE b. X (t) takes on only two values A and −A, and these are equally likely. Thus f X (x) = 1 2 δ (x −A) + 1 2 δ (x +A) Problem 5.4 a. The sample functions at the integrator output are of the form Y (i) (t) = Z t Acos (ω 0 λ) dλ = A ω 0 sinω 0 t where A is Gaussian. b. Y (i) (t 0 ) is Gaussian with mean zero and standard deviation σ Y = σ a |sin(ω 0 t 0 )| ω 0 c. Not stationary and not ergodic. Problem 5.5 a. By inspection of the sample functions, the mean is zero. Considering the time average correlation function, defined as R(λ) = lim T→∞ 1 2T Z T −T x(t) x(t +λ) dt = 1 T 0 Z T 0 x(t) x(t +λ) dt where the last expression follows because of the periodicity of x(t), it follows from a sketch of the integrand that R(τ) = A 2 (1 −4λ/T 0 ) , 0 ≤ λ ≤ T 0 /2 Since R(τ) is even and periodic, this defines the autocorrelation function for all λ. b. Yes, it is wide sense stationary. The random delay, λ, being over a full period, means that no untypical sample functions occur. 5.1. PROBLEM SOLUTIONS 3 Problem 5.6 a. The time average mean is zero. The statistical average mean is E[X (t)] = Z π/2 π/4 Acos (2πf 0 t +θ) dθ π/4 = 4A π sin(2πf 0 t +θ) ¯ ¯ ¯ ¯ π/2 π/4 = 4A π [sin(2πf 0 t +π/2) −sin(2πf 0 t +π/4)] = 4A π ·µ 1 − 1 √ 2 ¶ cos (2πf 0 t) − 1 √ 2 sin(2πf 0 t) ¸ The time average variance is A 2 /2. The statistical average second moment is E £ X 2 (t) ¤ = Z π/2 π/4 A 2 cos 2 (2πf 0 t +θ) dθ π/4 = 2A 2 π " Z π/2 π/4 dθ + Z π/2 π/4 cos (4πf 0 t + 2θ) dθ # = 2A 2 π " π 4 + 1 2 sin(4πf 0 t + 2θ) ¯ ¯ ¯ ¯ π/2 π/4 # = A 2 2 + A 2 π h sin(4πf 0 t +π) −sin ³ 4πf 0 t + π 2 í = A 2 2 − A 2 π [sin(4πf 0 t) + cos (4πf 0 t)] The statistical average variance is this expression minus E 2 [X (t)]. b. The time average autocorrelation function is hx(t) x(t +τ)i = A 2 2 cos (2πf 0 τ) The statistical average autocorrelation function is R(τ) = Z π/2 π/4 A 2 cos (2πf 0 t +θ) cos (2πf 0 (t +τ) +θ) dθ π/4 = 2A 2 π Z π/2 π/4 [cos (2πf 0 τ) + cos (2πf 0 (2t +τ) + 2θ)] dθ = A 2 2 cos (2πf 0 τ) + A 2 π sin(2πf 0 (2t +τ) + 2θ) ¯ ¯ ¯ ¯ π/2 π/4 = A 2 2 cos (2πf 0 τ) + A 2 π h sin(2πf 0 (2t +τ) +π) −sin ³ 2πf 0 (2t +τ) + π 2 í = A 2 2 cos (2πf 0 τ) − A 2 π [sin2πf 0 (2t +τ) + cos 2πf 0 (2t +τ)] 4 CHAPTER 5. RANDOM SIGNALS AND NOISE c. No. The random phase, not being uniform over (0, 2π), means that for a given time, t, certain phases will be favored over others. Problem 5.7 a. The mean is zero. The mean-square value is E £ Z 2 ¤ = E n [X (t) cos (ω 0 t)] 2 o = E £ X 2 (t) ¤ cos 2 (ω 0 t) = σ 2 X cos 2 (ω 0 t) The process is not stationary since its second moment depends on time. b. Again, the mean is clearly zero. The second moment is E £ Z 2 ¤ = E n [X (t) cos (ω 0 t +θ)] 2 o = E £ X 2 (t) ¤ E £ cos 2 (ω 0 t +θ) ¤ = σ 2 X 2 Problem 5.8 The pdf of the noise voltage is f X (x) = e −(x−2) 2 /10 √ 10π Problem 5.9 a. Suitable; b. Suitable; c. Not suitable because the Fourier transform is not everywhere nonnegative; d. Not suitable for the same reason as (c); e. Not suitable because it is not even. 5.1. PROBLEM SOLUTIONS 5 Problem 5.10 Use the Fourier transform pair 2Wsinc (2Wτ) ↔Π(f/2W) with W = 2 × 10 6 Hz to get the autocorrelation function as R X (τ) = N 0 W sinc (2Wτ) = ¡ 2 × 10 −8 ¢ ¡ 2 × 10 6 ¢ sinc ¡ 4 × 10 6 τ ¢ = 0.02 sinc ¡ 4 × 10 6 τ ¢ Problem 5.11 a. The r (τ) function, (5.55), is r (τ) = 1 T Z ∞ −∞ p (t) p (t +τ) dt = 1 T Z T/2−τ −T/2 cos (πt/T) cos · π (t +τ) T ¸ dt, 0 ≤ τ ≤ T/2 = 1 2 (1 −τ/T) cos ³ πτ T ´ , 0 ≤ τ ≤ T/2 Now r (τ) = r (−τ) and r (τ) = 0 for |τ| > T/2. Therefore it can be written as r (τ) = 1 2 Λ(τ/T) cos ³ πτ T ´ Hence, by (5.54) R a (τ) = A 2 2 Λ(τ/T) cos ³ πτ T ´ The power spectral density is S a (f) = A 2 T 4 ½ sinc 2 · T µ f − 1 2T ¶¸ + sinc 2 · T µ f + 1 2T ¶¸¾ b. The autocorrelation function now becomes R b (τ) = A 2 [2r (τ) +r (τ +T) +r (τ −T)] which is (5.62) with g 0 = g 1 = 1. The corresponding power spectrum is S b (f) = A 2 T £ sinc 2 (Tf −0.5) + sinc 2 (Tf + 0.5) ¤ cos 2 (πfT) 6 CHAPTER 5. RANDOM SIGNALS AND NOISE c. The plots are left for the student. Problem 5.12 The cross-correlation function is R XY (τ) = E[X (t) Y (t +τ)] = R n (τ) +E © A 2 cos (ω 0 t +Θ) × sin[ω 0 (t +τ) +Θ] ª = BΛ(τ/τ 0 ) + A 2 2 sin(ω 0 τ) Problem 5.13 a. By definition R Z (τ) = E[Z (t) Z (t +τ)] = E[X (t) X (t +τ) Y (t) Y (t +τ)] = E[X (t) X (t +τ)]E[Y (t) Y (t +τ)] = R X (τ) R Y (τ) b. Since a product in the time domain is convolution in the frequency domain, it follows that S Z (f) = S X (f) ∗ S Y (f) c. Use the transform pairs 2Wsinc (2Wτ) ←→Π(f/2W) and cos (2πf 0 τ) ←→ 1 2 δ (f −f 0 ) + 1 2 δ (f +f 0 ) Also, R Y (τ) = E{4 cos (50πt +θ) cos [50π (t +τ) +θ]} = 2 cos (50πτ) Using the first transform pair, we have R Y (τ) = 500 sinc (100τ) 5.1. PROBLEM SOLUTIONS 7 This gives R Z (τ) = [500 sinc (100τ)] [2 cos (50πτ)] = 1000 sinc (100τ) cos (50πτ) and S Z (f) = 5 × 10 3 · Y µ f −25 200 ¶ + Y µ f + 25 200 ¶¸ Problem 5.14 a. E £ X 2 (t) ¤ = R(0) = 5 W; E 2 [X (t)] = lim τ→∞ R(τ) = 4 W; σ 2 X = E £ X 2 (t) ¤ −E 2 [X (t)] = 5 −4 = 1 W. b. dc power = E 2 [X (t)] = 4 W. c. Total power = E £ X 2 (t) ¤ = 5 W. d. S (f) = 4δ (f) + 5 sinc 2 (5f). Problem 5.15 a. The autocorrelation function is R X (τ) = E[Y (t) Y (t +τ)] = E{[X (t) +X (t −T)][X (t +τ) +X (t +τ −T)]} = E[X (t) X (t +τ)] +E[X (t) X (t +τ +τ)] +E[X (t −T) X (t +τ)] +E[X (t −T) X (t +τ −T)] = 2R X (τ) +R X (τ −T) +R X (τ +T) b. Application of the time delay theorem of Fourier transforms gives S Y (f) = 2S X (f) +S X (f) [exp(−j2πfT) + exp(j2πfT)] = 2S X (f) + 2S X (f) cos (2πfT) = 4S X (f) cos 2 (πfT) 8 CHAPTER 5. RANDOM SIGNALS AND NOISE c. Use the transform pair R X (τ) = 6Λ(τ) ←→S X (f) = 6 sinc 2 (f) and the result of (b) to get S Y (f) = 24 sinc 2 (f) cos 2 (πf/4) Problem 5.16 a. The student should carry out the sketch. b. dc power = R 0+ 0− S (f) df = 5 W. c. Total power = R ∞ −∞ S (f) df = 5 + 10/5 = 7 W Problem 5.17 a. This is left for the student. b. The dc power is either lim τ→∞ R X (τ) or R 0+ 0− S X (f) df. Thus (1) has 0 W dc power, (2) has dc power K 1 W, and (3) has dc power K 2 W. c. Total power is given by R X (0) or the integral of R ∞ −∞ S X (f) df over all frequency. Thus (1) has total power K, (2) has total power 2K 1 +K 2 , and (3) has total power P 3 = K Z ∞ −∞ e −αf 2 df +K 1 +K 1 +K 2 = r π α K + 2K 1 +K 2 where the integral is evaluated by means of a table of definite integrals. d. (2) has a periodic component of frequency b Hz, and (3) has a periodic component of 10 Hz, which is manifested by the two delta functions at f = ±10 Hz present in the power spectrum. Problem 5.18 a. The output power spectral density is S out (f) = 10 −5 Π ¡ f/2 × 10 6 ¢ Note that the rectangular pulse function does not need to be squared because its amplitude squared is unity. 5.1. PROBLEM SOLUTIONS 9 b. The autocorrelation function of the output is R out (τ) = 2 × 10 6 × 10 −5 sinc ¡ 2 × 10 6 τ ¢ = 20 sinc ¡ 2 × 10 6 τ ¢ c. The output power is 20 W, which can be found from R out (0) or the integral of S out (f). Problem 5.19 a. By assuming a unit impulse at the input, the impulse response is h(t) = 1 T [u(t) −u(t −T)] b. Use the time delay theorem of Fourier transforms and the Fourier transform of a rectangular pulse to get H (f) = sinc (fT) e −jπfT c. The output power spectral density is S 0 (f) = N 0 2 sinc 2 (fT) d. Use F [Λ(τ/τ 0 )] = τ 0 sinc 2 (τ 0 f) to get R 0 (τ) = N 0 2T Λ(τ/T) e. By definition of the equivalent noise bandwidth E £ Y 2 ¤ = H 2 max B N N 0 The output power is also R 0 (0) = N 0 /2T. Equating the two results and noting that H max = 1, gives B N = 1/2T Hz. f. Evaluate the integral of the power spectral density over all frequency: E £ Y 2 ¤ = Z ∞ −∞ N 0 2 sinc 2 (fT) df = N 0 2T Z ∞ −∞ sinc 2 (u) du = N 0 2T = R 0 (0) 10 CHAPTER 5. RANDOM SIGNALS AND NOISE Problem 5.20 a. The output power spectral density is S 0 (f) = N 0 /2 1 + (f/f 3 ) 4 b. The autocorrelation function of the output, by inverse Fourier transforming the output power spectral density, is R 0 (τ) = F −1 [S 0 (f)] = Z ∞ −∞ N 0 /2 1 + (f/f 3 ) 4 e −j2πfτ df = N 0 Z ∞ 0 cos (2πfτ) 1 + (f/f 3 ) 4 df = N 0 f 3 Z ∞ 0 cos (2πf 3 τx) 1 +x 4 dx = πf 3 N 0 2 e − √ 2πf 3 τ cos ³ √ 2πf 3 τ −π/4 ´ c. Yes. R 0 (0) = πf 3 N 0 2 = N 0 B N , B N = πf 3 /2 Problem 5.21 We have |H (f)| 2 = (2πf) 2 (2πf) 4 + 5, 000 Thus |H (f)| = |2πf| q (2πf) 4 + 5, 000 This could be realized with a second-order Butterworth filter in cascade with a differentiator. Problem 5.22 a. The power spectral density is S Y (f) = N 0 2 Π(f/2B) The autocorrelation function is R Y (τ) = N 0 Bsinc (2Bτ) 5.1. PROBLEM SOLUTIONS 11 b. The transfer function is H (f) = A α +j2πf The power spectral density and autocorrelation function are, respectively, S Y (f) = |H (f)| 2 S X (f) = A 2 α 2 + (2πf) 2 B 1 + (2πβf) 2 = A 2 /α 2 1 + (2πf/α) 2 B 1 + (2πβf) 2 = A 2 B/α 2 1 −α 2 β 2 · 1 1 + (2πf/α) 2 − α 2 β 2 1 + (2πβf) 2 ¸ Use the transform pair exp(−|τ|/τ 0 ) ←→ 2τ 0 1+(2πfτ 0 ) 2 to find that the corresponding autocorrelation function is R Y (t) = A 2 B/α 2 h 1 −(αβ) 2 i h e −α|τ| −αβ e −|τ|/β i , α 6= 1/β Problem 5.23 a. E[Y (t)] = 0 because the mean of the input is zero. b. The output power spectrum is S Y (f) = 1 (10π) 2 1 1 + (f/5) 2 which is obtained by Fourier transforming the impulse response to get the transfer function and applying (5.90). c. Use the transform pair exp(−|τ|/τ 0 ) ←→ 2τ 0 1+(2πfτ 0 ) 2 to find the power spectrum as R Y (τ) = 1 10π e −10π|τ| d. Since the input is Gaussian, so is the output. Also, E[Y ] = 0 and var[Y ] = R Y (0) = 1/ (10π), so f Y (y) = e −y 2 /2σ 2 Y q 2πσ 2 Y 12 CHAPTER 5. RANDOM SIGNALS AND NOISE where σ 2 Y = 1 10π . e. Use (4.188) with x = y 1 = Y (t 1 ) , y = y 2 = Y (t 2 ) , m x = m y = 0, and σ 2 x = σ 2 y = 1/10π Also ρ (τ) = R Y (τ) R Y (0) = e −10π|τ| Set τ = 0.03 to get ρ(0.03) = 0.3897. Put these values into (4.188). Problem 5.24 Use the integrals I 1 = Z ∞ −∞ b 0 ds (a 0 s +a 1 ) (−a 0 s +a 1 ) = jπb 0 a 0 a 1 and I 2 = Z ∞ −∞ ¡ b 0 s 2 +b 1 ¢ ds (a 0 s 2 +a 1 s +a 2 ) (a 0 s 2 −a 1 s +a 2 ) = jπ −b 0 +a 0 b 1 /a 2 a 0 a 1 to get the following results for B N . Filter Type First Order Second Order Chebyshev πfc 2² πf c /² 2 √ 1+1/² 2 q √ 1+1/² 2 −1 Butterworth π 2 f c π 6 f c Problem 5.25 The transfer function is H (s) = ω 2 c ¡ s +ω c / √ 2 +jω c / √ 2 ¢ ¡ s +ω c / √ 2 −jω c / √ 2 ¢ = A ¡ s +ω c / √ 2 +jω c / √ 2 ¢ + A ∗ ¡ s +ω c / √ 2 −jω c / √ 2 ¢ where A = jω c /2 1/2 . This is inverse Fourier transformed, with s = jω, to yield h(t) = √ 2ω c e −ωct/ √ 2 sin µ ω c t √ 2 ¶ u(t) 5.1. PROBLEM SOLUTIONS 13 for the impulse response. Now 1 2 Z ∞ −∞ |h(t)| 2 dt = ω c 4 √ 2 = πf c 2 √ 2 after some effort. Also note that R ∞ −∞ h(t) dt = H (0) = 1. Therefore, the noise equivalent bandwidth, from (5.109), is B N = πf c 2 √ 2 Hz = 100π √ 2 Hz Problem 5.26 (a) Note that H max = 2, H (f) = 2 for −1 ≤ f ≤ 1, H (f) = 1 for −3 ≤ f ≤ −1 and 1 ≤ f ≤ 3 and is 0 otherwise. Thus B N = 1 H 2 max Z ∞ 0 |H (f)| 2 df = 1 4 µZ 1 0 2 2 df + Z 2 1 1 2 df ¶ = 1 4 (4 + 1) = 1.25 Hz (b) For this case H max = 2 and the frequency response function is a triangle so that B N = 1 H 2 max Z ∞ 0 |H (f)| 2 df = 1 4 Z 100 0 [2 (1 −f/100)] 2 df = Z 1 0 (1 −v) 2 dv = − 100 3 (1 −v) 3 ¯ ¯ ¯ ¯ 1 0 = 33.33 Hz Problem 5.27 By definition B N = 1 H 2 0 Z ∞ 0 |H (f)| 2 df = 1 4 Z 600 400 · 2Λ µ f −500 100 ¶¸ 2 df = 66.67 Hz Problem 5.28 a. Use the impulse response method. First, use partial fraction expansion to get the impulse response: H a (f) = 10 19 µ 1 j2πf + 1 − 1 j2πf + 20 ¶ ⇔h a (t) = 10 19 ¡ e −t −e −20t ¢ u(t) 14 CHAPTER 5. RANDOM SIGNALS AND NOISE Thus B N = R ∞ −∞ |h(t)| 2 dt 2 h R ∞ −∞ h(t) dt i 2 = ¡ 10 19 ¢ 2 R ∞ 0 ¡ e −t −e −20t ¢ 2 dt 2 £¡ 10 19 ¢ R ∞ 0 (e −t −e −20t ) dt ¤ 2 = 1 2 R ∞ 0 ¡ e −2t −2e −21t +e −40t ¢ dt £¡ −e −t + 1 20 e −20t ¢ ∞ 0 ¤ 2 = 1 2 ¡ − 1 2 e −2t + 2 21 e −21t − 1 40 e −40t ¢ ∞ 0 (1 −1/20) 2 = 1 2 µ 20 19 ¶ 2 µ 1 2 − 2 21 + 1 40 ¶ = 1 2 µ 20 19 ¶ 2 µ 361 21 × 40 ¶ = 0.238 Hz b. Again use the impulse response method to get B N = 0.625 Hz. Problem 5.29 (a) Choosing f 0 = f 1 moves −f 1 right to f = 0 and f 1 left to f = 0. (b) Choosing f 0 = f 2 moves −f 2 right to f = 0 and f 2 left to f = 0. Thus, the baseband spectrum can be written as S LP (f) = 1 2 N 0 Λ ³ f f 2 −f 1 ´ . (c) For this case both triangles (left and right) are centered around the origin and they add to give S LP (f) = 1 2 N 0 Π ³ f f 2 −f 1 ´ . (d) They are not uncorrelated for any case for an arbitrary delay. However, all cases give quadrature components that are uncorrelated at the same instant. Problem 5.30 a. By inverse Fourier transformation, the autocorrelation function is R n (τ) = α 2 e −α|τ| where K = α/2. b. Use the result from part (a) and the modulation theorem to get R p n (τ) = α 2 e −α|τ| cos (2πf 0 τ) c. The result is S nc (f) = S ns (f) = α 2 α 2 + (2πf) 2 5.1. PROBLEM SOLUTIONS 15 and S n c n s (f) = 0 Problem 5.31 (a) The equivalent lowpass power spectral density is given by S nc (f) = S ns (f) = N 0 Π ³ f f 2 −f 1 ´ . The cross-spectral density is S ncns (f) = 0. (b) For this case S nc (f) = S ns (f) = N 0 2 Π ³ f 2(f 2 −f 1 ) ´ and the cross-spectral density is S n c n s (f) = ½ − N 0 2 , −(f 2 −f 1 ) ≤ f ≤ 0 N 0 2 , 0 ≤ f ≤ (f 2 −f 1 ) (c) For this case, the lowpass equivalent power spectral densities are the same as for part (b) and the cross-spectral density is the negative of that found in (b). (d) for part (a) R ncns (τ) = 0 For part (b) R ncns(τ) = j " Z 0 −(f 2 −f 1 ) − N 0 2 e j2πfτ df + Z f 1 −f 2 0 N 0 2 e j2πfτ df # = j N 0 2 " −e −j2πfτ j2πτ ¯ ¯ ¯ ¯ 0 −(f 2 −f 1 ) + e −j2πfτ j2πτ ¯ ¯ ¯ ¯ (f 2 −f 1 ) 0 # = N 0 4πτ ³ −1 +e j2π(f 2 −f 1 )τ +e − j2π(f 2 −f 1 )τ −1 ´ = − N 0 4πτ {2 −2 cos [2π (f 2 −f 1 ) τ]} = − N 0 πτ sin 2 [π (f 2 −f 1 ) τ] = − h N 0 π (f 2 −f 1 ) 2 τ i sinc 2 [(f 2 −f 1 ) τ] For part (c), the cross-correlation function is the negative of the above. Problem 5.32 The result is S n 2 (f) = 1 2 S n (f −f 0 ) + 1 2 S n (f +f 0 ) 16 CHAPTER 5. RANDOM SIGNALS AND NOISE so take the given spectrum, translate it to the right by f 0 and to the left by f 0 , add the two translated spectra, and divide the result by 2. Problem 5.33 Define N 1 = A+N c . Then R = q N 2 1 +N 2 s Note that the joint pdf of N 1 and N s is f N 1 N s (n 1 , n s ) = 1 2πσ 2 exp ½ − 1 2σ 2 h (n 1 −A) 2 +n 2 s i ¾ Thus, the cdf of R is F R (r) = Pr (R ≤ r) = Z Z √ N 2 1 +N 2 s ≤r f N 1 Ns (n 1 , n s ) dn 1 dn s Change variables in the integrand from rectangular to polar with n 1 = αcos β and n 2 = αsinβ Then F R (r) = Z r 0 Z 2π 0 α 2πσ 2 e − 1 2σ 2 [(αcos β−A) 2 +(αsinβ) 2 ] dβdα = Z r 0 α σ 2 e − 1 2σ 2 (α 2 +A 2 ) I 0 µ αA σ 2 ¶ dα Differentiating with respect to r, we obtain the pdf of R: f R (r) = r σ 2 e − 1 2σ 2 (r 2 +A 2 ) I 0 µ rA σ 2 ¶ , r ≥ 0 Problem 5.34 The suggested approach is to apply S n (f) = lim T→∞ E n |=[x 2T (t)]| 2 o 2T where x 2T (t) is a truncated version of x(t). Thus, let x 2T (t) = N X k=−N n k δ (t −kT s ) 5.1. PROBLEM SOLUTIONS 17 The Fourier transform of this truncated waveform is =[x 2T (t)] = N X k=−N n k e −j2πkTs Thus, E n |=[x 2T (t)]| 2 o = E    ¯ ¯ ¯ ¯ ¯ N X k=−N n k e −j2πkT s ¯ ¯ ¯ ¯ ¯ 2    = E ( N X k=−N n k e −j2πkTs N X l=−N n l e j2πlTs ) = N X k=−N N X l=−N E[n k n l ] e −j2π(k−l)Ts = N X k=−N N X l=−N R n (0) δ kl e −j2π(k−l)T s = N X k=−N R n (0) = (2N + 1) R n (0) But 2T = 2NT s so that S n (f) = lim T→∞ E n |=[x 2T (t)]| 2 o 2T = lim N→∞ (2N + 1) R n (0) 2NT s = R n (0) T s Problem 5.35 To make the process stationary, assume that each sample function is displaced from the origin by a random delay uniform in (0, T]. Consider a 2nT chunk of the waveform and obtain its Fourier transform as =[X 2nT (t)] = n−1 X k=−n Aτ 0 sinc (fτ 0 ) e −j2πf(∆t k +τ 0 /2+kT) Take the magnitude squared and then the expectation to reduce it to E h |X 2nT (f)| 2 i = A 2 τ 2 0 sinc 2 (fτ 0 ) n−1 X k=−n n−1 X m=−n E n e −j2πf[∆t k −∆tm+(k−m)T] o 18 CHAPTER 5. RANDOM SIGNALS AND NOISE We need the expectation of the inside exponentials which can be separated into the product of the expectations for k 6= m, which is E{} = 4 T Z T/4 0 e −j2πf∆t k d∆t k = e −j2πf(k−m)T sinc 2 (fτ/4) For k = m, the expectation inside the double sum is 1. After considerable manipulation, the psd becomes S X (f) = A 2 τ 2 0 T sinc 2 (fτ 0 ) [1 −sinc (fT/4)] + 2A 2 τ 2 0 sinc 2 (τ 0 f) T cos 2 (πfT) · lim n→∞ sin 2 (nπfT) nsin 2 (πfT) ¸ Examine L = lim n→∞ sin 2 (nπfT) nsin 2 (πfT) It is zero for f 6= m/T, m an integer, because the numerator is finite and the denominator goes to infinity. For f = mT, the limit can be shown to have the properties of a delta function as n →∞. Thus S X (f) = A 2 τ 2 0 T sinc 2 (fτ 0 ) " 1 −sinc (fT/4) + 2 ∞ X m=−∞ δ (f −m/T) # Problem 5.36 The result is E £ y 2 (t) ¤ = 1 T 2 Z t+T t Z t+T t £ R 2 X (τ) +R 2 X (λ −β) +R X (λ −β −τ) +R X (λ −β +τ) ¤ dλdβ 5.1. PROBLEM SOLUTIONS 19 Problem 5.37 1. (a) In the expectation for the cross correlation function, write the derivative as a limit: R yy_dot p (τ) = E · y (t) dy (t +τ) dt ¸ = E ½ y (t) · lim ²→0 y (t +τ +²) −y (t +τ) ² ¸¾ = lim ²→0 1 ² {E[y (t) y (t +τ +²)] −E[y (t) y (t +τ)]} = lim ²→0 R y (τ +²) −R y (τ) ² = dR y (τ) dτ (b) The variance of Y = y (t) is σ 2 Y = N 0 B. The variance for Z = dy dt is found from its power spectral density. Using the fact that the transfer function of a differentiator is j2πf we get S Z (f) = (2πf) 2 N 0 2 Π(f/2B) so that σ 2 Z = 2π 2 N 0 Z B −B f 2 df = 2π 2 N 0 f 3 3 ¯ ¯ ¯ ¯ B −B = 4 3 π 2 N 0 B 3 The two processes are uncorrelated because R yy_dot p (0) = dRy(τ) dτ = 0 because the derivative of the autocorrelation function of y (t) exists at τ = 0 and therefore must be 0. Thus, the random process and its derivative are independent. Hence, their joint pdf is f Y Z (α, β) = exp ¡ −α 2 /2N 0 B ¢ √ 2πN 0 B exp ¡ −β 2 /2.67π 2 N 0 B 3 ¢ √ 2.67π 3 N 0 B 3 (c) No, because the derivative of the autocorrelation function at τ = 0 of such noise does not exist (it is a double-sided exponential). 20 CHAPTER 5. RANDOM SIGNALS AND NOISE 5.2 Computer Exercises Computer Exercise 5.1 % ce5_1.m: Computes approximations to mean and mean-square ensemble and time % averages of cosine waves with phase uniformly distributed between 0 and 2*pi % clf f0 = 1; A = 1; theta = 2*pi*rand(1,100); t = 0:.1:1; X = []; for n = 1:100 X = [X; A*cos(2*pi*f0*t+theta(n))]; end EX = mean(X); AVEX = mean(X’); EX2 = mean(X.*X); AVEX2 = mean((X.*X)’); disp(‘ ’) disp(‘Sample means (across 100 sample functions)’) disp(EX) disp(‘ ’) disp(‘Typical time-average means (sample functions 1, 20, 40, 60, 80, & 100)’) disp([AVEX(1) AVEX(20) AVEX(40) AVEX(60) AVEX(80) AVEX(100)]) disp(‘ ’) disp(‘Sample mean squares’) disp(EX2) disp(‘ ’) disp(‘Typical time-average mean squares’) disp([AVEX2(1) AVEX2(20) AVEX2(40) AVEX2(60) AVEX2(80) AVEX2(100)]) disp(‘ ’) for n = 1:5 Y = X(n,:); subplot(5,1,n),plot(t,Y),ylabel(‘X(t,\theta)’) if n == 1 title(‘Plot of first five sample functions’) end if n == 5 5.2. COMPUTER EXERCISES 21 xlabel(‘t’) end end A typical run follows: >> ce5_1 Sample means (across 100 sample functions) Columns 1 through 7 0.0909 0.0292 -0.0437 -0.0998 -0.1179 -0.0909 -0.0292 Columns 8 through 11 0.0437 0.0998 0.1179 0.0909 Typical time-average means (sample functions 1, 20, 40, 60, 80, & 100) -0.0875 -0.0815 -0.0733 -0.0872 0.0881 0.0273 Sample mean squares Columns 1 through 7 0.4960 0.5499 0.5348 0.4717 0.4477 0.4960 0.5499 Columns 8 through 11 0.5348 0.4717 0.4477 0.4960 Typical time-average mean squares 0.5387 0.5277 0.5136 0.5381 0.5399 0.4628 Computer Exercise 5.2 This is a matter of changing the statement theta = 2*pi*rand(1,100); in the program of Computer Exercise 5.1 to theta = (pi/2)*(rand(1,100) - 0.5); The time-average means will be the same as in Computer Exercise 5.1, but the ensembleaverage means will vary depending on the time at which they are computed. Computer Exercise 5.3 This was done in Computer exercise 4.2 by printing the covariance matrix. The diagonal terms give the variances of the X and Y vectors and the off-diagonal terms give the correlation coefficients. Ideally, they should be zero. 22 CHAPTER 5. RANDOM SIGNALS AND NOISE Figure 5.1: Computer Exercise 5.4 % ce5_4.m: plot of Ricean pdf for several values of K % clf A = char(‘-’,‘—’,‘-.’,‘:’,‘—.’,‘-..’); sigma = input(‘Enter desired value of sigma ’); r = 0:.1:15*sigma; n = 1; KdB = []; for KK = -10:5:15; KdB(n) = KK; K = 10.^(KK/10); fR = (r/sigma^2).*exp(-(r.^2/(2*sigma^2)+K)).*besseli(0, sqrt(2*K)*(r/sigma)); plot(r, fR, A(n,:)) if n == 1 hold on xlabel(’r’), ylabel(’f_R(r)’) grid on 5.2. COMPUTER EXERCISES 23 Figure 5.2: end n = n+1; end legend([‘K = ’, num2str(KdB(1)),‘ dB’],[‘K = ’, num2str(KdB(2)),‘ dB’], [‘K = ’, num2str(KdB(3)),‘ dB’], [‘K = ’, num2str(KdB(4)),‘ dB’],[‘K = ’, num2str(KdB(5)),‘ dB’], [‘K = ’, num2str(KdB(6)),‘ dB’]) title([‘Ricean pdf for \sigma = ’, num2str(sigma)]) A plot for several values of K is shown below: Chapter 6 Noise in Modulation Systems 6.1 Problems Problem 6.1 The signal power at the output of the lowpasss …lter is P T . The noise power is N 0 B N , where B N is the noise-equivalent bandwidth of the …lter. From (5.116), we know that the noise-equivalent bandwidth of an n th order Butterworth …lter with 3 dB bandwidth W is B n (n) = ¼W=2n sin (¼=2n) Thus, the signal-to-noise ratio at the …lter output is SNR = P T N 0 B N = sin (¼=2n) ¼=2n P T N 0 W so that f (n) is given by f (n) = sin (¼=2n) ¼=2n We can see from the form of f (n) that since sin (x) ¼ 1 for x ¿ 1, f (1) = 1. Thus for large n, the SNR at the …lter output is close to P T =N 0 W. The plot is shown In Figure 6.1. Problem 6.2 We express n(t) as n (t) =n c (t) cos · ! c t § 1 2 (2¼W) t +µ ¸ +n s (t) sin · ! c t § 1 2 (2¼W) t +µ ¸ where we use the plus sign for the USB and the minus sign for the LSB. The received signal is x r (t) =A c [m(t) cos (! c t +µ) ¨ b m(t) sin (! c t +µ)] 1 2 CHAPTER 6. NOISE IN MODULATION SYSTEMS 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 n f ( n ) Figure 6.1: Multiplying x r (t) +n(t) by 2 cos (! c t +µ) and lowpass …ltering yields y D (t) =A c m(t) +n c (t) cos (¼Wt) §n s (t) sin (¼Wt) From this expression, we can show that the postdetection signal and noise powers are given by S D = A 2 c m 2 N D =N 0 W S T = A c m 2 N T = N 0 W This gives a detection gain of one. The power spectral densities of n c (t) and n s (t) are illustrated in Figure 6.2. Problem 6.3 The received signal and noise are given by x r (t) =A c m(t) cos (! c t +µ) +n(t) At the output of the predetection …lter, the signal and noise powers are given by S T = 1 2 A 2 c m 2 N T =n 2 = N 0 B T The predetection SNR is (SNR) T = A 2 c m 2 2N 0 B T 6.1. PROBLEMS 3 1 2 W 1 2 W − ( ) ( ) , c s n n S f S f f 0 N Figure 6.2: 1 2 T B 1 2 T B − ( ) c n S f 0 f 0 N Figure 6.3: If the postdetection …lter passes all of the n c (t) component, y D (t) is y D (t) =A c m(t) +n c (t) The output signal power is A 2 c m 2 and the output noise PSD is shown in Figure 6.3. Case I: B D > 1 2 B T For this case, all of the noise, n c (t), is passed by the postdetection …lter, and the output noise power is N D = Z 1 2 BT ¡ 1 2 B D N 0 df = 2N 0 B T This yields the detection gain (SNR) D (SNR) T = A 2 c m 2 =N 0 B T A 2 c m 2 =2N 0 B T = 2 4 CHAPTER 6. NOISE IN MODULATION SYSTEMS Case II: B D < 1 2 B T For this case, the postdetection …lter limits the output noise and the output noise power is N D = Z BD ¡B D N 0 df = 2N 0 B D This case gives the detection gain (SNR) D (SNR) T = A 2 c m 2 =2N 0 B D A 2 c m 2 =2N 0 B T = B T B D Problem 6.4 This problem is identical to Problem 6.3 except that the predetection signal power is S T = 1 2 A 2 c h 1 +a 2 m 2 n i and the postdetection signal power is S D = A 2 c a 2 m 2 n The noise powers do not change. Case I: B D > 1 2 B T (SNR) D (SNR) T = A 2 c a 2 m 2 n =N 0 B T A 2 c h 1 +a 2 m 2 n i =2N 0 B T = 2a 2 m 2 n 1 +a 2 m 2 n Case II: B D < 1 2 B T (SNR) D (SNR) T = A 2 c a 2 m 2 n =2N 0 B D A 2 c h 1 +a 2 m 2 n i =2N 0 B T = a 2 m 2 n 1 +a 2 m 2 n = B T B D Problem 6.5 Since the message signal is sinusoidal m n (t) = cos (8¼t) 6.1. PROBLEMS 5 Thus, m 2 n = 0:5 The e¢ciency is therefore given by E ff = (0:5) (0:8) 2 1 + (0:5) (0:8) 2 = 0:2424 = 24:24% From (6.29), the detection gain is (SNR) D (SNR) T = 2E ff = 0:4848 = ¡3:14dB and the output SNR is, from (6.33), (SNT) D = 0:2424 P T N 0 W Relative to baseband, the output SNR is (SNR) D P T =N 0 W = 0:2424 =¡6:15 dB If the modulation index in increased to 0:9, the e¢ciency becomes E ff = (0:5) (0:9) 2 1 + (0:5) (0:9) 2 = 0:2883 = 28:83% This gives a detection gain of (SNR) D (SNR) T = 2E ff = 0:5765 = ¡2:39dB The output SNR is (SNR) D = 0:2883 P T N 0 W which, relative to baseband, is (SNR) D P T =N 0 W = 0:2883 =¡5:40 dB This represents an improvement of 0:75 dB. Problem 6.6 6 CHAPTER 6. NOISE IN MODULATION SYSTEMS The …rst step is to determine the value of M. First we compute P fX > Mg = Z 1 M 1 p 2¼¾ e ¡y 2 =2¾ 2 dy =Q µ M ¾ ¶ = 0:005 This gives M ¾ = 2:57 Thus, M = 2:57¾ and m n (t) = m(t) 2:57¾ Thus, since m 2 =¾ 2 m 2 n = m 2 (2:57¾) 2 = ¾ 2 6:605¾ 2 = 0:151 Since a = 1 2 , we have E ff = 0:151 ¡ 1 2 ¢ 2 1 + 0:151 ¡ 1 2 ¢ 2 = 3:64% and the detection gain is (SNR) D (SNR) T = 2E = 0:0728 Problem 6.7 The output of the predetection …lter is e(t) =A c [1 +am n (t)] cos [! c t +µ] +r n (t) cos [! c t +µ +Á n (t)] The noise function r n (t) has a Rayleigh pdf. Thus f R n (r n ) = r N e ¡r 2 =2N where N is the predetection noise power. This gives N = 1 2 n 2 c + 1 2 n 2 s = N 0 B T From the de…nition of threshold 0:99 = Z Ac 0 r N e ¡r 2 =2N dr 6.1. PROBLEMS 7 which gives 0:99 = 1 ¡e ¡A 2 c =2N Thus, ¡ A 2 c 2N = 1n(0:01) which gives A 2 c = 9:21 N The predetection signal power is P T = 1 2 A c h 1 +a 2 m 2 n i ¼ 1 2 A 2 c [1 +1] =A 2 c which gives P T N = A 2 c N = 9:21 ¼9:64dB Problem 6.8 Since m(t) is a sinusoid, m 2 n = 1 2 , and the e¢ciency is E ff = 1 2 a 2 1 + 1 2 a 2 = a 2 2 +a 2 and the output SNR is (SNR) D =E ff = a 2 2 +a 2 P T N 0 W In dB we have (SNR) D;dB = 10log 10 µ a 2 2 +a 2 ¶ + 10 log 10 P T N 0 W For a = 0:4, (SNR) D;dB = 10log 10 P T N 0 W ¡11:3033 For a = 0:5, (SNR) D;dB = 10 log 10 P T N 0 W ¡9:5424 For a = 0:7, (SNR) D;dB = 10 log 10 P T N 0 W ¡7:0600 For a = 0:9, (SNR) D;dB = 10 log 10 P T N 0 W ¡5:4022 8 CHAPTER 6. NOISE IN MODULATION SYSTEMS The plot of (SNR) dB as a function of P T =N 0 W in dB is linear having a slope of one. The modulation index only biases (SNR) D;dB down by an amount given by the last term in the preceding four expressions. Problem 6.9 Let the predetection …lter bandwidth be B T and let the postdetection …lter bandwidth be B D . The received signal (with noise) at the predetection …lter output is represented x r (t) =A c [1 +am n (t)] cos ! c t +n c (t) cos ! c t +n s (t) sin ! c t which can be represented x r (t) = fA c [1 +am n (t)] +n c (t)g cos ! c t ¡n s (t) sin ! c t The output of the square-law device is y (t) = fA c [1 +am n (t)] +n c (t)g 2 cos 2 ! c t ¡fA c [1 +am n (t)] +n c (t)gn s (t) cos ! c (t) +n 2 s (t) sin 2 cos ! c (t) Assuming that the postdetection …lter removes the double frequency terms, y D (t) can be written y D (t) = 1 2 fA c [1 +am n (t)] +n c (t)g 2 + 1 2 n 2 s (t) Except for a factor of 2, this is (6.50). We now let m n (t) = cos ! m t and expand. This gives y D (t) = 1 2 A 2 c [1 +acos ! m t] 2 +A c n c (t) +A c an c (t) cos ! m t + 1 2 n 2 c (t) + 1 2 n 2 s (t) We represent y D (t) by y D (t) =z 1 (t) +z 2 (t) +z 3 (t) +z 4 (t) +z 5 (t) where z i (t) is the i th term in the expression for y D (t). We now examine the power spectral density of each term. The …rst term is z 1 (t) = 1 2 A 2 c µ 1 + 1 2 a 2 ¶ +A 2 c acos ! m t + 1 4 A 2 c a 2 cos 2! m t z 2 (t) =A c n c (t) z 3 (t) =A c an c (t) cos ! m t 6.1. PROBLEMS 9 z 4 (t) = 1 2 n 2 c (t) and z 5 (t) = 1 2 n 2 s (t) A plot of the PSD of each of these terms is illiustrated in Figure 6.4. Adding these …ve terms together gives the PSD of y D (t). Problem 6.10 Assume sinusoidal modulation since sinusoidal modulation was assumed in the development of the square-law detector. For a = 1 and m n (t) = cos (2¼f m t) so that m 2 n = 0:5, we have, for linear envelope detection (since P T =N 0 W À1, we use the coherent result), (SNR) D;` = E ff P T N 0 W = a 2 m 2 n 1 +a 2 m 2 n P T N 0 W = 1 2 1 + 1 2 P T N 0 W = 1 3 P T N 0 W For square-law detection we have, from (6.61) with a = 1 (SNR) D;S` = 2 µ a 2 +a 2 ¶ 2 P T N 0 W = 2 9 P T N 0 W Taking the ratio (SNR) D;S` (SNR) D;` = 2 9 P T N 0 W 1 3 P T N 0 W = 2 3 = ¡1:76 dB This is approximately ¡1:8 dB. Problem 6.11 For the circuit given H(f) = R R +j2¼fL = 1 1 +j ³ 2¼fL R ´ and jH (f)j 2 = 1 1 + ³ 2¼fL R ´ 2 The output noise power is N = Z 1 ¡1 N 0 2 df 1 + ³ 2¼fL R ´ 2 = N 0 Z 1 0 1 1 +x 2 µ R 2¼L ¶ dx = N 0 R 4L 10 CHAPTER 6. NOISE IN MODULATION SYSTEMS 2 m f m f − m f − − 2 m f − 4 4 1 64 C A a 4 2 1 4 C A a 2 4 2 1 1 1 4 2 C A a   +     4 2 1 4 C A a 4 1 64 C A ( ) 1 Sz f 0 f 1 2 T B 1 2 T B − 2 0 C A N ( ) 2 Sz f 0 f 1 2 T m B f + 1 2 T m B f − 1 2 T m B f − + 1 2 T m B f − − ( ) 3 Sz f 0 f 2 0 1 4 T N B ( ) ( ) 4 5 , Sz f Sz f 0 f Figure 6.4: 6.1. PROBLEMS 11 The output signal power is S = 1 2 A 2 R 2 R 2 +(2¼f c L) 2 This gives the signal-to-noise ratio S N = 2A 2 RL N 0 h R 2 + (2¼f c L) 2 i Problem 6.12 For the RC …lter S N = 2A 2 RL N 0 h 1 +(2¼f c RC) 2 i Problem 6.13 The transfer function of the RC highpass …lter is H(f) = R R + 1 j2¼C = j2¼fRC 1 +j2¼fRC so that H (f) = (j2¼fRC) 2 1 +(j2¼fRC) 2 Thus, The PSD at the output of the ideal lowpass …lter is S n (f) = ( N0 2 (2¼fRC) 2 1+(2¼fRC) 2 , 0, jfj <W jfj >W The noise power at the output of the ideal lowpass …lter is N = Z W ¡W S n (f) df =N 0 Z W 0 (2¼fRC) 2 1 + (2¼fRC) 2 df with x = 2¼fRC, the preceding expression becomes N = N 0 2¼RC Z 2¼RCW 0 x 2 1 +x 2 dx Since x 2 1 +x 2 = 1 ¡ 1 1 +x 2 12 CHAPTER 6. NOISE IN MODULATION SYSTEMS we can write N = N 0 2¼RC ½Z 2¼RCW 0 df ¡ Z 2¼RCW 0 dx 1 +x 2 ¾ or N = N 0 2¼RC ¡ 2¼RCW ¡tan ¡1 (2¼RCW) ¢ which is N = N 0 W ¡ N 0 tan ¡1 (2¼RCW) 2¼RC The output signal power is S = 1 2 A 2 jH (f c )j 2 = A 2 2 (2¼f c RC) 2 1 + (2¼f c RC) 2 Thus, the signal-to-noise ratio is S N = A 2 2N 0 (2¼f c RC) 2 1 +(2¼f c RC) 2 2¼RC 2¼RCW ¡tan ¡1 (2¼RCW) Note that as W !1; S N !0. Problem 6.14 For the case in which the noise in the passband of the postdetection …lter is negligible we have ² 2 Q = ¾ 2 Á ; SSB and QDSB and ² 2 D = 3 4 ¾ 4 Á ; DSB Note that for reasonable values of phase error variance, namely ¾ 2 Á ¿1, DSB is much less sensitive to phase errors in the demodulation carrier than SSB or QDSB. Another way of viewing this is to recognize that, for a given acceptable level of ² 2 , the phase error variance for can be much greater for DSB than for SSB or QDSB. The plots follow by simply plotting the two preceding expressions. Problem 6.15 From the series expansions for sinÁ and cos Á we can write cos Á = 1 ¡ 1 2 Á 2 + 1 24 Á 4 sinÁ = Á ¡ 1 6 Á 3 6.1. PROBLEMS 13 Squaring these, and discarding all terms Á k for k >4, yields cos 2 Á = 1 ¡Á 2 + 1 3 Á 4 sin 2 Á = Á 2 ¡ 1 3 Á 4 Using (6.70) and recognizing that m 1 (t), m 2 (t), and Á(t) are independent, yields " 2 =¾ 2 m1 ¡2¾ 2 m1 cos Á +¾ 2 m1 cos 2 Á +¾ 2 m2 sin 2 Á +¾ 2 n Assuming Á(t) to be a zero-mean process and recalling that Á 4 = 3¾ 4 Á gives cos Á = 1 + 1 2 ¾ 2 Á + 1 8 ¾ 4 Á cos 2 Á = 1 ¡¾ 2 Á +¾ 4 Á sin 2 Á = ¾ 2 Á ¡¾ 4 Á This yields " 2 =¾ 2 m1 ¡2¾ 2 m1 µ 1 + 1 2 ¾ 2 Á + 1 8 ¾ 4 Á ¶ +¾ 2 m1 ¡ 1 ¡¾ 2 Á +¾ 4 Á ¢ +¾ 2 m2 ¡ ¾ 2 Á ¡¾ 4 Á ¢ +¾ 2 n which can be expressed " 2 = 3 4 ¾ 2 m1 ¾ 4 Á +¾ 2 m2 ¾ 2 Á ¡¾ 2 m2 ¾ 4 Á +¾ 2 n For QDSB we let ¾ 2 m1 = ¾ 2 m2 =¾ 2 m . This gives " 2 =¾ 2 m 2 µ ¾ 2 Á ¡ 1 4 ¾ 4 Á ¶ +¾ 2 n For ¾ 2 Á >>¾ 4 Á , we have " 2 =¾ 2 m ¾ 2 Á +¾ 2 n which yields (6.73). For DSB, we let ¾ 2 m1 =¾ 2 m and ¾ 2 m2 = 0. This gives (6.79) " 2 = 3 4 ¾ 2 m ¾ 4 Á +¾ 2 n 14 CHAPTER 6. NOISE IN MODULATION SYSTEMS Problem 6.16 From (6.73) and (6.79), we have 0:05 = ¾ 2 Á + ¾ 2 n ¾ 2 m ; SSB 0:05 = 3 4 ¡ ¾ 2 Á ¢ 2 + ¾ 2 n ¾ 2 m ; DSB Thus we have the following signal-to-noise ratios ¾ 2 m ¾ 2 n = 1 0:05 ¡¾ 2 Á ; SSB ¾ 2 m ¾ 2 n = 1 0:05 ¡ 3 4 ³ ¾ 2 Á ´ 2 ; DSB The SSB characteristic has an asymptote de…ned by ¾ 2 Á = 0:05 and the DSB result has an asymptote de…ned by 3 4 ¡ ¾ 2 Á ¢ 2 = 0:05 or ¾ 2 Á = 0:258 The curves are shown in Figure 6.5. The appropriate operating regions are above and to the left of the curves. It is clear that DSB has the larger operating region. Problem 6.17 Since we have a DSB system " 2 N = 3 4 ¾ 4 Á + ¾ 2 n ¾ 2 m Let the bandwidth of the predetection …lter be B T and let the bandwidth of the pilot …lter be B p = B T ® This gives ¾ 2 n ¾ 2 m = N 0 B T ¾ 2 m = B T B p N 0 B p ¾ 2 m = ® ½ 6.1. PROBLEMS 15 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 100 200 300 400 500 600 700 800 900 1000 Phase error variance SNR SSB DSB Figure 6.5: From (6.85) and (6.86), we have ¾ 2 Á = 1 k 2 1 2½ so that " 2 N = 3 16 1 k 4 ½ 2 + ® ½ For an SNR of 15 dB ½ = 10 1:5 = 31:623 Using this value for ½ and with k = 4, we have " 2 N = 7:32 ¡ 10 ¡7 ¢ + 0:032® The plot is obviously linear in ® with a slope of 0:032. The bias, 7:32 £10 ¡7 , is negligible. Note that for k >1 and reasonable values of the pilot signal-to-noise ratio, ½, the …rst term (the bias), which arises from pilot phase jitter, is negligible. The performance is limited by the additive noise. Problem 6.18 The mean-square error " 2 (A; ¿) =E © [y (t) ¡Ax(t ¡¿)] 2 ª 16 CHAPTER 6. NOISE IN MODULATION SYSTEMS can be written " 2 (A; ¿) =E © y 2 (t) ¡2Ax(t ¡¿)y (t) +A 2 x 2 (t ¡¿) ª In terms of autocorrelation and cross correlation functions, the mean-square error is " 2 (A; ¿) = R y (0) ¡2AR xy (¿) +A 2 R x (0) Letting P y =R y (0) and P x =R x (0) gives " 2 (A; ¿) =P y ¡2AR xy (¿) +A 2 P x In order to minimize " 2 we choose ¿ = ¿ m such that R xy (¿) is maximized. This follows since the crosscorrelation term is negative in the expression for " 2 . Therefore, " 2 (A; ¿ m ) =P y ¡2AR xy (¿ m ) +A 2 P x The gain A m is determined from d" 2 dA = ¡2AR xy (¿ m ) +A 2 P x = 0 which yields A m = R xy (¿ m ) P x This gives the mean-square error " 2 (A m ; ¿ m ) =P y ¡2 R 2 xy (¿ m ) P x + R 2 xy (¿ m ) P x which is " 2 (A m ; ¿ m ) = P y ¡ P 2 xy (¿ m ) P x The output signal power is S D =E © [A m x(t ¡¿ m )] 2 ª =A 2 m P x which is S D = R 2 xy (¿ m ) P x = R 2 xy (¿ m ) R x (0) Since N D is the error " 2 (A m ; ¿ m ) we have S D N D = R 2 xy (¿ m ) R x (0) R y (0) ¡R 2 xy (¿ m ) 6.1. PROBLEMS 17 Note: The gain and delay of a linear system is often de…ned as the magnitude of the transfer and the slope of the phase characteristic, respectively. The de…nition of gain and delay suggested in this problem is useful when the magnitude response is not linear over the frequency range of interest. Problem 6.19 The single-sided spectrum of a stereophonic FM signal and the noise spectrum is shown in Figure 6.6. The two-sided noise spectrum is given by S nF (f) = K 2 D A 2 C N 0 f, ¡1< f < 1 The predetection noise powers are easily computed. For the L+R channel P n;L+R = 2 Z 15;000 0 K 2 D A 2 C N 0 f 2 df = 2:25 ¡ 10 12 ¢ K 2 D A 2 C N 0 For the L¡R channel P n;L¡R = 2 Z 53;000 23;000 K 2 D A 2 C N 0 f 2 df = 91:14 ¡ 10 12 ¢ K 2 D A 2 C N 0 Thus, the noise power on the L¡R channel is over 40 times the noise power in the L+R channel. After demodulation, the di¤erence will be a factor of 20 because of 3 dB detection gain of coherent demodulation. Thus, the main source of noise in a stereophonic system is the L¡R channel. Therefore, in high noise environments, monophonic broadcasting is preferred over stereophonic broadcasting. Problem 6.20 The received FDM spectrum is shown in Figure 6.7. The kth channel signal is given by x k (t) =A k m k (t) cos 2¼kf 1 t Since the guardbands have spectral width 4W, fk = 6kW and the k th channel occupies the frequency band (6k ¡1) W · f · (6k +1) W Since the noise spectrum is given by S nF = K 2 D A 2 C N 0 f 2 , jfj < B T The noise power in the kth channel is N k =B Z (6k+1)W (6k¡1)W f 2 df = BW 3 3 h (6k +1) 3 ¡(6k ¡1) 3 i 18 CHAPTER 6. NOISE IN MODULATION SYSTEMS Noise Spectrum ( ) nF S f ( ) ( ) L f R f + ( ) ( ) L f R f + Pilot 53 23 15 0 ( ) f kHz Figure 6.6: where B is a constant. This gives N k = BW 3 3 ¡ 216k 2 + 2 ¢ » = 72BW 3 k 2 The signal power is proportional to A 2 k . Thus, the signal-to-noise ratio can be expressed as (SNR) D = ¸A 2 k k 2 =¸ µ A k k ¶ 2 where ¸ is a constant of proportionality and is a function of K D ,W,A C ,N 0 . If all channels are to have equal (SNR) D , A k =k must be a constant, which means that A k is a linear function of k. Thus, if A 1 is known, A k = k A 1 ; k = 1; 2; :::; 7. A 1 is …xed by setting the SNR of Channel 1 equal to the SNR of the unmodulated channel. The noise power of the unmodulated channel is N 0 =B Z W 0 f 2 df = B 3 W 3 yielding the signal-to-noise ratio (SNR) D = P 0 B 3 W 3 where P 0 is the signal power. Problem 6.21 6.1. PROBLEMS 19 7 f 6 f 5 f 4 f 3 f 2 f 1 f Noise PSD 0 f Figure 6.7: From (6.132) and (6.119), the required ratio is R = 2 K 2 D A 2 C N 0 f 3 3 ³ W f3 ¡tan ¡1 W f3 ´ 2 3 K 2 D A 2 C N 0 W 3 or R = 3 µ f 3 W ¶ 3 µ W f 3 ¡tan ¡1 W f 3 ¶ This is shown in Figure 6.8. For f 3 = 2:1kHz and W = 15kHz, the value of R is R = 3 µ 2:1 15 ¶ 2 µ 15 2:1 ¡tan ¡1 15 2:1 ¶ = 0:047 Expressed in dB this is R = 10log 10 (0:047) =¡13:3 dB The improvement resulting from the use of preemphasis and deemphasis is therefore 13:3 dB. Neglecting the tan ¡1 (W=f 3 ) gives an improvement of 21 ¡8:75 = 12:25 dB. The di¤erence is approximately 1 dB. Problem 6.22 From the plot of the signal spectrum it is clear that k = A W 2 20 CHAPTER 6. NOISE IN MODULATION SYSTEMS 0 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 W/f3 R Figure 6.8: Thus the signal power is S = 2 Z W 0 A W 2 f 2 df = 2 3 AW The noise power is N 0 B. This yields the signal-to-noise ratio (SNR) 1 = 2 3 AW N 0 B If B is reduced to W, the SNR becomes (SNR) 2 = 2 3 A N 0 This increases the signal-to-noise ratio by a factor of B=W. Problem 6.23 From the de…nition of the signal we have x(t) = Acos 2¼f c t dx dt = ¡2¼f c Asin2¼f c t d 2 x dt 2 = ¡(2¼f c ) 2 Acos 2¼f c t 6.1. PROBLEMS 21 The signal component of y (t) therefore has power S D = 1 2 A 2 (2¼f c ) 4 = 8A 2 ¼ 4 f 4 c The noise power spectral density at y (t) is S n (f) = N 0 2 (2¼f) 4 so that the noise power is N D = N 0 2 (2¼) 4 Z W ¡W f 4 df = 16 5 N 0 ¼ 4 W 5 Thus, the signal-to-noise ratio at y (t) is (SNR) D = 5 2 A 2 N 0 W µ f c W ¶ 4 Problem 6.24 Since the signal is integrated twice and then di¤erentiated twice, the output signal is equal to the input signal. The output signal is y s (t) =Acos 2¼f c t and the output signal power is S D = 1 2 A 2 The noise power is the same as in the previous problem, thus N D = 16 5 N 0 ¼ 4 W 5 This gives the signal-to-noise ratio (SNR) D = 5 32¼ 4 A 2 N 0 W 5 Problem 6.25 The signal-to-noise ratio at y (t) is (SNR) D = 2A N 0 µ f 3 W ¶ tan ¡1 W f 3 22 CHAPTER 6. NOISE IN MODULATION SYSTEMS 0 1 2 3 4 5 6 7 8 9 10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 W/f3 N o r m a l i z e d S N R Figure 6.9: The normalized (SNR) D , de…ned by (SNR) D =(2A?N 0 ); is illustrated in Figure 6.9. Problem 6.26 The instantaneous frequency deviation in Hz is ±f = f d m(t) =x(t) where x(t) is a zero-mean Gaussian process with variance ¾ 2 x =f 2 d ¾ 2 m . Thus, j±fj =jxj = Z 1 ¡1 jxj p 2¼¾ x e ¡x 2 =2¾ 2 x dx Recognizing that the integrand is even gives j±fj = r 2 ¼ 1 ¾ x Z 1 0 xe ¡x 2 =2¾ 2 x dx = r 2 ¼ ¾ x Therefore, j±fj = r 2 ¼ f d ¾ m 6.1. PROBLEMS 23 Substitution into (6.148) gives (SNR) D = 3 ³ f d W ´ 2 m 2 n P T N0W 1 + 2 p 3 B T W Q · q A 2 C N 0 B T ¸ + 6 q 2 ¼ f d ¾m W exp h ¡A 2 C 2N 0 B T i The preceding expression can be placed in terms of the deviation ratio, D, by letting D = f d W and B T = 2 (D+ 1) W Problem 6.27 (Note: This problem was changed after the …rst printing of the book. The new problem, along with the solution follow.) Assume a PCM system in which the the bit error probability P b is su¢ciently small to justify the approximation that the word error pobability is P w ¼ nP b . Also assume that the threshold value of the signal-to-noise ratio, de…ned by (6.175), occurs when the two denominator terms are equal, i.e., the e¤ect of quantizating errors and word errors are equivalent. Using this assumption derive the threshold value of P T =N 0 B p in dB for n = 4; 8; and 12. Compare the results with Figure 6.22 derived in Example 6.5. With P w =nP b , equating the two terms in the denominator of (6.175) gives 2 ¡2n = nP b (1 ¡2 ¡2n ) Solving for P b we have P n = 1 n 2 ¡2n 1 ¡2 ¡2n ¼ 1 n 2 ¡2n From (6.178) exp · ¡ P T 2N 0 B p ¸ = 2P b ¼ 1 n 2 (1¡2n) Solving for P T =N 0 B p gives, at threshold, P T N 0 B p ¼ ¡2 ln · 1 n 2 (1¡2n) ¸ = 2(2n ¡1) ln(2) + 2 ln(2) The values of P T =N 0 B p for n = 4; 8; and 12 are given in Table 6.1. Comparison with Figure 6.22 shows close agreement. 24 CHAPTER 6. NOISE IN MODULATION SYSTEMS Table 6.1: n Threshold value of P T =N 0 B p 4 10:96 dB 8 13:97 dB 12 15:66 dB Problem 6.28 The peak value of the signal is 7:5 so that the signal power is S D = 1 2 (7:5) 2 = 28:125 If the A/D converter has a wordlength n, there are 2 n quantizing levels. Since these span the peak-to-peak signal range, the width of each quantization level is S D = 15 2 n = 15 ¡ 2 ¡n ¢ This gives " 2 = 1 12 S 2 D = 1 12 (15) 2 ¡ 2 ¡2n ¢ = 18:75 ¡ 2 ¡2n ¢ This results in the signal-to-noise ratio SNR = S D " 2 = 28:125 18:75 ¡ 2 2n ¢ = 1:5 ¡ 2 2n ¢ Chapter 7 Binary Data Communication 7.1 Problem Solutions Problem 7.1 The signal-to-noise ratio is z = A 2 T N 0 = A 2 N 0 R Trial and error using the asymptotic expression Q(x) ≈ exp ¡ −x 2 /2 ¢ / ¡√ 2πx ¢ shows that P E = Q ³ √ 2z ´ = 10 −5 for z = 9.58 dB = 9.078 Thus A 2 N 0 R = 9.078 or A = p 9.078N 0 R = p 9.078 × 10 −7 × 20000 = 0.135 V Problem 7.2 The bandwidth should be equal to the data rate to pass the main lobe of the signal spectrum. By trial and error, solving the following equation for z, P E | desired = Q ³q 2 z| required ´ , z = E b /N 0 1 2 CHAPTER 7. BINARY DATA COMMUNICATION we find that z| required = 6.78 dB (4.76 ratio), 8.39 dB (6.90 ratio), 9.58 dB (9.08 ratio), 10.53 dB (11.3 ratio) for P E | desired = 10 −3 , 10 −4 , 10 −5 , 10 −6 , respectively. The required signal power is P s, required = E b, required /T symbol = E b, required R = z| required N 0 R For example, for P E = 10 −3 and R = 1000 bits/second, we get P s, required = z| required N 0 R = 4.76 × 10 −6 × 1000 = 4.76 × 10 −3 W Similar calculations allow the rest of the table to be filled in, giving the following results. R, bps P E = 10 −3 P E = 10 −4 P E = 10 −5 P E = 10 −6 1, 000 4.76 × 10 −3 W 6.9 × 10 −3 W 9.08 × 10 −3 W 11.3 × 10 −3 W 10, 000 4.76 × 10 −2 W 6.9 × 10 −2 W 9.08 × 10 −2 W 11.3 × 10 −2 W 100, 000 4.76 × 10 −1 W 6.9 × 10 −1 W 9.08 × 10 −1 W 1.13 W Problem 7.3 (a) For R = B = 5, 000 bps, we have signal power = A 2 = z| required N 0 R = 10 10.53/10 × 10 −6 × 5, 000 = 11.3 × 5 × 10 −3 = 0.0565 W where z| required = 10.53 dB from Problem 7.2. (b) For R = B = 50, 000 bps, the result is signal power = A 2 = 0.565 W (c) For R = B = 500, 000 bps, we get signal power = A 2 = 5.65 W (d) For R = B = 1, 000, 000 bps, the result is signal power = A 2 = 11.3 W Problem 7.4 The decision criterion is V > ε, choose +A V < ε, choose −A 7.1. PROBLEM SOLUTIONS 3 where V = ±AT +N N is a Gaussian random variable of mean zero and variance σ 2 = N 0 T/2. Because P (A sent) = P (−A sent) = 1/2, it follows that P E = 1 2 P (−AT +N > ε | −A sent) + 1 2 P (AT +N < ε | A sent) = 1 2 (P 1 +P 2 ) where P 1 = Z ∞ AT+ε e −η 2 /N 0 T √ πN 0 T dη = Q   s 2A 2 T N 0 + s 2ε 2 N 0 T   = Q ³ √ 2z + √ 2²/σ ´ , σ = N 0 T, z = A 2 T N 0 Similarly P 2 = Z −AT+ε −∞ e −η 2 /N 0 T √ πN 0 T dη = Z ∞ AT−ε e −η 2 /N 0 T √ πN 0 T dη = Q   s 2A 2 T N 0 − s 2ε 2 N 0 T   = Q ³ √ 2z − √ 2²/σ ´ Use the approximation Q(u) = e −u 2 /2 u √ 2π to evaluate 1 2 (P 1 +P 2 ) = 10 −6 by solving iteratively for z for various values of ε/σ. The results are given in the table below. ε/σ z dB for P E = 10 −6 ε/σ z dB for P E = 10 −6 0 10.54 0.6 11.34 0.2 10.69 0.8 11.67 0.4 11.01 1.0 11.98 4 CHAPTER 7. BINARY DATA COMMUNICATION Problem 7.5 Use the z| required values found in the solution of Problem 7.2 along with z| required = A 2 /N 0 R to get R = A 2 / z| required N 0 . Substitute N 0 = 10 −6 V 2 /Hz, A = 20 mV to get R = 400/ z| required . This results in the values given in the table below. P E z| required , dB (ratio) R, bps 10 −4 8.39 (6.9) 57.97 10 −5 9.58 (9.08) 43.05 10 −6 10.53 (11.3) 35.4 Problem 7.6 The integrate-and-dump detector integrates over the interval [0, T −δ] so that the SNR is z 0 = A 2 (T −δ) N 0 instead of z = A 2 T/N 0 . Thus z 0 z = 1 − T δ and the degradation in z in dB is D = −10 log 10 µ 1 − T δ ¶ Using δ = 10 −6 s and the data rates given, we obtain the following values of D: (a) for 1/T = R = 10 kbps, D = 0.04 dB; (b) for R = 50 kbps, D = 0.22 dB; (c) for R = 100 kbps, D = 0.46 dB. Problem 7.7 (a) For the pulse sequences (−A, −A) and (A, A), which occur half the time, no degradation results from timing error, so the error probability is Q ³ p 2E b /N 0 ´ . The error probability for the sequences (−A, A) and (A, −A), which occur the other half the time, result in the error probability Q h p 2E b /N 0 (1 −2 |∆T| /T) i (sketches of the pulse sequences with the integration interval offset from the transition between pulses is helpful here). Thus, the average error probability is the given result in the problem statement. Plots are shown in Figure 7.1 for the giving timing errors. (b) The plots given in Figure 7.1 for timing errors of 0 and 0.15 indicate that the degradation at P E = 10 −6 is about 2.8 dB. 7.1. PROBLEM SOLUTIONS 5 Figure 7.1: Problem 7.8 (a) The impulse response of the filter is h(t) = (2πf 3 ) e −2πf 3 t u(t) and the step response is the integral of the impulse response, or a (t) = ³ 1 −e −2πf 3 t ´ u(t) Thus, for a step of amplitude A, the signal out at t = T is s 0 (T) = A ³ 1 −e −2πf 3 T ´ The power spectral density of the noise at the filter output is S n (f) = N 0 /2 1 + (f/f 3 ) 2 so that the variance of the noise component at the filter output is E £ N 2 ¤ = Z ∞ −∞ S n (f) df = N 0 πf 3 2 = N 0 B N 6 CHAPTER 7. BINARY DATA COMMUNICATION where B N is the noise equivalent bandwidth of the filter. Thus SNR = s 2 0 (T) E[N 2 ] = 2A 2 ¡ 1 −e −2πf 3 T ¢ 2 N 0 πf 3 (b) To maximize the SNR, differentiate with respect to f 3 and set the result equal to 0. Solve the equation for f 3 . This results in the equation 4πf 3 Te −2πf 3 T −1 +e −2πf 3 T = 0 Let α = 2πf 3 T to get the equation (2α + 1) e −α = 1 A numerical solution results in α ≈ 1.25. Thus f 3, opt = 1.25 2πT = 0.199/T = 0.199R Problem 7.9 As in the derivation in the text for antipodal signaling, var [N] = 1 2 N 0 T The output of the integrator at the sampling time is V = ½ AT +N, A sent N, 0 sent The probabilities of error, given these two signaling events, are P (error | A sent) = P µ AT +N < 1 2 AT ¶ = P µ N < − 1 2 AT ¶ = Z −AT/2 −∞ e −η 2 /N 0 T √ πN 0 T dη = Q   s A 2 T 2N 0   7.1. PROBLEM SOLUTIONS 7 Similarly, P (error | 0 sent) = P µ N > 1 2 AT ¶ = Z ∞ AT/2 e −η 2 /N 0 T √ πN 0 T dη = Q   s A 2 T 2N 0   Thus P E = 1 2 P (error | A sent) + 1 2 P (error | 0 sent) = Q   s A 2 T 2N 0   But, the average signal energy is E ave = P (A sent) ×A 2 T +P (0 sent) × 0 = 1 2 A 2 T Therefore, the average probability of error is P E = Q " r E ave N 0 # Problem 7.10 Substitute (7.27) and (7.28) into P E = pP (E | s 1 (t)) + (1 −p) P (E | s 2 (t)) = p Z ∞ k exp h −(v −s 01 ) 2 /2σ 2 0 i p 2πσ 2 0 dv + (1 −p) Z k −∞ exp h −(v −s 02 ) 2 /2σ 2 0 i p 2πσ 2 0 dv Use Leibnitz’s rule to differentiate with respect to the threshold, k, set the result of the differentiation equal to 0, and solve for k = k opt . The differentiation gives −p exp h −(k −s 01 ) 2 /2σ 2 0 i p 2πσ 2 0 + (1 −p) exp h −(k −s 02 ) 2 /2σ 2 0 i p 2πσ 2 0 ¯ ¯ ¯ ¯ ¯ ¯ k=k opt = 0 8 CHAPTER 7. BINARY DATA COMMUNICATION or p exp h −(k opt −s 01 ) 2 /2σ 2 0 i p 2πσ 2 0 = (1 −p) exp h −(k opt −s 02 ) 2 /2σ 2 0 i p 2πσ 2 0 or −(k opt −s 01 ) 2 + (k opt −s 02 ) 2 = 2σ 2 0 ln µ 1 −p p ¶ (s 01 −s 02 ) k opt + s 2 02 −s 2 01 2 = σ 2 0 ln µ 1 −p p ¶ k opt = σ 2 0 s 01 −s 02 ln µ 1 −p p ¶ + s 02 −s 01 2 Problem 7.11 (a) Denote the output due to the signal as s 0 (t) and the output due to noise as n 0 (t). We wish to maximize s 2 0 (t) E £ n 2 0 (t) ¤ = 2 N 0 ¯ ¯ ¯ R ∞ −∞ S (f) H m (f) e j2πft 0 df ¯ ¯ ¯ 2 R ∞ −∞ H m (f) H ∗ m (f) df By Schwartz’s inequality, we have s 2 0 (t) E £ n 2 0 (t) ¤ ≤ 2 N 0 R ∞ −∞ S (f) S ∗ (f) df R ∞ −∞ H m (f) H ∗ m (f) df R ∞ −∞ H m (f) H ∗ m (f) e j2πft 0 df = 2 N 0 Z ∞ −∞ |S (f)| 2 df where the maximum value on the right-hand side is attained if H m (f) = S ∗ (f) e −j2πft 0 (b) The matched filter impulse response is h m (t) = s (t 0 −t) by taking the inverse Fourier transform of H m (f) employing the time reversal and time delay theorems. (c) The realizable matched filter has zero impulse response for t < 0. (d) By sketching the components of the integrand of the convolution integral, we find the following results: For t 0 = 0, s 0 (t 0 ) = 0; For t 0 = T/2, s 0 (t 0 ) = A 2 T/2; For t 0 = T, s 0 (t 0 ) = A 2 T; For t 0 = 2T, s 0 (t 0 ) = A 2 T. 7.1. PROBLEM SOLUTIONS 9 Problem 7.12 (a) These would be x(t) and y (t) reversed and then shifted to the right so that they are nonzero for t > 0. (b) A = √ 7B. (c) The outputs for the two cases are S o1 (t) = A 2 τΛ[(t −t 0 ) /τ] = 7B 2 τΛ[(t −t 0 ) /τ] S o2 (t) = 7B 2 τΛ[(t −t 0 ) /7τ] (d) The shorter pulse gives the more accurate time delay measurement. (e) Peak power is lower for y (t). Problem 7.13 (a) The matched filter impulse response is given by h m (t) = s 2 (T −t) −s 1 (T −t) (b) Using (7.55) and a sketch of s 2 (t) −s 1 (t) for an arbitrary t 0 , we obtain ζ 2 = 2 N 0 Z ∞ −∞ [s 2 (t) −s 1 (t)] 2 dt = 2 N 0 · A 2 t 0 + 4A 2 t 0 +A 2 µ T 2 −t 0 ¶¸ = 2 N 0 · A 2 T 2 + 4A 2 t 0 ¸ , 0 ≤ t 0 ≤ T 2 This result increases linearly with t 0 and has its maximum value for for t 0 = T/2, which is 5A 2 T/N 0 . (c) The probability of error is P E = Q · ζ 2 √ 2 ¸ To minimize it, use the maximum value of ζ, which is given in (b). (d) The optimum receiver would have two correlators in parallel, one for s 1 (t) and one for s 2 (t). The outputs for the correlators at t = T are differenced and compared with the threshold k calculated from (7.30), assuming the signals are equally likely. Problem 7.14 Use ζ 2 max = 2E s /N 0 for a matched filter, where E s is the signal energy. The required values for signal energy are: (a) A 2 T; (b) 3A 2 T/8; (c) A 2 T/2; (d) A 2 T/3 (Note that AΛ h (t−T/2) T i in the problem statement should be AΛ h 2(t−T/2) T i ). 10 CHAPTER 7. BINARY DATA COMMUNICATION Problem 7.15 The required formulas are E = 1 2 (E 1 +E 2 ) R 12 = 1 E Z T 0 s 1 (t) s 2 (t) dt k opt = 1 2 (E 2 −E 1 ) P E = Q   s (1 −R 12 ) E N 0   The energies for the three given signals are E A = A 2 T; E B = B 2 T/2; E C = 3C 2 T/8 (a) For s 1 = s A and s 2 = s B , the average signal energy is E = A 2 T 2 µ 1 + B 2 2A 2 ¶ The correlation coefficient is R 12 = 1 E Z T 0 ABcos · π (t −T/2) T ¸ dt = 1 E Z T 0 ABsin · πt T ¸ dt = 2ABT πE = 4AB π (A 2 +B 2 /2) The optimum threshold is k opt = 1 2 (E B −E A ) = 1 2 µ B 2 2 −A 2 ¶ T The probability of error is P E = Q " s T 2N 0 µ A 2 + B 2 2 − 4AB π ¶ # 7.1. PROBLEM SOLUTIONS 11 (b) s 1 = s A and s 2 = s C : E = 1 2 (E A +E C ) = 1 2 ¡ A 2 + 3C 2 /8 ¢ T R 12 = AC A 2 + 3C 2 /8 k opt = 1 2 (E C −E A ) = 1 2 µ 3C 2 8 −A 2 ¶ T P E = Q " s T 2N 0 µ A 2 + 3C 2 8 −AC ¶ # (c) s 1 = s B and s 2 = s C : E = 1 2 (E B +E C ) = 1 4 µ B 2 + 3C 2 4 ¶ T R 12 = 16BC 3π (B 2 + 3C 2 /4) k opt = 1 2 (E C −E B ) = 1 4 µ 3C 2 4 −B 2 ¶ T P E = Q " s T 4N 0 µ B 2 + 3C 2 4 − 16BC π ¶ # (d) s 1 = s B and s 2 = −s B : E = E B = B 2 T/2 R 12 = −1 k opt = 0 P E = Q   s B 2 T N 0   12 CHAPTER 7. BINARY DATA COMMUNICATION (e) s 1 = s C and s 2 = −s C : E = E C = 3C 2 T/8 R 12 = −1 k opt = 0 P E = Q   s 3C 2 T 8N 0   Problem 7.16 The appropriate equations to solve are P E = Q £√ z ¤ = 10 −5 , ASK P E = Q h p 2 (1 −m 2 ) z i = 10 −5 , PSK, m = 0.2 P E = Q £√ z ¤ = 10 −5 , FSK Trial and error shows that the Q-function has a value of 10 −5 for it argument equal to 4.265. Thus, we get the following results: For ASK and FSK, √ z = 4.265 or z = 18.19 or z dB = 12.6 dB For PSK with m = 0.2, p 2 (1 −m 2 ) z = 4.265 or z = 9.47 or z dB = 9.765 dB Problem 7.17 The program is given below. Curves corresponding to the numbers given in Table 7.2 are shown in Figure 7.2. % Program bep_ph_error.m; Uses subprogram pb_phase_pdf % bep_ph_error.m calls user-defined subprogram pb_phase_pdf.m % Effect of Gaussian phase error in BPSK; fixed variance clf sigma_psi2 = input(’Enter vector of variances of phase error in radians^2 ’); L_sig = length(sigma_psi2); Eb_N0_dB_min = input(’Enter minimum Eb/N0 desired ’); Eb_N0_dB_max = input(’Enter maximum Eb/N0 desired ’); A = char(’-.’,’:’,’—’,’-..’); a_mod = 0; for m = 1:L_sig 7.1. PROBLEM SOLUTIONS 13 Eb_N0_dB = []; Eb_N0 = []; PE = []; k = 1 for snr = Eb_N0_dB_min:.5:Eb_N0_dB_max Eb_N0_dB(k) = snr Eb_N0(k) = 10.^(Eb_N0_dB(k)/10); sigma_psi = sqrt(sigma_psi2(m)); Eb_N0 PE(k) = quad8(’pb_phase_pdf’,-pi,pi,[],[],Eb_N0(k),sigma_psi,a_mod) k = k+1; end PE semilogy(Eb_N0_dB, PE,A(m,:)),xlabel(’E_b/N_0, dB’),ylabel(’P_b’),... axis([Eb_N0_dB_min Eb_N0_dB_max 10^(-7) 1]),... if m == 1 hold on grid on end end P_ideal = .5*erfc(sqrt(Eb_N0)); semilogy(Eb_N0_dB, P_ideal),... legend([’BPSK; \sigma_\theta_e^2 = ’,num2str(sigma_psi2(1))], [’BPSK; \sigma_\theta_e^2 = ’,num2str(sigma_psi2(2))], [’BPSK; \sigma_\theta_e^2 = ’,num2str(sigma_psi2(3))], [’BPSK; \sigma_\theta_e^2 = 0’],3) % pb_phase_pdf.m; called by bep_ph_error.m and bep_ph_error2.m % function XX = pb_phase_pdf(psi,Eb_N0,sigma_psi,a) arg = Eb_N0*(1-a^2)*(cos(psi)-a/sqrt(1-a^2)*sin(psi)).^2; T1 = .5*erfc(sqrt(arg)); T2 = exp(-psi.^2/(2*sigma_psi^2))/sqrt(2*pi*sigma_psi^2); XX = T1.*T2; Problem 7.18 From the figure of Problem 7.17, the degradation for σ 2 φ = 0.01 is about 0.045 dB; for σ 2 φ = 0.05 it is about 0.6 dB; for σ 2 φ = 0.1 it is about 9.5 dB. For the constant phase error 14 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.2: model, the degradation is D const = −20 log 10 [cos (φ const )] As suggested in the problem statement, we set the constant phase error equal to the standard deviation of the Gaussian phase error for comparison putposes; thus, we consider constant phase errors of φ const = 0.1, 0.224, and 0.316 radians. The degradations corresponding to these phase errors are 0.044, 0.219, and 0.441 dB, respectively. The constant phase error degradations are much less serious than the Gaussian phase error degradations. Problem 7.19 (a) For ASK, P E = Q[ √ z] = 10 −5 gives z = 18.19 or z = 12.6 dB. (b) For BPSK, P E = Q £√ 2z ¤ = 10 −5 gives z = 9.1 or z = 9.59 dB. (c) Binary FSK is the same as ASK. (d) The degradation of BPSK with a phase error of 5 degrees is D const = −20 log 10 [cos (5 o )] = 0.033 dB, so the required SNR to give a bit error probability of 10 −5 is 9.59+0.033 = 9.623 dB. (e) For PSK with m = 1/ √ 2, P E = Q h p 2 (1 −1/2) z i = Q[ √ z] = 10 −5 gives z = 18.19 or z = 12.6 dB. (f) Assuming that the separate effects are additive, we add the degradation of part (d) to the SNR found in part (e) to get 12.6 + 0.033 = 12.633 dB. 7.1. PROBLEM SOLUTIONS 15 Figure 7.3: Problem 7.20 The degradation in dB is D const = −20 log 10 [cos (φ)]. It is plotted in Figure 7.3. The degradations for phase errors of 3, 5, 10, and 15 degrees are 0.0119, 0.0331, 0.1330, and 0.3011 dB, respectively. Problem 7.21 (a) Solve P E = Q ¡√ 2z ¢ = 10 −4 to get z = 8.4 dB. From (7.73), the additional SNR required due to a carrier component is −10 log 10 ¡ 1 −m 2 ¢ . A plot of SNR versus m is given in Fig. 7.4. (b) Solve P E = Q( √ z) = 10 −5 to get z = 9.57 dB. A plot of SNR versus m is given in Fig. 7.4. (c) Solve P E = Q( √ z) = 10 −6 to get z = 10.53 dB. A plot of SNR versus m is given in Fig. 7.4. Problem 7.22 (a) For BPSK, an SNR of 10.53 dB is required to give an error probability of 10 −6 . For ASK and FSK it is 3 dB more, or 13.54 dB. For BPSK and ASK, the required bandwidth is 2R Hz where R is the data rate in bps. For FSK with minimum tone spacing of 0.5R Hz, the required bandwidth is 2.5R Hz. Hence, for BPSK and ASK, the required bandwidth is 200 kHz. For FSK, it is 250 kHz. 16 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.4: (b) For BPSK, the required SNR is 9.6 dB. For ASK and FSK it is 12.6 dB. The required bandwidths are now 400 kHz for BPSK and ASK; it is 500 kHz for FSK. Problem 7.23 The correlation coefficient is R 12 = 1 E Z T 0 s 1 (t) s 2 (t) dt = 1 E Z T 0 A 2 cos (ω c t) cos [(ω c +∆ω) t] dt = sinc (2∆fT) where ∆f is the frequency separation between the FSK signals in Hz. Using a calculator, we find that the sinc-function takes on its first minimum at an argument value of about 1.4, or ∆f = 0.7/T. The minimum value at this argument value is about -0.216. The improvement in SNR is −10 log 10 (1 −R 12 ) = −10 log 10 (1 + 0.216) = 0.85 dB. Problem 7.24 The encoded bit streams, assuming a reference 1 to start with, are: (a) 1 100 001 000 010; (b) 1 100 110 011 001; (c) 1 111 111 111 111; (d)1 010 101 010 101; (e) 1 111 111 010 101; (f) 1 110 000 011 011; (g) 1 111 010 000 101; (h) 1 100 001 000 010. 7.1. PROBLEM SOLUTIONS 17 Problem 7.25 The encoded bit stream is 1 110 000 100 110 (1 assumed for the starting reference bit). The phase of the transmitted carrier is 000ππππ0ππ00π. Integrate the product of the received signal and a 1-bit delayed signal over a (0, T) interval. Each time the integrated product is greater than 0, decide 1; each time it is less than 0, decide 0. The resulting decisions are 110 111 001 010, which is the message signal. If the signal is inverted, the same demodulated signal is detected as if it weren’t inverted. Problem 7.26 Note that E[n 1 ] = E ·Z 0 −T n(t) cos (ω c t) dt ¸ = Z 0 −T E[n(t)] cos (ω c t) dt = 0 which follows because E[n(t)] = 0. Similarly for n 2 , n 3 , and n 4 . Consider the variance of each of these random variables: var [n 1 ] = E £ n 2 1 ¤ = E ·Z 0 −T Z 0 −T n(t) n(λ) cos (ω c t) cos (ω c λ) dtdλ ¸ = Z 0 −T Z 0 −T E[n(t) n(λ)] cos (ω c t) cos (ω c λ) dtdλ = Z 0 −T Z 0 −T N 0 2 δ (t −λ) cos (ω c t) cos (ω c λ) dtdλ = N 0 2 Z 0 −T cos 2 (ω c t) dt = N 0 T 4 where the fact that E[n(t) n(λ)] = N 0 2 δ (t −λ) has been used along with the sifting property of the δ-function to reduce the double integral to a single integral. Similarly for n 2 , n 3 , and n 4 . Therefore, w 1 has zero mean and variance var [w 1 ] = 1 4 var [n 1 ] + 1 4 var [n 2 ] = N 0 T 8 Similarly for w 2 , w 3 , and w 4 . Problem 7.27 The two error probabilities are P E, opt = 1 2 e −z and P E, delay-and-mult ≈ Q ¡√ z ¢ Comparative values are: 18 CHAPTER 7. BINARY DATA COMMUNICATION z, dB P E, opt P E, delay-and-mult 8 9.1 × 10 −4 6.8 × 10 −3 9 1.8 × 10 −4 2.7 × 10 −3 10 2.3 × 10 −5 8.5 × 10 −4 11 1.7 × 10 −6 2.1 × 10 −4 Problem 7.28 Noncoherent binary FSK has error probability P E, NCFSK = 1 2 e −z NCFSK /2 Solve for the SNR to get z NCFSK = −2 ln(2P E, NCFSK ) DPSK has error probability P E, DPSK = 1 2 e −z DPSK Solve for the SNR to get z DPSK = −ln(2P E, DPSK ) The values asked for are given in the table below: P E z NCFSK , dB z DPSK , dB 10 −3 10.94 7.93 10 −5 13.35 10.34 10 −7 14.89 11.88 Problem 7.29 Bandwidths are given by B BPSK = B DPSK = 2R; R BPSK = R BPSK = B DPSK /2 = 50 kbps B CFSK = 2.5R; R CFSK = B BPSK /2.5 = 40 kbps B NCFSK = 4R; R NCFSK = B BPSK /4 = 25 kbps 7.1. PROBLEM SOLUTIONS 19 Problem 7.30 Consider the bandpass filter, just wide enough to pass the ASK signal, followed by an envelope detector with a sampler and threshold comparator at its output. The output of the bandpass filter is u(t) = A k cos (ω c t +Θ) +n(t) = A k cos (ω c t +Θ) +n c (t) cos (ω c t +Θ) −n s (t) sin(ω c t +Θ) = x(t) cos (ω c t +Θ) −n s (t) sin(ω c t +Θ) where A k = A if signal plus noise is present at the receiver input, and A = 0 if noise alone is present at the input. In the last equation x(t) = A k +n c (t) Now u(t) can be written in envelope-phase form as u(t) = r (t) cos [ω c t +Θ+φ(t)] where r (t) = p x 2 (t) +n 2 s (t) and φ(t) = tan −1 · n s (t) x(t) ¸ The output of the envelope detector is r (t), which is sampled each T seconds and compared with the threshold. Thus, to compute the error probability, we need the pdf of the envelope for noise alone at the input and then with signal plus noise at the input. For noise alone, it was shown in Chapter 4 that the pdf of r (t) is Rayleigh, which is f R (r) = r R e −r 2 /2N , r ≥ 0, noise only where N is the variance (mean-square value) of the noise at the filter output. For signal plus noise at the receiver input, the pdf of the envelope is Ricean. Rather than use this complex expression, however, we observe that for large SNR at the receiver input r (t) ≈ A+n c (t) , large SNR, signal present which follows by expanding the argument of the square root, dropping the squared terms, and approximating the square root as √ 1 +² ≈ 1 + 1 2 ², |²| << 1 Thus, for large SNR conditions, the envelope detector output is approximately Gaussian for signal plus noise present. It has mean A and variance N (recall that the inphase and 20 CHAPTER 7. BINARY DATA COMMUNICATION quadrature lowpass noise components have variance the same as the bandpass process). The pdf of the envelope detector output with signal plus noise present is approximately f R (r) = e −(r−A) 2 /2N √ 2πN , large SNR For large SNR, the threshold is approximately A/2. Thus, for noise alone present, the probability of error is exactly P (E | 0) = Z ∞ A/2 r R e −r 2 /2N dr = e −A 2 /8N For signal plus noise present, the probability of error is approximately P (E | S +N present) = Z A/2 −∞ e −(r−A) 2 /2N √ 2πN dr = Q ³ p A 2 /2N ´ ≈ e −A 2 /4N p π/NA Now the average signal-to-noise ratio is z = 1 2 A 2 /2 N 0 B T = A 2 4N which follows because the signal is present only half the time. Therefore, the average probability of error is P E = 1 2 P (E | S +N) + 1 2 P (E | 0) ≈ e −z √ 4πz + 1 2 e −z/2 Problem 7.31 This is a matter of following the steps as outlined in the problem statement. Problem 7.32 Do the inverse Fourier transform of the given P (f) to show that p (t) results. The derivation is quite long and requires a fair amount of trigonometry. Only a sketch is given here. It is also useful to note that for an even function, the inverse Fourier transform integral reduces to p (t) = Z ∞ 0 P (f) exp(j2πft) df = 2 Z ∞ 0 P (f) cos (2πft) df 7.1. PROBLEM SOLUTIONS 21 because P (f) is even. Now substitute for P (f) from (7.124) to get p (t) = 2 Z 1−β 2T 0 T cos (2πft) df + 2 Z 1+β 2T 1−β 2T T 2 ½ 1 + cos · πT β µ f − 1 −β 2T ¶¸¾ cos (2πft) df = T sin h 2π ³ 1−β 2T ´ t i 2πt +T sin h 2π ³ 1+β 2T ´ t i 2πt + T 2 Z 1+β 2T 1−β 2T ½ cos · πT β µ 1 + 2β T t ¶ f − π 2β + π 2 ¸ + cos · πT β µ 1 − 2β T t ¶ f − π 2β + π 2 ¸¾ df = cos (πβt/T) sinc (t/T) − T 2 Z 1+β 2T 1−β 2T ½ sin · πT β µ 1 + 2β T t ¶ f − π 2β ¸ + sin · πT β µ 1 − 2β T t ¶ f − π 2β ¸¾ df = cos (πβt/T) sinc (t/T) + T 2   cos h πT β ³ 1 + 2β T t ´ f − π 2β i πT β ³ 1 + 2β T t ´ + cos h πT β ³ 1 − 2β T t ´ f − π 2β i πT β ³ 1 − 2β T t ´   1+β 2T 1−β 2T = cos (πβt/T) sinc (t/T) + T 2   cos h πT β ³ 1 + 2β T t ´ 1+β 2T − π 2β i πT β ³ 1 + 2β T t ´ + cos h πT β ³ 1 − 2β T t ´ 1+β 2T − π 2β i πT β ³ 1 − 2β T t ´   − T 2   cos h πT β ³ 1 + 2β T t ´ 1−β 2T − π 2β i πT β ³ 1 + 2β T t ´ + cos h πT β ³ 1 − 2β T t ´ 1−β 2T − π 2β i πT β ³ 1 − 2β T t ´   = cos (πβt/T) sinc (t/T) + T 2   cos h πT β ³ 1 + 2β T t ´ 1+β 2T − π 2β i −cos h πT β ³ 1 + 2β T t ´ 1−β 2T − π 2β i πT β ³ 1 + 2β T t ´   + T 2   cos h πT β ³ 1 − 2β T t ´ 1+β 2T − π 2β i −cos h πT β ³ 1 − 2β T t ´ 1−β 2T − π 2β i πT β ³ 1 − 2β T t ´   22 CHAPTER 7. BINARY DATA COMMUNICATION Expand the trigonometric function arguments to get p (t) = cos (πβt/T) sinc (t/T) + T 2   cos £ (1 +β) πt T + π 2 ¤ −cos £ (1 −β) πt T − π 2 ¤ πT β ³ 1 + 2β T t ´   + T 2   cos £ (1 +β) πt T − π 2 ¤ −cos £ (1 −β) πt T + π 2 ¤ πT β ³ 1 − 2β T t ´   Expand the cosines using to get p (t) = cos (πβt/T) sinc (t/T) − T 2   sin £ (1 +β) πt T ¤ + sin £ (1 −β) πt T ¤ πT β ³ 1 + 2β T t ´   + T 2   sin £ (1 +β) πt T ¤ + sin £ (1 −β) πt T ¤ πT β ³ 1 − 2β T t ´   = cos (πβt/T) sinc (t/T) −   cos (πβt/T) sin(πt/T) 2πt T ³ T 2βt + 1 ´   +   cos (πβt/T) sin(πt/T) 2πt T ³ T 2βt −1 ´   = T πt   1 − 1 2 ³ T 2βt + 1 ´ + 1 2 ³ T 2βt −1 ´   cos (πβt/T) sin(πt/T) = cos (πβt/T) sinc (t/T) 1 −(2βt/T) 2 Problem 7.33 For β = 0.25, P (f) =    T, 0 ≤ |f| ≤ 0.75 2T T 2 © 1 + cos £ πT 0.25 ¡ |f| − 0.75 2T ¢¤ª , 0.75 2T ≤ |f| ≤ 1.25 2T Hz 0, |f| > 0.75 2T The optimum transmitter and receiver transfer functions are |H T (f)| opt = A|P (f)| 1/2 G 1/4 n (f) |H C (f)| 1/2 7.1. PROBLEM SOLUTIONS 23 and |H R (f)| opt = K|P (f)| 1/2 G 1/4 n (f) |H C (f)| 1/2 Setting all scale factors equal to 1, these become |H T (f)| opt = |P (f)| 1/2 1/4 q 1 + (f/f C ) 2 1/4 q 1 + (f/f 3 ) 2 |H R (f)| opt = |P (f)| 1/2 1/4 q 1 + (f/f 3 ) 2 1/4 q 1 + (f/f C ) 2 A MATL:AB program for computing the optimum transfer functions is given below. Plots for the various cases are then given in Figures 7.5 - 7.7. % Problem 7.33: Optimum transmit and receive filters for raised-cosine pulse shaping, % first-order Butterworth channel filtering, and first-order noise spectrum % A = char(’-’,’:’,’—’,’-.’); clf beta00 = input(’Enter starting value for beta: >=0 and <= 1 ’); del_beta = input(’Enter step in beta: starting value + 4*step <= 1 ’); R = 1; f_3 = R/2; f_C = R/2; delf = 0.01; HT_opt = []; HR_opt = []; for n = 1:4 beta = (n-1)*del_beta+beta00; beta0(n) = beta; f1 = 0:delf:(1-beta)*R/2; f2 = (1-beta)*R/2:delf:(1+beta)*R/2; f3 = (1+beta)*R/2:delf:R; f = [f1 f2 f3]; P1 = ones(size(f1)); P2 = 0.5*(1+cos(pi/(R*beta)*(f2-0.5*(1-beta)*R))); P3 = zeros(size(f3)); Gn = 1./(1+(f/f_3).^2); 24 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.5: f 3 = f C = 1/2T HC = 1./sqrt(1+(f/f_C).^2); P = [P1 P2 P3]; HT_opt = sqrt(P).*(Gn.^.25)./sqrt(HC); HR_opt = sqrt(P)./((Gn.^.25).*sqrt(HC)); subplot(2,1,1),plot(f,HR_opt,A(n,:)),xlabel(’f, Hz’),ylabel(’|H_R_,_o_p_t(f)||’) title([’R = ’,num2str(R),’ bps; channel filter 3-dB frequency = ’,num2str(f_C),’ Hz; noise 3-dB frequency = ’,num2str(f_3),’ Hz’]) if n == 1 hold on end subplot(2,1,2),plot(f,HT_opt,A(n,:)),xlabel(’f, Hz’),ylabel(’|H_T_,_o_p_t(f)|’) if n == 1 hold on end end legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))], [’\beta = ’,num2str(beta0(3))],[’\beta = ’,num2str(beta0(4))],1) (a) f 3 = f C = 1/2T (b) f C = 2f 3 = 1/T 7.1. PROBLEM SOLUTIONS 25 Figure 7.6: f C = 2f 3 = 1/T (c) f 3 = 2f C = 1/T Problem 7.34 This a simple matter of sketching Λ(f/B) and Λ £¡ f − 1 2 B ¢ /B ¤ side by side and showing that they add to a constant in the overlap region. Problem 7.35 Follow the steps outlined in the problem statement. Problem 7.36 (a) The optimum transmitter and receiver transfer functions are |H T (f)| opt = A|P (f)| 1/2 G 1/4 n (f) |H C (f)| 1/2 and |H R (f)| opt = K|P (f)| 1/2 G 1/4 n (f) |H C (f)| 1/2 where P (f) is the signal spectrum and A and K are arbitrary constants which we set equal 26 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.7: f 3 = 2f C = 1/T to 1 for convenience. From the problem statement, G 1/4 n (f) = ¡ 10 −12 ¢ 1/4 = 10 −3 and |H C (f)| = 1 q 1 + (f/4800) 2 The signal spectrum is, for β = 1, P (f) = 1 2 × 4800 · 1 + cos µ π |f| 4800 ¶¸ , 0 ≤ |f| ≤ 4800 Hz = 1 4800 cos 2 µ π |f| 9600 ¶ , 0 ≤ |f| ≤ 4800 Hz Thus, |H T (f)| opt = |H R (f)| opt = ¯ ¯ ¯ ¯ cos µ π |f| 9600 ¶¯ ¯ ¯ ¯ q 1 + (f/4800) 2 , 0 ≤ |f| ≤ 4800 Hz where all scale factors have been set equal to 1. 7.1. PROBLEM SOLUTIONS 27 Problem 7.37 The MATLAB program is given below and Figure 7.8 shows the probability of error curves. % Plots for Prob. 7.37 % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-..’); delta = input(’ Enter value for delta: ’) taum_T0 = input(’ Enter vector of values for tau_m/T: ’) L_taum = length(taum_T0); z0_dB = 0:.1:15; z0 = 10.^(z0_dB/10); for ll = 1:L_taum ll taum_T = taum_T0(ll) P_E = 0.5*qfn(sqrt(2*z0)*(1+delta))+0.5*qfn(sqrt(2*z0)*((1+delta)-2*delta*taum_T)); semilogy(z0_dB, P_E,A(ll,:)) if ll == 1 hold on; grid on; axis([0, inf, 10^(-6), 1]); xlabel(’z_0, dB’); ylabel(’P_E’) end end title([’P_E versus z_0 in dB for \delta = ’, num2str(delta)]) if L_taum == 1 legend([’\tau_m/T = ’,num2str(taum_T0)],3) elseif L_taum == 2 legend([’\tau_m/T = ’,num2str(taum_T0(1))],[’\tau_m/T = ’,num2str(taum_T0(2))],3) elseif L_taum == 3 legend([’\tau_m/T = ’,num2str(taum_T0(1))],[’\tau_m/T = ’,num2str(taum_T0(2))], [’\tau_m/T = ’,num2str(taum_T0(3))],3) elseif L_taum == 4 legend([’\tau_m/T = ’,num2str(taum_T0(1))],[’\tau_m/T = ’,num2str(taum_T0(2))], [’\tau_m/T = ’,num2str(taum_T0(3))],[’\tau_m/T = ’,num2str(taum_T0(4))],3) end Problem 7.38 The program given in Problem 7.37 can be adapted for making the asked for plot. 28 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.8: Problem 7.39 (a) The output of the channel in terms of the input is y (t) = x(t) +βx(t −τ m ) Fourier transform both sides and take the ratio of output to input transforms to obtain the channel transfer function: H C (f) = Y (f) X (f) = 1 +βe −j2πfτm Using Euler’s theorem to expand the exponential and finding the maginitude gives |H C (f)| = q 1 + 2β cos (2πfτ m ) +β 2 Plots are shown in Fig. 7.9 for β = 0.5 and 1.0. % Plots for Prob. 7.39 % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-..’); tau_m = input(’ Enter value for tau_m: ’) 7.1. PROBLEM SOLUTIONS 29 beta0 = input(’ Enter vector of values for beta: ’) L_beta = length(beta0) f = -1:.005:1; for ll = 1:L_beta beta = beta0(ll); arg = 1+2*beta*cos(2*pi*tau_m*f)+beta^2; mag_H = sqrt(arg); plot(f, mag_H, A(ll,:)) if ll == 1 hold on; grid on; xlabel(’f’); ylabel(’|H(f)|’) end end title([’|H(f)| versus f for \tau_m = ’, num2str(tau_m)]) if L_beta == 1 legend([’\beta = ’,num2str(beta0)],3) elseif L_beta == 2 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))],3) elseif L_beta == 3 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))], [’\beta = ’,num2str(beta0(3))],3) elseif L_beta == 4 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))], [’\beta = ’,num2str(beta0(3))],[’\beta = ’,num2str(beta0(4))],3) end (b) The output of the tapped delay line filter in terms of the input is z (t) = N X n=0 β n y [t −(n −1) ∆] Fourier transform this equation and take the ratio of the output to the input transforms to obtain the transfer function of the equalizer as H eq (f) = Z (f) Y (f) = ∞ X n=0 β n e −j2π(n−1)∆f (c) From part (a) and using the geometric expansion, we obtain 1 H C (f) = 1 1 +βe −j2πfτm = ∞ X n=0 (−1) n β n e −j2πnτ m f 30 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.9: Equate this to the result of part (b) with τ m = ∆. Clearly, β n = (−β) n−1 . Problem 7.40 The appropriate equations are: BPSK: (7.73) with m = 0 for nonfading; (7.174) for fading; DPSK: (7.111) for nonfading; (7.176) for fading; CFSK: (7.87) for nonfading; (7.175) for fading; NFSK: (7.119) for nonfading; (7.177) for fading. Degradations for P E = 10 −2 are: Modulation type Fading SNR, dB Nonfaded SNR, dB Fading Margin, dB BPSK 13.85 4.32 9.52 DPSK 16.90 5.92 10.98 Coh. FSK 16.86 7.33 9.52 Noncoh. FSK 19.91 8.93 10.98 Problem 7.41 See Problem 7.40 for equation reference numbers. The inverse equations, expressing signal- to-noise ratio in dB, are: 7.1. PROBLEM SOLUTIONS 31 (a) BPSK z BPSK, NF = 10 log 10 ½ 1 2 £ Q −1 (P E ) ¤ 2 ¾ dB z BPSK, F = 10 log 10 ( (1 −2P E ) 2 1 −(1 −2P E ) 2 ) dB The margin in dB is the difference between these two results for a given P E . (b) DPSK z DPSK, NF = 10 log 10 {−ln(2P E )} dB z DPSK, F = 10 log 10 ½ 1 2P E −1 ¾ dB (c) Coherent FSK z BPSK, NF = 10 log 10 n £ Q −1 (P E ) ¤ 2 o dB z BPSK, F = 10 log 10 ( 2 (1 −2P E ) 2 1 −(1 −2P E ) 2 ) dB (d) Noncoherent FSK z NCFSK, NF = 10 log 10 {−2 ln(2P E )} dB z NCFSK, F = 10 log 10 ½ 1 P E −2 ¾ dB The fading margin for these various modulation cases are given in the table below for P E = 10 −3 . Modulation type Fading SNR, dB Nonfaded SNR, dB Fading Margin, dB BPSK 23.97 6.79 17.18 DPSK 26.98 7.93 19.05 Coh. FSK 26.98 9.80 17.18 Noncoh. FSK 29.99 10.94 19.05 Problem 7.42 The expressions with Q-functions can be integrated by parts (BPSK and coherent FSK). The other two cases involve exponential integrals which are easily integrated (DPSK and noncoherent FSK). 32 CHAPTER 7. BINARY DATA COMMUNICATION Problem 7.43 We need to invert the matrix [P C ] =   1 0.1 −0.01 0.2 1 0.1 −0.02 0.2 1   Using a calculator or MATLAB, we find the inverse matrix to be [P C ] −1 =   1.02 −0.11 0.02 −0.21 1.04 −0.11 0.06 −0.21 1.02   The optimum weights are the middle column. (a) The equation giving the equalized output is p eq (mT) = −0.11p C [(m+ 1) T] + 1.04p C [mT] −0.21p C [(m−1) T] (b) Calculations with the above equation using the given sample values result in p eq (−2T) = −0.02; p eq (−T) = 0; p eq (0) = 1; p eq (T) = 0; p eq (2T) = −0.06 Note that equalized values two samples away from the center sample are not zero because the equalizer can only provide three specified values. Problem 7.44 (a) The desired matrix is [R yy ] = N 0 T   ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2   (b) The following matrix is needed: [R yd ] =   R yd (−T) R yd (0) R yd (T)   =   0 A bA   Multiply the matrix given in (a) by the column matrix of unknown weights to get the equations N 0 T   ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2     a 0 −1 a 0 0 a 0 1   =   0 A bA   7.1. PROBLEM SOLUTIONS 33 or   ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2 bz + π 2 e −2π π 2 e −4π bz + π 2 e −2π ¡ 1 +b 2 ¢ z + π 2     a −1 a 0 a 1   =   0 A 2 T/N 0 bA 2 T/N 0   =   0 z bz   where the factor N 0 /T has been normalized out so that the right hand side is in terms of the signal-to-noise ratio, z (the extra factor of A needed mulitplies both sides of the matrix equation and on the left hand side is lumped into the new coefficients a −1 , a 0 , and a 1 . Invert the matrix equation above to find the weights. A MATLAB program for doing so is given below. For z = 10 dB and b = 0.1, the weights are a −1 = −0.2781, a 0 = 0.7821, and a 1 = 0.0773. % Solution for Problem 7.44 % z_dB = input(’Enter the signal-to-noise ratio in dB ’); z = 10^(z_dB/10); b = input(’Enter multipath gain ’); R_yy(1,1)=(1+b^2)*z+pi/2; R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=b*z+(pi/2)*exp(-2*pi); R_yy(1,3)=(pi/2)*exp(-4*pi); R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’R_yy’) disp(R_yy) B = inv(R_yy); disp(’R_yy^-1’) disp(B) R_yd = [0 z b*z]’ A = B*R_yd Typical output of the program: >> pr7_44 Enter the signal-to-noise ratio in dB 10 Enter multipath gain .5 R_yy R_yd = 34 CHAPTER 7. BINARY DATA COMMUNICATION 14.0708 5.0029 0.0000 0 5.0029 14.0708 5.0029 10 0.0000 5.0029 14.0708 5 R_yy^-1 A = 0.0831 -0.0338 0.0120 -0.2781 -0.0338 0.0951 -0.0338 0.7821 0.0120 -0.0338 0.0831 0.0773 7.2 Computer Exercises Computer Exercise 7.1 % ce7_1.m: Simulation BEP for antipodal signaling % clf n_sim = input(’Enter number of bits for simulation: ’); zdB = input(’Enter z = A^2*T/N_0 vector in dB: ’); z = 10.^(zdB/10); Lz = length(z); A = 1; T = 1; N0 = (A^2*T)./z; S = A*T*sign(rand(1,n_sim)-.5); PE = []; ET = []; for k = 1:Lz N = sqrt(0.5*N0(k)*T)*randn(1,n_sim); Y = S+N; Z = sign(Y); V = 0.5*(1+Z); W = 0.5*(1+sign(S)); E = xor(V,W); ET(k) = sum(E); PE(k) = ET(k)/length(Y); end PEth = 0.5*erfc(sqrt(z)); disp(’ ’) disp(’ SNR, dB Errors P_E est. P_E theory’) disp(’ ________________________________________’) 7.2. COMPUTER EXERCISES 35 disp(’ ’) disp([zdB’ ET’ PE’ PEth’]) disp(’ ’) semilogy(zdB, PE, ’o’), xlabel(’z in dB’), ylabel(’P_E’), grid hold on semilogy(zdB, PEth) title([’BER simulated with ’, num2str(n_sim), ’ bits per SNR value ’]) legend([’Simulated’], [’Theoretical’]) A typical run is given below: >> ce7_1 Enter number of bits for simulation: 20000 Enter z = A^2*T/N_0 vector in dB: [2 4 6 8] SNR, dB Errors P_E est. P_E theory ______________________________________ 2.0000 704.0000 0.0352 0.0375 4.0000 238.0000 0.0119 0.0125 6.0000 44.0000 0.0022 0.0024 8.0000 4.0000 0.0002 0.0002 A plot also appears showing the estimated and theoretical probabilities of error plotted versus signal-to-noise ratio. Computer Exercise 7.2 % ce7_2: Compute degradation in SNR due to bit timing error % clf A = char(’-’,’-.’,’:’,’—’); for m = 1:4 PE0 = 10^(-m); delT = []; deg = []; Eb1 = []; k = 0; for Y = 0:.025:.45; k = k+1; delT(k) = Y; Eb1(k) = fzero(@PE1, 2, [], delT(k), PE0); end Eb = fzero(@PE, 2, [], PE0); 36 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.10: deg = Eb1 - Eb; plot(delT, deg, A(m,:)), ... if m == 1 hold on axis([0 .45 0 20]) xlabel(’Fractional timing error, \DeltaT/T’), ylabel(’Degradation in dB’),... title([’Degradation in SNR for bit synchronization timing error’]), grid end end legend([’P_E = 10^-^1’], [’ = 10^-^2’], [’ = 10^-^3’], [’ = 10^-^4’], 2) The output plot is shown in Fig. 7.1. Computer Exercise 7.3 A MATLAB program for this computer exercise is given in the solution for Problem 7.17. 7.2. COMPUTER EXERCISES 37 Computer Exercise 7.4 % ce7_4: For a given data rate and error probability, find the required % bandwidth and Eb/N0 in dB; baseband, BPSK/DPSK, coherent FSK % and noncoherent FSK modulation may be specified. % I_mod = input(’Enter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK; 3 = CFSK; 4 = NCFSK: ’); I_BW_R = input(’Enter 1 if bandwidth specified; 2 if data rate specified: ’); if I_BW_R == 1 BkHz = input(’Enter vector of desired bandwidths in kHz: ’); elseif I_BW_R == 2 Rkbps = input(’Enter vector of desired data rates in kbps: ’); end PE = input(’Enter desired BEP: ’); N0 = 1; if I_BW_R == 1 if I_mod == 0 Rkpbs = BkHz; % Rate for baseband elseif I_mod == 1 | I_mod == 2 Rkbps = BkHz/2; % Rate in kbps for BPSK or DPSK elseif I_mod == 3 Rkbps = BkHz/3; % Rate in kbps for CFSK elseif I_mod == 4 Rkbps = BkHz/4; % Rate in kbps for NFSK end elseif I_BW_R == 2 if I_mod == 0 BkHz = Rkbps; % Transmission BW for baseband in kHz elseif I_mod == 1 | I_mod == 2 BkHz = 2*Rkbps; % Transmission BW for BPSK or DPSK in kHz elseif I_mod == 3 BkHz = 3*Rkbps; % Transmission BW for CFSK in kHz elseif I_mod == 4 BkHz = 4*Rkbps; % Transmission BW for NFSK in kHz end end R = Rkbps*1000; B = BkHz*1000; % PE = 0.5*erfc(sqrt(z)) MATLAB has an inverse erf but not an inverse Q-function 38 CHAPTER 7. BINARY DATA COMMUNICATION if I_mod == 0 | I_mod == 1 sqrt_z = erfinv(1-2*PE); z = sqrt_z^2; elseif I_mod == 2 z = -log(2*PE); elseif I_mod == 3 sqrt_z_over_2 = erfinv(1-2*PE); z = 2*sqrt_z_over_2^2; elseif I_mod == 4 z = -2*log(2*PE); end Eb_N0_dB = 10*log10(z); % This is the required Eb/N0 in dB if I_mod == 0 A = sqrt(R*N0*z); % z = A^2/(R*N0) for baseband elseif I_mod == 1 | I_mod == 2 | I_mod == 3 | I_mod == 4 A = sqrt(2*R*N0*z); % z = A^2/2*R*N0 for BPSK, DPSK, CFSK, NFSK end disp(’ ’) disp(’Desired P_E and required E_b/N_0 in dB: ’) format long disp([PE Eb_N0_dB]) format short disp(’ ’) disp(’ R, kbps BW, Hz A, volts’) disp(’ ___________________________’) disp(’ ’) disp([Rkbps’ BkHz’ A’]) disp(’ ’) 7.2. COMPUTER EXERCISES 39 A typical run is given below: >> ce7_4 Enter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK; 3 = CFSK; 4 = NCFSK: 3 Enter 1 if bandwidth specified; 2 if data rate specified: 2 Enter vector of desired data rates in kbps: [5 10 20 50] Enter desired BEP: 1e-3 Desired P_E and required E_b/N_0 in dB: 0.00100000000000 9.79982256904398 R, kbps BW, Hz A, volts ___________________________ 5.0000 15.0000 309.0232 10.0000 30.0000 437.0248 20.0000 60.0000 618.0465 50.0000 150.0000 977.2173 Computer Exercise 7.5 % ce7_5.m: Simulation of suboptimum bandpass filter/delay-and-multiply % demodulator with integrate-and-dump detection for DPSK; input bandpass % filter is Butterworth with selectable bandwidth-bit period product. % Simulation is broken up into contiguous blocks for memory management. % Eb_N0_dB_max = input(’Enter maximum Eb/N0 in dB: ’); Eb_N0_dB_min = input(’Enter minimum Eb/N0 in dB: ’); samp_bit = input(’Enter number of samples per bit used in simulation: ’); n_order = input(’Enter order of Butterworth detection filter: ’); BWT_bit = input(’Enter filter bandwidth normalized by bit rate: ’); N_bits = input(’Enter total number of bits in simulation: ’); N_bits_block = input(’Enter number of bits/sim. block: ’); N_blocks = floor(N_bits/N_bits_block); T_bit = 1; % Arbitrarily take bit time as 1 second BW = BWT_bit/T_bit; % Compute filter bandwidth from BW*T_bit [num,den] = butter(n_order,2*BW/samp_bit); % Obtain filter num/den coefficients e_tot = []; N_bits_sim = []; Eb_N0_dB = []; % Ensure that plotting arrays are empty Perror = []; clf % Clear any plots left over from previous runs k = 0; 40 CHAPTER 7. BINARY DATA COMMUNICATION for kk = Eb_N0_dB_min:Eb_N0_dB_max % Loop for simul. det. for each Eb/N0 k = k+1; Eb_N0_dB(k) = kk; Eb_N0 = 10^(Eb_N0_dB(k)/10); % Convert desired Eb/N0 from dB to ratio Eb = T_bit; % Bit energy is T_bit, if ampl. = 1 N0 = Eb/Eb_N0; % Compute noise PSD from Eb/N0 del_t = T_bit/samp_bit; % Compute sampling interval sigma_n = sqrt(N0/(2*del_t)); % Compute std deviations of noise samples e_sum_tot = 0; zi = []; nn = 1; while nn <= N_blocks & e_sum_tot < 200 ss = sign(rand(1,N_bits_block)-.5); % Generate sequence of random +-1s data = 0.5*(ss+1); % Logical data is sequence of 1s and 0s data_diff_enc = diff_enc(data); % Differentially encode data for DPSK s = 2*data_diff_enc-1; % Generate bipolar data for modulation sig = []; sig = s(ones(samp_bit,1),:); % Build array wth columns samp_bit long sig = sig(:); % Convert matrix of bit samples to vector bits_out = []; % Make sure various arrays are empty y_det = []; noise = sigma_n*randn(size(sig)); % Form sequence of Gauss. noise samples [y,zf] = filter(num,den,sig+noise,zi); % Filter S plus N with chosen filter zi = zf; % Save final values for future init. cond’ns y_ref = delay1(y,samp_bit); % Ref. signal is 1-bit delayed S + Ne y_mult = y.*y_ref; % Multiply rec’d S + N by 1 bit delay bits_out = int_and_dump(y_mult,samp_bit,N_bits_block); % I-&-D det. error_array = abs(bits_out-data); % Compare detected bits with input if nn == 1 error_array(1:10) = 0; % Don’t include 1st 5 bits due to transients elseif nn ~= 1 error_array(1:1) = 0; % Delete 1st bit due to y_ref = 0 end e_sum = sum(error_array); % Sum to get total number of errors/blk e_sum_tot = e_sum_tot + e_sum; % Running sum of total no. of errors nn = nn + 1; end e_tot(k) = e_sum_tot; N_bits_sim(k) = (nn-1)*N_bits_block-10-(nn-2); Perror(k) = e_sum_tot/N_bits_sim(k); 7.2. COMPUTER EXERCISES 41 end % End of Eb/N0 loop disp(’ ’) disp(’ Eb/N0, dB; PE errors’) % Display computed prob. of error disp(’ __________________________’) disp(’ ’) disp([Eb_N0_dB’ Perror’ e_tot’]) disp(’ ’) % Plot simulated bit error probabilities versus Eb/N0 semilogy(Eb_N0_dB, Perror,’—’), grid, xlabel(’E_b/N_0; dB’), ylabel(’P_E’), hold % Plot theoretical bit error probability for optimum DPSK detector semilogy(Eb_N0_dB, 0.5*exp(-10.^(Eb_N0_dB/10))) % Plot approximate theoretical result for suboptimum detector semilogy(Eb_N0_dB, qfn(sqrt(10.^(Eb_N0_dB/10))),’-.’) Title([’Comparison of optimum & delay-and-multiply detectors for DPSK; ’, num2str(n_order), ’-order Butterworth filter; BT = ’, num2str(BWT_bit)]) legend([’Simulation; BT = ’, num2str(BWT_bit),’; ’, num2str(N_bits),’ bits; ’, num2str(N_blocks), ’ blocks’],’Theory; optimum differential detector’, ’Theory; delay/multiply detector’, 3) %diff_enc(input); function to differentially encode a bit stream vector % function output = diff_enc(input) L_in = length(input); output = []; for k = 1:L_in if k == 1 output(k) = not(bitxor(input(k),1)); else output(k) = not(bitxor(input(k),output(k-1))); end end % int_and_dump(input,samp_bit,N_bits); % function bits_out = int_and_dump(input,samp_bit,N_bits) % Reshape input vector with each bit occupying a column samp_array = reshape(input, samp_bit, N_bits); integrate = sum(samp_array); % Integrate (sum) each bit (column) bits_out = (sign(integrate) + 1)/2; 42 CHAPTER 7. BINARY DATA COMMUNICATION A typical run is given below. A minimum of 200 errors are counted unless the total number of specified bits to be simulated is exceeded. A plot is also made comparing probabilities of error for the optimum detector, simulated suboptimum detector, and approximate theoretical result for the suboptimum detector. >> ce7_5 Enter maximum Eb/N0 in dB: 7 Enter minimum Eb/N0 in dB: 3 Enter number of samples per bit used in simulation: 5 Enter order of Butterworth detection filter: 2 Enter filter bandwidth normalized by bit rate: 1.5 Enter total number of bits in simulation: 15000 Enter number of bits/sim. block: 5000 Eb/N0, dB; PE errors __________________________ 3.0000 0.1040 519.0000 4.0000 0.0655 327.0000 5.0000 0.0485 242.0000 6.0000 0.0216 216.0000 7.0000 0.0101 152.0000 Computer Exercise 7.6 % ce7_6: program to plot Figure 7-28 % clf z0dB = 0:1:30; z0 = 10.^(z0dB/10); for delta = -0.9:0.3:0.6; if delta > -eps & delta < eps delta = 0; end for taum_over_T = 0:1:1 T1 = 0.5*qfn(sqrt(2*z0)*(1+delta)); T2 = 0.5*qfn(sqrt(2*z0)*(1+delta)-2*delta*taum_over_T); PE = T1 + T2; if taum_over_T == 0 semilogy(z0dB, PE, ’-’) elseif taum_over_T == 1 semilogy(z0dB, PE, ’—’) end 7.2. COMPUTER EXERCISES 43 Figure 7.11: if delta <= -.3 text(z0dB(10), PE(10), [’\delta = ’, num2str(delta)]) elseif delta > -0.3 & delta <= 0.3 text(z0dB(8), PE(8), [’\delta = ’, num2str(delta)]) elseif delta > 0.3 text(z0dB(5), PE(5), [’\delta = ’, num2str(delta)]) end if delta == -0.9 hold on ylabel(’P_E’), xlabel(’z_0 in dB’) axis([0 30 1E-5 1]) grid end end end legend([’\tau_m/T = 0’],[’\tau_m/T = 1’]) The output plot is shown in Fig. 7.2. Computer Exercise 7.7 Use Equations (7.73) (with m = 0), (7.87), (7.111), (7.119), and (7.174) - (7.177). Solve them for the SNR in terms of the error probability. Express the SNR in dB and subtract 44 CHAPTER 7. BINARY DATA COMMUNICATION the nonfading result from the fading result. The MATLAB program below does this for the modulation cases of BPSK, coherent FSK (CFSK), DPSK, and noncoherent (NFSK). % ce7_7.m: Program to evaluate degradation due to flat % Rayleigh fading for various modulation schemes % mod_type = input(’Enter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 = NFSK: ’); PE = input(’Enter vector of desired probabilities of error: ’); disp(’ ’) if mod_type == 1 disp(’ BPSK’) Z_bar = (0.25*(1 - 2*PE).^2)./(PE - PE.^2); z = (erfinv(1 - 2*PE)).^2; % MATLAB has an inv. error function elseif mod_type == 2 disp(’ CFSK’) Z_bar = (0.5*(1 - 2*PE).^2)./(PE - PE.^2); z = 2*(erfinv(1 - 2*PE)).^2; elseif mod_type == 3 disp(’ DPSK’) Z_bar = 1./(2*PE) -1; z = -log(2*PE); elseif mod_type == 4 disp(’ NFSK’) Z_bar = 1./PE -2; z = -2*log(2*PE); end Z_bar_dB = 10*log10(Z_bar); z_dB = 10*log10(z); deg_dB = Z_bar_dB - z_dB; disp(’ ’) disp(’ P_E Degradation, dB’) disp(’_________________________’) disp(’ ’) disp([PE’ deg_dB’]) disp(’ ’) A typical run is given below: >> ce7_7 Enter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 = NFSK: 4 7.2. COMPUTER EXERCISES 45 Enter vector of desired probabilities of error: [1e-2 5e-3 1e-3 5e-4 1e-4] NFSK P_E Degradation, dB _________________________ 0.0100 10.9779 0.0050 13.3239 0.0010 19.0469 0.0005 21.6023 0.0001 27.6859 Computer Exercise 7.8 The solution to this exercise consists of two MATLAB programs, one for the zero-forcing case and one for the MMSE case. % ce7_8a.m: Plots unequalized and equalized sample values for ZF equalizer % clf pc = input(’Enter sample values for channel pulse response (odd #): ’); L_pc = length(pc); disp(’Maximum number of zeros each side of decision sample: ’); disp((L_pc-1)/4) N = input(’Enter number of zeros each side of decision sample desired: ’); mid_samp = fix(L_pc/2)+1; Pc = []; for m = 1:2*N+1 row = []; for n = 2-m:2*N+2-m row = [row pc(mid_samp-1+n)]; end Pc(:,m) = row’; end disp(’ ’) disp(’Pc:’) disp(’ ’) disp(Pc) Pc_inv = inv(Pc); disp(’ ’) disp(’Inverse of Pc:’) disp(’ ’) disp(Pc_inv) 46 CHAPTER 7. BINARY DATA COMMUNICATION coef = Pc_inv(:,N+1); disp(’ ’) disp(’Equalizer coefficients:’) disp(’ ’) disp(coef) disp(’ ’) steps = (L_pc-length(coef))+1; p_eq = []; for m = mid_samp-floor(steps/2):mid_samp+floor(steps/2) p_eq(m) = sum(coef’.*fliplr(pc(m-N:m+N))); end p_eq_p = p_eq(N+1:mid_samp+floor(steps/2)); disp(’ ’) disp(’Equalized pulse train’) disp(’ ’) disp(p_eq_p) disp(’ ’) t=-5:.01:5; y=zeros(size(t)); subplot(2,1,1),stem(pc),ylabel(’Ch. output pulse’),... axis([1 L_pc -.5 1.5]),... title([’Ch. pulse samp. = [’,num2str(pc),’]’]) subplot(2,1,2),stem(p_eq_p),xlabel(’n’),ylabel(’Equal. output pulse’),... axis([1 L_pc -.5 1.5]),... title([’Output equalized with ’,num2str(N),’ zero samples either side’]) A typical run is given below: >> ce7_8a Enter sample values for channel pulse response (odd #): [-.05 .1 -.15 .25 -.3 1 -.2 .15 -.1 .05 -.02] Maximum number of zeros each side of decision sample: 2.5000 Enter number of zeros each side of decision sample desired: 2 Pc: 1.0000 -0.3000 0.2500 -0.1500 0.1000 -0.2000 1.0000 -0.3000 0.2500 -0.1500 0.1500 - 0.2000 1.0000 -0.3000 0.2500 -0.1000 0.1500 - 0.2000 1.0000 -0.3000 0.0500 - 0.1000 0.1500 -0.2000 1.0000 Inverse of Pc: 7.2. COMPUTER EXERCISES 47 Figure 7.12: 1.0877 0.2842 -0.1777 0.0366 -0.0107 0.1686 1.1317 0.2572 -0.1731 0.0366 -0.1095 0.1403 1.1476 0.2572 -0.1777 0.0588 -0.0956 0.1403 1.1317 0.2842 -0.0093 0.0588 -0.1095 0.1686 1.0877 Equalizer coefficients: -0.1777 0.2572 1.1476 0.1403 -0.1095 Equalized pulse train -0.0350 -0.0000 -0.0000 1.0000 0.0000 0 -0.0554 The MATLAB program for the MMSE case is given below: % Solution for ce 7.8(b) % z_dB = input(’Enter the signal-to-noise ratio in dB: ’); z = 10^(z_dB/10); b = input(’Enter multipath gain: ’); 48 CHAPTER 7. BINARY DATA COMMUNICATION R_yy(1,1)=(1+b^2)*z+pi/2; R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=b*z+(pi/2)*exp(-2*pi); R_yy(1,3)=(pi/2)*exp(-4*pi); R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’ ’) disp(’R_yy:’) disp(’ ’) disp(R_yy) B = inv(R_yy); disp(’ ’) disp(’R_yy^-1:’) disp(’ ’) disp(B) R_yd = [0 z b*z]’; disp(’ ’) disp(’R_yd:’) disp(’ ’) disp(R_yd) A = B*R_yd; disp(’ ’) disp(’Optimum coefficients:’) disp(A) disp(’ ’) A typical run is given below: >> ce7_8b Enter the signal-to-noise ratio in dB: 7 Enter multipath gain: .6 R_yy: 8.3869 3.0101 0.0000 3.0101 8.3869 3.0101 0.0000 3.0101 8.3869 R_yy^-1: 0.1399 -0.0576 0.0207 -0.0576 0.1606 -0.0576 7.2. COMPUTER EXERCISES 49 0.0207 -0.0576 0.1399 R_yd: 0 5.0119 3.0071 Optimum coefficients: -0.2267 0.6316 0.1319 Chapter 8 Advanced Data Communications Topics 8.1 Problem Solutions Problem 8.1 Use the relationship R b = (log 2 M) R s bps where R b is the data rate, R s is the symbol rate, and M is the number of possible signals per signaling interval. In this case, R s = 2000 symbols per second. For M = 4, R b = 2 × 2, 000 = 4, 000 bps, for M = 8, R b = 6, 000 bps, and for M = 64, R b = 12, 000 bps. Problem 8.2 (a) 5,000 symbols per second. (b) Recall that y (t) = A[d 1 (t) cos (ω c t) +d 2 (t) sin(ω c t)] = √ 2Acos [ω c t −θ i (t)] , θ i (t) = tan −1 · d 2 (t) d 1 (t) ¸ Alternating every other bit given in the problem statement between d 1 (t) and d 2 (t) gives d 1 (t) = 1, 1, −1, 1, 1, −1, 1, −1 and d 2 (t) = 1, −1, −1, −1, −1, 1, −1 which results in θ i (t) = π/4, 7π/4, 5π/4, 7π/4, 7π/4, 3π/4, 7π/4, · · ·. For QPSK the symbol switching points occur each T s seconds. (c) Now the switching instants are each T b seconds. Start with d 1 (t) = 1. Then d 2 (t) is staggered, or offset, by 1 bit time. So the first phase shift is for d 1 (t) = 1 and d 2 (t) = 1 or θ 2 (t) = π/4. After T s = 2T b seconds d 1 (t) takes on the second 1-value, but d 2 (t) is still 1, so θ 1 (t) = π/4. At 3T b seconds, d 2 (t) changes to −1, so θ 2 (t) = 7π/4. At 4T b seconds, d 1 (t) changes to −1, so θ 2 (t) = 7π/4, etc. 1 2 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Problem 8.3 For QPSK, P E, symbol = 2Q ³ p E s /N 0 ´ = 10 −5 Trial and error using the asymptotic approximation for the Q-function gives [E s /N 0 ] req’d ≈ 12.9 dB = 19.5. If the quadrature-carrier amplitudes are A, then the amplitude of the envelope-phase-modulated form of the carrier is √ 2A, and E s /N 0 = ¡√ 2A ¢ 2 T/ (2N 0 ) = A 2 /N 0 R. Hence, A req’d = q N 0 R[E s /N 0 ] req’d = p (10 −11 ) (19.5) R = 1.4 × 10 −5 √ R. The answers are as follows: (a) 0.0014 V; (b) 0.0031 V; (c) 0.0044 V; (d) 0.0099 V; (e) 0.014 V; (f) 0.0442 V. Problem 8.4 Take the expectation of the product after noting that the average values are 0 because E[n(t)] = 0: E[N 1 N 2 ] = E ·Z T s 0 n(t) cos (ω c t) dt Z T s 0 n(t) sin(ω c t) dt ¸ = Z T s 0 Z T s 0 E[n(t) n(λ)] cos (ω c t) sin(ω c λ) dtdλ = Z T s 0 Z T s 0 N 0 2 δ (t −λ) cos (ω c t) sin(ω c λ) dtdλ = N 0 2 Z T s 0 cos (ω c t) sin(ω c t) dt = N 0 4 Z T s 0 sin(2ω c t) dt = 0 Problem 8.5 (a) Use θ i (t) = tan −1 · d 2 (t) d 1 (t) ¸ (i) If θ i (t) = 45 o , d 1 (t) = 1 and d 2 (t) = 1; (ii) If θ i (t) = 135 o , d 1 (t) = −1 and d 2 (t) = 1; (iii) If θ i (t) = −45 o , d 1 (t) = 1 and d 2 (t) = −1; (iv) If θ i (t) = −135 o ; d 1 (t) = −1 and d 2 (t) = −1. (b) Error in detecting d 1 (t): (i) θ i (t) = 135 o ; (ii) θ i (t) = 45 o ; (iii) θ i (t) = −135 o ; (iv) θ i (t) = −45 o . 8.1. PROBLEM SOLUTIONS 3 (c) Error in detecting d 2 (t); (i) θ i (t) = −45 o ; (ii) θ i (t) = −135 o ; (iii) θ i (t) = 45 o ; (iv) θ i (t) = 135 o . Problem 8.6 (a) Both are equivalent in terms of symbol error probabilities. (b) QPSK is 3 dB worse in terms of symbol error probability for equal transmission bandwidths, but it handles twice as many bits per second. (c) Choose QPSK over BPSK in terms of performance; however, other factors might favor BPSK, such as simpler implementation. Problem 8.7 The exact result is P symbol = 1 −(1 −P E 1 ) 2 = 2P E 1 −P 2 E 1 The approximation is P symbol ≈ 2P E 1 , so the error term is ² = P 2 E 1 = " Q Ã r E s N 0 !# 2 Clearly, since the Q-function is monotonically decreasing with increasing E s /N 0 , the error term becomes negligible compared with P E 1 = Q ³q E s N 0 ´ as E s /N 0 becomes larger. Problem 8.8 For the data stream 11100 10111 00100 00011, the quadrature data streams are d 1 (t) = 11-1-11-11-1-11 d 2 (t) = 1-1111-1-1-1-11 Each 1 above means a positive rectangular pulse T s seconds in duration and each -1 above means a negative rectangular pulse T s seconds in duration. For QPSK, these pulse sequences are aligned and for OQPSK, the one corresponding to d 2 (t) is delayed by T s /2 = T b seconds with respect to d 1 (t). Type I MSK corresponds to the OQPSK waveforms d 1 (t) and d 2 (t) multiplied by cosine and sine waveforms with periods of 2T s , respectively, and type II MSK corresponds to d 1 (t) and d 2 (t) multiplied by absolute-value cosine and sine waveforms with periods of T s , respectively (one half cosine or sine per T s pulse). Waveforms for QPSK, OQPSK, and Type II MSK generated by a MATLAB program are shown in Figures 8.1 - 8.3 for a random (coin toss) serial bit sequence. 4 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.1: Figure 8.2: 8.1. PROBLEM SOLUTIONS 5 Figure 8.3: Problem 8.9 For QPSK, the excess phase corresponds to a stepwise function that may change values each T s seconds. The phase deviation is given by θ i (t) = tan −1 · d 2 (t) d 1 (t) ¸ so the sequence of phases for QPSK computed from d 1 (t) and d 2 (t) given in Problem 8.8 is π/4, −π/4, π/4, 3π/4, π/4, −3π/4, −π/4, −3π/4, −3π/4, π/4 radians. For OQPSK, the phase can change each T s /2 = T b seconds because d 2 (t) is delayed by T s /2 seconds with respect to d 1 (t). The maximum phase change is ±π/2 radians. For MSK, the excess phase trajectories are straight lines of slopes ±π/2T b radians/second. Problem 8.10 Write the sinc-functions as sin(x) /x functions. Use appropriate trigonometric identities to reduce the product of the sinc-functions to the given form. Problem 8.11 If d (t) is a sequence of alternating 1s and 0s, the instantaneous frequency is 1/4T b Hz above the carrier (with the definition of θ i (t) given by (8.17) and (8.18)). If it is a sequence of 6 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS 1s or 0s, the instantaneous frequency is 1/4T b Hz below the carrier. (a) The instantaneous frequency is 5, 000, 000 + 100, 000/4 = 1, 025, 000 Hz. (b) The instantaneous frequency is 5, 000, 000 −100, 000/4 = 975, 000 Hz. Problem 8.12 The signal points lie on a circle of radius √ E s centered at the origin equally spaced at angular intervals of 22.5 degrees (360/16). The decision regions are pie wedges centered over each signal point. Problem 8.13 The bounds on symbol error probability are given by P 1 ≤ P E, symbol ≤ 2P 1 where P 1 = Q " r 2E s N 0 sin(π/M) # For moderate signal-to-noise ratios and M ≥ 8, the actual error probability is very close to the upper bound. Assuming Gray encoding, the bit error probability is given in terms of symbol error probability as P E, bit ≈ P E, symbol log 2 (M) and the energy per bit-to-noise spectral density ratio is E b N 0 = 1 log 2 (M) E s N 0 Thus, we solve 2 log 2 (M) Q " r 2 log 2 (M) E b N 0 sin(π/M) # = 10 −5 or Q " r 2 log 2 (M) E b N 0 sin(π/M) # = 10 −5 2 × log 2 (M) = _ ¸ ¸ _ ¸ ¸ _ 1.5 × 10 −5 , M = 8 2 × 10 −5 , M = 16 2.5 × 10 −5 , M = 32 3 × 10 −5 , M = 64 8.1. PROBLEM SOLUTIONS 7 The argument of the Q-function to give these various probabilities are found, approximately, by trial and error (either using MATLAB and a program for the Q-function or a calculator and the asymptotic expansion of the Q-function): Q-function argument = _ ¸ ¸ _ ¸ ¸ _ 4.17, M = 8 4.1, M = 16 4.06, M = 32 4.02, M = 64 Thus, for M = 8, we have r 2 log 2 (8) E b N 0 sin(π/8) = 4.17 r 6 E b N 0 × 0.3827 = 4.17 E b N 0 = 10 log 10 " 1 6 µ 4.17 0.3827 ¶ 2 # = 12.96 dB For M = 16, we have r 2 log 2 (16) E b N 0 sin(π/16) = 4.1 r 8 E b N 0 × 0.1951 = 4.17 E b N 0 = 10 log 10 " 1 8 µ 4.17 0.1951 ¶ 2 # = 17.57 dB For M = 32, we have r 2 log 2 (32) E b N 0 sin(π/32) = 4.06 r 10 E b N 0 × 0.098 = 4.06 E b N 0 = 10 log 10 " 1 10 µ 4.06 0.098 ¶ 2 # = 22.35 dB 8 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS For M = 64, we have r 2 log 2 (64) E b N 0 sin(π/64) = 4.02 r 12 E b N 0 × 0.049 = 4.02 E b N 0 = 10 log 10 " 1 12 µ 4.02 0.049 ¶ 2 # = 27.49 dB Problem 8.14 The binary number representation for the decimal digits 0 −15 and their Gray code equiv- alents are given below: Bin. No. Gray Bin. No. Gray Bin. No. Gray Bin. No. Gray 0000 0000 0100 0110 1000 1100 1100 1010 0001 0001 0101 0111 1001 1101 1101 1011 0010 0011 0110 0101 1010 1111 1110 1001 0011 0010 0111 0100 1011 1110 1111 1000 Problem 8.15 Let the coordinates of the received data vector, given signal labeled 1111 was sent, be (X, Y ) = (a +N 1 , a +N 2 ) where N 1 and N 2 are zero-mean, uncorrelated Gaussian random variables with variances N 0 /2. Since they are uncorrelated, they are also independent. Thus, P (C | I) = P (0 ≤ X ≤ 2a) P (0 ≤ Y ≤ 2a) = Z 2a 0 e −(x−a) 2 /N 0 √ πN 0 dx Z 2a 0 e −(y−a) 2 /N 0 √ πN 0 dy Let u = √ 2 (x −a) N 0 and u = √ 2 (y −a) N 0 8.1. PROBLEM SOLUTIONS 9 This results in P (C | I) = Z r 2a 2 N 0 − r 2a 2 N 0 e −u 2 /2 √ 2π du Z r 2a 2 N 0 − r 2a 2 N 0 e −v 2 /2 √ 2π dv = 4 Z r 2a 2 N 0 0 e −u 2 /2 √ 2π du Z r 2a 2 N 0 0 e −v 2 /2 √ 2π dv = _ _ 1 −2Q _ _ s 2a 2 N 0 _ _ _ _ 2 Similar derivations can be carried out for the type II and III regions. Problem 8.16 The symbol error probability expression is P s = 1 − · 1 4 P (C | I) + 1 2 P (C | II) + 1 4 P (C | III) ¸ where P (C | I) = [1 −2Q(A)] 2 , A = s 2a 2 N 0 P (C | II) = [1 −2Q(A)] [1 −Q(A)] P (C | III) = [1 −Q(A)] 2 Thus P E = 1 − ½ 1 4 [1 −2Q(A)] 2 + 1 2 [1 −2Q(A)] [1 −Q(A)] + 1 4 [1 −Q(A)] 2 ¾ = 1 − ½ 1 4 £ 1 −4Q(A) + 4Q 2 (A) ¤ + 1 2 £ 1 −3Q(A) + 2Q 2 (A) ¤ + 1 4 £ 1 −2Q(A) +Q 2 (A) ¤ ¾ ≈ 1 −{1 + 3Q(A)} , neglecting all terms in Q 2 (A) = 3Q _ _ s 2a 2 N 0 _ _ 10 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Problem 8.17 This follows by counting type I, II, and III regions. There are √ M × √ M total regions, four of which are type III regions (the corner regions), 4 ³ √ M −2 ´ type II regions, and ³ √ M −2 ´ 2 type I regions. A sketch will help in determining this. Thus, the given expression for P E follows. The expression for a follows by computing the average symbol energy in terms of a. To accomplish this, the sums m X i=1 i = m(m+ 1) 2 and m X i=1 i 2 = m(m+ 1) (2m+ 1) 6 are convenient. The approximate probability of error expression follows in a manner similar to the derivation for M = 16 outlined in Problem 8.16. Problem 8.18 Use P E, bit = M 2 (M −1) P E, symbol and E b N 0 = 1 log 2 (M) E s N 0 A tight bound for the symbol error probability for coherent M-ary FSK is P E, symbol ≤ MQ Ã r E s N 0 ! Use the asymptotic expression for the Q-function and iteration on a calculator to get the following results: For M = 8, P E, bit = 10 −4 for E b /N 0 = 7.5 dB; For M = 16, P E, bit = 10 −4 for E b /N 0 = 6.5 dB; For M = 32, P E, bit = 10 −4 for E b /N 0 = 5.9 dB; For M = 64, P E, bit = 10 −4 for E b /N 0 = 5.3 dB. Problem 8.19 Use the same relations as in Problem 8.18 for relating signal-to-noise ratio per bit and symbol and for relating bit to symbol error probability, except now P E, symbol = M−1 X k=1 µ M −1 k ¶ (−1) k+1 k + 1 exp µ − k k + 1 E s N 0 ¶ Use MATLAB to perform the sum. The following results are obtained: For M = 2, P E, bit = 10 −4 for E b /N 0 = 12.3 dB; 8.1. PROBLEM SOLUTIONS 11 For M = 4, P E, bit = 10 −4 for E b /N 0 = 9.6 dB; For M = 8, P E, bit = 10 −4 for E b /N 0 = 8.2 dB. Problem 8.20 This is a normal cartesian coordinate system with the signal points located on the φ 1 and φ 2 axes at √ E s . The decision regions are formed by the positive coordinate axes and the 45-degree bisector of the first quadrant angle. Problem 8.21 The bandwidth efficiencies are given by R B = ( 0.5 log 2 (M) , M-PSK and M-QAM log 2 (M) M+1 , coherent M-FSK This gives the results given in the table below: PSK: R B = _ _ _ 1.5 bps/Hz, M = 8 2.0 bps/Hz, M = 16 2.5 bps/Hz, M = 32 ; B = _ _ _ 6.67 kHz, M = 8 5.0 kHz, M = 16 4.0 kHz, M = 32 16-QAM: R B = 2.0 bps/Hz, M = 16; B = 5.0 kHz FSK: R B = _ _ _ 0.33 bps/Hz, M = 8 0.24 bps/Hz, M = 16 0.15 bps/Hz, M = 32 ; B = _ _ _ 30.0 kHz, M = 8 42.6 kHz, M = 16 66.0 kHz, M = 32 Problem 8.22 Use Figure 8.19 noting that the abscissa is normalized baseband bandwidth. RF bandwidthis twice as large. The 90% power containment bandwidth is defined by the ordinate = -10 dB. (c) B 90, BPSK = 1.6/T b = 1.6R b ; (b) B 90, QPSK, OQPSK = 0.8/T b = 0.8R b ; (a) B 90, MSK = 0.8/T b = 0.8R b . Problem 8.23 Use Figure 8.19 noting that the abscissa is normalized baseband bandwidth. RF bandwidth is twice as great. The 99% power containment bandwidth is defined by the ordinate = -20 dB. (a) B 99, BPSK = 14/T b = 14R b ; (b) B 99, QPSK, OQPSK = 3.5/T b = 3.5R b ; (c) B 99, MSK = 1.23/T b = 1.23R b . Problem 8.24 The baseband power spectrum is G(f) = 2A 2 T b [log 2 (M)] sinc 2 [log 2 (M) T b f] 12 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS where A 2 = E s cos 2 θ i = E s sin 2 θ i = 1 M M X i=1 cos 2 · 2π (i −1) M ¸ = 1 M M X i=1 sin 2 · 2π (i −1) M ¸ The 10% power containment bandwidth is defined by (see Problem 8.22) log 2 (M) B 90, MPSK = 0.8/T b = 0.8R b B 90, MPSK = 0.8R b / log 2 (M) = _ _ _ 0.267R b Hz, M = 8 0.25R b Hz, M = 16 0.16R b Hz, M = 8 Problem 8.25 (a) For NRZ mark: R m = ½ (A) (A) 1 2 + (−A) (−A) 1 2 = A 2 , m = 0 (A) (A) 1 4 + (−A) (A) 1 4 + (A) (−A) 1 4 + (−A) (−A) 1 4 = 0, m 6= 0 S r (f) = T b sinc 2 (fT b ) S NRZ, mark (f) = A 2 T b sinc 2 (fT b ) (b) For polar RZ: R m = ½ A 2 , m = 0 0, m 6= 0 S r (f) = T b 2 sinc 2 µ fT b 2 ¶ S polar RZ (f) = A 2 T b 4 sinc 2 µ fT b 2 ¶ 8.1. PROBLEM SOLUTIONS 13 (c) Bipolar RZ: R m = _ ¸ _ ¸ _ (A) (A) 1 2 + (0) (0) 1 2 = A 2 2 , m = 0 (A) (A) ¡ 1 4 ¢ 2 + (A) (−A) ¡ 1 4 ¢ 2 + (−A) (A) ¡ 1 4 ¢ 2 + (−A) (−A) ¡ 1 4 ¢ 2 + (A) (0) ¡ 1 4 ¢ ¡ 1 2 ¢ +(0) (A) ¡ 1 2 ¢ ¡ 1 4 ¢ + (−A) (0) ¡ 1 4 ¢ ¡ 1 2 ¢ + (0) (−A) ¡ 1 2 ¢ ¡ 1 4 ¢ + (0) (0) ¡ 1 2 ¢ ¡ 1 2 ¢ = 0, m 6= 0 S r (f) = T b 2 sinc 2 µ fT b 2 ¶ S bipolar RZ (f) = A 2 T b 8 sinc 2 µ fT b 2 ¶ Problem 8.26 To show: cos µ πt 2T b ¶ Π µ t 2T b ¶ ←→ 4T b π cos (2πT b f) 1 −(4T b f) 2 Use the modulation theorem together with the Fourier transform of a rectangular pulse to get cos µ πt 2T b ¶ Π µ t 2T b ¶ ←→T b sinc [2T b (f −1/4T b )] +T b sinc [2T b (f + 1/4T b )] Rewrite the RHS as RHS = T b · sin(2πT b f −π/2) 2πT b f −π/2 + sin(2πT b f +π/2) 2πT b f +π/2 ¸ Note that sin(2πT b f ±π/2) = ±cos (2πT b f) Then RHS = 2T b π · 1 1 −4T b f + 1 1 + 4T b f ¸ cos (2πT b f) = 4T b π cos (2πT b f) 1 −(4T b f) 2 14 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.4: Problem 8.27 Apply (5.28). Problem 8.28 This is a matter of MATLAB programming and plotting. Problem 8.29 A plot is given in Figure 8.4. Problem 8.30 The states are (read from top to bottom, column by column): 1111 0100 1011 0111 0010 0101 0011 1001 1010 0001 1100 1101 1000 0110 1110 The output sequence is the last digit in each 4-element word. The autocorrelation function is a triangular pulse centered at the origin two units wide repeated every 15 units and a horzontal line between the triangles at −1/15 units. 8.1. PROBLEM SOLUTIONS 15 Problem 8.31 The states are (read from top to bottom, column by column): 11111 00101 10100 01101 01111 10010 01010 00110 00111 01001 11010 00011 10011 00100 11101 10001 11001 00010 01110 11000 01100 00001 01110 11100 10110 10000 10111 11110 01011 01000 11011 The output sequence is the last digit in each 5-element word. The autocorrelation function is a triangular pulse centered at the origin two units wide repeated every 31 units and a horzontal line between the triangles at −1/31 units. Problem 8.32 To find the aperiodic correlation function, we slide the sequence past itself and compute the number of alike bits minus the number of unalike bits for the overlap (the sequence is not periodically repeated). The result can be scaled by the length of the sequence if desired. (a) -1, 0, -1, 0, -1, 0, 7, 0, -1, 0, -1, 0, -1 (the maximum absolute value correlation is 1 - nonzero delay); (b) -1, 0, -1, 0, 3, 0, 1, -2, -1, -4, -1, 0, -1, 0, 15, . . . (the maximum absolute value correlation is 4 - nonzero delay). Problem 8.33 Note that the expectation of N g is zero because n(t) has zero mean. Write the expectation of the square of N g as an iterated integral. The expectation can be taken inside. Use the fact that E[n(t) n(λ)] = N 0 2 δ (t −λ) to reduce the double integral to a single integral. The integral of the resulting cosine squared is T b /2. The result for the variance (same as the mean square) is then found to be N 0 T b . Problem 8.34 The random variable to be considered is N I = Z T b 0 A I c (t) cos (∆ωt +θ −φ) dt 16 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS where θ − φ is assumed uniform in[0, 2π). Clearly, the expectation of N I is zero due to this random phase. The variance is the same as the mean square, which is given by var (N I ) = E ½Z T b 0 Z T b 0 A 2 I c (t) c (λ) cos (∆ωt +θ −φ) cos (∆ωλ +θ −φ) dtdλ ¾ = Z T b 0 Z T b 0 A 2 I E[c (t) c (λ)] cos (∆ωt +θ −φ) cos (∆ωλ +θ −φ) dtdλ = A 2 I T c Z T b 0 Z T b 0 1 T c Λ[(t −λ) /T c ] 1 2 cos [∆ω (t −λ)] dtdλ = A 2 I T c Z T b 0 1 2 dt = A 2 I T c T b 2 Problem 8.35 (a) The processing gain is G p = T b T c = R c R b or R c = G p R b For G p = 100 and a data rate of 1 Mbps, R c = 100 megachips per second; (b) B RF = 2R c = 200 MHz; (c) Use (8.99) to get the following: P E = 5.03 × 10 −5 for a JSR of 5 dB; P E = 8.34 × 10 −4 for a JSR of 10 dB; P E = 1.42 × 10 −2 for a JSR of 15 dB; P E = 0.34 for a JSR of 30 dB. Problem 8.36 The results are P E = 5.72 × 10 −6 for a JSR of 5 dB; P E = 1.09 × 10 −5 for a JSR of 10 dB; P E = 5.03 × 10 −5 for a JSR of 15 dB; P E = 0.0895 for a JSR of 30 dB. Problem 8.37 In the limit at SNR E b /N 0 → ∞, the argument of the Q-function for P E = Q ³ √ SNR ´ becomes √ SNR = r 3N K −1 ≈ 3.81 to give P E = 10 −4 8.1. PROBLEM SOLUTIONS 17 Thus, K = 3N (3.81) 2 + 1 = 53.7 Round this down to K = 53 users since the number of users must be an integer. Problem 8.38 The result is T acq = nT c T e 2 (1 −P FA ) 2 −P H P H = 2 (1000) ¡ 10 −3 ¢ 2 (1 −10 −3 ) 2 −0.9 0.9 = 1.22 seconds Problem 8.39 (a) Only the OBP case will be done. See the text for the bent-pipe case. Equation (8.121) becomes p d = 10 −5 −p u 1 −2p u For BPSK, the uplink and downlink probabilities of error are related to the uplink and downlink SNRs by p u = Q "s 2 µ E b N 0 ¶ u # = 1 2 " 1 −erf Ãs µ E b N 0 ¶ u !# p d = Q "s 2 µ E b N 0 ¶ d # = 1 2 " 1 −erf Ãs µ E b N 0 ¶ d !# where these probabilities have been put in terms of the error function because MATLAB includes an inverse error function. When these relations are inverted, these become µ E b N 0 ¶ u = £ erf −1 (1 −2p u ) ¤ 2 µ E b N 0 ¶ d = £ erf −1 (1 −2p d ) ¤ 2 Typical values are given in the table below. The MATLAB program for computing them is also given below along with a plot. Part (b) is just a matter of inputting the desired p_overall in the program. 18 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.5: p u (E b /N 0 ) u , dB p d (E b /N 0 ) d , dB 10 −9 12.5495 9.999 × 10 −6 9.5879 3 × 10 −9 12.3228 9.997 × 10 −6 9.5880 7 × 10 −9 12.0839 9.993 × 10 −6 9.5882 4.3 × 10 −8 11.5639 9.957 × 10 −6 9.5898 7.2 × 10 −7 10.6497 9.280 × 10 −6 9.6217 % Solution for Problem 8.39 % p_overall = input(’Enter desired end-to-end probability of error ’) pu = logspace(log10(p_overall)-4, log10(p_overall)+1e-20); pd = (p_overall - pu)./(1 - 2*pu); Eb_N0_u = 20*log10(erfinv(1 - 2*pu)); Eb_N0_d = 20*log10(erfinv(1 - 2*pd)); plot(Eb_N0_d, Eb_N0_u), xlabel(’(E_b/N_0)_d, dB’), ylabel(’(E_b/N_0)_d, dB’),... grid, axis square 8.1. PROBLEM SOLUTIONS 19 Problem 8.40 (a) This is similar to the previous problem except that the error probability expresseion for noncoherent FSK is P E = 1 2 exp µ − E b 2N 0 ¶ Solving for E b N 0 , we have E b N 0 = −2 ln(2P E ) Thus, p d = 10 −6 −p u 1 −2p u µ E b N 0 ¶ u = −2 ln(2p u ) µ E b N 0 ¶ d = −2 ln(2p d ) (b) For DPSK, the probability of error is P E = 1 2 exp µ − E b N 0 ¶ Therefore, the equations describing the link are p d = 10 −6 −p u 1 −2p u µ E b N 0 ¶ u = −ln(2p u ) µ E b N 0 ¶ d = −ln(2p d ) A MATLAB program for computing the performance curves for both parts (a) and (b) is given below. Typical performance curves are also shown. % Solution for Problem 8.40 % mod_type = input(’Enter 1 for NFSK; 2 for DPSK ’) p_overall = input(’Enter desired end-to-end probability of error ’) pu = logspace(log10(p_overall)-6, log10(1.000001*p_overall)); 20 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.6: pd = (p_overall - pu)./(1 - 2*pu); if mod_type == 1 Eb_N0_u = 10*log10(-2*log(2*pu)); Eb_N0_d = 10*log10(-2*log(2*pd)); elseif mod_type == 2 Eb_N0_u = 10*log10(-log(2*pu)); Eb_N0_d = 10*log10(-log(2*pd)); end plot(Eb_N0_d, Eb_N0_u), xlabel(’(E_b/N_0)_d, dB’), ylabel(’(E_b/N_0)_d, dB’),... grid, axis square if mod_type == 1 title([’OBP link performance for overall P_E = ’, num2str(p_overall), ’; NFSK modulation’]) elseif mod_type == 2 title([’OBP link performance for overall P_E = ’, num2str(p_overall), ’; DPSK modulation’]) end 8.1. PROBLEM SOLUTIONS 21 Figure 8.7: Problem 8.41 For noncoherent FSK, the bit error probability, using (8.51) and (8.54), is P b = M 2 2 (M −1) Q Ã r log 2 (M) E b N 0 ! ≈ M 2 Q Ã r log 2 (M) E b N 0 ! So, the equations for the OBP link are p d = 10 −6 −p u 1 −2p u p u ≈ M 2 Q Ãs log 2 (M) µ E b N 0 ¶ u ! = M 4 " 1 −erf Ãs log 2 (M) 2 µ E b N 0 ¶ u !# µ E b N 0 ¶ u = · 2 log 2 (M) erf −1 µ 1 − 4p u M ¶¸ 2 p d ≈ M 2 Q Ãs log 2 (M) µ E b N 0 ¶ d ! = M 4 " 1 −erf Ãs log 2 (M) 2 µ E b N 0 ¶ d !# µ E b N 0 ¶ d = · 2 log 2 (M) erf −1 µ 1 − 4p d M ¶¸ 2 22 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.8: A MATLAB program is given below and a plot for 16-FSK is also given in Fig. 8.8. % Solution for Problem 8.41 % p_overall = input(’Enter desired end-to-end probability of error ’) M = input(’Enter the desired M ’) pu = logspace(log10(p_overall)-4, log10(p_overall)+1e-20); pd = (p_overall - pu)./(1 - 2*pu); Eb_N0_u = 20*log10((2/log2(M))*erfinv(1 - (4/M)*pu)); Eb_N0_d = 20*log10((2/log2(M))*erfinv(1 - (4/M)*pd)); plot(Eb_N0_d, Eb_N0_u), xlabel(’(E_b/N_0)_d, dB’), ylabel(’(E_b/N_0)_d, dB’),... grid, axis square title([’OBP link performance for overall P_E = ’, num2str(p_overall), ’; coherent ’, num2str(M), ’-FSK modulation’]) 8.1. PROBLEM SOLUTIONS 23 Problem 8.42 (a) The number of users per reuse pattern remain the same at 120. For 20 dB minimum signal-to-interference ratio and a propagation power law of α = 3, we have 20 = 10(3) log 10 µ D co d A −1 ¶ −7.7815 or log 10 µ D co d A −1 ¶ = 20 + 7.7815 30 = 0.794 D co d A = 10 0.794 + 1 = 9.43 = √ 3N or N = d29.67e = 31 (i = 1, j = 5) The efficiency is η v = ¥ 120 31 ¦ 6 MHz = 3 6 = 1 2 voice circuits per base station per MHz (b) For 14 dB minimum signal-to-interference ratio and a propagation power law of α = 3, we have 14 = 10(3) log 10 µ D co d A −1 ¶ −7.7815 log 10 µ D co d A −1 ¶ = 14 + 7.7815 30 = 0.726 D co d A = 10 0.726 + 1 = 6.32 = √ 3N or N = d13.32e = 19 (i = 2, j = 3) The efficiency is η v = ¥ 120 19 ¦ 6 MHz = 6 6 = 1 voice circuits per base station per MHz 24 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS 8.2 Computer Exercises Computer Exercise 8.1 % ce8_1.m: BEPs for M-ary PSK in AWGN (uses upper bound for M > 4); % coherent FSK (uses union bound); noncoherent FSK (exact) % clf I_mod = input(’Enter type of mod: MPSK = 1; CMFSK = 2; NCMFSK = 3: ’); if I_mod == 1 z_dB = 3:.1:30; elseif I_mod == 2 | I_mod == 3 z_dB = 3:.1:15; end z = 10.^(z_dB/10); for j = 1:6 M=2^j; if I_mod == 1 A=2/log2(M); kk=(sin(pi/M))^2*log2(M); elseif I_mod == 2 A=M/2; kk=log2(M)/2; elseif I_mod == 3 A=M/(2*(M-1)); kk = 1; end if j == 1 & I_mod == 1 Pb = .5*erfc(sqrt(z)); elseif (j >= 2 & I_mod == 1) | (j >= 1 & I_mod == 2) Pb = (A/2)*erfc(sqrt(kk*z)); elseif I_mod == 3 Pb = zeros(size(z)); for k = 1:M-1 B = (prod(1:(M-1))/(prod(1:k)*prod(1:(M-1-k))))*(-1)^(k+1)/(k+1); alpha=k*log2(M)/(k+1); Pb = Pb+A*B*exp(-alpha*z); end end semilogy(z_dB,Pb),xlabel(’E_b/N_0, dB’),ylabel(’P_b’),... 8.2. COMPUTER EXERCISES 25 axis([min(z_dB) max(z_dB) 10^(-6) 1]),... if I_mod == 1 title(’BEP for MPSK’) if j <= 2 text(z_dB(35)+.2, Pb(35), ’M = 2 & 4’) elseif j >= 3 text(z_dB(105)+.2, Pb(105), [’M = ’, num2str(M)]) end elseif I_mod == 2 title(’BEP for coherent MFSK’) if j <= 3 text(z_dB(50)+.2, Pb(50), [’M = ’, num2str(M)]) elseif j > 3 text(z_dB(25)+.2, Pb(25), [’M = ’, num2str(M)]) end elseif I_mod == 3 title(’BEP for noncoherent MFSK’) if j <= 3 text(z_dB(50)+.2, Pb(50), [’M = ’, num2str(M)]) elseif j > 3 text(z_dB(25)+.2, Pb(25), [’M = ’, num2str(M)]) end end if j==1 hold on grid end end A typical run follows, with the resulting plot shown in Fig. 8.9: >> ce8_1 Enter type of modulation: MPSK = 1; coh. MFSK = 2; noncoh. MFSK = 3: 1 Computer Exercise 8.2 % ce8_2.m: Out-of-band power for M-ary PSK, QPSK (OQPSK), and MSK % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-..’); Tbf = 0:.025:7; 26 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.9: LTbf = length(Tbf); A_BPSK = []; A_QPSK = []; A_MSK = []; for k = 1:LTbf A_BPSK(k) = quadl(’G_BPSK’, Tbf(k), 14); A_QPSK(k) = quadl(’G_QPSK’, Tbf(k), 7); A_MSK(k) = quadl(’G_MSK’, Tbf(k), 7); end OBP_BPSK = 10*log10(A_BPSK/A_BPSK(1)); OBP_QPSK = 10*log10(A_QPSK/A_QPSK(1)); OBP_MSK = 10*log10(A_MSK/A_MSK(1)); plot(Tbf, OBP_BPSK, A(1,:)), ylabel(’Relative out-of-band power, dB’), xlabel(’T_bf’),... axis([0 5 -40 0]),grid,... hold on plot(Tbf, OBP_QPSK, A(2,:)) plot(Tbf, OBP_MSK, A(3,:)) legend(’BPSK’, ’QPSK/OQPSK’, ’MSK’) A typical plot generated by the program is shown in Fig. 8.10. 8.2. COMPUTER EXERCISES 27 Figure 8.10: Computer Exercise 8.3 % ce8_3.m: Computes spectrum of continuous phase FSK % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-..’); Tbf0 = input(’Enter f_0 times T_b ’) delTbf0 = .5:.25:1.5; L_delTbf = length(delTbf0) Tbf = 0:.025:10; for k = 1:L_delTbf delTbf = delTbf0(k); G_FSK = (S_CFSK(Tbf-Tbf0) + S_CFSK(Tbf-Tbf0-delTbf)).^2; area = sum(G_FSK)*0.025; G_FSKN = G_FSK/area; plot(Tbf, G_FSKN, A(k,:)) if k == 1 hold on grid xlabel(’T_bf’), ylabel(’Spectral amplitude’) end 28 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.11: end legend([’\DeltafT_b = ’,num2str(delTbf0(1))], [’\DeltafT_b = ’,num2str(delTbf0(2))], [’\DeltafT_b = ’,num2str(delTbf0(3))],[’\DeltafT_b = ’,num2str(delTbf0(4))], [’\DeltafT_b = ’,num2str(delTbf0(5))]) title([’Spectrum for continuous phase FSK; T_bf_0 = ’, num2str(Tbf0)]) % Function to approximate CFSK spectrum % function y = S_CFSK(Tbf) y = sinc(Tbf); A typical plot is shown in Fig. 8.11. Computer Exercise 8.4 % ce8_4.m: Computes bit error probability curves for jamming in DSSS. % For a desired P_E and given JSR and Gp, computes the required Eb/N0 % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-.x’,’-.o’); Gp_dB = input(’Enter desired processing gain in dB ’); 8.2. COMPUTER EXERCISES 29 Gp = 10^(Gp_dB/10); z_dB = 0:.5:30; z = 10.^(z_dB/10); k = 1; JSR0 = []; for JSR_dB = 5:5:25; JSR0(k) = JSR_dB; JSR = 10^(JSR_dB/10); arg = z./(1+z*JSR/Gp); PE = 0.5*erfc(sqrt(arg)); semilogy(z_dB, PE, A(k+1,:)) if k == 1 hold on axis([0 30 1E-15 1]) grid on xlabel(’E_b/N_0, dB’), ylabel(’P_E’) title([’BEP for DS BPSK in jamming for proc. gain = ’, num2str(Gp_dB),’ dB’]) end k = k+1; end PEG = 0.5*qfn(sqrt(2*z)); semilogy(z_dB, PEG), text(z_dB(30)-5, PEG(30), ’No jamming’) legend([’JSR = ’, num2str(JSR0(1))], [’JSR = ’, num2str(JSR0(2))], [’JSR = ’, num2str(JSR0(3))], [’JSR = ’, num2str(JSR0(4))], [’JSR = ’, num2str(JSR0(5))], 3) disp(’Strike ENTER to continue’); pause PE0 = input(’Enter desired value of P_E ’); JSR_dB_0 = input(’Enter given value of JSR in dB ’); Gp_dB_0 = input(’Enter given value of processing gain in dB ’); JSR0 = 10^(JSR_dB_0/10); Gp0 = 10^(Gp_dB_0/10); PELIM = 0.5*erfc(sqrt(Gp0/JSR0)); disp(’ ’) disp(’Infinite Eb/N0 BEP limit’) disp(PELIM) if PELIM >= PE0 disp(’For given choices of G_p, JSR, & desired P_E, a solution is not possible ’); 30 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Gp0_over_JSR0 = (erfinv(1-2*PE0))^2; Gp0_over_JSR0_dB = 10*log10(Gp0_over_JSR0); disp(’The minimum required value of G_p over JSR in dB is:’) disp(Gp0_over_JSR0_dB) else arg0 = (erfinv(1-2*PE0))^2; SNR0 = 1/(1/arg0 - JSR0/Gp0); SNR0_dB = 10*log10(SNR0); end disp(’ ’) disp(’To give BEP of:’) disp(PE0) disp(’Requires GP, JSR, and Eb/N0 in dB of:’) disp([Gp_dB JSR_dB SNR0_dB]) A typical run follows, with both a plot given in Fig. 8.12 and a specific output generated: >> ce8_4 Enter desired processing gain in dB: 30 Strike ENTER to continue Enter desired value of P_E: 1e-3 Enter given value of JSR in dB: 20 Enter given value of processing gain in dB: 30 Infinite Eb/N0 BEP limit 3.8721e-006 To give BEP of: 0.0010 Requires GP, JSR, and Eb/N0 in dB of: 30.0000 25.0000 9.6085 Computer Exercise 8.5 % ce8_5.m: Given a satellite altitude and desired spot diameter for % illumination, determines circular antenna aperture diameter and % maximum antenna gain. % h = input(’Enter satellite altitude in km: ’); d_spot = input(’Enter desired illuminated spot diameter at subsatellite point in km: ’); f0 = input(’Enter carrier frequency in GHz: ’); rho = input(’Enter desired antenna efficiency: 0 < \rho <= 1: ’); 8.2. COMPUTER EXERCISES 31 Figure 8.12: theta = d_spot/h; phi3dB = theta; lambda = 3E8/(f0*1E9); d = lambda/(phi3dB*sqrt(rho)); % G0 = rho*(pi*d/lambda)^2 G0 = (pi/phi3dB)^2; G0_dB = 10*log10(G0); disp(’ ’) disp(’3-dB beamwidth in degrees: ’) disp(phi3dB*57.3) disp(’Wavelength in meters:’) disp(lambda) disp(’Antenna diameter in meters:’) disp(d) disp(’Antenna gain in dB:’) disp(G0_dB) disp(’ ’) A typical run follows: 32 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS >> ce8_5 Enter satellite altitude in km: 200 Enter desired illuminated spot diameter at subsatellite point in km: 20 Enter carrier frequency in GHz: 8 Enter desired antenna efficiency: 0 < rho <= 1: .7 3-dB beamwidth in degrees: 5.7300 Wavelength in meters: 0.0375 Antenna diameter in meters: 0.4482 Antenna gain in dB: 29.9430 Computer Exercise 8.6 % ce8_6.m: Performance curves for bent-pipe relay and mod/demod relay % clf A = char(’—’,’-’,’-.’,’:’); I_mod = input(’Enter type of modulation: 1 = BPSK; 2 = coh. FSK; 3 = DPSK; 4 = noncoh. FSK ’); PE0 = input(’Enter desired value of P_E ’); for I_type = 1:2 if I_type == 1 if I_mod == 1 EbN0r = (erfinv(1 - 2*PE0))^2; elseif I_mod == 2 EbN0r = 2*(erfinv(1 - 2*PE0))^2; elseif I_mod == 3 EbN0r = -log(2*PE0); elseif I_mod == 4 EbN0r = -2*log(2*PE0); end EbN0r_dB = 10*log10(EbN0r) EbN0u_dB = EbN0r_dB+.000001:.01:35; EbN0u = 10.^(EbN0u_dB/10); den = 1/EbN0r-1./EbN0u; EbN0d = 1./den; EbN0d_dB = 10.*log10(EbN0d); 8.2. COMPUTER EXERCISES 33 elseif I_type == 2 d1 = log10(0.00000001*PE0); d2 = log10(0.9999999*PE0); pu = logspace(d1, d2, 5000); pd = (PE0 - pu)./(1 - 2*pu); if I_mod == 1 EbN0u = (erfinv(1 - 2*pu)).^2; EbN0d = (erfinv(1 - 2*pd)).^2; elseif I_mod == 2 EbN0u = 2*(erfinv(1 - 2*pu)).^2; EbN0d = 2*(erfinv(1 - 2*pd)).^2; elseif I_mod == 3 EbN0u = -log(2*pu); EbN0d = -log(2*pd); elseif I_mod == 4 EbN0u = -2*log(2*pu); EbN0d = -2*log(2*pd); end EbN0u_dB = 10*log10(EbN0u); EbN0d_dB = 10*log10(EbN0d); end plot(EbN0d_dB, EbN0u_dB,A(I_type,:)), if I_type == 1 axis square, axis([5 30 5 30]), grid on,... xlabel(’(E_b/N_0)_d, dB’), ylabel(’(E_b/N_0)_u, dB’) hold on end end legend([’Bent pipe’],[’Demod/remod’],1) if I_mod == 1 title([’Uplink E_b/N_0 versus downlink E_b/N_0, both in dB, to give P_E = ’,num2str(PE0),’; BPSK modulation’]) elseif I_mod == 2 title([’Uplink E_b/N_0 versus downlink E_b/N_0, both in dB, to give P_E = ’,num2str(PE0),’; coh. FSK modulation’]) elseif I_mod == 3 title([’Uplink E_b/N_0 versus downlink E_b/N_0, both in dB, to give P_E = ’,num2str(PE0),’; DPSK modulation’]) elseif I_mod == 4 title([’Uplink E_b/N_0 versus downlink E_b/N_0, both in dB, 34 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.13: to give P_E = ’,num2str(PE0),’; noncoh. FSK modulation’]) end A typical run follows: >> ce8_6 Enter type of modulation: 1 = BPSK; 2 = coh. FSK; 3 = DPSK; 4 = noncoh. FSK 2 Enter desired value of P_E 1e-3 EbN0r_dB = 9.7998 Computer Exercise 8.7 % ce8_7.m: plots waveforms or spectra for gmsk and msk % A = char(’-’,’-.’,’—’,’:.’); samp_bit = input(’Enter number of samples per bit used in simulation: ’); N_bits = input(’Enter total number of bits: ’); GMSK = input(’Enter 1 for GMSK; 0 for normal MSK: ’); if GMSK == 1 BT_bit = input(’Enter vector of bandwidth X bit periods: ’); N_samp = 50; end 8.2. COMPUTER EXERCISES 35 I_plot = input(’Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power ’); if I_plot == 1 f0 = 1; elseif I_plot == 2 f0 = 0; end clf kp = pi/2; T_bit = 1; LB = length(BT_bit); for k = 1:LB B = BT_bit(k)/T_bit; alpha = 1.1774/B; % Note that B is two-sided bandwidth del_t = T_bit/samp_bit; fs = 1/del_t; data = 0.5*(sign(rand(1,N_bits)-.5)+1); s = 2*data - 1; L = length(s); t=0:del_t:L*T_bit-del_t; s_t = s(ones(samp_bit,1),:); % Build array whose columns are samp_bit long s_t = s_t(:)’; % Convert matrix where bit samples occupy columns to vector L_s_t = length(s_t); tp=0:del_t:N_samp*T_bit-del_t; L_t = length(t); L_tp=length(tp); t_max = max(tp); if GMSK == 1 hG = (sqrt(pi)/alpha)*exp(-pi^2*(tp-t_max/2).^2/alpha^2); freq1 = kp*del_t*conv(hG,s_t); L_hG = length(hG); freq = freq1(L_hG/2:length(freq1)-L_hG/2); elseif GMSK == 0 freq = kp*s_t; end phase = del_t*cumsum(freq); if GMSK == 0 y_mod = exp(j*2*pi*f0*t+j*phase); elseif GMSK == 1 y_mod = exp(j*2*pi*f0*t+j*phase); end 36 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS if I_plot == 1 subplot(3,1,1),plot(t,s_t),axis([min(t), max(t), -1.2, 1.2]), ylabel(’Data’) if GMSK == 0 title(’MSK waveforms’) elseif GMSK == 1 title([’GMSK for BT_b = ’, num2str(BT_bit(k))]) end subplot(3,1,2),plot(t,phase), ylabel(’Excess phase, rad.’) for nn = 1:11 if nn == 1 hold on end subplot(3,1,2),plot(t,-nn*(pi/2)*ones(size(t)),’m’) subplot(3,1,2),plot(t,nn*(pi/2)*ones(size(t)),’m’) end subplot(3,1,3),plot(t,real(y_mod)), xlabel(’t’), ylabel(’Modulated signal’) elseif I_plot == 2 Z=PSD(real(y_mod),2048,fs); % MATLAB supplied function ZN = Z/max(Z); ET = sum(ZN); OBP = 1-cumsum(ZN)/ET; LZ = length(Z); del_FR = fs/2/LZ; FR = [0:del_FR:fs/2-del_FR]; subplot(1,2,1),semilogy(FR,ZN,A(k,:)), axis([0 3 10^(-8) 1]) if k == 1 hold on grid ylabel(’PSD, J/Hz’), xlabel(’fT_b_i_t’) title(’Power spectra and out-of-band power for GMSK’) end subplot(1,2,2),semilogy(FR,OBP,A(k,:)), axis([0 3 10^(-8) 1]) if k == 1 hold on grid ylabel(’Fraction of out-of-band power’), xlabel(’fT_b_i_t’) end if LB == 1 legend([’BT_b_i_t = ’,num2str(BT_bit(1))],1) elseif LB == 2 8.2. COMPUTER EXERCISES 37 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))],1) elseif LB == 3 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))], [’BT_b_i_t = ’,num2str(BT_bit(3))],1) elseif LB == 4 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))], [’BT_b_i_t = ’,num2str(BT_bit(3))], [’BT_b_i_t = ’,num2str(BT_bit(4))],1) end end end Two typical runs follow - one for plotting waveforms and one for plotting spectra. >> ce8_7 Enter number of samples per bit used in simulation: 20 Enter total number of bits: 50 Enter 1 for GMSK; 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: .5 Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 1 >> ce8_7 Enter number of samples per bit used in simulation: 10 Enter total number of bits: 2000 Enter 1 for GMSK; 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: [.5 1 1.5] Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 2 38 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.14: 8.2. COMPUTER EXERCISES 39 Figure 8.15: Chapter 9 Optimum Receivers and Signal Space Concepts 9.1 Problems Problem 9.1 a. Given H 1 , Z = N, so f Z (z|H 1 ) = f N (n) | z=n = 10e −10z u(z) Given H 2 , Z = S+N, where S and N are independent. Thus, the resulting pdf under H 2 is the convolution of the separate pdfs of S and N: f Z (z|H 2 ) = Z ∞ −∞ f s (z −λ) f N (λ) dλ = 20 Z z 0 e −2z e −8λ dλ, λ > 0 which integrates to the result given in the problem statement. b. The likelihood ratio is Λ(Z) = f Z (Z|H 2 ) f Z (Z|H 1 ) = 0.25 ¡ e 8Z −1 ¢ , Z > 0 c. The threshold is η = P 0 (c 21 −c 11 ) P 1 (c 12 −c 22 ) = 1 3 1 2 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS d. The likelihood ratio test becomes 0.25 ¡ e 8Z −1 ¢ H 2 > < H 1 1 3 This can be reduced to Z H 2 > < H 1 ln(7/3) 8 = 0.106 e. The probability of detection is P D = Z ∞ 0.106 2.5 ¡ e −2z −e −10z ¢ dz = 0.925 The probability of false alarm is P F = Z ∞ 0.106 10e −10z dz = 0.346 Therefore, the risk is Risk = 1 4 (5) + 3 4 (5) (1 −0.925) − 1 4 (5) (1 −0.346) = 0.714 f. Consider the threshold set at η. Then P F = e −10η ⇒η ≥ 0.921 to give P F ≤ 10 −4 . Also, from part (e), P D = 1.25e −2η −0.25e −10η For values of η ≥ 0.921, P D is approximately equal to the Þrst term of the above expression. For this range of ηP D is strictly decreasing with η , so the best value of P D occurs for η = 0.921 for P F ≤ 10 −4 . The resulting P D is 0.198. g. From part (f), P F = e −10η and P D = 1.25e −2η −0.25e −10η 9.1. PROBLEMS 3 Figure 9.1: From part (d), the likelihood ratio test is Z H 2 > < H 1 ln(4η + 1) 8 = γ A plot of P D versus P F as a function of η or γ constitutes the operating characteristic of the test. It is shown in Figure 9.1. Problem 9.2 a. The likelihood ratio is Λ(Z) = 1 2 e |Z| e − 1 2 Z 2 √ 2π = r π 2 e − 1 2 Z 2 −|Z| b. The conditional pdfs under each hypothesis are plotted below. The decision regions R 1 and R 2 are indicated in Figure 9.2 for η = 1. Problem 9.3 DeÞne A = (a 1 , a 2, a 3 ) and B = (b 1 , b 2 , b 3 ). DeÞne scalar multiplication by α(a 1 , a 2, a 3 ) = (αa 1 , αa 2, αa 3 ) and vector addition by (a 1 , a 2, a 3 ) +(b 1 , b 2 , b 3 ) = (a 1 +b 1 , a 2 +b 2 , a 3 +b 3 ). The properties listed under ”Structure of Signal Space” in the text then become: 4 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9.2: 1. α 1 A+α 2 B = (α 1 a 1 +α 2 b 1 , α 1 a 2 +α 2 b 2 , α 1 a 3 +α 2 b 3 ) ²R 3 ; 2. α(A+B) = α(a 1 +b 1, a 2 +b 2 , a 3 +b 3 ) = (αa 1 +αb 1, αa 2 +αb 2 , αa 3 +αb 3 ) = (αa 1 , αa 2, αa 3 ) + (αb 1, αb 2 , αb 3 ) = αA+αB² R 3 ; 3. α 1 (α 2 A) = (α 1 α 2 A) ² R 3 (follows by writing out in component form); 4. 1 • A = A (follows by writing out in component form); 5. The unique element 0 is (0, 0, 0) so that A+ 0 = A; 6. −A = (−a 1 , −a 2, −a 3 ) so that A+ (−A) = 0. Problem 9.4 Consider (x, y) = lim T→∞ Z T −T x(t) y ∗ (t) dt Then (x, y) = lim T→∞ Z T −T y (t) x ∗ (t) dt = · lim T→∞ Z T −T x(t) y ∗ (t) dt ¸ ∗ = (x, y) ∗ Also (αx, y) = lim T→∞ Z T −T αx(t) y ∗ (t) dt = α lim T→∞ Z T −T x(t) y ∗ (t) dt = α(x, y) 9.1. PROBLEMS 5 and (x +y, z) = lim T→∞ Z T −T [x(t) +y (t)] z (t) ∗ (t) dt = lim T→∞ Z T −T x(t) z ∗ (t) dt + lim T→∞ Z T −T y (t) z ∗ (t) dt = (x, z) + (y, z) Finally (x, x) = lim T→∞ Z T −T x(t) x ∗ (t) dt = lim T→∞ Z T −T |x(t)| 2 dt ≥ 0 The scalar product for power signals is considered in the same manner. Problem 9.5 a. This scalar product is (x 1 , x 2 ) = lim T→∞ Z T 0 2e −6t dt = 1 3 b. Use the energy signal scalar product deÞnition as in (a): (x 1 , x 2 ) = lim T→∞ Z T 0 e −(5+j)t dt = 1 5 +j c. Use the power signal product deÞnition: (x 1 , x 2 ) = lim T→∞ 1 2T Z T −T cos (2πt) sin 2 (2πt) dt = 0 d. Use the power signal scalar product deÞnition: (x 1 , x 2 ) = lim T→∞ 1 2T Z T 0 cos (2πt) dt = 0 Problem 9.6 Since the signals are assumed real, kx 1 +x 2 k 2 = lim T→∞ Z T −T [x 1 (t) +x 2 (t)] 2 dt = lim T→∞ Z T −T x 2 1 (t) dt + 2 lim T−→∞ Z T −T x 1 (t) x 2 (t) dt + lim T−→∞ Z T −T x 2 2 dt = kx 1 k 2 + 2 (x 1 , x 2 ) + kx 2 k 2 6 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS From this it is seen that kx 1 +x 2 k 2 = kx 1 k 2 + kx 2 k 2 if and only if (x 1 , x 2 ) = 0. Problem 9.7 By straight forward integration, it follows that kx 1 k 2 = 2 or kx 1 k = √ 2 kx 2 k 2 = 2 3 or kx 2 k = r 2 3 kx 3 k 2 = 8 3 or kx 3 k = r 8 3 Also (x 1 , x 2 ) = 0 and (x 3 , x 1 ) = 2 Since (x 1 , x 2 ) = 0, they are orthogonal. Choose them as basis functions (not normalized). Clearly, x 3 is their vector sum. Problem 9.8 It follows that kx 1 k 2 = N X n=1 |a n | 2 and kx 2 k 2 = N X n=1 |b n | 2 Also (x 1 , x 2 ) = N X n=1 a n b ∗ n The inequality can be reduced to X m X n |a n b m −a m b n | 2 ≥ 0 which is true because each term in the sum is nonnegative. Problem 9.9 a. A set of normalized basis functions is φ 1 (t) = 1 √ 2 s 1 (t) φ 2 (t) = r 2 3 · s 2 (t) − 1 2 s 1 (t) ¸ φ 3 (t) = √ 3 · s 3 (t) − 2 3 s 1 (t) − 2 3 s 2 (t) ¸ 9.1. PROBLEMS 7 b. Evaluate (s 1 , φ 1 ), (s 2 , φ 1 ), and (s 3 , φ 1 ) to get s 1 = √ 2φ 1 ; s 2 = 1 √ 2 φ 1 + r 3 2 φ 2 ; s 3 = √ 2φ 1 + r 2 3 (φ 2 +φ 3 ) Note that a basis function set could have been obtained by inspection. For example, three unit-height, unit-width, nonoverlapping pulses spanning the interval could have been used. Problem 9.10 A basis set is φ 1 (t) = Ks 1 (t) and φ 2 (t) = Ks 2 (t) where K = r f c NA 2 A signal point lies on each coordinate axis at ±1/K. Problem 9.11 First we set v 1 (t) = x 1 (t) = exp(−t)u(t) By deÞnition kv 1 k 2 = kx 1 k 2 = Z ∞ 0 exp(−2t)dt = 1 2 Thus kv 1 k = 1 √ 2 This leads to the Þrst basis function φ 1 (t) = √ 2 exp(−t)u(t) For the second basis function we compute <1 v 2 (t) = x 2 (t) −(x 2 , φ 1 )φ 1 (t) where (x 2 , φ 1 ) = Z ∞ 0 √ 2 exp(−t) exp(−2t)dt = √ 2 Z ∞ 0 exp(−3t)dt = √ 2 3 Thus v 2 (t) = exp(−2t)u(t) − √ 2 3 √ 2 exp(−t)u(t) = · exp(−2t) − 2 3 exp(−t) ¸ u(t) 8 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Next we compute kv 2 k 2 = Z ∞ 0 µ exp(−4t) − 4 3 exp(−3t) + 4 9 exp(−2t) ¶ dt = 1 36 This gives φ 2 (t) = [6 exp(−2t) −4 exp(−t)] u(t) We next consider v 3 (t) = x 3 (t) −(x 3 , φ 2 )φ 2 (t) −(x 3 , φ 1 )φ 1 (t) Using the same procedure as previously used (x 3 , φ 2 ) = Z ∞ 0 exp(−3t) [6 exp(−2t) −4 exp(−t)] dt = 1 5 so that (x 3 , φ 2 )φ 2 (t) = 6 5 exp(−2t) − 4 5 exp(−t) Also (x 3 , φ 1 ) = Z ∞ 0 exp(−3t) √ 2 exp(−t)dt = √ 2 4 so that (x 3 , φ 1 )φ 1 (t) = 1 2 exp(−t) For t ≥ 0 we then have v 3 (t) = exp(−3t) − 6 5 exp(−2t) + 3 10 exp(−t) This leads to kv 3 k 2 = 229 600 and φ 3 (t) = r 600 229 · exp(−3t) − 6 5 exp(−2t) + 3 10 exp(−t) ¸ u(t) Problem 9.12 First we set v 1 (t) = x 1 (t) = t By deÞnition kv 1 k 2 = kx 1 k 2 = Z 1 −1 t 2 dt = 2 3 9.1. PROBLEMS 9 Thus the Þrst basis function is φ 1 (t) = r 3 2 t, −1 < t < 1 For the next basis function we set v 2 (t) = x 2 (t) −(x 2 , φ 1 )φ 1 (t) where (x 2 , φ 1 ) = Z 1 −1 t 2 Ã r 3 2 t ! dt = 0 since the intergrand is odd and the limits are even. Thus kv 2 k 2 = Z 1 −1 t 4 dt = 2 5 from which φ 2 (t) = t 2 kv 2 k = r 5 2 t 2 For the third basis function we write v 3 (t) = x 3 (t) −(x 3 , φ 2 )φ 2 (t) −(x 3 , φ 1 )φ 1 (t) where (x 3 , φ 2 ) = Z 1 −1 t 3 Ã r 5 2 t 2 ! dt = 0 and (x 3 , φ 1 ) = Z 1 −1 t 3 Ã r 3 2 t ! dt = r 3 2 2 5 so that v 3 (t) = t 3 − r 3 2 2 5 r 3 2 t = t 3 − 3 5 t This gives kv 3 k 2 = Z 1 −1 · t 2 − 6 5 t 4 + 9 25 t 2 ¸ dt = 8 175 and φ 3 (t) = v 3 (t) kv 3 k = r 175 8 · t 3 − 3 5 t ¸ 10 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS For the next basis function we write v 4 (t) = x 4 (t) −(x 4 , φ 3 )φ 3 (t) −(x 4 , φ 2 )φ 2 (t) −(x 4 , φ 1 )φ 1 (t) or v 4 (t) = x 4 (t) −(x 4 , φ 2 )φ 2 (t) since (x 4 , φ 3 ) and (x 4 , φ 1 ) are zero. By deÞnition (x 4 , φ 2 ) = Z 1 −1 t 4 Ã r 5 2 t 2 ! dt = r 5 2 2 7 so that v 4 (t) = t 4 − r 5 2 2 7 r 5 2 t 2 = t 4 − 5 7 t 2 This gives kv 4 k 2 = Z 1 −1 · t 8 − 10 7 t 6 + 25 49 t 4 ¸ dt = 8 441 and φ 4 (t) = v 4 (t) kv 4 k = r 441 8 · t 4 − 5 7 t 2 ¸ Problem 9.13 a. A suitable basis is φ 1 (t) = r 2 T cos (2πf c t) and φ 2 (t) = r 2 T sin(2πf c t) for 0 ≤ t ≤ T The coordinates of the signal vectors are x i = Z T 0 s i (t) φ 1 (t) dt = √ Ecos µ 2πi M ¶ = √ Ecos ψ i and y i = Z T 0 s i (t) φ 2 (t) dt = − √ Esin µ 2πi M ¶ = − √ Esinψ i where E = A 2 T/2 and ψ i = 2πi M . Thus s i (t) = √ Ecos(ψ i )φ 1 (t) − √ Esin(ψ i )φ 2 (t) , i = 1, 2, 3, 4 9.1. PROBLEMS 11 b. The optimum partitioning of the signal space is a pie-wedge centered on each signal point with the origin as the vertex and having angle 2π/M. c. The optimum receiver consists of two parallel branches which correlate the incoming signal plus noise with φ 1 (t) and φ 2 (t). The correlator outputs are sampled at the end of each signaling interval. Call these samples x and y. The angle tan −1 (x/y) is formed to determine which pie-wedge decision region the received data vector is in. d. DeÞning z i (t) = s i (t) +n(t) gives z i (t) = ( √ Ecos(ψ i ) +N 1 )φ 1 (t) + (− √ Esin(ψ i ) +N 2 )φ 2 (t) , i = 1, 2, 3, 4 The probability of error can then be written in the form P[error |s i (t)] = 1 −Pr[correct recept.|s i (t)] = 1 − Z Z R i 1 πN 0 · exp · − 1 N 0 µ ³ x − √ Ecos ψ i ´ 2 + ³ y + √ Esin(ψ i ) ´ 2 ¶¸ dxdy where R i is the region for a correct decision given that s i (t) is transmitted. (Note that the noise variance is N 0 /2). The expression for the probability of error can be changed to polar coordinates by letting x = r p N 0 cos θ and y = r p N 0 sinθ With this transformation dxdy →N 0 rdrdθ This gives P[error |s i (t)] = 1 − 1 π Z Z R i r · exp · − 1 N 0 ³ ¡ r 2 N 0 ¢ −2r p EN 0 cos(θ +ψ i ) +E ´ ¸ drdθ where R i now represents the decision region for a correct decision in the R, θ plane. Since all decision regions have equal area and the noise has circular symmetry, the error probability is independent of the transmitted signal. Thus, let i = M so that ψ i = 2π and cos(θ+ψ i ) = cos θ. Since the error probability is independent of the signal chosen, the conditional symbol error probability is equal to the unconditional error probability for the case in which all signals are transmitted with equal probability. Therefore P[error] = 1 − 1 π Z π/M −π/M Z ∞ 0 r exp h − ³ r 2 −2r p E/N 0 cos(θ) +E/N 0 í drdθ 12 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS e. This was done in Chapter 8. Problem 9.14 Write P c = Z ∞ −∞ e −y 2 √ π " Z y+ √ E/N 0 −∞ e −x 2 √ π dx # dy and let z = x −y. This gives P c = Z √ E/N 0 −∞ e −z 2/2 √ π Z ∞ −∞ e −2(y+z/2) 2 √ π dydz Complete the square in the exponent of the inside integral and use a table of deÞnite integrals to show that is evaluates to (1/2) 1/2 . The result then reduces to P c = 1 −Q ³ p E/N 0 ´ which gives P c = Q h (E/N 0 ) 1/2 i , the desired result. Problem 9.15 a. The space is three-dimensional with signal points at the eight points ³ ± p E/3 ´ , ³ ± p E/3 ´ , ³ ± p E/3 ´ . The optimum partitions are planes determined by the coordinate axes taken two at a time (three planes). Thus the optimum decision regions are the eight quadrants of the signal space. b. Consider S 1 . Given the partitioning discussed in part (a), we make a correct decision only if kY −S 1 k 2 < kY −S 2 k 2 and kY −S 1 k 2 < kY −S 4 k 2 and kY −S 1 k 2 < kY −S 8 k 2 where S 2 , S 4 , and S 8 are the nearest-neighbor signal points to S 1 . These are the most probable errors. Substituting Y = (y 1 , y 2 , y 3 ) and S 1 = (1, 1, 1), S 2 (1, −1, 1, ), S 4 (−1, 1, 1), and S 8 = (1, 1, −1), the above conditions become (y 1 + 1) 2 < (y 1 −1) 2 , (y 2 + 1) 2 < (y 2 −1) 2 , and (y 3 + 1) 2 < (y 3 −1) 2 9.1. PROBLEMS 13 by canceling like terms on each side. For S 1 , these reduce to y 1 > 0, y 2 > 0, and y 3 > 0 to deÞne the decision region for S 1 . Therefore, the probability of correct decision is P [correct dec.|s 1 (t)] = Pr (y 1 > 0, y 2 > 0, y 3 > 0) = [Pr (y i > 0)] 3 because the noises along each coordinate axis are independent. Note that E[y i ] = (E/3) 1/2 , all i. The noise variances, and therefore the variances of the y i ’s are all N 0 . Thus P [correct dec.|s 1 (t)] = · 1 √ πN 0 Z ∞ 0 e − 1 N 0 ³ y− √ E/3 2 ´ dy ¸ 3 = h 1 −Q ³ p 2E/3N 0 í 3 Since this is independent of the signal chosen, this is the average probability of correct detection. The symbol error probability is 1 − P c . Generalizing to n dimensions, we have P e = 1 − h 1 −Q ³ p 2E/nN 0 í n where n = log 2 M. Note that E b = E/n since there are n bits per dimension. c. The symbol error probability is plotted in Figure 9.3. Problem 9.16 a. We need 1/2T Hz per signal. Therefore, M = W/ (1/2T) = 2WT. b. For vertices-of-a-hpercube signaling, M = 2 n where n is the number of dimensions. Hence M = 2 2WT . Problem 9.17 In the equation P (E|H 1 ) = Z ∞ 0 ·Z ∞ r 1 f R 2 (r 2 |H 1 ) dr 2 ¸ f R 1 (r 1 |H 1 ) dr 1 substitute f R 1 (r 1 |H 1 ) = 2r 1 2Eσ 2 +N 0 exp µ − r 2 1 2Eσ 2 +N 0 ¶ ; r 1 ≥ 0 and f R 2 (r 2 |H 1 ) = 2r 2 N 0 exp µ − r 2 2 N 0 ¶ , r 2 > 0 14 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9.3: The inside integral becomes I = exp µ − r 2 1 N 0 ¶ When substituted in the Þrst equation, it can be reduced to P (E|H 1 ) = Z ∞ 0 2r 1 2Eσ 2 +N 0 exp µ − r 2 1 K ¶ dr 1 where K = N 0 ¡ 2Eσ 2 +N 0 ¢ 2Eσ 2 +N 0 The remaining integral is then easily integrated since its integrand is a perfect differential if written as P (E|H 1 ) = K 2Eσ 2 +N 0 Z ∞ 0 2r 1 K exp µ − r 2 1 K ¶ dr 1 Problem 9.18 a. Decide hypothesis i is true if f Z|H i (z|H i ) ≥ f Z|H j (z|H j ), all j 9.1. PROBLEMS 15 where f Z|H i (z|H i ) = exp ³ − z 2 ci +z 2 si E i σ 2 +N 0 ´ π M (E i σ 2 +N 0 ) N M 0 exp   − 1 N 0 M X j=1,j6=i ¡ z 2 cj +z 2 sj ¢   This constitutes the maximum likelihood decision rule since the hypotheses are equally probable. Alternatively, we can take the natural log of both sides and use that as a test. This reduces to computing E i σ 2 E i σ 2 +N 0 ¡ z 2 ci +z 2 si ¢ and choosing the signal corresponding to the largest. If the signals have equal energy, then the optimum receiver is seen to be an M-signal replica of the one shown in Fig. 9.8a. b. The probability of correct decision can be reduced to P c = M−1 X i=0 µ m−1 i ¶ N 0 N 0 + (Eσ 2 +N 0 ) (M −1 −i) = 1 −P e Problem 9.19 a. It can be shown under hypothesis that H 1 , Y 1 is the sum of squares of 2N independent Gaussian random variables each having mean zero and variance σ 2 11 = 2 E N σ 2 + N 0 2 and Y 2 is the sum of squares of 2N independent Gaussian random variables each having zero mean and variance σ 2 21 = N 0 2 Therefore, from Problem 4.37, the pdfs under hypothesis H 1 are f Y i (y 1 |H 1 ) = y N−1 1 e −y 1 /2σ 2 11 2 N σ N 11 Γ(N) , y 1 ≥ 0 and f Y 2 (y 2 |H 1 ) = y N−1 2 e −y 2 /N 0 2 N (N 0 /2) N Γ(N) , y 2 ≥ 0 16 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9.4: b. The probability of error is P e = Pr (E|H 1 ) = Pr (E|H 2 ) = Pr (Y 2 > Y 1 |H 1 ) = Z ∞ 0 ·Z ∞ y i f Y 2 (y 2 |H 1 ) dy 2 ¸ f Y 1 (y 1 |H 1 ) dy 1 Using integration by parts, it can be shown that I (x; a) = Z ∞ x z n e −az dz = e −ax n X i=0 x i n! i!a n+1−i This can be used to get the formula given in the problem statement. c. Plots are given in Figure 9.4 for N = 1 (solid curve), 2, 3, and 4 (dash-dot curve). Problem 9.20 a. The characteristic function is found and used to obtain the moments. By deÞnition, Φ(jω) = E £ e jΛω ¤ = Z ∞ 0 e jλω β m Γ(m) e −βλ λ m−1 dλ 9.1. PROBLEMS 17 To integrate, note that Z ∞ 0 e jλω e −βλ dλ = 1 β −jω Differentiate both sides m−1 times with respect to β. This gives Z ∞ 0 e jλω (−1) m−1 λ m−1 e −βλ dλ = (−1) m−1 (m−1)! (β −jω) m Multiply both sides by β m and cancel like terms to get Z ∞ 0 e jλω β m Γ(m) e −βλ λ m−1 dλ = β m (β −jω) m Thus Φ(jω) = β m (β −jω) m Differentiating this with respect to ω, we get E[Λ] = jΦ 0 (0) − m β and E £ Λ 2 ¤ = j 2 Φ 00 (0) = m(m+ 1) β The variance is V ar (Λ) = E ¡ Λ 2 ¢ −E 2 (Λ) = m β 2 b. Use Bayes’ rule. But Þrst we need f Z (z), which is obtained as follows: f Z (z) = Z ∞ 0 f Z|Λ (z|λ) f Λ (λ) dλ = Z ∞ 0 λe −λz β m Γ(m) e −βλ λ m−1 dλ = mβ m (β +z) m+1 Z ∞ 0 e (β+z)λ (β +z) m+1 Γ(m+ 1) λ m dλ = mβ m (β +z) m+1 where the last integral is evaluated by noting that the integrand is a pdf and therefore integrates to unity. Using Bayes’ rule, we Þnd that f Λ|Z (λ|z) = λ m (β +z) m+1 e −(β+z)λ Γ(m+ 1) 18 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Note that this is the same pdf as in part (a) except that β has been replaced by β +z and m has been replaced by m + 1. Therefore, we can infer the moments from part (a) as well. They are E[Λ|Z] = m+ 1 β +Z and V ar [Λ|Z] = m+ 1 (β +Z) 2 c. Assume the observations are independent. Then f z 1 z 2 |Λ (z 1, z 2 |λ) = λ 2 e −λ(z 1 +z 2 ) , z 1 , z 2 ≥ 0 The integration to Þnd the joint pdf of Z 1 and Z 2 is similar to the procedure used to Þnd the pdf of Z above with the result that f z 1 z 2 (z 1, z 2 ) = m(m+ 1) β m (β +z 1 +z 2 ) m+2 Again using Bayes’ rule, we Þnd that f Λ|z 1 z 2 (λ|z 1 , z 2 ) = λ m+1 e −λ(β+z 1 +z 2 ) (β +z 1 +z 2 ) m+2 Γ(m+ 2) Since this is of the same form as in parts (a) and (b), we can extend the results there for the moments to E[Λ|Z 1 , Z 2 ] = m+ 2 β +Z 1 +Z 2 and V ar [Λ|Z 1 , Z 2 ] = m+ 2 (β +Z 1 +Z 2 ) 2 d. By examining the pattern developed in parts. (a), (b), and (c), we conclude that f Λ|z 1 z 2 ...z K (λ|z 1 , z 2, . . . z K ) = λ m+1+K e −λ(β+z 1 +z 2 +...+z K ) (β +z 1 +z 2 +. . . +z K ) m+K Γ(m+ 2) and E[Λ|Z 1 , Z 2 , . . . , Z K ] = m+ 2 β +Z 1 +Z 2 +. . . +Z K V ar [Λ|Z 1 , Z 2 , . . . , Z K ] = m+K (β +Z 1 +Z 2 +. . . +Z K ) 2 Problem 9.21 The conditional mean and Bayes’ estimates are the same if: 9.1. PROBLEMS 19 1. C (x) is symmetric; 2. C (x) is convex upward; 3. f A|Z is symmetric about the mean. With this, we Þnd the following to be true: a. Same - all conditions are satisÞed; b. Same - all conditions are satisÞed; c. Condition 3 is not satisÞed; d. Condition 2 is not satisÞed. Problem 9.22 a. The conditional pdf of the noise samples is f Z ¡ z 1 , z 2 , . . . , z K |σ 2 n ¢ = ¡ 2πσ 2 n ¢ −K/2 e − P K i=1 z 2 i /2σ 2 n The maximum likelihood estimate for σ 2 n maximizes this expression. It can be found by differentiating the pdf with respect to σ 2 n and setting the result equal to zero. Solving for σ 2 n , we Þnd the maximum likelihood estimate to be b σ 2 n = 1 K K X i=1 Z 2 i b. The variance of the estimate is V ar ¡ b σ 2 n ¢ = 2σ 4 n K c. Yes. d. P K i=1 Z 2 i is a sufficient statistic. 20 CHAPTER 9. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Problem 9.23 We base our estimate on the observation Z 0 , where Z 0 = M 0 + N where N is a Gaussian random variable of mean zero and variance N 0 /2. The sample value M 0 is Gaussian with mean m 0 and variance σ 2 m . The Þgure of merit is mean-squared error. Thus, we use a MAP estimate. It can be shown that the estimate for m 0 is the mean value of M 0 conditioned on Z 0 because the random variables involved are Gaussian. That is, b m 0 = E(m 0 |Z 0 = z 0 ) = ρz 0 where ρ = E(M 0 Z 0 ) p V ar (M 0 ) p V ar (Z 0 ) = σ m q σ 2 m + N 0 2 It is of interest to compare this with the ML estimate, which is b m 0,ML = Z 0 which results if we have no a priori knowledge of the parameter. Note that the MAP estimate reduces to the ML estimate when σ 2 m → ∞. Note also that for N 0 → ∞ (i.e., small signal-to-noise ratio) we don’t use the observed value of Z 0 , but instead estimate m 0 as zero (its mean). Problem 9.24 a. The results are f Z (z 1 , z 2 |θ, H 1,2 ) = 1 πN 0 e − 1 N 0 [(z 1 ∓A) 2 +(z 2 ±A) 2 ] where the top sign in the exponent goes with H 1 and the bottom sign goes with H 2 , b. The result is f Z|Θ (Z 1 , Z 2 |Θ) = 1 πN 0 e − 1 N 0 (Z 2 1 +Z 2 2 + T 2 A 2 ) cosh · 2 N 0 (A 1 Z 1 −A 2 Z 2 ) ¸ Since the log function is monotonic, we can take the natural log of both sides and differentiate it with respect to Θ, set the result to zero, and have a condition that the maximum likelihood estimate must satisfy. The result is tanh " r T 2 2A N 0 (Z 1 cos θ −Z 2 sinθ) # r T 2 2A N 0 (−Z 1 sinθ −Z 2 cos θ) = 0 9.1. PROBLEMS 21 Using deÞnitions for Z 1 and Z 2 , which are Z 1 = Z T 0 y (t) r 2 T cos ω c t dt and Z 2 = Z T 0 y (t) r 2 T sinω c t dt this can be put into the form tanh · 4z AT Z T 0 y (t) cos (ω c t +θ) dt ¸ 4z AT Z T 0 −y (t) sin(ω c t +θ) dt = 0 where z = A 2 T/2N 0 . This implies a Costas loop type structure with integrators in place of the low pass Þlters and a tanh-function in the leg with the cosine multiplication. Note that for x small tanh(x) ' x and the phase estimator becomes a Costas loop if the integrators are viewed as lowpass Þlters. c. The variance of the phase estimate is bounded by V ar ³ b θ ML ´ ≥ 1 4z (z + 1) Problem 9.24 This is similar to the previous problem. For (a) note that cos £ cos −1 m ¤ = m, sin £ cos −1 m ¤ = ¡ 1 −m 2 ¢ 1/2 The development of the log-likelihood function follows similarly to the development of the one for Problem 9.23. The block diagram of the phase estimator has another arm that adds into the feedback to the VCO that acts as a phase-lock loop for the carrier component. Chapter 10 Information Theory and Coding 10.1 Problem Solutions Problem 10.1 The information in the message is I (x) = −log 2 (0.8) = 0.3219 bits I (x) = −log e (0.8) = 0.2231 nats I (x) = −log 10 (0.8) = 0.0969 Hartleys Problem 10.2 (a) I (x) = −log 2 (52) = 5.7004 bits (b) I (x) = −log 2 [(52)(52)] = 11.4009 bits (c) I (x) = −log 2 [(52)(51)] = 11.3729 bits Problem 10.3 The entropy is I (x) = −0.3 log 2 (0.3) −0.25 log 2 (0.25) −0.25 log 2 (0.25) −0.1 log 2 (0.1) −0.05 log 2 (0.05) −0.05 log 2 (0.05) = 2.1477 bits The maximum entropy is I (x) = log 2 5 = 2.3219 bits Problem 10.4 1 2 CHAPTER 10. INFORMATION THEORY AND CODING 0.7 0.5 0.2 0.2 0.2 0.2 0.1 0.6 0.3 1 y 3 x 2 x 1 x 3 y 2 y Figure 10.1: I (x) = 2.471 bits Problem 10.5 (a) The channel diagram is shown is illustrated in Figure 10.1. (b) The output probabilities are p(y 1 ) = 1 3 (0.5 + 0.2 + 0.1) = 8 30 = 0.2667 p(y 2 ) = 1 3 (0.3 + 0.6 + 0.2) = 11 30 = 0.3667 p (y3) = 1 3 (0.2 + 0.2 + 0.7) = 11 30 = 0.3667 (c) Since [P (Y )] = [P (X)] [P (Y |X)] we can write [P (X)] = [P (Y )] [P (Y |X)] −1 which is [P (X)] = [0.333 0.333 0.333] _ _ 2.5333 −1.1333 −0.4000 −0.8000 2.2000 −0.4000 −0.1333 −0.4667 1.6000 _ _ 10.1. PROBLEM SOLUTIONS 3 This gives[P (X)] = [0.5333 0.2000 02.667] (d) The joint probability matrix is [P (X; Y )] = [0.333 0.333 0.333] _ _ 0.5333 0 0 0 0.2000 0 0 0 0.2667 _ _ _ _ 0.5 0.3 0.2 0.2 0.6 0.2 0.1 0.2 0.7 _ _ which gives [P (X; Y )] = _ _ 0.2667 0.1600 0.1067 0.0400 0.1200 0.0400 0.0267 0.0533 0.1867 _ _ Note that the column sum gives the output possibilities [P (Y )], and the row sum gives the input probabilities [P (X)] . Problem 10.6 For a noiseless channel, the transition probability matrix is a square matrix with 1s on the main diagonal and zeros elsewhere. The joint probability matrix is a square matrix with the input probabilities on the main diagonal and zeros elsewhere. In other words [P (X; Y )] = _ ¸ ¸ ¸ _ p (x 1 ) 0 · · · 0 0 p (x 2 ) · · · 0 . . . . . . . . . . . . 0 0 · · · p(x n ) _ ¸ ¸ ¸ _ Problem 10.7 This problem may be solved by raising the channel matrix A = · 0.999 0.001 0.001 0.999 ¸ which corresponds to an error probability of 0.001, to increasing powers n and seeing where the error probability reaches the critical value of 0.08. Consider the MATLAB program a = [0.999 0.001; 0.001 0.999]; % channel matrix n = 1; % initial value a1 = a; % save a while a1(1,2)<0.08 n=n+1; a1=a^n; end 4 CHAPTER 10. INFORMATION THEORY AND CODING n-1 % display result Executing the program yields n −1 = 87. Thus we compute (MATLAB code is given) À a^87 ans = 0.9201 0.0799 0.0799 0.9201 À a^88 ans = 0.9192 0.0808 0.0808 0.9192 Thus 87 cascaded channels meets the speciÞcation for P E < 0.08. However cascading 88 channels yields P E > 0.08 and the speciÞcation is not satisÞed. (Note: This may appear to be an impractical result since such a large number of cascaded channels are speciÞed. However, the channel A may represent a lengthly cable with a large number of repeaters. There are a number of other practical examples.) Problem 10.8 The Þrst step is to write H (Y |X) −H (Y ). This gives H (Y |X) −H (Y ) = − X i X j p (x i , y j ) log 2 p (y j |x i ) − X j p (y j ) log 2 p (y j ) which is H (Y |X) −H (Y ) = − X i X j p (x i , y j ) [log 2 p (y j |x i ) −log 2 p (y i )] or H (Y |X) −H (Y ) = − 1 ln2 X i X j p(x i , y j ) ln · p (x i , y j ) p (x i )p(x i ) ¸ or H (Y |X) −H (Y ) = 1 ln2 X i X j p (x i , y j ) ln · p (x i )(y j ) p (x i , y j ) ¸ Since lnx ≤ x −1, the preceding expression can be written H (Y |X) −H (Y ) ≤ 1 ln2 X i X j p (x i , y j ) · p (x i )p(y j ) p (x i , y j ) −1 ¸ 10.1. PROBLEM SOLUTIONS 5 2 y 1 y 3 x 2 x 1 x Figure 10.2: or H (Y |X) −H (Y ) ≤ 1 ln2 _ _ _ X i X j p (x i )p(y j ) − X i X j p (x i , y j ) _ _ _ The term in braces is zero since both double sums evaluate to one. Thus, H (Y |X) −H (Y ) ≤ 0 which gives H (Y ) ≥ H (Y |X) Problem 10.9 For this problem note that H (Y |X) = 0 so that I (Y |X) = H (X). It therefore follows that C = 1 bit/symbol and that the capacity is achieved for p (x 1 ) = p (x 2 ) = 1 2 Problem 10.10 For this problem, the channel diagram is illustrated in Figure 10.2. Note that p (y 1 |x 1 ) = 1 p (y 2 |x 1 ) = 0 p (y 1 |x 2 ) = 1 p (y 2 |x 2 ) = 0 p (y 1 |x 3 ) = 0 p (y 2 |x 3 ) = 1 6 CHAPTER 10. INFORMATION THEORY AND CODING Thus H (Y |X) = − X i X j p (x i , y j ) log 2 p(y j |x i ) = 0 since each term in the summand is zero. Thus I (X; Y ) = H (Y ) The capacity is therefore C = 1 bit/symbol and is achieved for source probabilities satisfying p (x 1 ) +p(x 2 ) = 1 2 p (x 3 ) = 1 2 Note that capacity is not achieved for equally likely inputs. Problem 10.11 For the erasure channel C = p bits/symbol and capacity is achieved for equally likely inputs. Problem 10.12 The capacity of the channel described by the transition probability matrix _ ¸ ¸ _ p q 0 0 q p 0 0 0 0 p q 0 0 q p _ ¸ ¸ _ is easily computed by maximizing I (X; Y ) = H (Y ) −H (Y |X) The conditional entropy H (Y |X) can be written H (Y |X) = − X i X j p (x i ) p (y j |x i ) log 2 p (y j |x i ) = X i p (x i ) H (Y |x i ) where H (Y |x i ) = − X j p (y j |x i ) log 2 p (y j |x i ) 10.1. PROBLEM SOLUTIONS 7 For the given channel H (Y |x i ) = −p log 2 p −q log 2 q = K so that H (Y |x i ) is a constant independent of i. This results since each row of the matrix contains the same set of probabilities although the terms are not in the same order. For such a channel H (Y |X) = X i p (x i ) K = K and I (X; Y ) = H (Y ) −K We see that capacity is achieved when each channel output occurs with equal probability. We now show that equally likely outputs result if the inputs are equally likely. Assume that p (x i ) = 1 4 for all i. Then p (y j ) = X i p (x i ) p(y j |x i ) For each j p (y j ) = 1 4 X i p (y j |x i ) = 1 4 (p +q) = 1 4 [p + (1 −p)] = 1 4 so that p (y j ) = 1 4 , all j Since p (y j ) = 1 4 for all j, H (Y ) = 2, and the channel capacity is C = 2 +p log 2 p + (1 −p) log 2 (1 −p) or C = 2 −H (p) This is shown in Figure 10.3. This channel is modeled as a pair of binary symmetric channels as shown. If p = 1 or p = 0, each input uniquely determines the output and the channel reduces to a noiseless channel with four inputs and four outputs. This gives a capacity of log 2 4 = 2 bits/symbol. If p = q = 1 2 , then each of the two subchannels (Channel 1 and Channel 2) have zero capacity. However, x 1 and x 2 can be viewed as a single input and x 3 and x 4 can be viewed as a single input as shown in Figure 10.4. The result is equivalent to a noiseless channel with two inputs and two outputs. This gives a capacity of log 2 2 = 1 bit/symbol. Note that this channel is an example of a general symmetric channel. The capacity of such channels are easily computed. See Gallager (1968), pages 91-94 for a discussion of these channels. 8 CHAPTER 10. INFORMATION THEORY AND CODING p 1 1/2 0 1 2 C Figure 10.3: 4 y 3 y 2 y 1 y 4 x 3 x 2 x 1 x q q q q p p p p Channel 1 Channel 2 Figure 10.4: 10.1. PROBLEM SOLUTIONS 9 Problem 10.13 To show (10.30), write p (x i , y j ) in (10.28) as p (x i |y j ) p (y j ) and recall that log 2 p (x i , y j ) = log 2 p (x i |y j ) + log 2 p (y j ) Evaluating the resulting sum yields (10.30). The same method is used to show (10.31) except that p (x i , y j ) is written p (y j |x i ) p (x i ) . Problem 10.14 The entropy is maximized when each quantizing region occurs with probability 0.25. Thus Z x 1 0 ae −ax dx = 1 −e ax 1 = 1 4 which gives e −ax 1 = 3 4 or x 1 = 1 a ln µ 4 3 ¶ = 0.2877/a In a similar manner Z x 2 x 1 ae −ax dx = e −ax 1 −e −ax 2 = 3 4 −e −ax 2 = 1 4 which gives e −ax 2 = 1 2 or x 2 = 1 a ln(2) = 0.6931/a Also Z x 3 x 2 ae −ax dx = 1 4 gives x 3 = 1.3863/a Problem 10.15 The thresholds are now parameterized in terms of σ. The results are x 1 = 0.7585σ x 2 = 1.1774σ x 3 = 1.6651σ 10 CHAPTER 10. INFORMATION THEORY AND CODING Problem 10.16 First we compute the probabilities of the six messages. Using symmetry, we only need compute three states; m 3 , m 4 , and m 5 . p (m 3 ) = 1 √ 2πσ Z σ 0 e −y 2 /2σ 2 dy = 1 2 −Q(1) = 0.3413 p (m 4 ) = 1 √ 2πσ Z 2σ σ e −y 2 /2σ 2 dy = Q(1) −Q(2) = 0.1359 p (m 5 ) = 1 √ 2πσ Z ∞ 2σ e −y 2 /2σ 2 dy = Q(2) = 0.0228 By symmetry p (m 0 ) = p(m 5 ) = 0.0228 p (m 1 ) = p(m 4 ) = 0.1359 p (m 2 ) = p(m 3 ) = 0.3413 the entropy at the quantizer output is H (X) = −2 [0.3413 log 2 0.3413 + 0.1359 log 2 0.1359 + 0.0228 log 2 0.0228] = 2.09 bits/symbol Since the symbol (sample) rate is 500 samples/s, the information rate is r = 500 H (X) = 1045 symbols/s Problem 10.17 Five thresholds are needed to deÞne the six quantization regions. The Þve thresholds are −k 2 , −k 1 , 0, k 1 , and k 2 . The thresholds are selected so that the six quantization regions are equally likely. Finding the inverse Q-function yields k 1 = 0.4303 k 2 = 0.9674 This gives the quantizing rule Quantizer Input Quantizer Output −∞< x i < −0.9674σ m 0 −0.9674σ < x i < −0.4303σ m 1 −0.4303σ < x i < 0 m 2 0 < x i < 0.4303σ m 3 0.4303σ < x i < 0.9674σ m 4 0.9674σ < x i < ∞ m 5 10.1. PROBLEM SOLUTIONS 11 Problem 10.18 The transition probabilities for the cascade of Channel 1 and Channel 2 are · p 11 p 12 p 21 p 22 ¸ = · 0.9 0.1 0.1 0.9 ¸ · 0.75 0.25 0.25 0.75 ¸ = · 0.7 0.3 0.3 0.7 ¸ Therefore, the capacity of the overall channel is C = 1 + 0.7 log 2 0.7 + 0.3 log 2 0.3 = 0.1187 bits/symbol The capacity of Channel 1 is C = 1 + 0.9 log 2 0.9 + 0.1 log 2 0.1 = 0.5310 bits/symbol and the capacity of Channel 2 is C = 1 + 0.75 log 2 0.75 + 0.25 log 2 0.25 = 0.1887 bits/symbol It can be seen that the capacity of the cascade channel is less than the capacity of either channel taken alone. Problem 10.19 For BPSK, the error probability is Q ¡√ 2z ¢ . From the data given Uplink error probability = 0.000191 Downlink error probability = 0.005954 Using the techniques of the previous problem, or equivalently, Equation (10.24), yields the overall error probability P E = 0.006143 Note that for all practical purposes the performance of the overall system is established by the downlink performance. Problem 10.20 The source entropy is H (X) = − 3 4 log 2 3 4 − 1 4 log 2 1 4 = 0.8113 and the entropy of the fourth-order extension is H ¡ X 4 ¢ = 4H (X) = 4 (0.3113) = 3.2451 12 CHAPTER 10. INFORMATION THEORY AND CODING The second way to determine the entropy of the fourth-order extension, is to determine the probability distribution of the extended source. We have 1 symbol with probability 81 256 4 symbols with probability 27 256 6 symbols with probability 9 256 4 symbols with probability 3 256 1 symbol with probability 1 256 Thus H ¡ X 4 ¢ = − 81 256 log 2 81 256 −4 µ 27 256 log 2 27 256 ¶ −6 µ 9 256 log 2 9 256 ¶ −4 µ 3 256 log 2 3 256 ¶ − 1 256 log 2 1 256 = 3.2451 bits/symbol Problem 10.21 For the fourth-order extension we have Source Symbol P (·) Codeword AAAA 0.6561 0 BAAA 0.0729 100 ABAA 0.0729 101 AABA 0.0729 110 AAAB 0.0729 1110 AABB 0.0081 111100 ABAB 0.0081 1111010 BAAB 0.0081 1111011 ABBA 0.0081 1111100 BABA 0.0081 1111101 BBAA 0.0081 1111110 ABBB 0.0009 111111100 BABB 0.0009 111111101 BBAB 0.0009 111111110 BBBA 0.0009 1111111110 BBBB 0.0001 1111111111 10.1. PROBLEM SOLUTIONS 13 The average wordlength: L = 1.9702 which is L/4 = 1.9702/4 = 0.4926. The efficiencies are given in the following table: n L H (X n ) Efficiency 1 1 0.4690 46.90% 2 1.29 0.9380 72.71% 3 1.598 1.4070 88.05% 4 1.9702 1.8760 95.22% Problem 10.22 For the Shannon-Fano code we have Source Symbol P (·) Codeword m 1 0.2 00 m 2 0.2 01 m 3 0.2 10 m 4 0.2 110 m 5 0.2 111 The entropy is H (X) = −5 (0.2 log 2 0.2) = 2.3219 The average wordlength is I = (0.2) (2 + 2 + 2 + 3 + 3) = 2.4 The efficiency is therefore given by H (X) L = 2.319 2.4 = 0.9675 = 96.75% The diagram for determining the Huffman code is illustrated in Figure 10.5. This diagram yields the codewords m 1 01 m 2 10 m 3 11 m 4 000 m 5 001 The average wordlength is L = 0.2 (2 + 2 + 2 + 3 + 3) = 2.4 14 CHAPTER 10. INFORMATION THEORY AND CODING 0.4 0.6 0.4 0.4 0.2 1 1 1 1 0 0 0 0 5 m 4 m 2 m 0.2 0.2 0.2 0.2 0.2 1 m 3 m Figure 10.5: Since the average wordlength is the same as for the Shannon-Fano code, the efficiency is also the same. Problem 10.23 The codewords for the Shannon-Fano and Huffman codes are summarized in the following table: Codewords Source Symbol P (·) Shannon-Fano Huffman m 1 0.40 00 1 m 2 0.19 01 000 m 3 0.16 10 001 m 4 0.15 110 010 m 5 0.10 111 011 The average wordlengths are: L = 2.25 (Shannon-Fano code) L = 2.22 (Huffman code) Note that the Huffman code gives the shorter average wordlength and therefore the higher efficiency. 10.1. PROBLEM SOLUTIONS 15 Problem 10.24 The entropy is H (X) = −0.85 log 2 0.85 −0.15 log 2 0.15 = 0.60984 bits/symbol We know that lim n→∞ r L n ≤ 300 and that lim n→∞ L n = H (X) Therefore, for n large rH (X) ≤ 300 or r ≤ 300 H (X) = 300 0.60984 which gives r ≤ 491.932 symbols/s Problem 10.25 The Shannon-Fano and Huffman codes are as follows. (Note that the codes are not unique.) Codewords Source Symbol Shannon-Fano Huffman m 1 000 001 m 2 001 010 m 3 010 011 m 4 011 100 m 5 100 101 m 6 101 110 m 7 110 111 m 8 1110 0000 m 9 1111 0001 In both cases, there are seven codewords of length three and two codewords of length four. Thus, both codes have the same average wordlengths and therefore have the same efficiencies. The average wordlength is L = 1 9 [7 (3) + 2 (4)] = 29 9 = 3.2222 16 CHAPTER 10. INFORMATION THEORY AND CODING Since the source entropy is H (X) = log 2 9 = 3.1699 the efficiency is E = H (X) L = 98.38% Problem 10.26 For the Shannon-Fano code we have Source Symbol P (·) Codeword m 1 1/12 000 m 2 1/12 0010 m 3 1/12 0011 m 4 1/12 010 m 5 1/12 0110 m 6 1/12 0111 m 7 1/12 100 m 8 1/12 1010 m 9 1/12 1011 m 10 1/12 110 m 11 1/12 1110 m 12 1/12 1111 The average wordlength is L = 1 12 [4 (3) + 8 (4)] = 44 12 = 3.6667 and the entropy of the source is H (X) = log 2 12 = 3.5850 This gives the efficiency E = H (X) L = 0.9777 = 97.77% The diagram for determining the Huffman code is illustrated in Figure 10.6. This yields 10.1. PROBLEM SOLUTIONS 17 1 0 1 0 1/3 2/3 1/3 1/3 1/3 1 0 1 0 1 0 1/6 1/6 1/6 1/6 1/6 1/6 1 0 1 0 12 m 11 m 10 m 9 m 8 m 7 m 6 m 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1 1 1 1 0 0 0 0 5 m 4 m 2 m 1/12 1/12 1/12 1/12 1/12 1 m 3 m Figure 10.6: the codewords: Source Symbol Codeword m 1 100 m 2 101 m 3 110 m 4 111 m 5 0000 m 6 0001 m 7 0010 m 8 0011 m 9 0100 m 10 0101 m 11 0110 m 12 0111 As in the case of the Shannon-Fano code, we have four codewords of length three and eight codewords of length four. The average wordlength is therefore 3.6667 and the efficiency is 18 CHAPTER 10. INFORMATION THEORY AND CODING 97.77%. Problem 10.27 The probability of message m k is P (m k ) = Z 0.1k 0.1(k−1) 2xdx = 0.01 (2k −1) A Huffman code yields the codewords in the following table: Source Symbol P (·) Codeword m 10 0.19 11 m 9 0.17 000 m 8 0.15 010 m 7 0.13 011 m 6 0.11 100 m 5 0.09 101 m 4 0.07 0011 m 3 0.05 00100 m 2 0.03 001010 m 1 0.01 001011 The source entropy is given by H (X) = 3.0488 bits/source symbol and the average wordlength is L = 3.1000 binary symbols/source symbol A source symbol rate of 250 symbols (samples) per second yields a binary symbol rate of rL = 250(3.1000) = 775 binary symbols/second and a source information rate of R s = rH (X) = 250(3.0488) = 75.2 bits/second Problem 10.28 The source entropy is H (X) = −0.4 log 2 0.4 −0.3 log 2 0.3 −0.2 log 2 0.2 −0.1 log 2 0.1 = 1.84644 bits/source symbol 10.1. PROBLEM SOLUTIONS 19 and the entropy of the second-order extension of the source is H ¡ X 2 ¢ = 2H (X) = 3.69288 The second-order extension source symbols are shown in the following table. For a D = 3 alphabet we partition into sets of 3. The set of code symbols is (0, 1, 2). Source Symbol P (·) Codeword m 1 m 1 0.16 00 m 1 m 2 0.12 01 m 2 m 1 0.12 02 m 2 m 2 0.09 10 m 1 m 3 0.08 11 m 3 m 1 0.08 12 m 2 m 3 0.06 200 m 3 m 2 0.06 201 m 3 m 3 0.04 202 m 1 m 4 0.04 210 m 4 m 1 0.04 211 m 2 m 4 0.03 212 m 4 m 2 0.03 220 m 3 m 4 0.02 221 m 4 m 3 0.02 2220 m 4 m 4 0.01 2221 The average wordlength of the extended source is L = 2 (0.16 + 0.12 + 0.12 + 0.09 + 0.08 + 0.08) +3 (0.06 + 0.06 + 0.04 + 0.04 + 0.03 + 0.03 + 0.02) +4 (0.02 + 0.01) = 2.38 Thus the efficiency is E = H ¡ X 2 ¢ L log 2 D = 3.69288 (2.38) log 2 3 = 97.9% Problem 10.29 The set of wordlengths from Table 10.3 is {1, 3, 3, 3, 5, 5, 5, 5}. The Kraft inequality gives 8 X i=1 2 l i = 2 −1 + 3 ¡ 2 −3 ¢ + 4 ¡ 2 −5 ¢ = 1 2 + 3 8 + 4 32 = 1 20 CHAPTER 10. INFORMATION THEORY AND CODING 10 2 10 3 10 4 10 5 10 6 10 7 10 8 0 1 2 3 4 5 6 x 10 5 Frequency C a p a c i t y - b i t s / s Figure 10.7: Thus the Kraft inequality is satisÞed. Problem 10.30 From (10.82), the Shannon-Hartley law with S = 40 and N = 10 −4 B gives C c = B log 2 µ 1 + S N ¶ = B log 2 Ã 1 + 40 ¡ 10 4 ¢ B ! This is illustrated in Figure 10.7. It can be seen that as B →∞ the capacity approaches a Þnite constant. Problem 10.31 From the previous problem C c = B log 2 Ã 1 + 40 ¡ 10 4 ¢ B ! 10.1. PROBLEM SOLUTIONS 21 Since ln(1 +u) ≈ u for ¿ 1, we can write that, as B →∞, C c = 40 ¡ 10 4 ¢ ln(2) = 5.7708 ¡ 10 5 ¢ bits/s Problem 10.32 For a rate 1 5 repetition code we have, assuming an error probability of (1 −p) on a single transmission, P E = µ 5 3 ¶ p 2 (1 −p) 3 + µ 5 4 ¶ p (1 −p) 4 + µ 5 5 ¶ (1 −p) 5 which is P E = 10p 2 (1 −p) 3 + 5p (1 −p) 4 + (1 −p) 5 or P E = 10 (1 −q) 2 q 3 + 5 (1 −q) q 4 +q 5 where q = 1 − p. For a rate 1 3 code we have the corresopnding result (See (10.94) with q = 1 −p) P E = 3q 2 (1 −q) +q 3 The results, generated by the MATLAB code, q = 0:0.01:0.1; p5 = 10*((1-q).^2).*(q.^3)+5*(1-q).*(q.^4)+(q.^5); p3 = 3*(1-q).*(q.^2)+(q.^3); plot(q,p3,q,p5) xlabel(’Error probability - (1-p)’) ylabel(’Word error probability’) are illustrated in Figure 10.8. The top curve (poorest performance) is for the rate 1 3 code and the bottom curve is for the rate 1 5 repetition code. Note that these results are not typically compariable since the values of p and q change as the code rate changes. Problem 10.33 The parity-check matrix for a (15, 11) Hamming code is [H] = _ ¸ ¸ _ 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 1 0 1 0 0 0 1 _ ¸ ¸ _ 22 CHAPTER 10. INFORMATION THEORY AND CODING 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0 0.005 0.01 0.015 0.02 0.025 0.03 Error probability - q=(1-p) W o r d e r r o r p r o b a b i l i t y Figure 10.8: Since the columns of [H] are distinct and all nonzero vectors of length three are included in [H], all single errors can be corrected. Thus, e = 1 2 (d m −1) = 1 or d m = 3 Note that it is not necessary to Þnd the set of codewords. Problem 10.34 The parity-check matrix will have 15 − 11 = 4 rows and 15 columns. Since rows may be 10.1. PROBLEM SOLUTIONS 23 permuted, the parity-check matrix is not unique. One possible selection is [H] = _ ¸ ¸ _ 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 1 0 1 0 0 0 1 _ ¸ ¸ _ The corresponding generator matrix is [G] = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1 1 1 0 1 1 0 1 1 0 0 1 1 1 1 0 1 1 0 1 0 1 0 1 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ = · I 11 H 11 ¸ where I 11 is the 11×11 identity matrix and H 11 represents the Þrst 11 columns of the parity- check matrix. From the structure of the parity-check matrix, we see that each parity symbol is the sum of 7 information symbols. Since 7 is odd, the all-ones information sequence yields the parity sequence 1111. This gives the codeword [T] t = [111111111111111] A single error in the third position yields a syndrome which is the third column of the parity-check matrix. Thus, for the assumed parity-check matrix, the syndrome is [S] = _ ¸ ¸ _ 0 1 1 0 _ ¸ ¸ _ Problem 10.35 24 CHAPTER 10. INFORMATION THEORY AND CODING For the given parity-check matrix, the generator matrix is [G] = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ This gives the 16 codewords (read vertically) a 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 a 2 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 a 3 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 a 4 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 c 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 c 2 0 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 c 3 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 Problem 10.36 We Þrst determine [T i ] = [G] [A i ] for i = 1, 2. The generator matrix for the previous problem is used. For [A 1 ] we have [G] [A 1 ] = [G] _ ¸ ¸ _ 0 1 1 0 _ ¸ ¸ _ = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 0 1 1 0 1 0 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ = [T 1 ] and for [A 2 ] we have [G] [A 2 ] = [G] _ ¸ ¸ _ 1 1 1 0 _ ¸ ¸ _ = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 1 1 0 0 1 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ = [T 2 ] 10.1. PROBLEM SOLUTIONS 25 We note that [A 1 ] ⊕[A 2 ] = _ ¸ ¸ _ 1 0 0 0 _ ¸ ¸ _ and [G] {[A 1 ] ⊕[A 2 ]} = [G] _ ¸ ¸ _ 1 0 0 0 _ ¸ ¸ _ = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 1 1 0 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ which is [T 1 ] ⊕[T 2 ]. Problem 10.37 For a rate 1 3 repetition code [G] = _ _ 1 1 1 _ _ and for a rate 1 5 repetition code [G] = _ ¸ ¸ ¸ ¸ _ 1 1 1 1 1 _ ¸ ¸ ¸ ¸ _ The generator matrix for a rate 1 n repetition code has one column and n rows. All elements are unity. Problem 10.38 For a Hamming code with r = n −k parity check symbols, the wordlength n is n = 2 r −1 and k = 2 r −r −1 The rate is R = k n = 2 r −r −1 2 r −1 26 CHAPTER 10. INFORMATION THEORY AND CODING If n is large, r is large. As r →∞ both k and n approach 2 r . Thus lim n→∞ R = lim r→∞ R = 2 r 2 r = 1 Problem 10.39 The codeword is of the form (a 1 a 2 a 3 a 4 c 1 c 2 c 3 ) where c 1 = a 1 ⊕a 2 ⊕a 3 c 2 = a 2 ⊕a 3 ⊕a 4 c 3 = a 1 ⊕a 2 ⊕a 4 This gives the generator matrix [G] = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 1 1 1 1 1 0 1 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ Using the generator matrix, we can determine the codewords. The codewords are (read vertically) a 1 0 1 1 1 1 0 0 0 1 1 1 0 0 0 0 1 a 2 0 0 1 1 1 0 0 1 0 1 0 1 1 0 1 0 a 3 0 0 0 1 1 0 1 1 1 0 0 1 0 1 0 1 a 4 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 c 1 0 1 0 1 1 0 1 0 0 0 1 0 1 1 1 0 c 2 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 c 3 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 1 Let α i denote the i-th code word and let α i →α j indicate that a cyclic shift of α i yields α j . The cyclic shifts are illustrated in Figure 10.9. Problem 10.40 The parity-check matrix for the encoder in the previous problem is [H] = _ _ 1 1 1 0 1 0 0 0 1 1 1 0 1 0 1 1 0 1 0 0 1 _ _ 10.1. PROBLEM SOLUTIONS 27 10 α 16 α 13 α 11 α 7 α 8 α 6 α 14 α 15 α 9 α 12 α 3 α 2 α 5 α 1 α 4 α Figure 10.9: For the received sequence [R] t = [1101001] the syndrome is [S] = [H] [R] = _ _ 0 0 0 _ _ This indicates no errors, which is in agreement with Figure 10.17. For the received sequence [R] t = [1101011] the syndrome is [S] = [H] [R] = _ _ 0 1 0 _ _ This syndrome indicates an error in the 6th position. Thus the assumed transmitted codeword is [T] t = [1101001] which is also in agreement with Figure 10.17. Problem 10.41 28 CHAPTER 10. INFORMATION THEORY AND CODING For the modiÞed encoder we can write Step S 1 S 2 S 3 1 a 1 0 0 2 a 1 ⊕a 2 a 1 0 3 a 1 ⊕a 2 ⊕a 3 a 1 ⊕a 2 a 1 4 a 2 ⊕a 3 ⊕a 4 a 1 ⊕a 2 ⊕a 3 a 1 ⊕a 2 5 0 a 2 ⊕a 3 ⊕a 4 a 1 ⊕a 2 ⊕a 3 6 0 0 a 2 ⊕a 3 ⊕a 4 7 0 0 0 Thus, the output for each step is Step Output 1 a 1 2 a 2 3 a 3 4 a 4 5 a 1 ⊕a 3 ⊕a 4 6 a 1 ⊕a 2 ⊕a 3 7 a 2 ⊕a 3 ⊕a 4 This gives the generator matrix [G] = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ Using the generator matrix, the codewords can be constructed. The 16 codewords are (read vertically) a 1 0 1 1 1 1 0 0 0 1 1 1 0 0 0 0 1 a 2 0 0 1 1 1 0 0 1 0 1 0 1 1 0 1 0 a 3 0 0 0 1 1 0 1 1 1 0 0 1 0 1 0 1 a 4 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 c 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 1 0 c 2 0 1 0 1 1 0 1 0 0 0 1 0 1 1 1 0 c 3 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 10.1. PROBLEM SOLUTIONS 29 11 α 14 α 15 α 9 α 8 α 4 α 6 α 7 α 12 α 10 α 16 α 13 α 2 α 5 α 1 α 3 α Figure 10.10: As in Problem 10.39, let α i denote the i-th code word and let α i →α j indicate that a cyclic shift of α i yields α j . These cyclic shifts are illustrated in Figure 10.10. Problem 10.42 For a (15, 11) Hamming code, the uncoded symbol error probability is given by q u = Q h p 2z/k i where z = ST w /N 0 and k = 11. This gives the uncoded word error probability P eu = 1 −(1 −q u ) 11 For the coded case, the symbol error probability is q c = Q h p 2z/15 i and the coded word error probability is P ec = 15 X i=2 µ 15 i ¶ q i c (1 −q c ) 15−i The performance curves are generated using the MATLAB M-Þles given in Computer Ex- ercise 10.2 in Section 10.2 (Computer Exercises) of this solutions manual. The results are shown in Figure 10.11. The solid curves represent the symbol error probability with the 30 CHAPTER 10. INFORMATION THEORY AND CODING 10 12 14 16 18 20 22 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 STw/No in dB P r o b a b i l i t y Figure 10.11: coded case on top (worse performance) and the uncoded case on bottom. The dashed curves represent the word error probability with the coded case (better performance) on bottom. Problem 10.43 The probability of two errors in a 7 symbol codeword is P {2 errors} = µ 7 2 ¶ (1 −q c ) 5 q 2 c = 21q 2 c (1 −q c ) 5 and the probability of three errors in a 7 symbol codeword is P {3 errors} = µ 7 3 ¶ (1 −q c ) 4 q 3 c = 35q 3 c (1 −q c ) 4 The ratio of P {3 errors} to P {2 errors} is R = Pr {3 errors} Pr {2 errors} = 35q 3 c (1 −q c ) 4 21q 2 c (1 −q c ) 5 = 5 3 q c 1 −q c 10.1. PROBLEM SOLUTIONS 31 We now determine the coded error probability q c . Since BPSK modulation is used q c = Q Ã r 2z 7 ! where z is ST w /N 0 . This gives us the following table z q c R = Pr{3 errors} Pr{2 errors} 8dB 0.0897 0.1642 9dB 0.0660 0.1177 10dB 0.0455 0.0794 11dB 0.0290 0.0497 12dB 0.0167 0.0278 13dB 0.0085 0.0141 14dB 0.0037 0.0062 It is clear that as the SNR, z, increases, the value of R becomes negligible. Problem 10.44 The parity-check matrix for a (7, 4) Hamming code is [H] = _ _ 0 0 0 1 1 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 1 _ _ The parity checks are in positions 1,2, and 4. Moving the parity checks to positions 5, 6, and 7 yields a systematic code with the generator matrix [G] = _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 1 _ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ _ which has the partity-check matrix [H] = _ _ 0 1 1 1 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 _ _ 32 CHAPTER 10. INFORMATION THEORY AND CODING Note that all columns of [H] are still distinct and therefore all single errors are corrected. Thus, the distance three property is preserved. Since the Þrst four columns of [H] can be in any order, there are 4! = 24 different (7, 4) systematic codes. If the code is not required to be systematic, there are 7! = 5040 different codes, all of which will have equivalent performance in an AWGN channel. Problem 10.45 For the encoder shown, the constraint span is k = 4. We Þrst need to compute the states for the encoder. The output is v 1 v 2 where v 1 = s 1 ⊕s 2 ⊕s 3 ⊕s 4 and v 2 = s 1 ⊕s 2 ⊕s 4 . We now compute the state transitions and the output using the state diagram shown in Figure 10.12. This gives the following table. Previous Current State s 1 s 2 s 2 Input s 1 s 2 s 3 s 4 State Output A 000 0 0000 A 00 1 1000 E 11 B 001 0 0001 A 11 1 1001 E 00 C 010 0 0010 B 10 1 1010 F 01 D 011 0 0011 B 01 1 1011 F 10 E 100 0 0100 C 11 1 1100 G 00 F 101 0 0101 C 00 1 1101 G 11 G 110 0 0110 D 01 1 1110 H 10 H 111 0 0111 D 10 1 1111 H 01 Problem 10.46 For the encoder shown, we have v 1 = s 1 ⊕ s 2 ⊕ s 3 and v 2 = s 1 ⊕ s 2 . The trellis diagram is illustrated in Figure 10.13. The state transitions, and the output corresponding to each transition appears in the following table. 10.1. PROBLEM SOLUTIONS 33 H G F D E C B A (0 0) (1 0) (1 0) (0 1) (1 0) (1 1) (1 0) (0 1) (0 1) (0 1) (1 1) (1 1) (1 1) (0 0) (0 0) (0 0) Figure 10.12: State s 1 s 2 Input s 1 s 2 s 3 State Output A 00 0 000 A 00 1 100 C 11 B 01 0 001 A 11 1 101 C 00 C 10 0 010 B 10 1 110 D 01 D 11 0 011 B 01 1 111 D 10 Problem 10.47 34 CHAPTER 10. INFORMATION THEORY AND CODING (01) (01) (01) (01) (10) (10) (10) (10) (11) (11) (11) (11) (11) (11) (00) (00) (00) (00) (00) (00) Termination Steady-State Transitions Figure 10.13: The source rate is r s = 2000 symbols/second. The channel symbol rate is r c = r s n k = 2000 µ 7 4 ¶ = 3500 If the burst duration is T b , then r c T b = r s T b n k channel symbols will be affected. Assuming a (7, 4) block code, we have r c T b = (2000) (0.15) µ 7 4 ¶ = 525 symbols affected by the burst. Let the table contain l = 525 codewords. This corresponds to (525) (7) = 3675 symbols in the table. Thus 3675 − 525 = 3150 must be transmitted without error. The transmission time for 3150 symbols is 3150 symbols 3500 symbols/second = 0.9 seconds 10.2. COMPUTER EXERCISES 35 Figure 10.14: Problem 10.48 The required plot is shown in Figure 10.14. 10.2 Computer Exercises Computer Exercise 10.1 For two messages the problem is easy since there is only a single independent probability. The MATLAB program follows a = 0.001:0.001:0.999; 36 CHAPTER 10. INFORMATION THEORY AND CODING 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 a E n t r o p y Figure 10.15: b =1-a; lna = length(a); H = zeros(1,lna); H=-(a.*log(a)+b.*log(b))/log(2); plot(a,H); xlabel(’a’) ylabel(’Entropy’) The plot is illustrated in Figure 10.15. For three source messages we have two independent probabilities. Two of them will be speciÞed and the remaining probability can then be determined. The resulting MATLAB code follows: n = 50; % number of points in vector nn = n+1; % increment for zero 10.2. COMPUTER EXERCISES 37 H = zeros(nn,nn); % initialize entropy array z = (0:n)/n; % probability vector z(1) = eps; % prevent problems with log(0) for i=1:nn for j=1:nn c = 1-z(i)-z(j); % calculate third probability if c>0 xx = [z(i) z(j) c]; H(i,j)= -xx*log(xx)’/log(2); % compute entropy end end end colormap([0 0 0]) % b&w colormap for mesh mesh(z,z,H); % plot entropy view(-45,30); % set view angle xlabel(’Probability of message a’) ylabel(’Probability of message b’) zlabel(’Entropy’) figure % new figure contour(z,z,H,10) % plot contour map xlabel(’Probability of message a’) ylabel(’Probability of message b’) The resulting plot is illustrated in Figure 10.16. One should experiment with various viewing locations. Next we draw a countour map with 10 countour lines. The result is illustrated in Figure 10.17. Note the peak at 0.33, 0.33, indicating that the maximum entropy occurs when all three source messages occur with equal probability. Computer Exercise 10.2 The performance curves for a (15,11) Hamming code, on the basis of word energy and word error probability were derived in Problem, 10.42. The equivalent results for a (31,26) Hamming code are given in Figure 10.18. As in the case of the (7,4) Hamming code, the solid curves represent the symbol error probability with the coded case on top (worse performance) and the uncoded case on bottom. The dashed curves represent the word error probability with the coded case (better performance) on bottom. The computer code for the (31,26) code is as follows: n = 31; k = 26; zdB = 0:10; % Es/No in dB 38 CHAPTER 10. INFORMATION THEORY AND CODING Figure 10.16: z = 10.^(zdB/10); % Es/No in linear units arg1 = sqrt(2*z); % Q-function argument for PSK psu = 0.5*(1-erf(arg1/sqrt(2.0))); % PSK symbol error probability pwu = 1-(1-psu).^k; % uncoded word error probability arg2 = sqrt(2*k*z/n); % Q-function argument for coded PSK psc = 0.5*(1-erf(arg2/sqrt(2.0))); % coded PSK symbol error probability pwc = werrorc(n,k,1,psc); % coded word error probability zz = zdB+10*log10(k); % scale axis for plot (STw/No) % semilogy(zz,psu,’k-’,zz,psc,’k-’,zz,pwu,’k--’,zz,pwc,’k--’) xlabel(’STw/No in dB’) ylabel(’Probability’) The results for the (7,4) Hamming code are generated by changing the Þrst line of the program to n = 7; k = 4. The main program calls two functions werrorc.m and nkchoose.m. 10.2. COMPUTER EXERCISES 39 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Probability of message a P r o b a b i l i t y o f m e s s a g e b Figure 10.17: These follow. function [y] = werrorc(n,k,t,psc) lx = length(psc); y = zeros(1,lx); % initialize y for i=1:lx % do for each element of psc sum = 0; for j=(t+1):n nkc = nkchoose(n,j); term = log(nkc)+j*log(psc(i))+(n-j)*log(1-psc(i)); sum = sum+exp(term); end y(i) = sum; end 40 CHAPTER 10. INFORMATION THEORY AND CODING 14 16 18 20 22 24 26 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 STw/No in dB P r o b a b i l i t y Figure 10.18: function out=nkchoose(n,k) % Computes n!/k!/(n-k)! a = sum(log(1:n)); % ln of n! b = sum(log(1:k)); % ln of k! c = sum(log(1:(n-k))); % ln of (n-k)! out = round(exp(a-b-c)); % result % End of function file. Note: The preceding function is called nkchoose rather than nchoosek in order to avoid conßict with the function nchoosek in the communications toolbox. We prefer to compute the binomial coefficient using logorithms in order to obtain the large dynamic range needed for some very long codes. Computer Exercise 10.3 The MATLAB code for implementing the given equation is 10.2. COMPUTER EXERCISES 41 function [ber] = cer2ber(q,n,d,t,ps) % Converts channel symbol error rate to decoded BER. % Based on (1.7.30) in ’’Principles of Secure Communication Systems’’ % by D.J. Torrieri (2nd Edition), Artech House, 1992. % Parameters: % q - alphabet size (2 for binary) % n - block length % d - minimum distance of code % t - correctable errors per block (usually t=(d-1)/2) % ps - vector of channel symbol error rates % ber - vector of bit error rates % lnps = length(ps); % length of error vector ber = zeros(1,lnps); % initialize output vector for k=1:lnps % iterate error vector cer = ps(k); % channel symbol error rate sum1 = 0; sum2 = 0; % initialize sums % first loop evaluates first sum for i=(t+1):d term = nkchoose(n,i)*(cerî)*((1-cer))^(n-i); sum1 = sum1+term; end % second loop evaluates second sum for i=(d+1):n term = i*nkchoose(n,i)*(cerî)*((1-cer)^(n-i)); sum2 = sum2+term; end % comput BER (output) ber(k) = (q/(2*(q-1)))*((d/n)*sum1+(1/n)*sum2); end In order to demonstrate the preceding function with a (7,4) Hamming code, we use the following MATLAB program: zdB = 0:0.1:10; % set Eb/No axis in dB z = 10.^(zdB/10); % convert to linear scale beru = qfn(sqrt(2*z)); % PSK result cerham = qfn(sqrt(4*z*2/7)); % CSER for (7,4) Hamming code berham = cer2ber(2,7,3,1,cerham); % BER for Hamming code semilogy(zdB,beru,zdB,berham) % plot results 42 CHAPTER 10. INFORMATION THEORY AND CODING 0 1 2 3 4 5 6 7 8 9 10 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 E b /N o in dB B i t E r r o r P r o b a b i l i t y Figure 10.19: xlabel(’E_b/N_o in dB’) % label x axis ylabel(’Bit Error Probability’) % label y axis The MATLAB function for the Gaussian Q-function follows: function out=qfn(x) % Gaussian Q function out =0.5*erfc(x/sqrt(2)); Executing the code yields the result illustrated in Figure 10.19. It can be seen that the (7,4) Hamming code yields only a moderate performance improvement at high SNR. Computer Exercise 10.4 The MATLAB program is zdB = 0:0.1:10; % set Eb/No axis in dB 10.2. COMPUTER EXERCISES 43 z = 10.^(zdB/10); % convert to linear scale ber1 = qfn(sqrt(2*z)); % PSK result ber2 = qfn(sqrt(12*2*z/23)); % CSER for (23,12) Golay code ber3 = qfn(sqrt(4*z*2/7)); % CSER for (15,11) Hamming code bergolay = cer2ber(2,23,7,3,ber2); % BER for Golay code berhamming = cer2ber(2,7,3,1,ber3); % BER for Hamming code semilogy(zdB,ber1,zdB,bergolay,zdB,berhamming) xlabel(’E_b/N_o in dB’) ylabel(’Bit Error Probability’) The funcrions Q and cer2ber are given in the previous computer exercise. The results are illustrated in Figure 10.20. The order of the curves, for high signal-tonoise ratio, are in order of the error correcting capability. The top curve (worst performance) is the uncoded case. The middle curve is for the (7,4) Hamming code and the bottom curve is for the (23,12) Golay code. Computer Exercise 10.5 The MATLAB code for generating the performance curves illustrated in Figure 10.18 in the text follows. zdB = 0:30; z = 10.^(zdB/10); qa = 0.5*exp(-0.5*z); qa3 = 0.5*exp(-0.5*z/3); qa7 = 0.5*exp(-0.5*z/7); pa7 = 35*((1-qa7).^3).*(qa7.^4)+21*((1-qa7).^2).*(qa7.^5)... +7*(1-qa7).*(qa7.^6)+(qa7.^7); pa3 = 3*(1-qa3).*(qa3.^2)+(qa3.^3); qr = 0.5./(1+z/2); qr3 = 0.5./(1+z/(2*3)); qr7 = 0.5./(1+z/(2*7)); pr7 = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5)... +7*(1-qr7).*(qr7.^6)+(qr7.^7); pr3 = 3*(1-qr3).*(qr3.^2)+(qr3.^3); semilogy(zdB,qr,zdB,pr3,zdB,pr7,zdB,qa,zdB,pa3,zdB,pa7) axis([0 30 0.0001 1]) xlabel(’Signal-to-Noise Ratio, z - dB’) ylabel(’Probability ’) Executing the code yields the results illustrated in Figure 10.21. The curves can be identiÞed from Figure 10.18 in the text. 44 CHAPTER 10. INFORMATION THEORY AND CODING 0 1 2 3 4 5 6 7 8 9 10 10 -10 10 -8 10 -6 10 -4 10 -2 10 0 E b /N o in dB B i t E r r o r P r o b a b i l i t y Figure 10.20: Computer Exercise 10.6 For this Computer Exercise we use the uncoded Rayleigh fading result from the previous Computer Exercise and also use the result for a Rayleigh fading channel with a rate 1/7 repetition code. The new piece of information needed for this Computer Exercise is the result from Problem 9.17. The MATLAB code follows: zdB = 0:30; z = 10.^(zdB/10); qr1 = 0.5./(1+z/2); qr7 = 0.5./(1+z/(2*7)); pr7 = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5)... +7*(1-qr7).*(qr7.^6)+(qr7.^7); % Calculations for diversity system pd7 = zeros(1,31); N = 7; 10.2. COMPUTER EXERCISES 45 0 5 10 15 20 25 30 10 -4 10 -3 10 -2 10 -1 10 0 Signal-to-Noise Ratio, z - dB P r o b a b i l i t y Figure 10.21: for i=1:31 sum = 0; a = 0.5/(1+z(i)/(2*N)); for j=0:(N-1) term = nkchoose(N+j-1,j)*((1-a)^j); sum = sum+term; end pd7(i) = (a^N)*sum; end % semilogy(zdB,qr1,zdB,pr7,zdB,pd7) axis([0 30 0.0001 1]) xlabel(’Signal-to-Noise Ratio, z - dB’) ylabel(’Probability’) 46 CHAPTER 10. INFORMATION THEORY AND CODING 0 5 10 15 20 25 30 10 -4 10 -3 10 -2 10 -1 10 0 Signal-to-Noise Ratio, z - dB P r o b a b i l i t y Figure 10.22: The function nkchoose follows: function out=nkchoose(n,k) % Computes n!/k!/(n-k)! a = sum(log(1:n)); % ln of n! b = sum(log(1:k)); % ln of k! c = sum(log(1:(n-k))); % ln of (n-k)! out = round(exp(a-b-c)); % result % End of function file. Executing the MATLAB program yields the result illustrated in Figure 10.22. The curves can be identiÞed by comparison with Figure 10.20 inthe text. Appendi x A Physi cal Noi se Sour ces and Noi se Cal cul at i ons Pr obl em A.1 All parts of the problem are solved using the relation V rms = √ 4kTRB where k = 1.38 ×10 −23 J/K B = 30 MHz = 3 ×10 7 Hz (a) For R = 10, 000 ohms and T = T 0 = 290 K V rms = p 4 (1.38 ×10 −23 ) (290) (10 4 ) (3 ×10 7 ) = 6.93 ×10 −5 V rms = 69.3 µV rms (b) V rms is smaller than the result in part (a) by a factor of √ 10 = 3.16. Thus V rms = 21.9 µV rms (c) V rms is smaller than the result in part (a) by a factor of √ 100 = 10. Thus V rms = 6.93 µV rms (d) Each answer becomes smaller by factors of 2, √ 10 = 3.16, and 10, respectively. 1 2 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Pr obl em A.2 Use I = I s · exp µ eV kT ¶ −1 ¸ We want I > 20I s or exp ¡ eV kT ¢ −1 > 20. (a) At T = 290 K, e kT ' 40, so we have exp(40V ) > 21 giving V > ln(21) 40 = 0.0761 volts i 2 rms ' 2eIB ' 2eBI s exp µ eV kT ¶ or i 2 rms B = 2eI s exp µ eV kT ¶ = 2 ¡ 1.6 ×10 −19 ¢¡ 1.5 ×10 −5 ¢ exp(40 ×0.0761) = 1.0075 ×10 −22 A 2 /Hz (b) If T = 29 K, then e kT = 400, and for I > 20I s , we need exp(400V ) > 21 or V > ln(21) 400 = 7.61 ×10 −3 volts Thus i 2 rms B = 2 ¡ 1.6 ×10 −19 ¢¡ 1.5 ×10 −5 ¢ exp ¡ 400 ×7.61 ×10 −3 ¢ = 1.0075 ×10 −22 A 2 /Hz as before. Pr obl em A.3 (a) Use Nyquist’s formula to get the equivalent circuit of R eq in parallel with R 3 , where R eq is given by R eq = R L (R 1 +R 2 ) R 1 +R 2 +R L The noise equivalent circuit is then as shown in Figure A.1. The equivalent noise voltages are V 1 = p 4kTR eq B V 2 = p 4kTR 3 B 3 V 1 V 2 R eq R 3 V 0 Figure A.1: Adding noise powers to get V 0 we obtain V 2 0 = µ R 3 R eq +R 3 ¶ 2 ¡ V 2 1 +V 2 2 ¢ = µ R 3 R eq +R 3 ¶ 2 (4kTB) (R eq +R 3 ) = (4kTB) R 2 3 (R eq +R 3 ) (b) With R 1 = 2000 Ω, R 2 = R L = 300 Ω, and R 3 = 500 Ω, we have R eq = 300 (2000 + 300) 300 + 2000 + 300 = 265.4 Ω and V 2 0 B = (4kT) R 2 3 (R eq +R 3 ) = 4 ¡ 1.38 ×10 −23 ¢ (290) (500) 2 500 + 265.4 = 5.23 ×10 −18 V 2 /Hz 4 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Pr obl em A.4 Find the equivalent resistance for the R 1 , R 2 , R 3 combination and set R L equal to this to get R L = R 3 (R 1 +R 2 ) R 1 +R 2 +R 3 Pr obl em A.5 Using Nyquist’s formula, we Þnd the equivalent resistance looking backing into the terminals with V rms across them. It is R eq = 50k k 20 k k (5 k + 10 k + 5 k) = 50k k 20 k k 20 k = 50k k 10 k = (50k) (10 k) 50k + 10 k = 8, 333 Ω Thus V 2 rms = 4kTR eq B = 4 ¡ 1.38 ×10 −23 ¢ (400) (8333) ¡ 2 ×10 6 ¢ = 3.68 ×10 −10 V 2 which gives V rms = 19.18 µV rms Pr obl em A.6 To Þnd the noise Þgure, we Þrst determine the noise power due to a source at the output, and then due to the source and the network. Initally assume unmatched conditions. The results are V 2 0 ¯ ¯ due to R S , only = µ R 2 k R L R S +R 1 +R 2 k R L ¶ 2 (4kTR S B) V 2 0 ¯ ¯ due to R 1 and R 2 = µ R 2 k R L R S +R 1 +R 2 k R L ¶ 2 (4kTR 1 B) + µ R L k (R 1 +R S ) R 2 + (R 1 +R S ) k R L ¶ 2 (4kTR 2 B) 5 V 2 0 ¯ ¯ due to R S , R 1 and R 2 = µ R 2 k R L R S +R 1 +R 2 k R L ¶ 2 [4kT (R S +R 1 ) B] + µ R L k (R 1 +R S ) R 2 + (R 1 +R S ) k R L ¶ 2 (4kTR 2 B) The noise Þgure is F = 1 + R 1 R S + µ R L k (R 1 +R S ) R 2 + (R 1 +R S ) k R L ¶ 2 µ R S +R 1 +R 2 k R L R 2 k R L ¶ 2 R 2 R S In the above, R a k R b = R a R b R a +R b Note that the noise due to R L has been excluded because it belongs to the next stage. Since this is a matching circuit, we want the input matched to the source and the output matched to the load. Matching at the input requires that R S = R in = R 1 +R 2 k R L = R 1 + R 2 R L R 2 +R L and matching at the output requires that R L = R out = R 2 k (R 1 +R S ) = R 2 (R 1 +R S ) R 1 +R 2 +R S Next, these expressions are substituted back into the expression for F. After some simpli- Þcation, this gives F = 1 + R 1 R S + µ 2R 2 L R S (R 1 +R 2 +R S ) / (R S −R 1 ) R 2 2 (R 1 +R S +R L ) +R 2 L (R 1 +R 2 +R S ) ¶ 2 R 2 R S Note that if R 1 >> R 2 we then have matched conditions of R L = R 2 and R S = R 1 . Then, the noise Þgure simpliÞes to F = 2 + 16 R 1 R 2 Note that the simple resistive pad matching circuit is very poor from the standpoint of noise. The equivalent noise temperature is found by using T e = T 0 (F −1) = T 0 · 1 + 16 R 1 R 2 ¸ 6 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Pr obl em A.7 The important relationships are F l = 1 + T e l T 0 T e l = T 0 (F l −1) T e 0 = T e 1 + T e 2 G a 1 + T e 3 G a 1 G a 2 Completion of the table gives Ampl. No. F T e i G a i 1 not needed 300 K 10 dB = 10 2 6 dB 864.5 K 30 dB = 1000 3 11 dB 3360.9 K 30 dB = 1000 Therefore, T e 0 = 300 + 864.5 10 + 3360.9 (10) (1000) = 386.8 K Hence, F 0 = 1 + T e 0 T 0 = 2.33 = 3.68 dB (b) With ampliÞers 1 and 2 interchanged T e 0 = 864.5 + 300 10 + 3360.9 (10) (1000) = 865.14 K This gives a noise Þgure of F 0 = 1 + 865.14 290 = 3.98 = 6 dB (c) See part (a) for the noise temperatures. 7 (d) For B = 50 kHz, T S = 1000 K, and an overall gain of G a = 10 7 , we have, for the conÞguration of part (a) P na, out = G a k (T 0 +T e 0 ) B = 10 7 ¡ 1.38 ×10 −23 ¢ (1000 + 386.8) ¡ 5 ×10 4 ¢ = 9.57 ×10 −9 watt We desire P sa, out P na, out = 10 4 = P sa, out 9.57 ×10 −9 which gives P sa, out = 9.57 ×10 −5 watt For part (b), we have P na, out = 10 7 ¡ 1.38 ×10 −23 ¢ (1000 + 865.14) ¡ 5 ×10 4 ¢ = 1.29 ×10 −8 watt Once again, we desire P sa, out P na, out = 10 4 = P sa, out 1.29 ×10 −8 which gives P sa, out = 1.29 ×10 −4 watt and P sa, in = P sa, out G a = 1.29 ×10 −11 watt Pr obl em A.8 (a) The noise Þgure of the cascade is F overall = F 1 + F 2 −1 G a 1 = L + F −1 (1/L) = LF (b) For two identicalattenuator-ampliÞer stages F overall = L + F −1 (1/L) + L −1 (1/L) L + F −1 (1/L) L(1/L) = 2LF −1 ≈ 2LF, L >> 1 (c) Generalizing, for N stages we have F overall ≈ NFL 8 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Pr obl em A.9 The data for this problem is Stage F i G i 1 (preamp) 2 dB = 1.58 G 1 2 (mixer) 8 dB = 6.31 1.5 dB = 1.41 3 (ampliÞer) 5 dB = 3.16 30 dB = 1000 The overall noise Þgure is F = F 1 + F 2 −1 G 1 + F 3 −1 G 1 G 2 which gives 5 dB = 3.16 ≥ 1.58 + 6.31 −1 G 1 + 3.16 −1 1.42G 1 or G 1 ≥ 5.31 −(2.16/1.41) 1.58 = 4.33 = 6.37 dB (b) First, assume that G 1 = 15 dB = 31.62. Then F = 1.58 + 6.31 −1 31.62 + 3.16 −1 (1.41) (31.62) = 1.8 = 2.54 dB Then T es = T 0 (F −1) = 290 (1.8 −1) = 230.95 K and T overall = T es +T a = 230.95 + 300 = 530.95 K Now use G 1 as found in part (a): F = 3.16 T es = 290 (3.16 −1) = 626.4 K T overall = 300 + 626.4 = 926.4 K 9 (c) For G 1 = 15 dB = 31.62, G a = 31.62 (1.41) (1000) = 4.46 ×10 4 . Thus P na, out = ¡ 4.46 ×10 4 ¢¡ 1.38 ×10 −23 ¢ (530.9) ¡ 10 9 ¢ = 3.27 ×10 −9 watts For G 1 = 6.37 dB = 4.33, G a = 4.33 (1.41) (1000) = 6.11 ×10 3 . Thus P na, out = ¡ 6.11 ×10 3 ¢¡ 1.38 ×10 −23 ¢ (926.4) ¡ 10 7 ¢ = 7.81 ×10 −10 watts Note that for the second case, we get less noise power out even wth a larger T overall . This is due to the lower gain of stage 1, which more than compensates for the larger input noise power. (d) A transmission line with loss L = 2 dB connects the antenna to the preamp. We Þrst Þnd T S for the transmission line/preamp/mixer/amp chain: F S = F TL + F 1 −1 G TL + F 2 −1 G TL G 1 + F 3 −1 G TL G 1 G 2 , where G TL = 1/L = 10 −2/ 10 = 0.631 and F TL = L = 10 2/ 10 = 1.58 Assume two cases for G 1 : 15 dB and 6.37 dB. First, for G 1 = 15 dB = 31.6, we have F S = 1.58 + 1.58 −1 0.631 + 6.31 −1 (0.631) (31.6) + 3.16 −1 (0.631) (31.6) (1.41) = 2.84 This gives T S = 290 (2.84 −1) = 534 K and T overall = 534 + 300 = 834 K Now, for G 1 = 6.37 dB = 4.33, we have F S = 1.58 + 1.58 −1 0.631 + 6.31 −1 (0.631) (4.33) + 3.16 −1 (0.631) (4.33) (1.41) = 5 This gives T S = 290 (5 −1) = 1160 K 10 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS and T overall = 1160 + 300 = 1460 K Pr obl em A.10 (a) Using P na, out = G a kT S B with the given values yields P na, out = 7.45 ×10 −6 watts (b) We want P sa, out P na, out = 10 5 or P sa, out = ¡ 10 5 ¢¡ 7.45 ×10 −6 ¢ = 0.745 watts This gives P sa, in = P sa, out G a = 0.745 ×10 −8 watts = −51.28 dBm Pr obl em A.11 (a) For ∆A = 1 dB, Y = 1.259 and the effective noise temperature is T e = 600 −(1.259) (300) 1.259 −1 = 858.3 K For ∆A = 1.5 dB, Y = 1.413 and the effective noise temperature is T e = 600 −(1.413) (300) 1.413 −1 = 426.4 K For ∆A = 2 dB, Y = 1.585 and the effective noise temperature is T e = 600 −(1.585) (300) 1.585 −1 = 212.8 K (b) These values can be converted to noise Þgure using F = 1 + T e T 0 11 With T 0 = 290 K, we get the following values: (1) For ∆A = 1 dB, F = 5.98 dB; (2) For ∆A = 1.5 dB, F = 3.938 dB; (3) For ∆A = 2 dB, F = 2.39 dB. Pr obl em A.12 (a) Using the data given, we can determine the following: λ = 0.039 m µ λ 4πd ¶ 2 = −202.4 dB G T = 39.2 dB P T G T = 74.2 dBW This gives P S = −202.4 + 74.2 + 6 −5 = −127.2 dBW (b) Using P n = kT e B for the noise power, we get P n = 10 log 10 · kT 0 µ T e T 0 ¶ B ¸ = 10 log 10 [kT 0 ] + 10 log 10 µ T e T 0 ¶ + 10 log 10 (B) = −174 + 10 log 10 µ 1000 290 ¶ + 10 log 10 ¡ 10 6 ¢ = −108.6 dBm = −138.6 dBW (c) µ P S P n ¶ dB = −127.2 −(−138.6) = 11.4 dB = 10 1.14 = 13.8 ratio (d) Assuming the SNR = z = E b /N 0 = 13.8, we get the results for various digital signaling techniques given in the table below: Modulation type Error probability BPSK Q ¡ √ 2z ¢ = 7.4 ×10 −8 DPSK 1 2 e −z = 5.06 ×10 −7 Noncoh. FSK 1 2 e −z/ 2 = 5.03 ×10 −4 QPSK Same as BPSK 2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY Single-sided amplitude 10 5 f, Hz 0 2 4 6 0 π/4 π/8 Single-sided phase, rad. f, Hz 2 4 6 Figure 2.1: Problem 2.3 (a) Not periodic. (b) Periodic. To find the period, note that 20π 6π = n1 f0 and = n2 f0 2π 2π Therefore n2 10 = 3 n1 Hence, take n1 = 3, n2 = 10, and f0 = 1 Hz. (c) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 7, and f0 = 1 Hz. (d) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 3, n3 = 11, and f0 = 1 Hz. Problem 2.4 (a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6 Hz, and the phase spectrum consists of a single line of height -π/6 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of 6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -π/6 radians at frequency 6 Hz and a line of height π/6 at frequency -6 radians Hz. (b) Write the signal as xb (t) = 3 cos(12πt − π/2) + 4 cos(16πt) From this it is seen that the single-sided amplitude spectrum consists of lines of height 3 and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists 2.1. PROBLEM SOLUTIONS 3 Double-sided amplitude 5 f, Hz -6 -4 -2 0 2 4 6 Double-sided phase, rad. π/4 π/8 -6 -4 -2 f, Hz 0 -π/8 -π/4 2 4 6 Figure 2.2: 4 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY of a line of height -π/2 radians at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of height 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrum consists of a line of height -π/2 radians at frequency 6 Hz and a line of height π/2 radians at frequency -6 Hz. Problem 2.5 (a) This function has area Area = ∞ Z ² −1 = −∞ ∞ Z −∞ · sin(πt/²) (πt/²) ¸2 ¸2 dt · sin(πu) (πu) du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the central lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. (b) The area for the function is Area = ∞ Z 1 exp(−t/²)u (t) dt = ² −∞ ∞ Z 0 exp(−u)du = 1 A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. R1 R² (c) Area = −² 1 (1 − |t| /²) dt = −1 Λ (t) dt = 1. As ² → 0, the function becomes narrower ² and higher, so it approximates a delta function in the limit. Problem 2.6 (a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9. Problem 2.7 (a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively. The waveform of part (c) is a periodic train of impulses extending from -∞ to ∞ spaced by 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). The waveform of part (e) is a doubly-infinite train of square pulses, each of which is one unit high and one unit wide, centered at · · ·, −6, −3, 0, 3, 6, · · ·. Waveform (f) is a raised cosine of minimum and maximum amplitudes 0 and 2, respectively. 2.1. PROBLEM SOLUTIONS Problem 2.8 (a) The result is ³ ´ h i ¢ ¡ x(t) = Re ej6πt + 6 Re ej(12πt−π/2) = Re ej6πt + 6ej(12πt−π/2) 1 1 x(t) = ej6πt + e−j6πt + 3ej(12πt−π/2) + 3e−j(12πt−π/2) 2 2 5 (b) The result is (c) The single-sided amplitude spectrum consists of lines of height 1 and 6 at frequencies of 3 and 6 Hz, respectively. The single-sided phase spectrum consists of a line of height −π/2 at frequency 6 Hz. The double-sided amplitude spectrum consists of lines of height 3, 1/2, 1/2, and 3 at frequencies of −6, −3, 3, and 6 Hz, respectively. The double-sided phase spectrum consists of lines of height π/2 and −π/2 at frequencies of −6 and 6 Hz, respectively. Problem 2.9 (a) Power. Since it is a periodic signal, we obtain 1 P1 = T0 Z T0 0 1 4 sin (8πt + π/4) dt = T0 2 Z T0 0 2 [1 − cos (16πt + π/2)] dt = 2 W where T0 = 1/8 s is the period. (b) Energy. The energy is Z ∞ Z E2 = e−2αt u2 (t)dt = −∞ ∞ e−2αt dt = 0 1 2α (c) Energy. The energy is E3 = (d) Neither energy or power. E4 = lim P4 = 0 since limT →∞ 1 T Z ∞ e2αt u2 (−t)dt = −∞ Z 0 e2αt dt = −∞ 1 2α T →∞ −T (α2 Z T dt + t2 )1/4 =∞ Since it is the sum of x1 (t) and x2 (t), its energy is the sum of the energies of these two signals, or E5 = 1/α. RT dt −T (α2 +t2 )1/4 = 0.(e) Energy. 6 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (f) Power. Since it is an aperiodic signal (the sine starts at t = 0), we use P6 = Z T 1 1 [1 − cos (10πt)] dt T →∞ 2T 0 2 0 · ¸ 1 T 1 1 sin (20πt) T = W − = lim T →∞ 2T 2 2 20π 4 0 1 T →∞ 2T lim T Z sin2 (5πt) dt = lim Problem 2.10 (a) Power. Since the signal is periodic with period π/ω, we have ω π Z π/ω P = 0 A2 |sin (ωt + θ)|2 dt = ω π Z π/ω 0 A2 A2 {1 − cos [2 (ωt + θ)]} dt = 2 2 (b) Neither. The energy calculation gives E = lim Z T T →∞ −T (Aτ )2 dt √ √ = lim τ + jt τ − jt T →∞ Z T −T (Aτ )2 dt √ →∞ τ 2 + t2 The power calculation gives 1 P = lim T →∞ 2T (c) Energy: E= (d) Energy: E=2 Z ∞ Z T −T (Aτ )2 dt (Aτ )2 √ = lim ln τ 2 + t2 T →∞ 2T Ã ! p 1 + T 2 /τ 2 p =0 −1 + 1 + T 2 /τ 2 1+ 0 3 A2 t4 exp (−2t/τ ) dt = A2 τ 5 (use table of integrals) 4 ÃZ Z ! τ /2 22 dt + τ 12 dt = 5τ 0 τ /2 Problem 2.11 (a) This is a periodic train of “boxcars”, each 3 units in width and centered at multiples of 6: µ ¶ Z Z 1 3 2 t 1 1 1.5 Π dt = W dt = Pa = 6 −3 3 6 −1.5 2 P = ∞. (c) The energy is Z Z ∞ 2 {2 [u (t) − u (t − 7)]} dt = E= −∞ 7 4dt = 28 J 0 . P = 0 W. (c) E = ∞. PROBLEM SOLUTIONS 7 (b) This is a periodic train of unit-high isoceles triangles.5 2 t 22 4 ¯ dt = 1− W 1− Pb = Λ dt = − ¯ = 5 −2. (b) E = 5 J. each 2 units wide and spaced by 3 units: ¶ µ ¶ ¯2 Z µ t 2 1 2 12 2 t 3¯ ¯ 1− dt = − 1− Pc = ¯ = W 3 0 2 33 2 ¯ 9 0 Problem 2.13 (a) The energy is E= (b) The energy is Z E= Z 6 cos (6πt) dt = 2 2 −6 Z 0 6· ¸ 1 1 + cos (12πt) dt = 6 J 2 2 ∞ ∞ −∞ Z h i2 −|t|/3 e cos (12πt) dt = 2 e −2t/3 0 · ¸ 1 1 + cos (24πt) dt 2 2 where the last integral follows by the eveness of the integrand of the first one. P = 49 W.5 2 5 0 2 53 2 ¯ 15 0 (d) This is a full-wave rectified cosine wave of period 1/5 (the width of each cosine pulse): ¸ Z 1/10 Z 1/10 · 1 1 1 2 |cos (5πt)| dt = 2 × 5 Pd = 5 + cos (10πt) dt = W 2 2 2 −1/10 0 (c) This is a backward train of sawtooths (right triangles with the right angle on the left). P = 2 W. each 4 units wide and centered at multiples of 5: ¶ µ ¶ ¯2 µ ¶ Z Z µ 2 2 t 3¯ t 2 1 2.1.12 (a) E = ∞.2. Problem 2. Use a table of definte integrals to obtain Z ∞ Z ∞ 3 2/3 −2t/3 e dt + e−2t/3 cos (24πt) dt = + E= 2 (2/3)2 + (24π)2 0 0 Since the result is finite. this is an energy signal. (d) E = ∞. and X4 = 7/8. They are centered at t = T /8. Problem 2. (d) Using the expression for the generalized Fourier series coefficients.083 × 10−2 T . integrate term by term. we use ²N = Z |x (t)| dt − 2 N X T n=1 RT R Note that T |x (t)|2 dt = 0 (t/T )2 dt = T /3. 5T /8. T ].34) and simplify. the ramp signal is approximated by t 3 5 7 1 = φ1 (t) + φ2 (t) + φ3 (t) + φ4 (t) .208 × 10−3 T . X3 = 5/8. they are adjacent to each other and fill the interval [0. ISEe = T − T 16 + 16 = 2. this is an energy signal. t ≥ 0 −∞ which is called the unit ramp. (e) These are unit-high rectangular pulses of width T /2 and centered at t = T /4 and 3T /4. and simplify making use of the orthogonality property of the orthonormal functions. we find that X1 = 1/8. Thus. for (d). 3 4 ¡ 1 64 9 64 ¢ For (e). Since they are spaced by T /4. ¡1 ¢ 9 ISEd = T − T 64 + 64 + 25 + 49 = 5. (c) These are unit-high rectangular pulses of width T /4. (f) To compute the ISE. this is an energy signal. Also. The energy is E= Z ∞ −∞ [r (t) − 2r (t − 10) + r (t − 20)] dt = 2 2 Z 0 10 µ t 10 ¶2 dt = 20 J 3 where the last integral follows because the integrand is a symmetrical triangle about t = 10. (d) Note that ½ Z t 0. t < 0 u (λ) dλ = r (t) = t. cn = T /4. Hence. and 7T /8. 3T /8. (b) Add and subtract the quantity suggested right above (2.14 (a) Expand the integrand. SIGNAL AND LINEAR SYSTEM THEORY Since the result is finite. 3 2 ¯ 2¯ cn ¯Xn ¯ . X2 = 3/8. 0 ≤ t ≤ T T 8 8 8 8 where the φn (t)s are given in part (c). We find that X1 = 1/4 and X2 = 3/4. Since the result is finite.8 CHAPTER 2. 15 (a) The Fourier coefficients are (note that the period = 9 1 2π 2 ω0 ) 1 1 X−1 = X1 = .16 The expansion interval is T0 = 4 and the Fourier coefficients are ¶ µ Z Z 1 2 2 −jn(π/2)t 2 2 2 nπt dt 2t e dt = 2t cos Xn = 4 −2 4 0 2 which follows by the eveness of the integrand.1. Problem 2. n6=0 16 jn(π/2)t e nπ . the integral for the coefficients is Z 1 2 2 8 2t dt = X0 = 4 −2 3 The Fourier series is therefore 8 x (t) = + 3 ∞ X (−1)n n=−∞. (d) The Fourier coefficients for this case are 1 3 X−3 = X3 = . X0 = 4 2 All other coefficients are zero. X−1 = X1 = 8 8 All other coefficients are zero. X0 = − 8 4 4 All other coefficients are zero. PROBLEM SOLUTIONS Problem 2. (c) The Fourier coefficients for this case are (note that the period is 1 1 1 X−2 = X2 = . Let u = nπt/2 to obtain the form µ ¶3 Z nπ 2 16 n Xn = 2 u2 cos u du = 2 (−1) nπ (nπ) 0 If n is odd. If n = 0. (b) The Fourier coefficients for this case are ∗ X−1 = X1 = 1 (1 + j) 2 2π 2ω0 ) All other coefficients are zero. X−1 = X1 = . the Fourier coefficients are zero as is evident from the eveness of the function being represented.2. integral for the Fourier coefficients is (note that the period really is T0 /2) Z A T0 /2 Xn = sin (ω 0 t) e−jnω0 t dt T0 0 ¯T0 /2 ¯ Ae−jnω0 t = − [jn sin (ω 0 t) + cos (ω0 t)]¯ ¯ ω0 T0 (1 − n2 ) 0 ¡ ¢ −jnπ A 1+e . the integral is Z A T0 /2 jA ∗ sin (ω 0 t) [cos (jnω 0 t) − j sin (jnω 0 t)] dt = − = −X−1 X1 = T0 0 4 This is the same result as given in Table 2. Problem 2. Problem 2. Also. For (d).1. in this case. Ptotal = A2 /2. This is the same as requiring that |n| ≤ 1/τ f0 = T0 /τ = 2. for a pulse train.18 This is a matter of integration. Then use the odd half-wave symmetry contition.19 (a) Use Parseval’s theorem to get P|nf0 | ≤ 1/τ = N X N X µ Aτ ¶2 |Xn | = sinc2 (nf0 τ ) T0 2 n=−N The n=−N where N is an appropriately chosen limit on the sum. Thus 2 µ ¶ P|nf0 | ≤ 1/τ 2 X A 2 = sinc2 (nf0 τ ) Ptotal A2 n=−2 2 = = = 2 1 X sinc2 (nf0 τ ) 2 n=−2 ¡ ¢¤ 1£ 1 + 2 sinc2 (1/2) + sinc2 (1) 2" µ ¶2 # 2 1 1+2 = 0.17 Parts (a) through (c) were discussed in the text. We are given that only frequences for which |nf0 | ≤ 1/τ are to be included. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.91 2 π . n 6= ±1 = ω 0 T0 (1 − n2 ) For n = 1. Ptotal = A2 τ /T0 and. break the integral for x (t) up into a part for t < 0 and a part for t > 0.10 CHAPTER 2. Only the solution for part (b) will be given here. t0 in the theorem proved in part (a) here is −π/2ω 0 . By Euler’s theorem.7568) + (0.5046) + (0. a sine wave can be expressed as 1 1 sin (ω0 t) = ejω0 t − e−jω0 t 2j 2j 1 1 Its Fourier coefficients are therefore X1 = 2j and X−1 = − 2j . which results in · Z T0 −t0 ¸ ¡ 0 ¢ −jnω0 t0 0 −jnω0 t0 1 x t e dt e = Xn e−jnω0 t0 Yn = T0 −t0 y (t) = A cos ω 0 t = A sin (ω 0 t + π/2) = A sin [ω 0 (t + π/2ω 0 )] Thus.1. According to the theorem proved in part (a).2. we obtain Y1 = For n = −1. |n| ≤ 5. Ptotal = A2 /5. PROBLEM SOLUTIONS (b) In this case. we obtain 1 jπ/2 1 = e 2j 2 1 −jπ/2 1 = e 2j 2 which gives the Fourier series representation of a cosine wave as Y−1 = − 1 1 y (t) = ejω 0 t + e−jω0 t = cos ω 0 t 2 2 .2339) 5 = 0. we multiply these by the factor e−jnω0 t0 = e−jnω0 (−π/2ω 0 ) = ejnπ/2 For n = 1.90 11 Problem 2. and P|nf0 | ≤ 1/τ Ptotal = 5 1 X sinc2 (n/5) 5 n=−5 io h 1n 2 2 2 2 = 1 + 2 (0.20 (a) The integral for Yn is Yn = 1 T0 Z y (t) e−jnω 0 t dt = 1 T0 Z T0 T0 0 x (t − t0 ) e−jnω0 t dt (b) Note that Let t0 = t − t0 .9355) + (0. 12 CHAPTER 2.1). Problem 2. Therefore. let t0 = −T0 /8 and τ = T0 /4 to get ¶ µ ³n´ A πnf0 Xn = sinc exp j 4 4 4 . its sum is 8 π2 . otherwise it looks like a squarewave spectrum.1 with A = 1 and t = 0 to obtain the series 1=···+ 2 4 4 4 4 4 4 + 2+ 2+ 2+ 2+ +··· 2 25π 9π π π 9π 25π 2 Multiply both sides by π to get the series in given in the problem. (b) Note that dxB (t) xA (t) = K dt where K is a suitably chosen constant. Hence. The relationship between spectral components is therefore A B Xn = K (jnω0 ) Xn where the superscript A refers to xA (t) and B refers to xB (t).22 (a) In the expression for the Fourier series of a pulse train (Table 2. the sum is π . Note that 4 4 this can be viewed as having a sinc-function envelope with zeros at multiples of 3T0 . Problem 2. Problem 2. 4 (b) The amplitude spectrum is the same as for part (a) except that X0 = 3A .23 (a) There is no line at dc. SIGNAL AND LINEAR SYSTEM THEORY We could have written down this Fourier representation directly by using Euler’s theorem. The phase spectrum can be obtained from that of part (a) by adding a phase shift of π for negative frequencies and subtracting π for postitive frequencies (or vice versa). 8 (b) Use the Fourier series of a square wave (specialize the Fourier series of a pulse train) with A = 1 and t = 0 to obtain the series µ ¶ 1 1 4 1− + −··· 1= π 3 5 Multiply both sides by π 4 to get the series in the problem statement.21 (a) Use the Fourier series of a triangular wave as given in Table 2. since x4 (t) is really a triangle function.25 (a) Using a table of Fourier transforms and the time reversal theorem. PROBLEM SOLUTIONS 13 Problem 2.2. that X3 (f ) = X1 (f) − X2 (f) jA = sinc (2f τ ) πf (d) Since x4 (t) = x1 (t) + x2 (t) we have. that X4 (f ) = X1 (f) + X2 (f) sin2 (πfτ ) = Aτ (πfτ )2 = Aτ sinc2 (fτ ) This is the expected result. Therefore. after some simplification. we deduce that F [sgn (t)] = 1 1 1 − = j2πf −j2πf jπf . after some simplification. (b) Since x2 (t) = x1 (−t) we have. transform as α → 0. the Fourierr transform of the given signal is 1 1 X (f ) = − α + j2πf α − j2πf Taking the limit of the above Fourier Note that x (t) → sgn(t) in the limit as α → 0. Problem 2. or A (1 − t/τ ). that ∗ X2 (f) = X1 (f) = X1 (−f) · ´¸ 1 ³ A j2πf τ 1+ 1−e = −j2πf j2πf τ (c) Since x3 (t) = x1 (t) − x2 (t) we have.1.24 (a) This is the right half of a triangle waveform of width τ and height A. the Fourier transform is Z τ X1 (f) = A (1 − t/τ ) e−j2πf t dt 0 · ´¸ 1 ³ A −j2πf τ 1− 1−e = j2πf j2πfτ where a table of integrals has been used. by the time reversal theorem. we obtain X2 (f) = (c) Two differentiations give d2 x3 (t) = δ (t) − δ (t − 1) − δ (t − 2) + δ (t − 3) dt2 1 − 2e−j2πf + e−j4πf = sinc2 (f ) e−j2πf (j2πf )2 .14 CHAPTER 2. we obtain F [u (t)] = = 1 1 F [sgn (t)] + F [1] 2 2 1 1 + δ (f) j2πf 2 (c) The same result as obtained in part (b) is obtained.26 (a) Two differentiations give dδ (t) d2 x1 (t) = − δ (t − 2) + δ (t − 3) dt2 dt Application of the differentiation theorem of Fourierr transforms gives (j2πf)2 X1 (f ) = (j2πf) (1) − 1 · e−j4πf + 1 · e−j6πf where the time delay theorem and the Fourier transform of a unit impulse have been used. we obtain X1 (f) = 1 1 e−j4πf − e−j6πf e−j5πf = − − sinc (2f) j2πf j2πf j2πf (j2πf)2 (b) Two differentiations give d2 x2 (t) = δ (t) − 2δ (t − 1) + δ (t − 2) dt2 Application of the differentiation theorem gives (j2πf )2 X2 (f) = 1 − 2e−j2πf + e−j4πf Dividing both sides by (j2πf )2 . Problem 2. SIGNAL AND LINEAR SYSTEM THEORY (b) Using the given relationship between the unit step and the signum function and the linearity property of the Fourier transform. Dividing both sides by (j2πf )2 . so its Fourier transform is real and even. we obtain X3 (f) = (d) Two differentiations give dδ (t − 2) d2 x4 (t) = 2Π (t − 1/2) − 2δ (t − 1) − 2 2 dt dt Application of the differentiation theorem gives (j2πf)2 X4 (f ) = 2sinc (f) e−jπf − 2e−j2πf − 2 (j2πf) e−j4πf Dividing both sides by (j2πf )3 . so its Fourier transform is even and purely real. time delay. (d) This signal is neither even nor odd signal. we obtain X2 (f ) = ej12πt − e−j12πt = 2j sin (12πf ) The Fourier transform is odd and imaginary because the signal is odd. so its Fourier transform is odd and purely imaginary. we obtain X1 (f) = ej16πt + 2 + e−j16πt = 4 cos2 (6πt) The Fourier transform is even and real because the signal is even. we obtain X4 (f) = 2e−j2πf + (j2πf) e−j4πf − sinc (f) e−jπf 2 (πf)2 1 − e−j2πf − e−j4πf + e−j6πf (j2πf)2 15 Problem 2. and the Fourierr transform of an impulse. (c) This is an odd signal.28 (a) Using superposition. so its Fourier transform is odd and purely imaginary. . so its Fourier transform is even and purely real. (b) Using superposition. (b) This is an even signal.27 (a) This is an odd signal.2. so its Fourier transform is complex.1. PROBLEM SOLUTIONS Application of the differentiation theorem gives (j2πf)2 X3 (f ) = 1 − e−j2πf − e−j4πf + e−j6πf Dividing both sides by (j2πf )2 . (e) This is an even signal. Problem 2. and the Fourier transform of an impulse. (f) This signal is even. time delay. the energy spectral density is 4 X2 (f ) = Π2 9 (c) The Fourier transform of this signal is µ ¶ 4 f X3 (f) = sinc 5 5 so the energy spectral density is 16 G3 (f) = sinc2 25 µ ¶ f 5 µ f 30 ¶ µ ¶ 4 f = Π 9 30 (d) The Fourier transform of this signal is · µ ¶ µ ¶¸ 2 f − 20 f + 20 sinc + sinc X4 (f) = 5 5 5 . SIGNAL AND LINEAR SYSTEM THEORY X3 (f) = n=0 4 X¡ ¢ n2 + 1 e−j4πnf It is complex because the signal is neither even nor odd. the energy spectral density is G1 (f) = ½ 2/3 1 + [f / (3/2π)]2 ¾2 (b) The Fourier transform of this signal is µ ¶ 2 f X2 (f) = Π 3 30 Thus.29 (a) The Fourier transform of this signal is X1 (f) = 2 (1/3) 2/3 2 = 1 + (2πf/3) 1 + [f/ (3/2π)]2 Thus.16 (c) The Fourier transform is CHAPTER 2. Problem 2. is ¸2 Z ∞ Z ∞· 1 2 |X3 (f)| df = df I3 = 2 2 −∞ −∞ α + (2πf ) Z ∞ Z ∞ 1 1 1 −2α|t| e dt = e−2αt dt = = 2 2α2 0 4α3 (2α) −∞ (d) Use the transform pair µ ¶ 1 t Λ ←→ sinc2 (τ f ) τ τ . PROBLEM SOLUTIONS so the energy spectral density is · µ ¶ µ ¶¸ 4 f − 20 f + 20 2 G4 (f) = sinc + sinc 25 5 5 Problem 2.2.1. we obtain the integral relationship Z ∞ Z ∞ Z ∞ Z ∞ df 1 2 2 |X1 (f )| df = df = |x1 (t)| dt = e−2αt dt = 2 + (2πf)2 2α −∞ −∞ α −∞ 0 (b) Use the transform pair µ ¶ t 1 ←→ sinc (τ f) = X2 (f) x2 (t) = Π τ τ Rayleigh’s energy theorem gives Z ∞ |X2 (f )|2 df −∞ = = Z ∞ sinc (τ f ) df = 1 2 Π τ2 2 −∞ Z ∞ −∞ µ ¶ t dt = τ Z ∞ −∞ Z τ /2 |x2 (t)|2 dt dt 1 = 2 τ τ −τ /2 (c) Use the transform pair x3 (t) = e−α|t| ←→ α2 2α + (2πf )2 The desired integral.30 (a) Use the transform pair x1 (t) = e−αt u (t) ←→ 1 α + j2πf 17 Using Rayleigh’s energy theorem. by Rayleigh’s energy theorem. t ≤ −1/2 ¤ 1 −α(t−1/2) − e−α(t+1/2) . by Rayleigh’s energy theorem. t ¤ τ − 1/2 ≤  £ 1 1 − e−α(t−τ +1/2) . t > 1/2 t−1/2 Integration of these three cases gives £ ¤  1  £ α eα(t+1/2) − eα(t−1/2) . −3/2 ≤ t < −1/2   3/2 − t. 1/2 ≤ t < 3/2 ∞ (c) The convolution results in Z y3 (t) = e −α|λ| −∞ Π (λ − t) dλ = Z t+1/2 e−α|λ| dλ t−1/2 Sketches of the integrand for various values of t gives the following cases:  R t+1/2 αλ   t−1/2 e dλ. τ − 1/2 ¤ t ≤ τ + 1/2 < y1 (t) = α  1 £ −α(t−τ −1/2) −α(t−τ +1/2) .31 (a) The convolution operation gives  0. −1/2 ≤ t ≤ 1/2 tr (t) =  3/2 + t. t > τ + 1/2 −e α e (b) The convolution of these two signals gives y2 (t) = Λ (t) + tr (t) where tr(t) is a trapezoidal function given by   0. −1/2 < t ≤ 1/2 y3 (t) = t−1/2 e 0  R t+1/2 −αλ   e dλ. t ≤ −1/2  R R 0 αλ dλ + t+1/2 e−αλ dλ. −1/2 < t ≤ 1/2 y3 (t) = α e £ ¤  1 e−α(t−1/2) − e−α(t+1/2) . t < −3/2 or t > 3/2   1.18 CHAPTER 2. t > 1/2 α . SIGNAL AND LINEAR SYSTEM THEORY The desired integral. is Z ∞ Z ∞ 2 I4 = |X4 (f)| df = sinc4 (τ f ) df −∞ −∞ Z ∞ Z τ 2 1 2 Λ (t/τ ) dt = 2 [1 − (t/τ )]2 dt = τ 2 −∞ τ 0 Z 2 2 1 [1 − u]2 du = = τ 0 3τ Problem 2. 1.32 (a) Using the convolution and time delay theorems. PROBLEM SOLUTIONS (d) The convolution gives y4 (t) = Z 19 t x (λ) dλ −∞ Problem 2. that the integration theorem can be applied directly) Y4 (f) = F [x (t) ∗ u (t)] ¸ · 1 1 + δ (f) = X (f) j2πf 2 X (f) 1 = + X (0) δ (f ) j2πf 2 Problem 2. also.33 (a) The normalized inband energy is 2 E1 (|f | ≤ W ) = tan−1 Etotal π µ 2πW α ¶ . we obtain £ ¤ Y1 (f) = F e−αt u (t) ∗ Π (t − τ ) ¤ £ = F e−αt u (t) F [Π (t − τ )] 1 sinc (f) e−j2πf τ = α + j2πf (b) The superposition and convolution theorems give Y2 (f) = F {[Π (t/2) + Π (t)] ∗ Π (t)} = [2sinc (2f) + sinc (f )] sinc (f) (c) By the convolution theorem h i Y3 (f) = F e−α|t| ∗ Π (t) = 2α sinc (f) α2 + (2πf )2 (d) By the convolution theorem (note.2. It makes no difference whether the original square wave is even or odd or neither.35 Combine the exponents of the two factors in the integrand of the Fourier transform integral. Problem 2.34 (a) By the modulation theorem X (f) = = ½ · ¸ · ¸¾ T0 T0 AT0 sinc (f − f0 ) + sinc (f + f0 ) 4 2 2 ¶¸ · µ ¶¸¾ ½ · µ AT0 1 f 1 f − 1 + sinc +1 sinc 4 2 f0 2 f0 (b) Use the superposition and modulation theorems to get ½ · · µ µ ¸ ¶ ¶¸¾ f 1 f 1 1 f AT0 sinc sinc + − 2 + sinc +2 X (f) = 4 2f0 2 2 f0 2 f0 Problem 2. Rename variables to obtain Z T 1 R (τ ) = lim x (β) x (t + β) dβ T →∞ 2T −T Problem 2.36 Consider the development below: Z Z ∞ x (−λ) x (t − λ) dλ = x (t) ∗ x (−t) = −∞ ∞ x (β) x (t + β) dβ −∞ where β = −λ has been substituted. Problem 2. SIGNAL AND LINEAR SYSTEM THEORY Z E1 (|f | ≤ W ) =2 Etotal τW sinc2 (u) du 0 The integration must be carried out numerically.37 The result is an even triangular wave with zero average value of period T0 .20 (b) The result is CHAPTER 2. complete the square. . and use the given definite integral. write H (f) as H (f) = 1 − Inverse Fourier transforming gives h (t) = δ (t) − 5e−5t u (t) (b) Use the transform pair Ae−αt u (t) ←→ A α + j2πf 5 5 + j2πf and the time delay theorem to find the unit impulse as 2 8 h (t) = e− 15 (t−3) u (t − 3) 5 . Problem 2.39 (a) The find the unit impulse response.1.2. PROBLEM SOLUTIONS 21 Problem 2.38 Fourier transform both sides of the differential equation using the differentiation theorem of Fourier transforms to get [j2πf + a] Y (f) = [j2πbf + c] X (f ) Therefore. the transfer function is H (f) = The amplitude response function is q c2 + (2πbf)2 |H (f)| = q a2 + (2πf)2 −1 c + j2πbf Y (f) = X (f) a + j2πf and the phase response is arg [H (f )] = tan µ 2πbf c ¶ − tan −1 µ 2πf a ¶ Amplitude and phase responses for various values of the constants are plotted below. Problem 2. will work. Q. we obtain µ ¶ R1 R1 + R2 h (t) = δ (t) − exp − t u (t) L L . and the definitions for ω 0 . SIGNAL AND LINEAR SYSTEM THEORY Problem 2.42 (a) By long division H (f) = 1 − R1 +R2 L R1 /L + j2πf Using the transforms of a delta function and a one-sided exponential. Call the junction at the negative input of the operational amplifier 3.41 (a) Replace the capacitors with 1/jωC which is their ac-equivalent impedance. Use the constraint equation for the operational amplifier. Call the junction of the input resistor. (b) If W > B.40 Use the transform pair for a sinc function to find that µ ¶ µ ¶ f f Y (f) = Π Π 2B 2W (a) If W < B.22 CHAPTER 2. it follows that µ ¶ f Y (f ) = Π 2B ³ ´ ³ ´ f f because Π 2W = 1 throughout the region where Π 2B is nonzero. Call the junction at the positive input of the operational amplifier 2.3 × 10−4 seconds and Ra = 2.5757 Rb ³ ´ ³ ´ f f because Π 2B = 1 throughout the region where Π 2W is nonzero. and K to get the given transfer function. feedback resistor. (d) Combinations of components giving RC = 2. which is V2 = V3 . Write down the KCL equations at these three junctions. it follows that f Y (f) = Π 2W µ ¶ Problem 2. and capacitors 1. (c) The condition for stability is Z ∞ Z ∞ 1 u (t − 1) dt |h3 (t)| dt = t −∞ Z−∞ ∞ dt = = ln (t)|∞ → ∞ 1 t 1 .43 The Payley-Wiener criterion gives the integral Z ∞ βf 2 df I= 2 −∞ 1 + f which does not converge. PROBLEM SOLUTIONS 23 (b) Substituting the ac-equivalent impedance for the inductor and using voltage division.1. (b) The condition for stability is Z ∞ Z ∞ |h2 (t)| dt = |cos (2πf0 t) u (t)| dt −∞ −∞ Z ∞ = |cos (2πf0 t)| dt → ∞ 0 which follows by integrating one period of |cos (2πf0 t)| and noting that the total integral is the limit of one period of area times N as N → ∞. the impulse response is · µ ¶ ¸ R2 R1 R2 R1 R2 h (t) = δ (t) − exp − t u (t) R1 + R2 (R1 + R2 ) L (R1 + R2 ) L Problem 2.2. the given function is not suitable as the transfer function of a causal LTI system. Hence. Problem 2.44 (a) The condition for stability is Z ∞ Z ∞ |h1 (t)| dt = |exp (−αt) cos (2πf0 t) u (t)| dt −∞ Z−∞ Z ∞ ∞ 1 = exp (−αt) |cos (2πf0 t)| dt < exp (−αt) dt = < ∞ α 0 0 which follows because |cos (2πf0 t)| ≤ 1. the transfer function is ¶ µ j2πfL R2 (R1 k R2 ) /L R2 = 1− H (f ) = R R1 + R2 R 1 R2 + j2πfL R1 + R2 (R1 k R2 ) /L + j2πf +R 1 2 Therefore. 46 Using the Fourier coefficients of a half-rectified sine wave from Table 2. using X2 (f ) = Π (f/2W ) .85/τ .45 The energy spectral density of the output is Gy (f) = |H (f )|2 |X (f)|2 where X (f) = Hence 1 2 + j2πf 100 ih i Gy (f ) = h 9 + (2πf)2 4 + (2πf)2 Problem 2. we obtain B90 = 0.0055α 2π 3A 3A − cos (20πt) π 2 (b) For this case. The ideal rectangular filter passes all frequencies less than 13 Hz and rejects all frequencies greater than 13 Hz. Therefore y (t) = Problem 2.9W (c) Numerical integration gives B90 = 0. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.24 CHAPTER 2.47 (a) The 90% energy containment bandwidth is given by B90 = α tan (0.1 and noting that those of a half-rectified cosine wave are related by Xcn = Xsn e−jnπ/2 The fundamental frequency is 10 Hz.45π) = 1. no phase distortion. no phase distortion. Tg (f) = Tp (f) = 0. no phase distortion.49 (a) Amplitude distortion. (c) No amplitude distortion.667πf)] 2πf . Problem 2.1.2πf ) 1 + (0.667πf )2 1 [tan (0.50 The transfer function corresponding to this impulse response is ¶¸ · µ 2 2πf 2 =q exp −j tan H (f) = 3 + j2πf 3 2 9 + (2πf) · µ ¶¸ 1 d 2πf Tg (f ) = − − tan 2π df 3 3 = 9 + (2πf)2 The phase delay is Tp (f ) = − θ (f) = 2πf tan ³ 2πf 3 The group delay is 2πf ´ Problem 2.51 The group and phase delays are.2.2πf ) − tan (0. (d) No amplitude distortion.333 2 − 1 + (0. respectively. PROBLEM SOLUTIONS Problem 2. we find its output in response to the above input to be · ³ ¸ ³ π´ 1 π´ A 2A cos ω 0 t − − cos 3ω 0 t − +··· y (t) = + 2 π 2 3 2 Problem 2.7 · ¸ 1 A 2A cos (ω 0 t) − cos (3ω 0 t) + · · · x (t) = + 2 π 3 25 From the transfer function of a Hilbert transformer. phase distortion.1 0.48 From Example 2. (b) No amplitude distortion. 2 6Λ 6 6 6 where Λ (t) is an isoceles triangle of unit height going from -1 to 1.2 Π +Π ∗ Π +Π 6 6 6 6 · µ ¶ µ ¶¸ f − 20 f + 20 = 4 Π +Π 6 6 ¶ µ ¶ µ ¶¸ · µ f f + 40 f − 40 + 12Λ + 6Λ +3.52 In terms of the input spectrum. the output spectrum is Y (f) = X (f ) + 0..26 CHAPTER 2.00005.00005 1 + 16 × 106 Q2 . The student should sketch the output spectrum given the above analytical result.2X (f ) ∗ X (f) ¶ µ ¶¸ · µ f + 20 f − 20 +Π = 4 Π 6 6 · µ ¶ µ ¶¸ · µ ¶ µ ¶¸ f − 20 f + 20 f − 20 f + 20 +3.53 (a) The output of the nonlinear device is y (t) = 1.075 cos (2000πt) + 0. Problem 2. SIGNAL AND LINEAR SYSTEM THEORY Problem 2.025 cos (6000πt) The steadtstate filter output in response to this input is z (t) = 1.005% = 0.075 |H (1000)| cos (2000πt) + 0.025 |H (3000)| cos (6000πt) so that the THD is THD = (1. (b) For THD = 0.025)2 |H (3000)|2 432 = 1 + 4Q2 (1000 − 3000)2 1849 = 1 + 16 × 106 Q2 where H (f ) is the transfer function of the filter. the equation for Q becomes 1849 = 0.075)2 |H (1000)|2 (0. PROBLEM SOLUTIONS or 6 × 106 Q2 = 1849 × 2 × 104 − 1 Q = 1. the delay theorem. and the inverse Fourier transform of a rectangular pulse function to get h (t) = H0 δ (t − t0 ) − 2BH0 sinc [2B (t − t0 )] Problem 2. 1 |X (f)| df = √ 2 2πτ −∞ 2 √ 2πτ = 1 2W T = √ 2 2πτ .2. To use this as a frequency multiplier.52 27 Problem 2.55 Write the transfer function as H (f) = H0 e −j2πf t0 f − H0 Π 2B µ ¶ e−j2πf t0 Use the inverse Fourier transform of a constant. Z ∞ √ 1 |x (t)| dt = 2πτ T = x (0) −∞ W= Therefore.3ω 1 .56 (a) The Fourier transform of this signal is √ ¡ ¢ X (f ) = A 2πb2 exp −2π 2 τ 2 f 2 By definition. or at 3ω1 or 3ω 2 to use as a tripler. ω 1 − ω 2 . 2ω 1 . 2ω 2 . ω 1 . 2ω 1 + ω 2 .54 Frequency components are present in the output at radian frequencies of 0. 2ω 1 − ω 2 . ω1 + 2ω2 . 1 2X (0) Z ∞ Similarly. ω 2 . using a table of integrals. 3ω 2 .1. pass the components at 2ω 1 or 2ω 2 to use as a doubler. Problem 2. ω 1 + ω 2 . 2ω 2 − ω 1 . the group delay is Tg (f) = − 1 d [θ (f)] 2π df For the second-order Butterworth filter considered here. function is H (s) = h s+ ω2 3 ih ω √3 (1 − j) s + 2 ω √3 2 Its s-domain transfer (1 + j) Letting ω 3 = 2πf3 and s = jω = j2πf. we obtain H (j2πf) = i= ω2 √ 3 s2 + 2ω 3 s + ω 2 3 2 4π 2 f3 f2 √ √3 = 2 2 −4π 2 f 2 + 2 (2πf3 ) (j2πf) + 4π 2 f3 −f 2 + j 2f3 f + f3 (b) If the phase response function of the filter is θ (f). 2W T = 2 ³α´ µ 2 ¶ 4 α Problem 2.28 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY (b) The Fourier transform of this signal is X (f ) = The pulse duration is T = The bandwidth is 1 x (0) Z 2A/α 1 + (2πf/α)2 ∞ −∞ ∞ |x (t)| dt = 2 α α 4 1 W = 2X (0) Z −∞ |X (f)| df = =1 Thus. Ã√ ! 2f3 f −1 θ (f) = − tan 2 f3 − f 2 .57 (a) The poles for a second order Butterworth filter are given by ω3 s1 = s∗ = − √ (1 − j) 2 2 where ω 3 is the 3-dB cutoff frequency of the Butterworth filter. 34/f3 seconds.08 and f3 t90% ≈ 0. (b) and (c) .3.use sketches to show. (b) The spectrum is Y (f) = Xδ (f) H (f ) . From the MATLAB plot of Fig. the group delay is Tg (f ) = = " Ã√ !# 2f3 f 1 d tan−1 2 2π df f3 − f 2 2 1 1 + (f/f3 )2 f3 f 3 + f 2 √ =√ 4 2π f3 + f 4 2πf3 1 + (f/f3 )4 (c) Use partial fraction expansion of H (s) and then inverse Laplace transform H (s) /s to get the given step response.2.3: Therefore.5 seconds. PROBLEM SOLUTIONS 29 Figure 2. f3 t10% ≈ 0. Problem 2.58 (a) 0.59 (a) The leading edges of the flat-top samples follow the waveform at the sampling instants.1.42 so that the 10-90 % rise time is about 0. Problem 2. 2. Plot it and estimate the 10% and 90% times from the plot. 30 where CHAPTER 2. (b) The output spectrum of the zero-order hold circuit is Y (f) = sinc (Ts f) ∞ X n=−∞ X (f − nfs ) exp (−jπf Ts ) −1 where fs = Ts .63 The Fourier transform is Y (f) = 1 1 X (f − f0 ) + X (f + f0 ) 2 2 ¸ · 1 1 −jπ/2 jπ/2 + δ (f + f0 ) e + [−jsgn (f) X (f)] ∗ δ (f − f0 ) e 2 2 1 1 = X (f − f0 ) [1 − sgn (f − f0 )] + X (f + f0 ) [1 + sgn (f + f0 )] 2 2 . we want Ts << W −1 . For small distortion. very little distortion will result if τ −1 >> W . only (f) will work. all but (b) and (e) will work theoretically. Problem 2. For lowpass sampling and recovery.62 For bandpass sampling and recovery. Problem 2.1− kHz. although an ideal filter with bandwidth exactly equal to the unsampled signal bandwidth is necessary. Problem 2. If W is the bandwidth of X (f). Problem 2.60 (a) The sampling frequency should be large compared with the bandwidth of the signal.9+ kHz to 2. SIGNAL AND LINEAR SYSTEM THEORY Xδ (f ) = fs and n=−∞ ∞ X X (f − nfs ) H (f) = τ sinc (f τ ) exp (−jπfτ ) The latter represents the frequency response of a filter whose impulse response is a square pulse of width τ and implements flat top sampling.61 The lowpass recovery filter can cut off in the range 1. Multiply x (t) and x (t) together term by term.65 (a) Note that F [jb(t)] = j [−jsgn (f)] X (f). Hence x x1 (t) = 3 3 1 1 x (t) + jb (t) → X1 (f) = X (f) + j [−jsgn (f)] X (f) x 4 4 4 4 ¸ · 3 1 + sgn (f ) X (f) = 4 4 ½ 1 2 X (f) . Problem 2. integrate over time to get b Z ∞ −∞ 1 sin (2ω0 t) dt −T 2 ¯ 1 cos (2ω 0 t) ¯T ¯ =0 = lim T →∞ 2T 4ω 0 ¯−T lim 1 2T Z T sin (ω0 t) cos (ω 0 t) dt −T Z T x (t) x (t) dt = A2 b = A2 = A2 Z Π (t/τ ) −∞ "Z Z τ /2 ∞ Z −τ /2 τ /2 −τ /2 where the result is zero by virtue of the integrand of the last integral being odd. use trigonometric identities for the product of sines and cosines. f < 0 = X (f) . Then find the b Hilbert transform of x (t) by phase shifting by −π/2.1. PROBLEM SOLUTIONS Problem 2.4. take the product. f > 0 1 dλ dt −τ /2 π (t − λ) ¯ ¯ 1 ¯ t − τ /2 ¯ ¯ ¯ dt = 0 ln π ¯ t + τ /2 ¯ τ /2 ·Z ∞ −∞ ¸ Π (λ/τ ) dλ dt π (t − λ) # A sketch is shown in Figure 2.2. The integrand will be a sum of terms similar to that of part (a). (c) Use the integral definition of x (t). then integrate. . so b 1 lim T →∞ 2T Z T −T 31 1 x (t) x (t) dt = lim b T →∞ 2T = T →∞ (b) Use trigonometric identities to express x (t) in terms of sines and cosines.64 (a) xa (t) = cos (ω 0 t − π/2) = sin (ω 0 t). The limit as T → ∞ will be zero term-by-term. Problem 2. f > f0 2 · A sketch is shown in Figure 2. (d) For this signal · ¸ 3 1 x4 (t) = x (t) − jb (t) exp (jπW t) x 4 4 · ¸ 3 1 − sgn (f − W/2) X (f − W/2) → X4 (f) = 4 4 A sketch is shown in Figure 2.66 (a) The spectrum is Xp (f) = X (f) + j [−jsgn (f)] X (f) = [1 + sgn (f)] X (f) The Fourier transform of x (t) is µ ¶ µ ¶ 1 f − f0 1 f + f0 X (f) = Π + Π 2 2W 2 2W Thus.4. (c) This case has the same spectrum as part (a). f < f0 = 3 X (f − f0 ) . SIGNAL AND LINEAR SYSTEM THEORY ¸ 3 3 x (t) + jb (t) exp (j2πf0 t) x x2 (t) = 4 4 3 → X2 (f ) = [1 + sgn (f − f0 )] X (f − f0 ) 4 ½ 0.32 (b) It follows that CHAPTER 2. · ¸ 3 1 x (t) + jb (t) exp (j2πW t) x x3 (t) = 4 4 · ¸ 3 1 + sgn (f − W ) X (f − W ) → X3 (f ) = 4 4 A sketch is shown in Figure 2. except that it is shifted right by W Hz. Xp (f) = Π (b) The complex envelope is defined by µ f − f0 2W ¶ if f0 > 2W ˜ xp (t) = x (t) ej2πf0 t . That is.4.4. Hz 0 W 2W 0 W X2(f) 3A/2 f. Hz 0 X4(f) 2A A f.4: .W Figure 2.W/2 0 W/2 3W/2 f0 f0 . Hz . PROBLEM SOLUTIONS 33 X1(f) 2A A f.2. Hz -W X3(f) 2A A f.1. exp(-t)u(t). or sinewave.2 Computer Exercises Computer Exercise 2. sin(2*pi*t) % tau = input(’Enter approximating pulse width: ’). type_wave = input(’Enter waveform type: 1 = decaying exponential. For |t| ≤ τ /2. 2 = sinewave ’). the result is In the above equations. θ is given by (α/2) e−αt y (t) = q α2 + (2π∆f)2 n o × eατ /2 cos [2π (f0 + ∆f ) t − θ] − e−ατ /2 cos [2π (f0 + ∆f) t + θ] θ = − tan−1 µ 2π∆f α ¶ 2. the result is y (t) = α/2 q α2 + (2π∆f)2 n o × cos [2π (f0 + ∆f ) t − θ] − e−α(t+τ /2) cos [2π (f0 + ∆f ) t + θ] For t > τ /2.67 For t < τ /2. if type_wave == 1 . the output is zero.1 % ce2_1: Computes generalized Fourier series coefficients for exponentially % decaying signal. SIGNAL AND LINEAR SYSTEM THEORY x (t) = xp (t) e−j2πf0 t ˜ Hence F [˜ (t)] = F [xp (t)]|f →f +f0 x (c) The complex envelope is x (t) = F ˜ −1 ¶¸ · µ f = 2W sinc (2W t) Π 2W µ ¶¯ ¶ f − f0 ¯ f ¯ =Π =Π ¯ 2W 2W f →f +f0 µ Problem 2.34 Therefore CHAPTER 2. elseif type_wave == 2 MSE = quad(@integrand_sine2. elseif type_wave == 2 a(n) = (1/cn)*quad(@integrand_sine. elseif type_wave == 2 t_max = 1. n_max = floor(t_max/tau) for n = 1:n_max LL = (n-1)*tau.2. x_a_p_p_r_o_x(t)’) 35 . ylabel(’x(t). UL).01:t_max. end if type_wave == 1 MSE = quad(@integrand_exp2. axis([0 t_max -inf inf]). end clf a = []. end end disp(’ ’) disp(’c_n’) disp(cn) disp(’ ’) disp(’Expansion coefficients (approximating pulses not normalized):’) disp(a) disp(’ ’) x_approx = zeros(size(t)). end disp(’Mean-squared error:’) disp(MSE) disp(’ ’) if type_wave == 1 plot(t. LL. COMPUTER EXERCISES t_max = input(’Enter duration of exponential: ’). t_max) .cn*a*a’. t = 0:.cn*a*a’.2. 0.n-1). if type_wave == 1 a(n) = (1/cn)*quad(@integrand_exp. t_max) . LL. cn = tau. xlabel(’t’). UL). for n = 1:n_max x_approx = x_approx + a(n)*phi(t/tau. UL = n*tau. 0. x_approx). exp(-t)) title([’Approximation of exp(-t) over [0. ylabel(’x(t). x_approx). SIGNAL AND LINEAR SYSTEM THEORY hold plot(t.num2str(tau)]) elseif type_wave == 2 plot(t.num2str(tau)]) end % Decaying exponential function for ce_1 % function z = integrand_exp(t) z = exp(-t).^2.num2str(t_max). % Sine function for ce_1 % function z = integrand_sine(t) z = sin(2*pi*t). sin(2*pi*t)) title([’Approximation of sin(2*pi*t) over [0. % Sin^2 function for ce_1 % function z = integrand_sine2(t) z = (sin(2*pi*t)). axis([0 t_max -inf inf]). A typical run follows: >> ce2_1 Enter approximating pulse width: 0.’] with contiguous rectangular pulses of width ’.36 CHAPTER 2. 2 = sinewave: 2 n_max = 8 c_n 0.125 Enter waveform type: 1 = decaying exponential. ’. x_a_p_p_r_o_x(t)’) hold plot(t. xlabel(’t’).’] with contiguous rectangular pulses of width ’.1250 Expansion coefficients (approximating pulses not normalized): . ’.num2str(t_max). % Decaying exponential squared function for ce_1 % function z = integrand_exp2(t) z = exp(-2*t). elseif type_wave == 2 t_max = 1.2 % ce2_1: Computes generalized Fourier series coefficients for exponentially % decaying signal.2.3729 0.01:t_max. type_wave = input(’Enter waveform type: 1 = decaying exponential. COMPUTER EXERCISES 37 Figure 2. exp(-t)u(t). or sinewave. sin(2*pi*t) % tau = input(’Enter approximating pulse width: ’). 2 = sinewave ’).9003 0. .3729 -0.9003 -0. t = 0:.0252 0.3729 -0.5: Columns 1 through 7 0.2. end clf a = [].3729 Mean-squared error: 0. if type_wave == 1 t_max = input(’Enter duration of exponential: ’).9003 Column 8 -0.9003 Computer Exercise 2. num2str(t_max). elseif type_wave == 2 MSE = quad(@integrand_sine2. if type_wave == 1 a(n) = (1/cn)*quad(@integrand_exp.n-1).38 CHAPTER 2. x_a_p_p_r_o_x(t)’) hold plot(t. elseif type_wave == 2 a(n) = (1/cn)*quad(@integrand_sine. xlabel(’t’). UL = n*tau. UL). 0. exp(-t)) title([’Approximation of exp(-t) over [0.cn*a*a’. end if type_wave == 1 MSE = quad(@integrand_exp2. t_max) . x_approx).’] with contiguous rectangular pulses of width ’. ylabel(’x(t). end end disp(’ ’) disp(’c_n’) disp(cn) disp(’ ’) disp(’Expansion coefficients (approximating pulses not normalized):’) disp(a) disp(’ ’) x_approx = zeros(size(t)). 0. x_approx). axis([0 t_max -inf inf]). UL). end disp(’Mean-squared error:’) disp(MSE) disp(’ ’) if type_wave == 1 plot(t.cn*a*a’.num2str(tau)]) elseif type_wave == 2 plot(t. LL. for n = 1:n_max x_approx = x_approx + a(n)*phi(t/tau. x_a_p_p_r_o_x(t)’) . n_max = floor(t_max/tau) for n = 1:n_max LL = (n-1)*tau. SIGNAL AND LINEAR SYSTEM THEORY cn = tau. t_max) . ’. ylabel(’x(t). axis([0 t_max -inf inf]). LL. xlabel(’t’). 3 % ce2_4. COMPUTER EXERCISES hold plot(t. full-rectified sinewave.2. 3 = square. del_t = 0. sine. % and triangular waveforms % clf I_wave = input(’Enter type of waveform: 1 = positive squarewave. % Decaying exponential squared function for ce_1 % function z = integrand_exp2(t) z = exp(-2*t). A typical run follows: >> ce2_2 Enter type of waveform: 1 = HR sine. ’.001. 3 = half-rect.’] with contiguous rectangular pulses of width ’. t = 0:del_t:T. % Sin^2 function for ce_1 % function z = integrand_sine2(t) z = (sin(2*pi*t)). 2 = 0-dc level triangular. .m: FFT plotting of line spectra for half-rectified. 4 = triangle: 1 39 Computer Exercise 2. 4 = full-wave sine ’). sin(2*pi*t)) title([’Approximation of sin(2*pi*t) over [0. 2 = FR sine. square wave.2. T = 2.^2. % Sine function for ce_1 % function z = integrand_sine(t) z = sin(2*pi*t).num2str(tau)]) end % Decaying exponential function for ce_1 % function z = integrand_exp(t) z = exp(-t).num2str(t_max). 1 level squarewave’) elseif I_wave == 2 x = 2*trgl_fn(2*(t-T/2)/T)-1. SIGNAL AND LINEAR SYSTEM THEORY Figure 2.6: L = length(t). fs = (L-1)/T. X_th(9) = 0. disp(’ ’) disp(’ 0 . n = [0 1 2 3 4 5 6 7 8 9].^2)./(pi^2*n.^2))). X_th(7) = 0. X_th(5) = 0. % Set n = even coeficients to zero (odd indexed because of MATLAB) X_th(3) = 0./(pi*(1-n. if I_wave == 1 x = pls_fn(2*(t-T/4)/T). . del_f = 1/T.*pls_fn(2*(t-T/4)/T). X_th = abs(1. X_th(1) = 0.5*sinc(n/2)). X_th = abs(0. X_th = 4.40 CHAPTER 2. disp(’ ’) disp(’ 0-dc level triangular wave’) elseif I_wave == 3 x = sin(2*pi*t/T). plot(f. disp(’ ’) disp(’ Full-rectified sinewave’) end X = 0.2. X_th(10) = 0.2).5 because of 1/T_0 with T_0 = 2 f = 0:del_f:fs. COMPUTER EXERCISES 41 X_th(2) = 0.. ylabel(’x(t)’) subplot(2.’o’).5*fft(x)*del_t. % Multiply by 0. 4 = full-wave sine 3 Half-rectified sinewave Magnetude of the Fourier coefficients n FFT Theory __________________________ . sine. xlabel(’n’).. Y = abs(X(1:10)). disp(’ ’) disp(’ Half-rectified sinewave’) elseif I_wave == 4 x = abs(sin(2*pi*t/T)). ylabel(’|X_n|’) A typical run follows: >> ce2_3 Enter type of waveform: 1 = positive squarewave.1. X_th(8) = 0./(pi*(1-n.1). X_th = abs(2.^2))).plot(t. disp(’Magnetude of the Fourier coefficients’).axis([0 10 0 1]). 3 = half-rect. X_th(8) = 0.25. disp(’ ’) disp(’ n FFT Theory’). x).2.1. subplot(2. X_th(6) = 0. abs(X). % Set n = odd coeficients to zero (even indexed because of MATLAB) X_th(4) = 0. 2 = 0-dc level triangular. xlabel(’t’). disp(’ __________________________’) disp(’ ’) disp(Z). X_th(4) = 0. % Set n = odd coeficients to zero (even indexed because of MATLAB) X_th(6) = 0. X_th(10) = 0. X_th(2) = 0.. Z = [n’ Y’ X_th’]. ’).0000 6.2501 0.3183 0.42 CHAPTER 2.0212 0 0. Computer Exercise 2. 3 = half-rect.0000 5. per_cent = input(’Enter percent total desired energy: ’).0000 0.m: Finding the energy ratio in a preset bandwidth % I_wave = input(’Enter type of waveform: 1 = positive squarewave.1062 0. tau = input(’Enter pulse width ’).0091 0.0000 0.4 Make the time window long compared with the pulse width.0000 0.0051 0 Computer Exercise 2.0001 0.3183 0.7: 0 1.0000 8.0051 0.0000 7. SIGNAL AND LINEAR SYSTEM THEORY Figure 2.0000 9.1061 0 0. .0001 0.0000 3.0000 2.0091 0 0. sine.0212 0.0000 4. 4 = raised cosine. 2 = triangular.2500 0.5 % ce2_5. *pls_fn((t-tau/2)/tau)). L_test = length(test). t = 0:del_t:T.2.01. ff = f1(1:NN). k = 1. f = []. end 43 . G = G1(1:NN). f = f1(1:NN+1). else tau1 = tau. test = E_W . elseif I_wave == 4 x = abs(sin(pi*t/tau).^2. G = []. elseif I_wave == 3 x = sin(pi*t/tau).per_cent/100. if I_wave == 1 x = pls_fn((t-tau/2)/tau). del_t = 0.*conj(X). E_f = cumsum(G). NN = floor(length(G1)/2).*pls_fn((t-tau/2)/tau). while test(k) <= 0 k = k+1. L = length(t). end B = k*del_f. del_f = 1/T. fs = (L-1)/T. end X = fft(x)*del_t. f1 = 0:del_f*tau:fs*tau. E_tot = sum(G). elseif I_wave == 2 x = trgl_fn(2*(t-tau/2)/tau). G1 = X. if I_wave == 2 tau1 = tau/2.2. E_W = [0 E_f]/E_tot. COMPUTER EXERCISES clf T = 20. 7 Use Computer Example 2.4) A typical run follows: >> ce2_5 Enter type of waveform: 1 = positive squarewave. 4 = raised cosine.6 The program for this exercise is similar to that for Computer Exercise 2. E_W.2]) if I_wave == 1 title([’Energy containment bandwidth for square pulse of width ’.5. 2 = triangular.2” should be “Computer Example 2.2). except that the waveform is used in the energy calculation. axis([0 2 0 1. abs(G.1.2”). ’ bandwidth % X (pulse width) = ’. ylabel(’G’).semilogy(ff*tau1.plot(f*tau1.’ seconds’]) elseif I_wave == 4 title([’Energy containment bandwidth for raised cosine pulse of width ’. x).1)./max(G))).1. Computer Exercise 2.’—’). num2str(tau).2 as a pattern for the solution (note that in earlier printings of the book “Computer Excercise 2. axis([0 4 0 1.44 CHAPTER 2. ylabel(’x(t)’).2]) legend([num2str(per_cent). ylabel(’E_W’). 3 = half-rect. num2str(tau).. axis([0 10 -inf 1]) subplot(3. xlabel(’f\tau’). num2str(tau). SIGNAL AND LINEAR SYSTEM THEORY subplot(3. num2str(B*tau)].1. xlabel(’f\tau’). sine.plot(t/tau. .’ seconds’]) elseif I_wave == 2 title([’Energy containment bandwidth for triangular pulse of width ’. xlabel(’t/\tau’). ..3).’ seconds’]) end subplot(3. num2str(tau). 3 Enter pulse width 2 Enter percent total desired energy: 95 Computer Exercise 2.’ seconds’]) elseif I_wave == 3 title([’Energy containment bandwidth for half-sine pulse of width ’. 2. COMPUTER EXERCISES 45 Figure 2.8: .2. in general. the demodulated output becomes yD (t) = Ac m (t) cos [θ0 − φ0 ] Letting Ac = 1 gives the error ε (t) = m (t) [1 − cos (θ0 − φ0 )] The mean-square error is E 2 ® D 2 ε (t) = m (t) [1 − cos (θ0 − φ0 )]2 1 .1 The demodulated output.Chapter 3 Basic Modulation Techniques 3. With xc (t) = Ac m (t) cos [ω c t + φ0 ] the demodulated output becomes yD (t) = Lp {2Ac m (t) cos [ω c t + φ0 ] cos [ω c t + θ (t)]} Performing the indicated multiplication and taking the lowpass portion yields yD (t) = Ac m (t) cos [θ (t) − φ0 ] If θ(t) = θ0 (a constant). is yD (t) = Lp{xc (t) 2 cos[ω c t + θ (t)]} where Lp {•} denotes the lowpass portion of the argument.1 Problems Problem 3. for convenience. For this case E 2 ® 2 ®D ε (t) = m (t) [1 − cos (ω 0 t − φ0 )]2 Problem 3. and the error is zero. Taking the average over an integer number of periods yields E D ® = 1 − 2 cos (ω 0 t − φ0 ) + cos2 (ω 0 t − φ0 ) [1 − cos (ω 0 t − φ0 )]2 = 1+ 3 1 = 2 2 Thus 2 ® 3 2 ® m (t) ε (t) = 2 In many cases. we have 2 ® 2 ® ε (t) = m (t) [1 − cos (θ0 − φ0 )]2 Note that for θ0 = φ0 . the average of a product is the product of the averages. (We will say more about this in Chapters 4 and 5).2 Multiplying the AM signal xc (t) = Ac [1 + amn (t)] cos ω c t by xc (t) = Ac [1 + amn (t)] cos ω c t and lowpass Þltering to remove the double frequency (2ω c ) term yields yD (t) = Ac [1 + amn (t)] cos θ (t) . the demodulation carrier is phase coherent with the original modulation carrier. BASIC MODULATION TECHNIQUES where h·i denotes the time-average value. Since the term [1 − cos (θ0 − φ0 )] is a constant. gives the error ε (t) = m (t) [1 − cos (ω 0 t − φ0 )] giving the mean-square error E 2 ® D 2 ε (t) = m (t) [1 − cos (ω 0 t − φ0 )]2 Note that 1 − cos (ω 0 t − φ0 ) is periodic.2 CHAPTER 3. For θ (t) = ω0 t we have the demodulated output yD (t) = Ac m (t) cos (ω0 t − φ0 ) Letting Ac = 1. 5 By inspection. the normalized message signal is as shown in Figure 3.3 A full-wave rectiÞer takes the form shown in Figure 3.1: For negligible demodulation phase error. The waveforms are shown in Figure 3.4 Part a b c 2 ® mn (t) 1/3 1/3 1 a = 0.2.1.7% Ef f = 10.1% Ef f = 13.3.8% a = 0. 1% Ef f = 5. Thus 2 T 0≤t≤ mn (t) = t. It is clear that the full-wave rectiÞed xc (t) deÞnes the message better than the half-wave rectiÞed xc (t) since the carrier frequency is effectively doubled. This process is not generally used in AM since the reason for using AM is to avoid the necessity for coherent demodulation. m (t).1.7% Ef f = 26. this becomes yD (t) = Ac + Ac amn (t) The dc component can be removed resulting in Ac amn (t). with the half-wave rectiÞer on top and the full-wave rectiÞer on the bottom.5% a=1 Ef f = 25% Ef f = 25% Ef f = 50% Problem 3. PROBLEMS 3 xC ( t ) yD ( t ) R C Figure 3.6 Ef f = 10. Problem 3.3.3(b) in the text. Problem 3. T 2 . Decreasing exponentials can be drawn from the peaks of the waveform as depicted in Figure 3. which is a signal proportional to the message. The message signal is the envelopes.4 Ef f = 5. θ(t) ≈ 0. 1 0.2 0.2: mn ( t ) 1 0 T -1 t Figure 3.5 1 0.3 0.5 0 0 0.5 0.6 0.1 0.7 0.7 0.2 0.8 0.9 1 0.9 1 Figure 3.5 0.5 1 0.6 0.4 0.3: .5 0 0 2 1.8 0.3 0.4 CHAPTER 3.4 0. BASIC MODULATION TECHNIQUES 2 1. 6)2 1 3 ¡ ¢ + (0.5 Watts c 2 2 The efficiency is Ef f = Thus (0.1.107 Pc + Psb where Psb represents the power in the sidebands and Pc represents the power in the carrier.48 Watts 1.6 Since the index is 0.107 = 10.107 + 0.7% 3.107Psb This gives Psb = 0. The above expression can be written Psb = 0.0 − 0. PROBLEMS and Z T /2 µ 5 ¶ µ ¶ µ ¶ 2 2 2 2 21 T 3 1 = t dt = T T T 3 2 3 Also 2 ® 2 mn (t) = T 0 Ac [1 + a] = 40 Ac [1 − a] = 10 This yields or 1+a 40 = =4 1−a 10 1 + a = 4 − 4a 5a = 3 Thus a = 0.6 ¡1¢ = 0.6] = 40 This gives Ac = This carrier power is 1 1 Pc = A2 = (25)2 = 312.107 Pc = 97. we can write Ac [1 + 0.36 Psb = 0.6)2 1 3 40 = 25 1.6.3.36 = 0.107 . and the minimum falls at t = 0. Thus the normalized message signal is mn (t) = 1 [9 cos 20πt − 7 cos 60πt] 11.9523 With the given value of c (t) and the index a.9523. 6 mn (t) = m (t) and Ef f = 81.5)2 (0.77% T . 6 1 mn (t) = m (t) 5 and Ef f = 14. we have xc (t) = 100 [1 + 0.0352 and t = 0. BASIC MODULATION TECHNIQUES Problem 3.455 2 (c)This gives the efficiency E= (0.6415 cos 140πt + 18.25% Problem 3.6 For the Þrst signal τ= and for the second signal τ= 5T . or use a root-Þnding algorithm in order to determine the minimum value of m (t).7 (a) The Þrst step in the solution to this problem is to plot m (t). ® (b) The value of m2 (t) is n 2 ® mn (t) = µ 1 11. We Þnd that the minimum of m (t) = −11.5)2 (0.9523 ¶2 µ ¶ h i 1 (9)2 + (7)2 = 0.6415 cos 260πt We will need thislater to plot the spectrum.6 CHAPTER 3.5mn (t)] cos 200πt This yields xc (t) = −14.0648.8248 cos 220πt − 14.455) .8428 cos 180πt +100 cos 200πt +18.213% 1 + (0.455) = 10. 8248 = 9.4: (d) The two-sided amplitude spectrum is shown in Figure 3.4615 = 7. (e) By inspection the signal is of the form xc (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass.3.2307 A= 2 and 18.1.8 . This the envelope is q e(t) = 100 a2 (t) cos2 200πt + a2 (t) sin2 200πt = 100a(t) a(t) = [1 + 0. PROBLEMS 7 50 50 A B B A A B B A f -130 –110 –100 –90 -70 70 90 100 110 130 Figure 3.5mn (t)] Problem 3.4. where 14. Thus the Hilbert transform of c(t) is c xc (t) = a(t)b(t) b where in which c(t) = 100 cos 200πt. The spectra of a(t) and c(t) are do not overlap.4124 B= 2 The phase spectrum results by noting that A is negative and all other terms are positive. 03125 2 (e) As in the previous problem.0625) = 7.969% (d) The expression for xc (t) is · ¸ 1 (9 cos 20πt + cos 60πt) cos 200πt xc (t) = 100 1 + 32 = 10.46875 2 1 (14. This the envelope is q e(t) = 100 a2 (t) cos2 200πt + a2 (t) sin2 200πt ¸ · 1 (9 cos 20πt + cos 60πt) = 100a(t) = 100 1 + 32 Problem 3.25(0. Thus the Hilbert transform of c(t) is c xc (t) = a(t)b(t) b where c(t) = 100 cos 200πt.05969 = 5.2539) = 0.25(0.8 CHAPTER 3.9 The modulator output xc (t) = 40 cos 2π (200) t + 4 cos 2π (180) t + 4 cos 2π (220) t can be written xc (t) = [40 + 8 cos 2π (20) t] cos 2π (200) t . the signal is of the form xc (t) = a(t)c(t) where a(t) is lowpass and c(t( is highpass.2539) (c) Ef f = 1+0.9375 cos 140πt + 14.2539 2 0. The amplitude spectrum is identical to that shown in the previous problem except that A = B = 1 (10.0625 cos 180πt +100 cos 200πt +14.0625 cos 220πt + 10.9375) = 5. BASIC MODULATION TECHNIQUES 1 (a) mn (t) = 16 [9 cos 20πt + 7 cos 60πt] i 2 ® ¡ 1 ¢2 ¡ 1 ¢ h 2 (b) mn (t) = 16 (9) + (7)2 = 0.9375 cos 260πt Note that all terms are positive so the phase spectrum is everywhere zero. The spectra of a(t) and c(t) are do not overlap. 1.3.96% Pc + Psb 800 + 16 1 1 (4)2 + (4)2 = 16 Watts 2 2 Problem 3.10 A = 14.2 40 9 By inspection.1547 Problem 3. is .11 The modulator output xc (t) = 20 cos 2π (150) t + 6 cos 2π (160) t + 6 cos 2π (140) t is · ¸ 12 cos 2π (10) t cos 2π (150) t xc (t) = 20 1 + 20 a= The carrier power is Pc = 1 (20)2 = 200 Watts 2 12 = 0.14 B = 8. the modulation index is a= Since the component at 200 Hertz represents the carrier. a. PROBLEMS or · ¸ 8 cos 2π (20) t cos 2π (200) t xc (t) = 40 1 + 40 8 = 0. Thus the sideband power is Psb = Thus. the carrier power is Pc = 1 (40)2 = 800 Watts 2 The components at 180 and 220 Hertz are sideband terms.16 a = 1. the modulation index.0196 = 1. the efficiency is Ef f = Psb 16 = = 0.6 20 Thus. 102Ac 0.102Ac 0.432.1020Ac cos 2π (fc − 2fm ) t +0.051Ac 0. BASIC MODULATION TECHNIQUES 1 1 (6)2 + (6)2 = 36 Watts 2 2 36 = 0.1525 200 + 36 Ef f = Problem 3.8%.2040Ac cos 2π (fc − fm ) t (b) The efficiency is 15. Thus mn (t) = 0.5Ac 0. the efficiency is CHAPTER 3.102Ac 0.102Ac 0.051Ac 0.7mn (t)] cos 2πfc t = 0.102Ac 0.2040Ac cos 2π (fc + fm ) t +0.2040Ac cos 2π (fc − 5fm ) t +Ac cos 2πfc t +0.5828 cos (2πfm t) + 0.13 .1020Ac cos 2π (fc + 2fm ) t +0.5Ac 0.102Ac +0.102Ac 0. It contains 14 discrete components as shown Comp 1 2 3 4 5 6 7 Freq −fc − 5fm −fc − 2fm −fc − fm −fc −fc + fm −fc + 2fm −fc + 5fm Amp 0.2040Ac cos 2π (fc + 5fm ) t The spectrum is drawn from the expression for xc (t).12 (a) By plotting m (t) or by using a root-Þnding algorithm we see that the minimum value of m (t) is M = −3.051Ac 0.2914 cos (4πfm t) + 0.5828 cos (10πfm t) The AM signal is xc (t) = Ac [1 + 0. Problem 3.10 and the sideband power is Psb = Thus.102Ac Comp 8 9 10 11 12 13 14 Freq fc − 5fm fc − 2fm fc − fm fc fc + fm fc + 2fm fc + 5fm Amp 0.051Ac 0. PROBLEMS 11 ℑ{m ( t )} ℑ m2 (t ) { } Filter Characteristic 0 W 2W fc − W fc fc + W 2 fc f Figure 3.75 x (t) = m (t) + cos ω c t With the given relationship between x (t) and y (t) we can write y (t) = 4 {m (t) + cos ω c t} + 10 {m (t) + cos ω c t}2 which can be written y (t) = 4m (t) + 4 cos ω c t + 10m2 (t) + 20m (t) cos ω c t + 5 + 5 cos 2ω c t The equation for y (t) is more conveniently expressed y (t) = 5 + 4m (t) + 10m2 (t) + 4 [1 + 5m (t)] cos ω c t + 5 cos 2ωc t (b) The spectrum illustrating the terms involved in y (t) is shown in Figure 3. and fc + W < 2fc .5. fc − W > 2W or fc > 3W .1. (c) From the deÞnition of m (t) we have m (t) = Mmn (t) so that g (t) = 4 [1 + 5Mmn (t)] cos ω c t It follows that a = 0.3. The center frequency of the Þlter is fc and the bandwidth must be greater than or equal to 2W . In addition.8 = 5M .5: (a) From Figure 3. The last inequality states that fc > W . which is redundant since we know that fc > 3W . BASIC MODULATION TECHNIQUES 0. θ1 must equal−φ. the modulator output is a sinusoid of frequency fc + fm . yD (t) must be of the form m (t − τ ).8 = 0. while for the LSB SSB case.16 5 (d) This method of forming a DSB signal avoids the need for a multiplier. the modulator output is a sinusoid of frequency of fc − fm .12 Thus CHAPTER 3.15 For the USB SSB case. This gives yD (t) = A cos (ω 1 t + φ) + B cos (ω 2 t + θ2 ) which we can write µ ¶ µ ¶ θ1 θ2 + B cos ω 2 t + yD (t) = A cos ω1 t + ω1 ω2 θ1 θ2 = ω1 ω2 For no distortion.13 and the phases given in the problem statement. Problem 3.16 Using Figure 3. Thus . M= Problem 3. the modulator output becomes xc (t) = Aε cos [(ω c − ω1 ) t + φ] 2 A (1 − ε) cos [(ω c + ω1 ) t + θ 1 ] + 2 B + cos [(ω c + ω 2 ) t + θ2 ] 2 Multiplying xc (t) by 4 cos ω c t and lowpass Þltering yields the demodulator output yD (t) = Aε cos (ω 1 − φ) + A (1 − ε) cos (ω 1 t + θ1 ) + B cos (ω 2 t + θ2 ) For the sum of the Þrst two terms to equal the desired output with perhaps a time delay.14 1 sgn (f + fc ) 2 1 1 xU SB (t) = Ac m (t) cos ω c t − Ac m (t) sin ω c t b 2 2 HU (f ) = 1 − Problem 3. The fact that φ = −θ1 tells us that the Þlter phase response must have odd symmetry about the carrier frequency. Thus y (t) = xc (t) + K cos ω c t It can be shown that y (t) can be written y (t) = y1 (t) cos ω c (t) + y2 (t) sin ω c t where y1 (t) = A B cos ω 1 t + cos ω 2 t + K 2 µ ¶ 2 A B Aε + sin ω 1 t − sin ω 2 t y2 (t) = 2 2 y (t) = R (t) cos (ω c t + θ) where R (t) is the envelope and is therefore the output of an envelope detector. which is 1 m (t) + K. 2 Problem 3. θ2 = Problem 3.3. the phase must be linear.18 The required Þgure appears in Figure 3.19 . It follows that q 2 2 R (t) = y1 (t) + y2 (t) Also For K large. K is removed and the output y (t) is m (t) scaled by 1 .6. R (t) = |y1 (t)|. Thus if the detector 2 is ac coupled. PROBLEMS so that 13 ω2 θ1 ω1 Thus.1. where K is a dc bias.17 We assume that the VSB waveform is given by xc (t) = 1 Aε cos (ω c − ω 1 ) t 2 1 + A (1 − ε) cos (ω c + ω 1 ) t 2 1 + B cos (ω c + ω 2 ) t 2 We let y (t) be xc (t) plus a carrier. Problem 3. BASIC MODULATION TECHNIQUES Desired Signal ω1 Local Oscillator ω ω1 − ω 2 Signal at Mixer Output ω ω 2 − ω IF Image Signal 2ω1 − ω 2 ω ω 2ω1 − ω 2 Image Signal at Mixer Output ω 3ω 2 − 2ω1 ω2 Figure 3.6: .14 CHAPTER 3. 5 0.0 R 4.MHz 0.08 3.21 For high-side tuning fLO = fi + fIF = 1120 + 2500 = 3620 kHz fIM AGE = fi + 2fIF = 1120 + 5000 = 6120 kHz For low-side tuning fLO = fi − fIF = 1120 − 2500 = −1380 kHz fLO = 1380 kHz fIM AGE = fi − 2fIF = 1120 − 5000 = −3880 kHz fLO = 3880 kHz .20 For high-side tuning we have fLO = fi + fIF = 1120 + 455 = 1575 kHz fIM AGE = fi + 2fIF = 1120 + 910 = 2030 kHz For low-side tuning we have fLO = fi − fIF = 1120 − 455 = 665 kHz fIM AGE = fi − 2fIF = 1120 − 910 = 210 kHz Problem 3.7 1.70 4. We make the following table fIF . the carrier frequency of the input signal. PROBLEMS Since high-side tuning is used.33 4. the local oscillator frequency is fLO = fi + fIF 15 where fi .4 0.1.86 A plot of R as a function of fIF is the required plot.51 4.3.5 2. varies between 5 and 25 M Hz.63 4.0 1. Problem 3. The ratio is fIF + 25 R= fIF + 5 where fIF is the IF frequency expressed in MHz. 6 0.23 Let φ (t) = β cos ω m t where ωm = 2πfm .6 1.8 2 0 -1 0 1 0.6 1. This gives o n xc2 (t) = Ac Re ejω c t ejβ cos ωm t Expanding a Fourier series gives ejβ cos ωm t = ∞ X Cn ejnω m t n=−∞ .4 1. the middle pane is for kp = −π/2.2 0.8 1 1.8 1 1.2 1. The top pane is for kp = π.4 1.6 0.8 1 1. recall that the spectra are symmetrical about f = 0.6 0.4 0.4 0.4 1. BASIC MODULATION TECHNIQUES 1 0 -1 0 1 0.2 1.7.6 1. and the bottom pane is for kp = π/4.8 2 Figure 3. Problem 3. Problem 3.7: In the preceding development for low-side tuning.8 2 0 -1 0 0.22 By deÞnition xc (t) = Ac cos [ω c t + kp m (t)] = Ac cos [ω c t + kp u(t − t0 )] The waveforms for the three values of kp are shown in Figure 3.4 0.2 0.2 1.2 0.16 CHAPTER 3. the preceding becomes 2 1 Cn = 2π This gives Cn = e j nπ 2 With x = ω m t.24 Since sin(x) = cos(x − π ) we can write 2 which is ³ ´ π xc3 (t) = Ac sin(ω c t + β sin ωm t) = Ac cos ωc t − + β sin ω m t 2 o n xc3 (t) = Ac Re ej(ωc t−π/2) ej sin ωm t . The phase spectrum is the same as in the preceding problem except that nπ is added to each term. the Fourier coefficients become Z π 1 ejβ cos x e−jnx dx Cn = 2π −π ¢ ¡ Since cos x = sin x + π 2 Z π π 1 ej [β sin(x+ 2 )−nx] dx Cn = 2π −π Z 3π/2 ej [β sin y−ny+n 2 ] dy ¾ π −π/2 where the limits have been adjusted by recognizing that the integrand is periodic with period 2π. 2 Problem 3.1. Thus nπ Cn = ej 2 Jn (β) and xc2 (t) = Ac Re ejωc t Taking the real part yields xc2 (t) = Ac ∞ X ½ 1 2π Z π e j[β sin y−ny] dy −π ( n=−∞ ∞ X Jn (β) ej ( nω m t+ nπ 2 ) ) h nπ i Jn (β) cos (ω c + nω m ) t + 2 n=−∞ The amplitude spectrum is therefore the same as in the preceding problem. PROBLEMS where ωm Cn = 2π Z 17 π/ωm ejβ cos ω m t e−jnωm t dt −π/ω m With x = x + π .3. we see that the only difference is in the phase spectrum. xc2 (t) and xc3 (t). The phase spectrum of xc3 (t) is formed from the spectrum of xc1 (t) by adding −π/2 to each term.25 From the problem statement xc (t) = Ac cos [2π (40) t + 10 sin (2π (5) t)] .18 Since CHAPTER 3. BASIC MODULATION TECHNIQUES ej sin ωm t = we have ( n=−∞ ∞ X ∞ X Jn (β)ejnωm t ) xc3 (t) = Ac Re e Thus xc3 (t) = Ac Thus xc3 (t) = Ac ∞ X j(ωc t−π/2) Jn (β)e jnωm t n=−∞ n=−∞ ∞ X o n Jn (β) Re ej(ωc t+nωm t−π/2) h πi Jn (β) cos (ω c + nωm ) t − 2 n=−∞ Note that the amplitude spectrum of xc3 (t) is identical to the amplitude spectrum for both xc1 (t) and xc2 (t). Problem 3. For xc4 (t) we write ³ ´ π xc4 (t) = Ac sin(ω c t + β cos ωm t) = Ac cos ωc t − + β cos ω m t 2 Using the result of the preceding problem we write ( ) ∞ X j (nω m t+ nπ ) j(ωc t−π/2) 2 Jn (β) e xc4 (t) = Ac Re e n=−∞ This gives xc4 (t) = Ac Thus xc4 (t) = Ac n=−∞ ∞ X ∞ X o n π nπ Jn (β) Re ej(ωc t+nω m t− 2 + 2 ) i h π Jn (β) cos (ω c + nωm ) t + (n − 1) 2 n=−∞ Compared to xc1 (t). 3391) + 0.26 We are given J0 (3) = −0.4862 3 2n Jn (β) − Jn−1 (β) β . Using Jn+1 (β) = with β = 3 we have 2 jn+1 (3) = nJn (3) − Jn−1 (3) 3 With n = 1. Where terms overlap. The fact that xc (t) has a nonzero dc value is obvious.8036 The waveform xc (t) is illustrated in the top pane of Figure 3.101). The terms resulting from the portion of the spectrum centered at f = 40 are denoted S40 and the terms resulting from the portion of the spectrum centered at f = −40 are denoted S−40 .1 gives the total power 0.2601 = 0. Problem 3.8. With the exception of the dc term.2601 and J1 (3) = −0. In developing the Table 3.3391.1 for f ≤ 0 (positive frequency terms and the term at dc).1 be sure to remember that J−n (β) = −Jn (β) for odd n. normalized with respect to Ac . Given the percision of the Table of Bessel coefficients (Page 131 in the textbook). The sum is denoted ST in Table 3. are given in Table 3. J2 (3) = 2 J1 (3) − J0 (3) 3 2 = (0.1. Since fc = 40 and fm = 5 (it is important to note that fc is an integer multiple of fm ) there is considerable overlap of the portion of the spectrum centered about f = −40 with the portion of the spectrum centered about f = 40 for large β (such as 10). we have overlap for |f| ≤ 45.0552 0.1.8036A2 = 40 c which is Ac = r 40 = 7.3. The power at dc is ST A2 = (0. PROBLEMS 19 The spectrum is determined using (3. c c Carring out these operations with the aid of Table 3.636)2 A2 . The terms in the spectral plot. they must be summed. normalized with respect to Ac . The spectrum. 2 the power at each frequency is ST A2 /2 (the power for negative frequencies is equal to the c power at positive frequencies and so the positive frequency power can simply be doubled to 2 give the total power with the dc term excepted). is illustrated in the bottom frame. The power must now be determined in order to Þnd Ac . 002 0.075 0.043 −0.357 −0.234 −0.029 0.318 ST 0.014 −0.001 0.001 0.220 0.005 0.123 0.001 0.002 0.044 −0.207 0.255 0.002 0.318 0.012 0.058 0.005 0.063 0.005 0.292 0.255 −0. BASIC MODULATION TECHNIQUES Table 3.157 0.058 −0.292 0.217 0.246 −0.207 0.318 0.244 −0.063 0.014 −0.20 CHAPTER 3.012 0.029 0.014 −0.207 0.058 0.234 −0.038 0.123 0.123 0.063 0.220 0.193 0.029 0.012 0.220 0.219 −0.318 S−40 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.029 −0.292 0.1: Data for Problem 3.234 −0.219 −0.043 0.636 .255 0.067 −0.25 f 125 120 115 110 105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 0 S40 0. .1 Time 0.2 0.5 -1 0 0.06 0. J3 (3) = = Finally. The magnitude spectrum is plotted assuming Ac = 1.2 0 -150 -100 -50 0 Frequency 50 100 150 Figure 3.5 x c (t) 0 -0.8: With n = 2.14 0.04 0.1.8 0. PROBLEMS 21 1 0. with n = 3 we have J4 (3) = 2J3 (3) − J2 (3) = 2 (0.02 0.9.3391 = 0.4 0.3091) + 0.3.18 0.3091 3 Problem 3.1320 4 J2 (3) − J1 (3) 3 4 (0. The single-sided magnitude and phase spectra are shown in Figure 3.4862 = 0.27 The amplitude and phase spectra follow directly from the table of Fourier-Bessel coefficients.08 0.16 0.12 0.6 M agnitude 0.04862) + 0. 2 0. BASIC MODULATION TECHNIQUES 0.4 0.22 CHAPTER 3.1 0 600 700 800 900 1000 1100 Frequency 1200 1300 1400 0 -1 P hase -2 -3 -4 600 700 800 900 1000 1100 Frequency 1200 1300 1400 Figure 3.9: .3 M agnitude 0. we can write 2 or where h n io π xc (t) = Re ej200πt 5e−j40πt + 10 + 3ej (40πt− 2 ) © ª xc (t) = Re a (t) ej200πt a (t) = (5 cos 40πt + 10 + 3 sin 40πt) +j (5 sin 40πt − 3 cos 40πt) .1. we can write e xc (t) = [10 + 3 cos 2π (20t) + 5 cos 2π (20) t] e +j [3 sin 2π (20t) − 5 sin 2π (20) t] = [10 + 8 cos 2π (20t)] − j [2 sin 2π (20) t] It follows from the deÞnition of xc (t) that xc (t) = R (t) ejφ(t) e R2 (t) = [10 + 8 cos 2π (20t)]2 + 4 sin2 2π (20) t = 134 + 160 cos 2π (20) t + 30 cos 2π (40) t This gives R (t) = Also p 134 + 160 cos 2π (20) t + 30 cos 2π (40) t φ (t) = tan−1 −2 sin 2π (20) t 10 + 8 cos 2π (20) t Thus Problem 3. PROBLEMS Problem 3. Denoting the complex envelope by xc (t).28 The modulated signal can be written i h xc (t) = Re 10 + 3ej2π(20)t + 5e−j2π(20)t ej2π(100)t 23 We will concentrate on the term in brackets.3.29 ¡ ¢ Since sin x = cos x − π . which is the complex envelope described in Chapter 2. φ (t).30 (a) Since the carrier frequency is 1000 Hertz.24 Thus if we write xc (t) as CHAPTER 3. φ (t). is given by The envelope. can be simpliÞed and expressed in several different forms. is therefore given by φ (t) = 20t2 The frequency deviation is dφ = 40t dt or 1 dφ 20 = t 2π dt π rad/sec Hertz rad (b) The phase deviation is φ (t) = 2π (500) t2 − 2π (1000) t and the frequency deviation is dφ dt or = 4π (500) t − 2π (1000) = 2000π (t − 1) 1 dφ = 1000 (t − 1) 2π dt rad/sec rad/sec rad Hertz . the general form of xc (t) is xc (t) = Ac cos [2π (1000) t + φ (t)] The phase deviation. R (t). is given by i1 h 2 2 2 R (t) = (5 cos 40πt + 10 + 3 sin 40πt) + (5 sin 40πt − 3 cos 40πt) φ (t) = tan−1 5 sin 40πt − 3 cos 40πt 10 + 5 cos 40πt + 3 sin 40πt and the phase deviation. R (t). Problem 3. BASIC MODULATION TECHNIQUES xc (t) = R (t) cos (200πt + φ (t)) the envelope. in Hz. (d) The phase deviation is √ φ (t) = 200πt + 10 t rad and the phase deviation is 1 dφ 5 1 = 200π + (10) t− 2 = 200π + √ dt 2 t rad/sec or 5 1 dφ = 100 + √ 2π dt 2π t Hertz Problem 3.3. t≥4 (b) The frequency deviation.1. 2π dt 0 t<0 0≤t<4 t≥4 . 1 dφ = 30m(t) = 240.31 (a) The phase deviation is φ(t) = 2π(30) Z t (8)dt = 480πt. φ(t) = 480πt. 0≤t<4  1920π. PROBLEMS (c) The phase deviation is φ (t) = 2π (100) t = 200πt and the frequency deviation is dφ = 200π dt or rad/sec rad 25 1 dφ = 100 Hertz 2π dt which should be obvious from the expression for xc (t). is 0. The required plot is simply  t<0  0. 0 t≤4 The maximum phase deviation is φ(4) = 480π (4) = 1920π. 6 ≤ t ≤ 8 φ (t) = 240π. m (t) =  8−t   0. BASIC MODULATION TECHNIQUES The required sketch follows simply.32 (a) The message signal is The phase deviation is   0. 4 ≤ t ≤ 6 φ (t) = 2πfd ¡ ¢ = 120π + 60π (8) (t − 6) − 30π t2 − 36 ¡ ¢ = 120π − 30π t2 − 16t + 60 . Z t 4 t<4 4≤t≤6 6≤t≤8 t>8 φ (t) = 2πfd ¡ ¢ = 60π (−4) (t − 4) + 30π t2 − 16 ¢ ¡ = 30π t2 − 8t + 16 . (b) The frequency deviation in Hertz is 30m(t). (c) The peak frequency deviation is 8fd = 8 (30) = 240 Hertz (d) The peak phase deviation is Z 4 (8) dt = 2π (30) (8) (4) = 1920π 2π (30) 0 rad The modulator output power is 1 1 P = A2 = (100)2 = 5000 c 2 2 Watts Problem 3. t < 4 The sketches follow immediately from the equations.   t − 4. (d) The peak frequency deviation = 120 Hertz. (c) The peak phase deviation = 240π rad. t > 8 φ (t) = 0. Z t 6 (t − 4) dt (8 − t) dt + φ (6) Also .26 CHAPTER 3. assuming a sufficiently high carrier frequency. 3.76 with the ordinate values multiplied by 25. we have φ (t) = 20π Z t Z 3 αdα = 10πt2 0 . PROBLEMS (e) The carrier power is. 1 1 P = A2 = (100)2 = 5000 2 c 2 Watts 27 Problem 3. for t > 4 we recognize that φ (t) = φ (4) = 475π.34 The frequency deviation in Hertz is the plot shown in Fig.77 with the ordinate values multiplied by 10.The phase deviation in radians is given Z t Z t m (α) dα = 50π m (α) dα φ (t) = 2πfd For 0 ≤ t ≤ 1. Problem 3.33 The frequency deviation in Hertz is the plot shown in Fig. 3.3. The phase deviation is given by Z t Z t m (α) dα = 20π m (α) dα φ (t) = 2πfd For 0 ≤ t ≤ 1. we have φ (t) = 50π For 1 ≤ t ≤ 2 φ (t) = φ(1) + 50π Z t 2αdα = 50πt2 0 = −175π + 250πt − 25πt2 For 2 ≤ t ≤ 3 φ (t) = φ(2) + 50π For 3 ≤ t ≤ 4 φ (t) = φ(3) + 50π Z t Z t 1 ¡ ¢ (5 − α) dα = 50π + 250π (t − 1) − 25π t2 − 1 2 t 3dα = 225π + 150π (t − 2) 2dα = 375π + 100π(t − 3) Finally. The required Þgure results by plotting these curves.1. 5 ≤ t ≤ 3 Z t 2 ¡ ¢ (10 − 4α)dα = 10π + 10π (10) (t − 2) − 10π(2) t2 − 4 φ (t) = φ(2. BASIC MODULATION TECHNIQUES For 2 ≤ t ≤ 4 t ¡ ¢ (α − 2) dα = 10π + 10π t2 − 1 − 40π(t − 1) φ (t) = φ(1) + 20π 1 ¢ ¡2 = 10π t − 4t + 4 = 10π (t − 2)2 Z φ (t) = φ(2) + 20π 2 = −20π(t − 6t + 8) Z t 2 ¡ ¢ (6 − 2α)dα = 0 + 20π (6) (t − 2) − 20π t2 − 4 Finally. Problem 3.5) Z t 2.78 with the ordinate values multiplied by 5. 3.5) − 10π = 10π(−2t + 5. we have φ (t) = 10π For 1 ≤ t ≤ 2 φ (t) = φ(1) + 10π For 2 ≤ t ≤ 2.5) . for t > 4 we recognize that φ (t) = φ(4) = 0.5 2dα = 15π − 20π(t − 2. The required Þgure follows by plotting these expressions. The phase deviation in radians is given by φ (t) = 2πfd For 0 ≤ t ≤ 1.5 φ (t) = φ(2) + 10π 2 Z t m (α) dα = 10π Z Z t m (α) dα t 0 (−2α)dα = −10πt2 Z t 1 2dα = −10π + 20π (t − 1) = 10π (2t − 3) = 10π(−2t + 10t − 11) For 2.35 The frequency deviation in Hertz is the plot shown in Fig.28 For 1 ≤ t ≤ 2 CHAPTER 3. the modulation index is 50 10 = 5. for t > 4 we recognize that φ (t) = φ(4) = −5π. (c) Since β is not ¿ 1.see Problem 3.04 0. The sketch is that of Figure 3.25.3.37 The results are given in the following table: Part a b c d fd 20 200 2000 20000 D = 5fd /W 0. kp (4) = 5 or kp = 1.1. (b) The magnitude spectrum is a Fourier-Bessel spectrum with β = 5. PROBLEMS For 3 ≤ t ≤ 4 φ (t) = φ(3) + 10π 29 = 10π(t2 − 8t + 15.4 4 B = 2 (D + 1) W 50. The required Þgure follows by plotting these expressions.38 From xc (t) = Ac we obtain n=−∞ ∞ 2 ® 1 2 X 2 xc (t) = Ac J (β) 2 n=−∞ n ∞ X Jn (β) cos (ω c + ω m ) t We also know that (assuming that xc (t) does not have a signiÞcant dc component .2 kHz 52 kHz 70 kHz 250 kHz Z t 3 (2α − 8) dα = 5π + 10π(t2 − 9) − 10π(8)(t − 3) Problem 3.36 (a) The peak deviation is (12. Thus. (d) For phase modulation.24 in the text. The n = 0 term falls at 1000 Hz and the spacing between components is 10 Hz. Problem 3.004 0.24) ® 2 ® 2 xc (t) = Ac cos2 [ω c t + φ (t)] . Problem 3. this is not narrowband FM.5) Finally. The bandwidth exceeds 2fm .5)(4) = 50 and fm = 10. we have Z 1 π Jn (β) = cos [β sin (π − λ) − n (π − λ)] (−1) dλ π 0 Z 1 π = cos [β sin (π − λ) − nπ + nλ] dλ π 0 Since sin (π − λ) = sin λ. is 2 ® 1 2 xc (t) = Ac 2 ∞ 1 2 1 2 X 2 Ac = Ac J (β) 2 2 n=−∞ n ∞ X 2 Jn (β) = 1 This gives from which n=−∞ Problem 3. we can write Z 1 π Jn (β) = cos [β sin λ + nλ − nπ] dλ π 0 .39 Since 1 Jn (β) = 2π Z π we can write 1 Jn (β) = 2π Z π e −j(nx−β sin x) −π 1 dx = 2π Z Z π ej(β sin x−nx) dx −π π 1 cos (β sin x − nx) dx + j 2π −π −π sin (β sin x − nx) dx The imaginary part of Jn (β) is zero. Thus Z π 1 cos (β sin x − nx) dx Jn (β) = 2π −π Since the integrand is even Jn (β) = 1 π Z π 0 cos (β sin x − nx) dx which is the Þrst required result. With the change of variables λ = π − x. BASIC MODULATION TECHNIQUES which. assuming that ω c À 1 so that xc (t) has no dc component.30 CHAPTER 3. since the integrand is an odd function of x and the limits (−π. π) are even. The output spectrum consistes of the n = 0 term and three terms each side of the n = 0 term. the second integral in the preceding expression is zero. Thus the output spectrum has terms at 470.1. P0 = 16.76 Watts (e) The spectrum of the input signal is a Fourier_Bessel spectrum with β = 8. 480. The n = 0 term is at the carrier frequency of 500 Hz and the spacing between components is 10 Hz. 520 and 530 Hz.3.40 (a) Peak frequency deviation = 80 Hz (b) φ (t) = 8 sin (20πt) (c) β = 8 (d) Pi = 50 Watts. Also cos (nπ) = (−1)n Thus 1 Jn (β) = (−1) π n However Z π π cos [β sin λ + nλ] dλ 0 1 J−n (β) = π Thus or equivalently Z cos [β sin λ + nλ] dλ 0 Jn (β) = (−1)n J−n (β) J−n (β) = (−1)n Jn (β) Problem 3. . 500. 490. 510. PROBLEMS Using the identity cos (u − ν) = cos u cos ν + sin u sin ν with u = β sin λ + nλ and ν = nπ yields Jn (β) = 1 π + Z π 31 cos [β sin λ + nλ] cos (nπ) dλ Z π 0 1 π sin [β sin λ + nλ] sin (nπ) dλ 0 Since sin (nπ) = 0 for all n. 10: . BASIC MODULATION TECHNIQUES Figure 3.32 CHAPTER 3. 42 We wish to Þnd k such that 2 Pr = J0 (10) + 2 k X 2 J0 (10) ≥ 0. The modulation indices are.05 20 = 400 0. we have fc1 = 110 This gives n= and fc1 = n (100) kHz = 44 MHz The two permissible local oscillator frequencies are f`0. β = 5. and β = 10.80 n=1 This gives k = 9.1. from top to bottom.2 = 100 + 44 = 144 MHz The center frequency of the bandpass Þlter must be fc = 100 MHz and the bandwidth is ¡ ¢ B = 2 (D + 1) W = 2 (20 + 1) (10) 103 or B = 420 kHz kHz fd1 = 0. Problem 3.41 The required spectra are given in Figure 3. The bandwidth is therefore B = 2kfm = 2 (9) (150) = 2700 Hz For Pr ≥ 0.9603. we have k = 10 for a power ratio of 0. β = 2.5.8747. yielding a power ratio of Pr = 0. β = 1. β = 0. PROBLEMS 33 Problem 3.05) = 20 .3.10.1 = 100 − 44 = 56 MHz f`0.43 From the given data.9. This gives B = 2kfm = 2 (10) (150) = 3000 Hz Problem 3.05 fd2 = n (0. 46 ¢ ¡ = 1. With these values. the discriminator constant is approximately ¢ ¡ KD ≈ 8.45 We can solve this problem by determining the peak of the amplitude response characteristic. The result is ¢ ¡ KD ∼ 8 10−6 = Problem 3. This gives L= 1 2 (2π) fp C 2 We Þnd the value of R by trial and error using plots of the amplitude response.001 10−12 . An appropriate value for R is found to be 1 MΩ. R 103 ¢ ¡ ¡ ¢ = RC = 103 10−9 = 10−6 The slope of the operating characteristic at the operating point is measured from the amplitude response. Thus an appropriate carrier frequency is fc = 118 + 54 = 86 2 kHz 10−3 L = = 10−6 .44 For the circuit shown H (f ) = or CHAPTER 3.5 10−9 Problem 3. Let fp = 150 MHz and let C = 0. This peak falls at 1 fp = √ 2π LC ¢ ¡ It is clear that fp > 100 MHz. BASIC MODULATION TECHNIQUES R E (f) = X (f ) R + j2πfL + 1 ³ 1 + j 2πfτ L − 1 j2πf C H (f) = where τL = τC 1 2πf τ C ´ A plot of the amplitude response shows that the linear region extends from approximately 54 kHz to118 kHz.126 10−3 .34 Problem 3. φ (t) = kf t≥0 . from (3.5. for Ai À Ac we see from the phasor diagram that ψ (t) ≈ θ (t) = ω i t and yD (t) = KD d (2πfi t) = fi 2π dt h ³ π x í x π − = − ψ(t) = tan−1 − tan 2 2 2 2 µ ¶ 1 d 2πfi t π 1 yD (t) = − = fi 2π dt 2 2 2 ³π x´ sin x x = cot = tan − 1 − cos x 2 2 2 Problem 3.3.47 From Example 3. xr (t) = Ac [cos ω c t + cos (ω c + ω i ) t] which is xr (t) = Ac [(1 + cos ω i t) cos ω c t − sin ω i t sin ω c t] This yields xr (t) = R (t) cos [ω c t + ψ (t)] where ψ (t) = tan This gives −1 35 · · ¸ ¸ sin ω i t ωi t ωi t −1 tan = tan = 1 + cos ω i t 2 2 µ 2πfi t 2 ¶ 1 = fi 2 1 d yD (t) = 2π dt For Ai = −Ac . PROBLEMS For Ai = Ac we can write.1.176). Thus Z t Au (α) dα = Akf t. we get ψ(t) = tan Since −1 · ¸ ¸ · − sin ω i t sin ω i t −1 = − tan 1 − cos ω i t 1 + cos ω i t we have Thus Finally. m (t) = Au (t). The required sketch follows. BASIC MODULATION TECHNIQUES φ (t ) θ (t ) Ak f KT t 0 ψ (t ) Figure 3.48 For m (t) = A cos ω m t and φ (t) = Akf Z t cos ω m αdα = Akf sin ω m t ωm .2 = and KT = 5Akf The VCO constant is contained in KT .36 CHAPTER 3. Problem 3. This gives ψ (t) = The maximum phase error occurs as t → ∞ and is clearly Akf /KT .11: Also Θ (s) = Taking the inverse transform gives θ (t) = Akf µ ¶ 1 1 −KT t e − t+ u (t) KT KT ¢ Akf ¡ 1 − e−KT t u (t) KT Akf KT AKT kf 2 (s + K ) s T The phase error is ψ (t) = φ (t) − θ (t). Thus we require 0. If kf = Kν . since KT is large.57.1. This gives m eν (t) ≈ Akf cos ω m t Kν and we see that eν (t) is proportional to m (t). The output of the top multiplier is m (t) cos ωc t [2 cos (ω c t + θ)] = m (t) cos θ + m (t) cos (2ω c t + θ) which. KT + ω 2 ≈ KT . we have eν (t) ≈ m (t) Problem 3.49 The Costas PLL is shown in Figure 3. is m (t) cos θ. is m (t) sin θ. PROBLEMS and Φ (s) = The VCO output phase is Θ (s) = Φ (s) Akf KT KT = s + KT (s + KT ) (s2 + ω 2 ) m s2 37 Akf + ω2 m Using partial fraction expansion. Thus. only the third term is signiÞcant. after lowpass Þltering. after lowpass Þltering. For large KT .Θ (s) can be expressed as · ¸ Akf KT 1 s KT Θ (s) = − 2 + 2 KT 2 + ω 2 s + KT s + ω2 s + ω2 m m m This gives. The quadrature multiplier output is m (t) cos ω c t [2 sin (ω c t + θ)] = m (t) sin θ + m (t) sin (2ω c t + θ) which. Akf K 2 1 dθ ¡ 2 T ¢ cos ω m t = eν (t) = Kν dt Kν KT + ω 2 m 2 2 Also. The multiplication of the lowpass Þlter outputs is m (t) cos θm (t) sin θ = m2 (t) sin 2θ .3. for t ≥ 0 · ¸ Akf KT KT −KT t θ (t) = e − cos ω m t + sin ω m t KT 2 + ω 2 ωm m The Þrst term is the transient response. we desire e0 (t) = A cos 2π 7 f0 t.13: as indicated.12: Phase Detector Loop Filter and Amplifier VCO ev ( t ) Bandpass Filter e0 ( t ) Figure 3. Thus the VCO is deÞned by dθ dψ = = −Kν eν (t) dt dt This is shown below. Problem 3. The spectrum of the VCO output consists of components separated by . Since the dψ intersection is on a portion of the curve with negative slope. the phase error is θ. BASIC MODULATION TECHNIQUES dψ / dt ψ A Figure 3. The pulse should be narrow so that the seventh harmonic 3 is relatively large. Note that with the assumed input m (t) cos ω c t and VCO output 2 cos (ω c t + θ). the point A dt at the origin is a stable operating point.38 CHAPTER 3. Assume that the VCO output is a 3 pulse train with frequency 1 fo . Thus the loop locks with zero phase error and zero frequency error.50 ¡ ¢ With x (t) = A2πf0 t. where τ is the pulse width. PROBLEMS This component (the fundamental) tracks the input signal 39 Bandpass filter passband 0 1 f0 3 7 f0 3 f Figure 3.13 defrees .3.51 The phase plane is deÞned by ψ = ∆ω − Kt sin ψ (t) at ψ = 0.1.14: with an envelope of sinc(τ f). The center frequency of the bandpass Þlter is 7 f0 and the bandwidth is on the order of 1 f0 as shown. ψ = ψ ss . Thus µ ¶ µ ¶ ∆ω −1 ∆ω −1 ψss = sin = sin Kt 2π (100) For ∆ω = 2π (30) ψ ss = sin For ∆ω = 2π (50) ψ ss = sin For ∆ω = 2π (80) ψ ss = sin −1 −1 1 3 f0 µ 30 100 ¶ = 17. the steady-state phase error.46 degrees −1 µ 50 100 ¶ = 30 degrees µ 80 100 ¶ = 53. 3 3 Problem 3. 1 ε = 0.228) Kt which is Kt F (s) Θ (s) ³ ´ = = Φ (s) s + Kt F (s) s + Kt s+a s+ε Kt (s + a) Kt (s + a) Θ (s) = = 2 Φ (s) s (s + ε) + Kt (s + a) s + (Kt + ε) s + Kt a s2 + 2ζω n s + ω 2 = s2 + (Kt + ε) s + Kt a n ³ s+a s+ε ´ Therefore This gives ωn = and Kt + ε ζ= √ 2 Kt a p Kt a Problem 3.13 degrees For ∆ω = 2π (120).1Kt = 80.40 For ∆ω = −2π (80) CHAPTER 3.52 From (3. BASIC MODULATION TECHNIQUES µ ¶ ψ ss = sin −1 −80 100 = −53.78 we have Thus .53 Since ω n = 2π (100) we have or Since p Kt a = 2π (100) ¡ ¢ Kt a = 4π2 104 1. Problem 3.1Kt Kt + ε 1 = ς=√ = √ 2 (2π) (100) 2 Kt a 2 √ 2 (2π) (100) = 807.8 Kt = 1. there is no stable operating point and the frequency error and the phase error oscillate (PLL slips cycles continually). 54 1 From (3.17.15: and ¡ ¢ 4π2 104 a= = 488. .15. we see that the phase deviation is reduced by 1+ 2π KD Kν . Z 1 i (t) dt ν (t) = C Multiplication by the pulse train can be realized by sampling the capacitor voltage.46 Problem 3.5 = 1 + KD (2π) (25) D2 0.7 807.8 Problem 3.247). the simple circuit is as illustrated in Figure 3. The integrator can be realized by using a capacitor since.3. The operation should be clear from the waveforms shown in Figure 3.5 Thus KD = 0.16. The VCO constant is 25 Hz /volt. for a capacitor.1.55 A system converting PWM to PAM can be realized as illustrated in Figure 3.4 2π which gives 1 + 25KD = 12. Thus. PROBLEMS 41 x PWM (t ) z (⋅)dt x1 (t ) xPAM ( t ) x2 ( t ) Pulse train Figure 3. Thus rad/s/volt Kν = 2π (25) we may then write 5 D1 1 = = 12. 42 CHAPTER 3. BASIC MODULATION TECHNIQUES x PWM (t ) t 0 x1 (t ) T 2T 3T 4T t 0 x2 ( t ) T 2T 3T 4T t 0 x PAM (t ) T 2T 3T 4T t 0 T 2T 3T 4T Figure 3.16: . 56 Let A be the peak-to-peak value of the data signal.3. The peak error is 0. PROBLEMS 43 xPWM ( t ) = i ( t ) sampling switch is closed for xPAM ( t ) switch closes at t = nT to restart integration nT − τ < t < nT C Figure 3. Each sample is encoded into n = 7 1 pulses.57 . For our case τ= Thus the bandwidth is 1 1 = nfs 2W n 1 = 2W n = 2W log2 q τ B = 2 (8. The required number of quantizating levels is A = 100 ≤ 2n = q 0.01A so we choose q = 128 and n = 7.1. For k = 1 Problem 3.5% and the peakto-peak error is 0.17: Problem 3. Then the sampling frequency is fs = 2W = 8 kHz. 000) (7) = 112 kHz and so k = 1.01 A. The bandwidth is B = 2W k log2 q = 2W k(7) The value of k is estimated by assuming that the speech is sampled at the Nyquist rate. Let each pulse be τ with corresponding bandwidth τ . 05π . BASIC MODULATION TECHNIQUES Let the maximum error be λA where A is the peak-to-peak message signal.5 ¡1¢ log 2λ 8.322 0 BN 9 7 6 4 3 2 1 0 ¸ · 1 n = log2 2λ A plot of BN as a function of λ gives the required plot.58 The message signal is m(t) = 4 sin 2π(10)t + 5 sin 2π(20)t The derivative of the message signal is dm (t) = 80π cos 2π (10) t + 200π cos 2π (20t) dt The maximum value of dm (t) /dt is obviously 280π and the maximum occurs at t = 0.05 0.322 1.966 6. this gives a normalized bandwidth of · ¸ 1 B = log2 BN = 2W 2λ We make the following table λ 0.644 3. The peak-topeak error is 2λA and the minimum number of quantizing levels qmin = The wordlength is given by A 1 = 2λA 2λ where [x] is the smallest integer greater than or equal to x.1 0. Thus δ0 ≥ 280π Ts or 280π 280π = 5600 fs ≥ = δ0 0.322 0. Problem 3.005 0.322 2.01 0.4 0.44 CHAPTER 3.644 5.2 0.001 0. With k = 1 (as in the previous problem). 60 The single-sided spectrum for x (t) is shown in Figure 3. After taking the sample at point 12 the commutator moves to point 1. The minimum commutator speed is 2W revolutions per second.18. Problem 3. 2 samples of x3 .18: Thus. PROBLEMS 45 x1 (BW = W ) x4 (BW = 4W ) x2 (BW = W ) x3 (BW = 2W ) x5 (BW = 4W ) 1 2 3 4 5 6 7 8 9 10 11 12 Output Figure 3. From the deÞnition of y (t) we have Y (s) = a1 X (f) + a2 X (f) ∗ X (f) . the illustration should be viewed as it would appear wrapped around a cylinder. Thus. The signal at the point labeled “output” is the baseband signal.59 One possible commutator conÞguration is illustrated in Figure 3. the minimum sampling frequency is 5600 Hz. and one sample of x1 and x2 .1.19. The minimum transmission bandwidth is X B = Wi = W + W + 2W + 4W + 4W i = 12W Problem 3. On each revolution. the commutator collects 4 samples of x4 and x5 . For simplicity the commutator is laid out in a straight line.3. the problem becomes more difficult.46 CHAPTER 3. assume that f2 = 2f1 . the harmonic distortion arising from the spectrum centered about f1 falls exactly on top of the spectrum centered about f2 .19: The spectrum for Y (f ) is given in Figure 3. As more signals are included in x (t).20. For this case. For example. . The difficulty with harmonically related carriers is that portions of the spectrum of Y (f) are sure to overlap. BASIC MODULATION TECHNIQUES W W f f1 f2 Figure 3. Demodulation can be a problem since it may be difficult to Þlter the desired signals from the harmonic and intermodulation distortion caused by the nonlinearity. 3.1. PROBLEMS 47 Y(f) a1 2a2W a2W f f 2 − f1 f1 f2 2 f2 f1 + f 2 2 f2 Figure 3.20: . .1 Problem Solutions Problem 4.2 to show that ½ · ¸ · ¸ · ¸ · ¸¾ 4 3 3 2 3 4 3 2 3 4 3 2 4 3 3 2 4 3 + 3 + + 3 P (3K.2 except that we now look at outcomes that give 4 of a kind.Chapter 4 Probability and Random Variables 4. or 25) = 15/25 = 3/5. 2. 2A) = 52 51 50 49 48 51 50 49 48 51 50 49 48 51 50 49 48 = 9. 12. Thus P (Ai ) = 1/25 (a) P (even number) = P (2 or 4 or · · · or 24) = 12/25. Let Ai = event that the pointer stops on the ith part. Problem 4. we multiply by 13 and get ¶ ¸ ¾ ½ · µ 2 1 48 48 3 2 1 48 4 3 2 1 4 3 3 + + P (4 of a kind) = 13 52 51 50 49 48 51 50 49 48 52 51 50 49 48 = 0. 4. 25..0002401 1 .2 (a) Use a tree diagram similar to Fig. Since the tree diagram is for a particular denomination. (d) P (number > 10) = P (11.1 S = sample space is the collection of the 25 parts. .2345 × 10−6 (b) Use the tree diagram of Fig. These events are exhaustive and mutually exclusive. 4. · · · . (c) P (4 or 5 or 9) = 3/25.. i = 1. (b) P (A25 ) = 1/25. C) P (B. on the fourth card it is 10/49. J. K.3 P (A. B.001981 51 50 49 48 Problem 4. The only way that both of these conditions can be true is for P (A) = P (B) = 0.4 A and B being mutually exclusive implies that P (A|B) = P (B|A) = 0. Therefore. on the third card it is 11/50. A and B being statistically independent implies that P (A|B) = P (A) and P (B|A) = P (B). C) P (A. Given a particular suit on the first card.5 (a) The result is ¢2 ¡ P (AB) = 1 − P (1 or more links broken) = 1 − 1 − q 2 (1 − q) ¡ ¢ P (AB| link 4 removed) = 1 − 1 − q 2 (1 − q) (b) If link 4 is removed.833 48 4 1 1 1 1 = 1. PROBABILITY AND RANDOM VARIABLES (c) The first card can be anything. K. the result is .2 CHAPTER 4. C) = = = = P (A) P (B) P (A) P (C) P (B) P (C) P (A) P (B) P (C) Problem 4.283 × 10−8 52 51 50 49 48 12 11 10 9 = 0. 10 of same suit) = (e) The desired probability is P (Q|A. 10 not all of same suit) = 4 = 0. the probability of suit match on the second card is 12/51. Problem 4. on the fifth and last card it is 9/48. B) P (A. P (all same suit) = 1 (d) For a royal flush P (A. Q. J. 9) (0. by total probability P (B|A) P (A) P (B) ¢ ¡ ¢ £ ¡ ¢¤ ¡ ¢ ¡ P (B) = P (B|A) P (A) + P B|A P A = P (B|A) P (A) + 1 − P B|A P A = (0.6 with ¡ ¢ ¢ P B|A P (A) ¡ ¡ ¢ P A|B = P B Thus ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ P B = P B|A P (A) + P B|A P A = (0.6 0. . P (B1 ) = 0.6.6 Therefore P (A|B) = Similarly.091. P (B3 ) = 0.4) + (0.4) (0.333.6) (0. P (A2 .083.4) ¡ = 0. P (B3 |A2 ) = 0.1.6) = 0.1) (0.4) + (0.7 (a) P (A2 ) = 0.05.15. the result is ¡ ¢2 P (AB| link 2 removed) = 1 − 1 − q 2 3 (d) Removal of link 4 is more severe than removal of link 2.4 ¢ (0.3.6 Using Bayes’ rule P (A|B) = where. B2 ) = 0.4 Problem 4. Problem 4. P (A1 . B1 ) = 0.4) = 0.05. B3 ) = 0. P (B2 |A1 ) = 0. (b) P(A3 |B3 ) = 0.1) (0. (0.6) = 0.4. PROBLEM SOLUTIONS (c) If link 2 is removed.05.8 See the tables below for the results. Problem 4.25. P (A3 . P (B2 ) = 0.1 P A|B = 0.9) (0. 9 The cdf is zero for x < 0. Fx (10) = 1 also.4 P spots up 2 3 4 5 6 7 8 9 10 11 12 CHAPTER 4. which says A × 103 = 1 or A = 10−3 .10 (a) As x → ∞. (d) D = τ /π . jumps another 3/8 at x = 2. the quadratic 3 × 10−3 x2 for 0 ≤ x ≤ 10. jumps by 1/8 at x = 0. (c) C = γ/2. the cdf approaches 1.343 = 0.316 Problem 4. Assuming continuity of the cdf. (b) The pdf is given by fX (x) = dFX (x) = 3 × 10−3 x2 u (x) u (10 − x) dx The graph of this pdf is 0 for t < 0. and jumps another 1/8 at x = 3. jumps another 3/8 at x = 1.657 (d) The desired probability is P (3 ≤ X < 7) = FX (7) − FX (3) = 0. Therefore B = 1. PROBABILITY AND RANDOM VARIABLES X1 4 9 16 25 36 49 64 81 100 121 144 P (X1 = xi ) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 P spots up 2 3 4 5 6 7 8 9 10 11 12 X2 1 1 1 1 1 0 0 0 0 0 0 P (X2 = xi ) 15/36 (a) (b) 21/36 Problem 4. (b) B = β.11 (a) A = α. (c) The desired probability is ¢ ¡ P (X > 7) = 1 − FX (7) = 1 − 10−3 (7)3 = 1 − 0. and an impulse of weight 1/8 at x = 3. and 0 for t > 10. impulses of weights 3/8 at x = 1 and 2. The pdf consists of an impulse of weight 1/8 at x = 0. Problem 4. y) dxdy = 1 = Axye−(x+y) dxdy −∞ −∞ 0 Z ∞0 Z ∞ −x = A xe dx ye−y dy 0 0 = A .1. 3) = (1+3x)/255. Next find fX (x) and fY (y) as ( Z 1−x 3 (1 − x)2 . 0 ≤ x ≤ 6. otherwise 0 and fY (y) = Z 1−y 0 6 (1 − x − y) dx = ( 3 (1 − y)2 . PROBLEM SOLUTIONS Problem 4.13 RR (a) fXY (x. 1. y) 1 + xy = . fY (y) = 6(1+3y) . Problem 4. Problem 4. the two separate factors are marginal pdfs.0098 n 255 (c) fXY (x. y) dxdy = 1 gives C = 1/255. Thus X and Y are statistically independent.5 = 0. y) dxdy = 1 or C (1 − x − y) dxdy = 1 −∞ −∞ 0 0 This gives C = 6.4.14 (a) To find A. otherwise integration. First find C from the integral Z 1 Z 1−y Z ∞Z ∞ fXY (x. 255 (d) By 0 ≤ y ≤ 5. (b) fXY (0. evaluate the double integral Z ∞Z ∞ Z ∞Z ∞ fXY (x. otherwise Since the joint pdf is not equal to the product of the two marginal pdfs. so the desired conditional pdf is fX|Y (x|y) = fXY (x. 0 ≤ y ≤ 5 fY (y) 6 (1 + 3y) Substitute y = 3 to get the asked for result. X and Y are not statistically independent. y) = √ √ A exp [− |x|] A exp [− |y|] 5 With proper choice of A.5) = 1+1. (b) Note that the pdf is the volume between a plane which intersects the x and y coordinate axes one unit out and the fXY coordinate axis at C. 0≤x≤6 0. 0 ≤ x ≤ 1 6 (1 − x − y) dy = fX (x) = 0. 0 ≤ y ≤ 1 0.1.12 (a) Factor the joint pdf as fXY (x. 15 (a) Use normalization of the pdf to unity: Z Z ∞ αx−2 u (x − α) dx = −∞ ∞ αx−2 dx = 2x−1 α Z ∞ =1 α Hence. the joint pdf factors into the product of the marginal pdfs. Problem 4. y ≥ 0 √ 2  2πσ y 0.6 CHAPTER 4. (b) Clearly. PROBABILITY AND RANDOM VARIABLES Thus A = 1.16 The result is fY (y) =  2  exp(−y/2σ ) . y>0 fY (y) = √ 2πa2 σ2 . Therefore ´ ³ 2 exp − 2ay σ2 2 . this is the pdf for any value of α. α > 10 α/10. x < α (1 − α/x) . y < 0 Problem 4. α < 10 Problem 4. (b) The cdf is ½ Z x −2 αz u (z − α) dz = FX (x) = −∞ 0. we have g −1 (y) = y/a. transformation of variables gives ¯ −1 ¯ ¯ dg (y) ¯ ¯ ¯ fY (y) = fX (x) ¯ dy ¯ x=g −1 (y) Since y = g (x) = ax.17 First note that P (Y = 0) = P (X ≤ 0) = 1/2 For y > 0. fX (x) = xe−x u (x) and fY (y) = ye−y u (y) (c) Yes. x > α ½ (c) The desired probability is P (X ≥ 10) = 1 − P (X < 10) = 1 − FX (10) = 1. for all y. The mean of the hyperbolic pdf is zero by virtue of the evenness of the pdf. and using a tabulated definite integral. The mean and variance for the Poisson and geometric distributions follow similarly. For the variance of the hyperbolic pdf.18 (a) The normalization of the pdf to unity provides the relationship Z ∞ Z ∞ e−b|x| dx = 2A e−bx dx = 2A/b = 1 A −∞ 0 where evenness of the integrand has again been used. Problem 4. Z ∞ Z ∞ © ª b 2 −b|x| x e σ2 = E X 2 = dx = b x2 e−bx dx = 2/b2 X 2 −∞ 0 Problem 4. Its variance follows by defining the random variable Y = X − m. Expanding. (b) E [X] = 0 because the pdf is an even function of x. see Example 4. The mean and variance of the Rayleigh random variable can be obtained by using tabulated definite integrals. The mean and variance of the single-sided exponential pdf are easily found in terms of tabulated integrals.162). The mean of a Laplacian random variable is zero by the evenness of the pdf. Hence.20 For the uniform distribution. and its variance is worked out below. £ ¤ we have E X 2 − {E [X]}2 > 0 if X 6= 0. The mean for the Gaussian pdf follows by a change of variables and the normalization property of the Gaussian pdf. The mean and variance of the Nakagami-m pdf can be put in terms of tabulated definite integrals.18. Thus A = b/2. and the last integral can be found in an integral table. consider Z ∞ 2 Z ∞ 2 £ 2¤ x (m − 1) hm−1 dx x (m − 1) hm−1 dx dx = dx E X = 2 (|x| + h)m 2 (x + h)m −∞ 0 . The mean and variance of the binomial random variable is worked out beginning with (4.4. where the second integral follows because of evenness of the integrand. PROBLEM SOLUTIONS 7 For y = 0. Problem 4.5.1.19 o n Use the fact that E [X − E (X)]2 ≥ 0 (0 only if X = 0 with probability one). which has zero mean. we need to add 0. the result is ´ ³ 2 exp − 2ay σ2 2 1 u (y) fY (y) = δ (y) + √ 2 2πa2 σ2 where u (y) is the unit step function.5δ (y) to reflect the fact that Y takes on the value 0 with probability 0.21. The variance was obtained in Problem 4. (c) Since the expectation of X is zero. 8 CHAPTER 4. the result is (A/b) 1 − e−bx . Problem 4. and the absolute value on x is unnecessary because the integration is for positive x. subtract the result of part (c) squared from the result of part (d). (e) For the variance.22 (a) The integral evaluates as follows: Z ∞ Z ∞³ √ Z ´2n e−y2 −x2/2σ2 © 2n ª 2n+1 σ2n ∞ 2n −y 2n e √ √ dy = √ = 2 2σy x dx = 2 y e dy E X π π 2πσ2 −∞ 0 0 ¢ ¡ where y = x/ 21/2 σ . Integrate by parts twice to obtain #∞ Z (" ) ∞ £ 2¤ x2 (h + h)−m−1 2x (x + h)−m+1 m−1 + dx E X = (m − 1) h −m + 1 m−1 0 0 The first term is zero if m ≥ 3. (c) The mean is ¸ · e−bB 1 1 − bB E [X] = b 1 − e−bB (d) The mean-square is ¸ · £ ¤ 2A 1 e−bB AB 2 −bB − 2 (1 + bB) − e E X2 = 2 b b b b where A must be substituted from part (a. (b) The integral is zero by virtue of the oddness of the integrand.23 Z ¾ 1 1 f (x − 4) + [u (x − 3) − u (x − 7)] dx x 2 8 −∞ Z ∞ Z 1 1 7 9 xf (x − 4) dx + xdx = 2 −∞ 8 3 2 ∞ E [X] = = ½ . PROBABILITY AND RANDOM VARIABLES where the second integral follows by evenness of the integrand in the first integral. m≥4 (m − 2) (m − 3) Problem£4. Therefore Z ∞ £ 2¤ m−1 x (x + h)−m+1 dx = E X = 2h 0 2h2 .). Problem 4. the given result is obtained. Using a table of definite integrals.21 ¤−1 . (a) A = b 1 − e−bB ¢ ¡ (b) The cdf is 0 for x < 0 and 1 for x > B. For 0 ≤ x ≤ B. 76. the mean of Z is E {Z} = E {3X − 4Y } = 3E {X} − 4E {Y } = 3 (2) − 4 (1) = 2 The variance of Z is o n var (Z) = E [Z − E (Z)]2 o n = E [3X − 4Y − 3E(X) + 4E(Y )]2 n£ ¡ ¢ ¤2 o = E 3 X − X − 4(Y − Y ) ¢ ¡ ¢ © ¡ ª = E 9 X − X − 24 X − X (Y − Y ) + 16(Y − Y )2 = 9σ 2 + 16σ2 − 24σX σY σ XY X Y Putting in numbers.y) . (d) 14. The Jacobian is 1 − ρ2 . v) = y = u Thus X and Y are equivalent to the random variables in the problem statement and which proves the desired result. Since the transforma¡ ¢−1/2 tion is linear. To see this. the new random variables are Gaussian.1.23.25 Consider independent Gaussian random variables U and V and define a transformation p g1 (u. (b) 83.05. PROBLEM SOLUTIONS £ ¤ E X2 = Z ∞ 9 x2 −∞ α2 X ¾ 1 1 127 f (x − 4) + [u (x − 3) − u (x − 7)] dx = 2 8 6 £ ¤ 127 81 11 = E X 2 − E 2 [X] = − = 6 4 12 ½ Problem 4. (c) 51. y) = p ¯ −1 2πσ 2 (1 − ρ2 ) 1 − ρ2 2πσ 2 ¯ u=g1 (x. note that ´ o n³ p E {XY } = E ρU + 1 − ρ2 V U = ρσ2 ¢1/2 ¡ The inverse transformation is u = y and v = (x − ρy) / 1 − ρ2 .24 Regardless of the correlation coefficient. Problem 4.4. v) = x = ρu + 1 − ρ2 u and g2 (u. Thus the joint pdf of the new random variables X and Y is i h 2 ¯ 2 2 −2ρxy+y2 − u +v ¯ exp − x2σ2 (1−ρ2 ) 2 ¯ e 2σ ¯ 1 = fXY (x.y) −1 v=g2 (x. the results for the variance of Z are: (a) 107. 5 ≤ z ≤ 3.5 ≤ z ≤ 0.5 ≤ z < 2.29 (a) The characteristic function is MX (jv) = a a − jv .26 Divide the joint Gaussian pdf of two random variables by the Gaussisn pdf of Y .5) /3.27 Convolve the two component pdfs. 2. z > 3. h i E [Y ] = E (2 + 3X)2 ¤ £ ¤ £ = E 4 + 12X + 9X 2 = 4 + 12E [X] + 9E X 2 = 4 + 12 + 9/32 = 4. and the overlap ends when z > 3. Problem 4. The result.5) /3. and the desired result is a Gaussian pdf with ¡ ¢ ρσx (Y − mY ) and var (X|Y ) = σ2 1 − ρ2 E {X|Y } = mx + X σY Problem 4. − 0. z < −0. either obtained graphically or analytically.5    0. collect all exponential terms in a common exponent.28125 − 22 = 0. is given by   0.5  1/3.28125 Problem 4. complete the square of this exponent which will be quadratic in x and y. 2 σ = 4.28 £ ¤ (a) E [X] = 0. E X 2 = 1/32 =var[X].5    (z + 0.5.5 Problem 4.10 CHAPTER 4.28125. (b) The pdf of Y is (c) 4 fY (y) = e−8|y−3|/3 3 E [Y ] = E [2 + 3X] = 2. 0.5 fZ (z) =   − (z − 3. A sketch of the two component pdfs making up the integrand show that there is no overlap until z > −1/2. PROBABILITY AND RANDOM VARIABLES which proves the desired result. 099 0.384 0. Problem 4.090 0.30 The required distributions are µ ¶ n k P (k) = p (1 − p)n−k (Binomial) k P (k) = P (k) = A comparison is given below: n.302 0.231 0.4.033 0.0×10−5 4.0014 Laplace 0. (c) var[X] = 1/a2 .012 3.222 0.729 0.1/10 (b) n.741 0.002 1.1.243 0. p 3.096 0.090 0.6×10−4 1.003 Poisson 0.512 0.9×10−4 e−(k−np) /[2np(1−p)] p (Laplace) 2πnp (1 − p) 2 11 (np)k −np (Poisson) e k! n.008 Binomial 0.090 0.026 0.201 0.135 0.027 0.001 Binomial 0.487 0.650 0.231 0.1×10−6 Laplace 0.268 0.329 0.3×10−4 4.107 0.1/5 (a) k 0 1 2 3 k 0 1 2 3 k 0 1 2 3 4 5 6 7 8 9 Binomial 0.1/5 (c) . p 10.396 0.020 Poisson 0.4×10−3 8.310 0.088 0.315 0.180 0.1×10−8 Poisson 0. p 3.075 0.005 7. PROBLEM SOLUTIONS £ ¤ (b) The mean and mean-square values are E [X] = 1/a and E X 2 = 2/a2 .9×10−4 7.1×10−6 7.036 0.549 0.271 0.271 0.004 1×10−6 Laplace 0.019 0. The Laplace-approximated probability is P (< 3 heads in 10 tosses) 2 X e−(k−5)2 /5 √ 5π k=0 ´ ³ 1 e−5 + e−16/5 + e−9/5 = 0.241 0.003 6×10−5 3.057 0.194 0.6×10−7 9.0518 (b) P (first head at trial 15) = (1/2)(1/2)14 = 3.31 P (a) P (5 or 6 heads in 20 trials) = k=5.0×10−9 1×10−10 Laplace 0.241 0.5×10−16 1.0537 = √ 5π = (c) The percent error is % error = 0. (b) Np = (26)4 = 456.1/10 k 0 1 2 3 4 5 6 7 8 9 CHAPTER 4.052 × 10−5 P 100! (c) P (50 to 60 heads in 100 trials) = 60 k!(100−k)! 2−100 k=50 Problem 4. 800. p 10.33 (a) The desired probability is P (fewer than 3 heads in 10 tosses) = 2 X µ 10 ¶ µ 1 ¶10 k=0 k 2 = (1 + 10 + 45) 2−10 = 0.12 n.183 0.368 0.9×10−7 6.4×10−4 3. 0547 − 0.4×10−3 1×10−6 1×10−7 (d) Problem 4.1×10−4 3.3×10−13 1.046 0.6 [20!/k! (20 − k)!] 2−20 = 0.0547 (b) npq = 10 (1/2) (1/2) = 2.32 (a) Np = 26 (25) (24) (21) = 358.387 0.0547 . (c) The answers are the reciprocals of the numbers found in (a) and (b).061 0. Problem 4.011 1.5×10−3 1.1×10−3 5.84% 0.015 3. 976.25.420 0.2×10−20 Poisson 0.0537 × 102 = 1. np = 5.368 0.348 0. PROBABILITY AND RANDOM VARIABLES Binomial 0. k = 2πσX σ Y 1 − ρ2 σX σY · 2 ¸ u − 2ρuv + v 2 (x.35 First note that Z ∞ e − 1 au2 2 du = −∞ Z ∞ e − 1 a(u−b)2 2 du = −∞ r 2π a (a) In the joint Gaussian pdf.34 P (3 or more errors in 106 ) = 0.v= . PROBLEM SOLUTIONS 13 Problem 4. (b) First find the characteristic function: ª © MX (jv) = E ejvX = Z Put the exponents together and complete the square in the exponent to get ¸ · ¸ · Z ∞ ¢ 1 1 2 2 1 ¡ 2 2 q MX (jv) = dx exp − σX v + jmX v exp − 2 x − mX − jvσ X 2 2σX −∞ 2πσ2 X 2σX e ejvx q dx 2 −∞ 2πσX ∞ − (x−mX )2 . y) = k exp − 2 (1 − ρ2 ) " Then the joint Gaussian pdf becomes fXY Complete the square in v to obtain fXY # ¡ ¢ (v − 2ρu)2 exp −u2 /2 (x.4.08 Problem 4.1. y) = k exp − 2) 2 (1 − ρ Use the definite integral above to integrate over v with the result " # q ¡ 2 ¢ 1 (x − mX )2 exp − fX (x) = 2π (1 − ρ2 ) σ2 exp −u /2 = q Y 2σ2 X 2πσ2 X The pdf of Y can be found in a similar manner. let u= ´−1 ³ p x − mX y − mY . X £ ¤ q E X 2 = (−j)2 MX (jv) |v=0 = σ2 + m2 X X Problem 4.36 (a) K = 1/π. (d) Compare the form of the characteristic function with the answer given in (c) and do a suitable redefinition of variables. Thus ¢−N/2 ¡ MY (jv) = 1 − j2vσ2 e−(1/2σ −jv)x √ dx = 2πσ2 −∞ ¡ ¢−1/2 = 1 − j2vσ2 Z (b) The given pdf follows easily by letting α = 2σ2 in the given Fourier transform pair. . Problem 4. but one could argue that it is zero from the oddness of the integrand for this expectation. it clearly does not converge (the integrand approaches a constant as |x| → ∞). PROBABILITY AND RANDOM VARIABLES Use the definite integral given above to get ¸ · 1 MX (jv) = exp jvmX − σ 2 v2 2 X It is then easy to differentiate and get p E [X] = −jMX (jv) |v=0 = mX and from which it follows that the variance is σ2 .35. (c) The answer is given. If one writes down the integral for the second moment. (b) E [X] is not defined.37 (a) The characteristic function is found from n Pn o © ª 2 MY (jv) = E ejvY = E ejv i=1 Xi = E But Z n o 2 jvXi E e = −x2 /2σ2 jvx2 e √ e dx 2πσ2 −∞ ∞ ∞ 2 2 (n Y i=1 e 2 jvXi ) which follows by using the definite integral defined at the beginning of Problem 4.14 CHAPTER 4. Note the extreme difference between exact and approximation due to the central limit theorem not being valid for the former case. (e) Evident by direct substitution.1.4. h¡ ¢ i 2 = 3σ4 − σ4 = 2σ 4 Xi2 . Since the terms in the sum defining Y are independent. var[Y ] = 2N σ4 . (d) A comparison of the cases N = 4 and N = 16 is shown in Figure 4.1: (c) The mean of Y is E [Y ] = N σ2 by taking the sum of the expectations of the separate terms of the sum defining Y . The mean-square value of Xi2 is h¡ ¢ i d2 2 2 2 = (−j) E Xi var h¡ ¢−1/2 i¯ 1 − j2vσ2 ¯ ¯ ¯ 2 dv = 3σ4 v=0 Thus.1. PROBLEM SOLUTIONS 15 Figure 4. The mean and variance of Y can be put into the definition of a Gaussian pdf to get the desired approximation. kσx > 1 1 . Problem 4.k>0 k2 P (|X| ≤ kσ X ) = P (−kσX Chebyshev’s bound is P (|X| ≤ kσX ) ≥ 1 − A plot is left to the student.40 The exact probability is A plot may easily be obtained by a MATLAB program. It is suggested that you use the ¡ R∞ ¢ 2 MATLAB function erfc(x) = √π x exp −t2 dt. the cdf of a Gaussian random variable is Z x FX (x) = Change variables to −∞ e− 2σ2 √ du = 1 − 2πσ 2 u−m σ (u−m)2 Z ∞ x e− 2σ2 √ du 2πσ2 (u−m)2 v= with the result that Z x x−m σ FX (x) = 1 − µ ¶ v2 e− 2 x−m √ du = 1 − Q σ 2π Problem 4. exp(−t2 /2) √ dx 2π and Qa (x) = exp(−x2 /2) √ 2πx on the same Problem 4.39 By definition.38 R∞ This is a matter of plotting Q (x) = x set of axes.41 (a) The probability is P (X ≤ 11) = = Z 2/30 ³ √ ´ e−(x−12) √ dx = Q 1/ 15 30π −∞ ³ √ ´ 1 erfc 1/ 30 = 0.398 2 11 . ½ kσx .16 CHAPTER 4. PROBABILITY AND RANDOM VARIABLES Problem 4. kσx ≤ 1 ≤ X ≤ kσ X ) = 1. 281 2 12 Problem 4.42 Observe that σ 2 = ≥ ≥ Z ∞ = k2 σ2 P {|X − m| > kσ} = k2 σ2 {1 − P [|X − m| > kσ]} or P [|X − m| < kσ] ≥ 1 − 1 k2 Z Z −∞ (x − m)2 fX (x) dx (x − m)2 fX (x) dx k 2 σ2 fX (x) dx |x−m|>kσ |x−m|>kσ .1.197 2 17 12 (d) The result is 2/30 ³ √ ´ e−(x−12) √ dx = 1 − 2Q 1/ 15 P (11 < X ≤ 13) = 30π 11 ³ √ ´ = erf 1/ 30 = 0.4.204 Z 13 P (9 < X ≤ 12) = = Z 2/30 ³ √ ´ e−(x−12) 1 √ dx = − Q 3/ 15 2 30π 9 1 ³ √ ´ erf 3/ 30 = 0. PROBLEM SOLUTIONS (b) This probability is P (10 < X ≤ 12) = = (c) The desired probability is Z 2/30 ³ √ ´ e−(x−12) 1 √ dx = − Q 2/ 15 2 30π 10 1 ³ √ ´ erf 2/ 30 = 0. ylabel(‘f_X(x)’).1 % ce4_1.. . ‘—’). N = input(‘Enter number of exponential random variables to be generated: ’). U = rand(1.43 The mean is 0 by even symmetry of the pdf. the variance is £ ¤ var [X] = E X 2 Z ∞ ³a´ x2 = exp (−a1 × 1) dx 2 −∞ Z ∞ ³a´ x2 exp (−ax) dx = 2 2 0 Z ∞ ³ ´2 y dy = 2a exp (−y) a a µ0 ¶ µ ¶ 1 1 = 1/a2 = 2a 3 a 2 4. ’o’) hold plot(X. PROBABILITY AND RANDOM VARIABLES Problem 4. plot(X. norm_hist. X] = hist(V). 1) % clf a = input(‘Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x): ’). Thus.m: Generate an exponentially distributed random variable % with parameter a from a uniform random variable in [0. V = -(1/a)*log(U). [M.. xlabel(‘x’).18 CHAPTER 4. of random variables generated’) disp(N) disp(‘ ’) disp(‘Bin centers’) disp(X) disp(‘ ’) disp(‘No. of random variable counts in each bin’) disp(M) disp(‘ ’) norm_hist = M/(N*(X(2)-X(1))).2 Computer Exercises Computer Exercise 4.N). disp(‘ ’) disp(‘No. a*exp(-a*X).. 4.*cos(2*pi*U)+m.*sin(2*pi*U)+m.9375 1. R = sqrt(-2*log(V)). a = ’.4375 Columns 8 through 10 2.8125 3. of random variables generated 5000 Bin centers Columns 1 through 7 0. [MY. norm_MX = MX/(N*(X_bin(2)-X_bin(1))). ‘ computer generated exponential random variables’]) legend([‘Histogram points’]. U = rand(1.5626 0.2. N = input(‘Enter number of Gaussian random variable pairs to be generated: ’). X_bin] = hist(X. X = sigma*R.N). of random variable counts in each bin Columns 1 through 6 2677 1212 590 263 129 71 Columns 7 through 10 34 9 11 4 Current plot held 19 Computer Exercise 4.6875 2.num2str(N). 20).5624 No. disp(‘ ’) disp(‘Covarance matrix of X and Y vectors:’) disp(cov(X. num2str(a)]) A typical run follows: >> ce4_1 Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x): 2 Enter number of exponential random variables to be generated: 5000 No.1876 0.3125 1.N). V = rand(1. sigma = input(‘Enter the standard deviation of the Gaussian random variables: ’).0625 2.Y)) disp(‘ ’) [MX.[‘Exponential pdf. 20).1875 3.2 % ce4_2. Y_bin] = hist(Y. random variables from Rayleigh & uniform RVs % clf m = input(‘Enter the mean of the Gaussian random variables: ’). . COMPUTER EXERCISES title([‘Theoretical pdf and histogram for ’.m: Generate pairs of Gauss. Y = sigma*R. norm_MY.1).2). plot(Y_bin. ylabel(‘f_X(x)’).1.^2/(2*sigma^2))/sqrt(2*pi*sigma^2).. ‘—’)..1. norm_MX.m).1. plot(X_bin. num2str(m)]) subplot(2. xlabel(‘y’). PROBABILITY AND RANDOM VARIABLES Figure 4.num2str(N). ’o’) hold subplot(2..2: norm_MY = MY/(N*(Y_bin(2)-Y_bin(1))).^2/(2*sigma^2))/sqrt(2*pi*sigma^2). ’. ‘ computer generated independent Gaussian RV pairs’]) legend([‘Histogram points’]. num2str(sigma). title([‘Theoretical pdfs and histograms for ’.m). gauss_pdf_Y.[‘Gauss pdf. \mu = ’.‘. gauss_pdf_X. ylabel(‘f_Y(y)’) A typical run follows: >> ce4_2 Enter the mean of the Gaussian random variables: 1 Enter the standard deviation of the Gaussian random variables: 3 Enter number of Gaussian random variable pairs to be generated: 2000 Covarance matrix of X and Y vectors: 8.1). subplot(2. ’o’) hold subplot(2.5184 0. gauss_pdf_X = exp(-(X_bin . xlabel(‘x’).0828 . gauss_pdf_Y = exp(-(Y_bin . \sigma. plot(X_bin.20 CHAPTER 4. ‘—’).2).1. plot(Y_bin. X = []. rho = input(‘Enter correlation coefficient between adjacent Gauss.rho^2). for k = 2:N m12 = m + rho*(X(k-1) .0828 8.9416 Current plot held Current plot held Computer Exercise 4. COMPUTER EXERCISES 21 Figure 4.3 % ce4_3. X(1) = sigma*randn(1)+m.m). end . X(k) = sqrt(var12)*randn(1) + m12.3: 0. var12 = sigma^2*(1 . sigma = input(‘Enter the standard deviation of the Gaussian random variables: ’).m: Generate a sequence of Gaussian random variables with specified correlation % between adjacent RVs % clf m = input(‘Enter the mean of the Gaussian random variables: ’). N = input(‘Enter number of Gaussian random variables to be generated: ’).2. RVs: ’).4. . norm_MX = MX/(N*(X_bin(2)-X_bin(1))).^2/(2*sigma^2))/sqrt(2*pi*sigma^2). PROBABILITY AND RANDOM VARIABLES [MX. plot(Z. ‘o’) hold subplot(2.1).num2str(N). 20). ’. gauss_pdf_X.m: Testing the validity of the Central Limit Theorem with sums of uniform % random numbers % clf N = input(‘Enter number of Gaussian random variables to be generated: ’).2) Z = X(1:50).. \mu = ’.m).N)-0. plot(X_bin.4 % ce4_4.2). subplot(2. title([‘Theoretical pdf and histogram for ’. ylabel(‘f_X(x)’). X_bin] = hist(X. norm_MX = MX/(N*(X_bin(2)-X_bin(1))). ‘—’). norm_MX.22 CHAPTER 4. subplot(2.1. plot(X_bin. ‘ between samples’]) A typical run follows: >> ce4_3 Enter the mean of the Gaussian random variables: 0 Enter the standard deviation of the Gaussian random variables: 2 Enter correlation coefficient between adjacent Gaussian random variables: .1).[‘Gauss pdf.5. num2str(m)]. gauss_pdf_X = exp(-X_bin.. disp(‘ ’) disp(‘Estimated variance of the Gaussian RVs:’) disp(cov(X_gauss)) disp(‘ ’) [MX. xlabel(‘k’). ‘o’) hold .^2/2)/sqrt(2*pi).1. xlabel(‘x’). % Mean of uniform = 0. X_bin] = hist(X_gauss.1. num2str(rho). num2str(sigma). variance = 1/12 Y = rand(M. X_gauss = sum(Y)/(sqrt(M/12)).‘. 20). \sigma. ylabel(‘X(k)’) title([‘Gaussian sample values with correlation ’.7 Enter number of Gaussian random variables to be generated: 2000 Current plot held Computer Exercise 4. gauss_pdf_X = exp(-(X_bin . ‘x’). plot(X_bin. norm_MX. ‘ computer generated Gaussian RVs’]) legend([‘Histogram points’]. M = input(‘Enter number of uniform RVs to be summed to generate one GRV: ’). ‘ uniform RVs’]) legend([‘Histogram points’].. title([‘Theoretical pdf & histogram for ’.4.. \sigma = 1. gauss_pdf_X.2.. ‘—’).4: plot(X_bin.2) A typical run follows: >> ce4_4 Enter number of Gaussian random variables to be generated: 10000 Enter number of uniform random variables to be summed to generate one GRV: 30 Estimated variance of the Gaussian RVs: 1. COMPUTER EXERCISES 23 Figure 4.0075 Current plot held . xlabel(‘x’). [‘Gauss pdf. num2str(M).num2str(N). \mu = 0’]. ylabel(‘f_X(x)’). ‘ Gaussian RVs each generated as the sum of ’. 5: . PROBABILITY AND RANDOM VARIABLES Figure 4.24 CHAPTER 4. A. and 1/2. 3A. −3A. −4A. For case (a) of problem 5. The probabilities are 1. and 5/6. Sample functions for case (c) are horizontal lines at levels 4A. the probability is again 2/3 because 4 out of 6 of the sample functions will be less than 2A at t = 4. respectively.2 a. 2/3. −5A. 0.Chapter 5 Random Signals and Noise 5. each case of which occurs equally often (with probability 1/3). b. since the sample functions are constant with time and are less than or equal 2A.1. For case (b) of problem 5. −2A. or oblique straight lines of slope A or −A.1 Problem Solutions Problem 5.3 a. the probability is one. respectively. For case (c) of problem 5.1 The various sample functions are as follows.1. Problem 5. −A. since the sample functions are constant with time and 4 out of 6 are less than or equal to 2A. each case of which occurs equally often (with probability 1/6). 2A. The probabilities are now 2/3. The sketches would consist of squarewaves of random delay with respect to t = 0.1. −A. the probability is 2/3. each case of which occurs equally often (with probability 1/6). c. 1/2. 1 . Sample functions for case (b) are horizontal lines at levels 5A. Problem 5. Sample functions for case (a) are horizontal lines at levels A. Y (i) (t0 ) is Gaussian with mean zero and standard deviation σY = σa c. Considering the time average correlation function. this defines the autocorrelation function for all λ.5 a. the mean is zero. b. Not stationary and not ergodic. it follows from a sketch of the integrand that R (τ ) = A2 (1 − 4λ/T0 ) . λ. The sample functions at the integrator output are of the form Z t A (i) Y (t) = A cos (ω 0 λ) dλ = sin ω 0 t ω0 where A is Gaussian. X (t) takes on only two values A and −A. Thus 1 1 fX (x) = δ (x − A) + δ (x + A) 2 2 Problem 5. The random delay. being over a full period. 0 ≤ λ ≤ T0 /2 Since R (τ ) is even and periodic. it is wide sense stationary.2 CHAPTER 5. RANDOM SIGNALS AND NOISE b. . By inspection of the sample functions. |sin (ω 0 t0 )| ω0 Problem 5. defined as R (λ) = lim 1 T →∞ 2T Z T x (t) x (t + λ) dt = −T 1 T0 Z x (t) x (t + λ) dt T0 where the last expression follows because of the periodicity of x (t).4 a. and these are equally likely. b. Yes. means that no untypical sample functions occur. 1. PROBLEM SOLUTIONS Problem 5.5. The statistical average second moment is Z π/2 £ 2 ¤ dθ E X (t) = A2 cos2 (2πf0 t + θ) π/4 π/4 "Z " # ¯π/2 # Z π/2 π/2 ¯ 1 2A2 2A2 π + sin (4πf0 t + 2θ)¯ = dθ + cos (4πf0 t + 2θ) dθ = ¯ π π 4 2 π/4 π/4 π/4 í ³ 2 2 h A A π + sin (4πf0 t + π) − sin 4πf0 t + = 2 π 2 2 2 A A [sin (4πf0 t) + cos (4πf0 t)] = − 2 π The statistical average variance is this expression minus E 2 [X (t)].6 a. The time average autocorrelation function is A2 cos (2πf0 τ ) 2 The statistical average autocorrelation function is Z π/2 dθ A2 cos (2πf0 t + θ) cos (2πf0 (t + τ ) + θ) R (τ ) = π/4 π/4 2 Z π/2 2A = [cos (2πf0 τ ) + cos (2πf0 (2t + τ ) + 2θ)] dθ π π/4 ¯π/2 ¯ A2 A2 cos (2πf0 τ ) + sin (2πf0 (2t + τ ) + 2θ)¯ = ¯ 2 π π/4 ³ 2 2 h A A π í cos (2πf0 τ ) + = sin (2πf0 (2t + τ ) + π) − sin 2πf0 (2t + τ ) + 2 π 2 2 2 A A cos (2πf0 τ ) − [sin 2πf0 (2t + τ ) + cos 2πf0 (2t + τ )] = 2 π hx (t) x (t + τ )i = . The time average mean is zero. b. The statistical average mean is ¯π/2 Z π/2 ¯ 4A dθ E [X (t)] = = sin (2πf0 t + θ)¯ A cos (2πf0 t + θ) ¯ π/4 π π/4 π/4 = = 4A [sin (2πf0 t + π/2) − sin (2πf0 t + π/4)] π ·µ ¸ ¶ 1 1 4A cos (2πf0 t) − √ sin (2πf0 t) 1− √ π 2 2 3 The time average variance is A2 /2. Suitable. Not suitable for the same reason as (c). Suitable. means that for a given time. the mean is clearly zero. Not suitable because it is not even. Again. d. The mean-square value is n o £ ¤ E Z 2 = E [X (t) cos (ω 0 t)]2 £ ¤ = E X 2 (t) cos2 (ω 0 t) = σ 2 cos2 (ω0 t) X The process is not stationary since its second moment depends on time. The second moment is n o £ ¤ E Z 2 = E [X (t) cos (ω0 t + θ)]2 £ ¤ £ ¤ σ2 = E X 2 (t) E cos2 (ω 0 t + θ) = X 2 Problem 5.7 a. e. b. b. not being uniform over (0.4 CHAPTER 5. The random phase.8 The pdf of the noise voltage is e−(x−2) /10 √ fX (x) = 10π 2 Problem 5. Problem 5.9 a. No. . c. The mean is zero. t. certain phases will be favored over others. RANDOM SIGNALS AND NOISE c. 2π). Not suitable because the Fourier transform is not everywhere nonnegative. 54) Now r (τ ) = r (−τ ) and r (τ ) = 0 for |τ | > T /2.5) + sin c2 (T f + 0. 0 ≤ τ ≤ T /2 T −T /2 T ³ πτ ´ 1 (1 − τ /T ) cos .5) cos2 (πfT ) .10 Use the Fourier transform pair 2W sinc (2W τ ) ↔ Π (f/2W ) with W = 2 × 106 Hz to get the autocorrelation function as ¢¡ ¢ ¡ ¡ ¢ RX (τ ) = N0 W sinc (2W τ ) = 2 × 10−8 2 × 106 sinc 4 × 106 τ ¡ ¢ = 0.1. PROBLEM SOLUTIONS Problem 5. (5. Therefore it can be written as ³ πτ ´ 1 r (τ ) = Λ (τ /T ) cos 2 T Ra (τ ) = ³ πτ ´ A2 Λ (τ /T ) cos 2 T The power spectral density is · µ · µ ½ ¶¸ ¶¸¾ 1 1 A2 T 2 2 sin c T f − + sinc T f + Sa (f ) = 4 2T 2T b.11 a. 0 ≤ τ ≤ T /2 = 2 T 5 Hence. The r (τ ) function.5. The corresponding power spectrum is £ ¤ Sb (f ) = A2 T sinc2 (T f − 0.55). The autocorrelation function now becomes Rb (τ ) = A2 [2r (τ ) + r (τ + T ) + r (τ − T )] which is (5. by (5.02 sinc 4 × 106 τ Problem 5.62) with g0 = g1 = 1. is Z 1 ∞ p (t) p (t + τ ) dt r (τ ) = T −∞ · ¸ Z π (t + τ ) 1 T /2−τ cos (πt/T ) cos = dt. RY (τ ) = E {4 cos (50πt + θ) cos [50π (t + τ ) + θ]} = 2 cos (50πτ ) Using the first transform pair. Since a product in the time domain is convolution in the frequency domain. By definition RZ (τ ) = E [Z (t) Z (t + τ )] = E [X (t) X (t + τ ) Y (t) Y (t + τ )] = E [X (t) X (t + τ )]E[Y (t) Y (t + τ )] = RX (τ ) RY (τ ) b. we have RY (τ ) = 500 sinc (100τ ) . The plots are left for the student. RANDOM SIGNALS AND NOISE © ª RXY (τ ) = E [X (t) Y (t + τ )] = Rn (τ ) + E A2 cos (ω0 t + Θ) × sin [ω0 (t + τ ) + Θ] = BΛ (τ /τ 0 ) + A2 sin (ω0 τ ) 2 Problem 5.6 c.12 The cross-correlation function is CHAPTER 5.13 a. Use the transform pairs 2W sinc (2W τ ) ←→ Π (f/2W ) and 1 1 cos (2πf0 τ ) ←→ δ (f − f0 ) + δ (f + f0 ) 2 2 Also. Problem 5. it follows that SZ (f ) = SX (f ) ∗ SY (f ) c. E 2 [X (t)] = limτ →∞ R (τ ) = 4 W.15 a.5.1. The autocorrelation function is RX (τ ) = E [Y (t) Y (t + τ )] = E {[X (t) + X (t − T )][X (t + τ ) + X (t + τ − T )]} = E [X (t) X (t + τ )] + E[X (t) X (t + τ + τ )] +E [X (t − T ) X (t + τ )] + E[X (t − T ) X (t + τ − T )] = 2RX (τ ) + RX (τ − T ) + RX (τ + T ) b.14 ¤ £ a. Problem 5. S (f ) = 4δ (f ) + 5 sinc2 (5f ). ¤ £ c. Application of the time delay theorem of Fourier transforms gives SY (f ) = 2SX (f ) + SX (f ) [exp (−j2πfT ) + exp (j2πfT )] = 2SX (f ) + 2SX (f ) cos (2πf T ) = 4SX (f ) cos2 (πf T ) . dc power = E 2 [X (t)] = 4 W. £ ¤ σ 2 = E X 2 (t) − E 2 [X (t)] = 5 − 4 = 1 W. X d. Total power = E X 2 (t) = 5 W. E X 2 (t) = R (0) = 5 W. b. PROBLEM SOLUTIONS This gives RZ (τ ) = [500 sinc (100τ )] [2 cos (50πτ )] = 1000 sinc (100τ ) cos (50πτ ) and ·Y µ ¶ Yµ ¶¸ f − 25 f + 25 + SZ (f ) = 5 × 10 200 200 3 7 Problem 5. 8 c. Use the transform pair CHAPTER 5. RANDOM SIGNALS AND NOISE RX (τ ) = 6Λ (τ ) ←→ SX (f ) = 6 sinc2 (f ) and the result of (b) to get SY (f ) = 24 sinc2 (f ) cos2 (πf /4) Problem 5.16 a. The student should carry out the sketch. R 0+ b. dc power = 0− S (f ) df = 5 W. R∞ c. Total power = −∞ S (f ) df = 5 + 10/5 = 7 W Problem 5.17 a. This is left for the student. R 0+ b. The dc power is either limτ →∞ RX (τ ) or 0− SX (f ) df . Thus (1) has 0 W dc power, (2) has dc power K1 W, and (3) has dc power K2 W. R∞ c. Total power is given by RX (0) or the integral of −∞ SX (f ) df over all frequency. Thus (1) has total power K, (2) has total power 2K1 + K2 , and (3) has total power r Z ∞ π −αf 2 K + 2K1 + K2 e df + K1 + K1 + K2 = P3 = K α −∞ where the integral is evaluated by means of a table of definite integrals. d. (2) has a periodic component of frequency b Hz, and (3) has a periodic component of 10 Hz, which is manifested by the two delta functions at f = ±10 Hz present in the power spectrum. Problem 5.18 a. The output power spectral density is ¢ ¡ Sout (f ) = 10−5 Π f /2 × 106 Note that the rectangular pulse function does not need to be squared because its amplitude squared is unity. 5.1. PROBLEM SOLUTIONS b. The autocorrelation function of the output is ¡ ¡ ¢ ¢ Rout (τ ) = 2 × 106 × 10−5 sinc 2 × 106 τ = 20 sinc 2 × 106 τ 9 c. The output power is 20 W, which can be found from Rout (0) or the integral of Sout (f ). Problem 5.19 a. By assuming a unit impulse at the input, the impulse response is h (t) = 1 [u (t) − u (t − T )] T b. Use the time delay theorem of Fourier transforms and the Fourier transform of a rectangular pulse to get H (f ) = sinc (fT ) e−jπf T c. The output power spectral density is S0 (f ) = d. Use F [Λ (τ /τ 0 )] = τ 0 sinc2 (τ 0 f ) to get R0 (τ ) = N0 Λ (τ /T ) 2T N0 sinc2 (fT ) 2 e. By definition of the equivalent noise bandwidth £ ¤ 2 E Y 2 = Hmax BN N0 The output power is also R0 (0) = N0 /2T . Equating the two results and noting that Hmax = 1, gives BN = 1/2T Hz. f. Evaluate the integral of the power spectral density over all frequency: Z ∞ Z £ 2¤ N0 N0 ∞ N0 2 = R0 (0) E Y = sinc (f T ) df = sinc2 (u) du = 2T −∞ 2T −∞ 2 10 Problem 5.20 CHAPTER 5. RANDOM SIGNALS AND NOISE a. The output power spectral density is S0 (f ) = N0 /2 1 + (f /f3 )4 b. The autocorrelation function of the output, by inverse Fourier transforming the output power spectral density, is Z ∞ N0 /2 −1 −j2πf τ df R0 (τ ) = F [S0 (f )] = 4e −∞ 1 + (f /f3 ) Z ∞ Z ∞ cos (2πf τ ) cos (2πf3 τ x) = N0 df = N0 f3 dx 4 1 + x4 1 + (f /f3 ) 0 0 ³√ ´ πf3 N0 −√2πf3 τ = e cos 2πf3 τ − π/4 2 c. Yes. R0 (0) = Problem 5.21 We have |H (f )|2 = Thus |2πf | |H (f )| = q (2πf )4 + 5, 000 (2πf )2 (2πf )4 + 5, 000 πf3 N0 2 = N0 BN , BN = πf3 /2 This could be realized with a second-order Butterworth filter in cascade with a differentiator. Problem 5.22 a. The power spectral density is SY (f ) = The autocorrelation function is RY (τ ) = N0 B sinc (2Bτ ) N0 Π (f /2B) 2 5.1. PROBLEM SOLUTIONS b. The transfer function is H (f ) = A α + j2πf 11 The power spectral density and autocorrelation function are, respectively, SY (f ) = |H (f )|2 SX (f ) = = B A2 2 + (2πf )2 1 + (2πβf )2 α · ¸ 2 /α2 B 1 α2 β 2 A A2 B/α2 = − 1 − α2 β 2 1 + (2πf/α)2 1 + (2πβf )2 1 + (2πf/α)2 1 + (2πβf )2 2τ 0 Use the transform pair exp (−|τ |/τ 0 ) ←→ 1+(2πf τ )2 to find that the corresponding 0 autocorrelation function is h i A2 B/α i e−α|τ | − αβ e−|τ |/β , α 6= 1/β RY (t) = h 2 1 − (αβ)2 Problem 5.23 a. E [Y (t)] = 0 because the mean of the input is zero. b. The output power spectrum is SY (f ) = 1 1 2 (10π) 1 + (f /5)2 which is obtained by Fourier transforming the impulse response to get the transfer function and applying (5.90). c. Use the transform pair exp (−|τ |/τ 0 ) ←→ RY (τ ) = 2τ 0 1+(2πf τ 0 )2 to find the power spectrum as 1 −10π|τ | e 10π d. Since the input is Gaussian, so is the output. Also, E [Y ] = 0 and var[Y ] = RY (0) = 1/ (10π), so e−y /2σY fY (y) = q 2πσ 2 Y 2 2 12 where σ 2 = Y 1 10π . CHAPTER 5. RANDOM SIGNALS AND NOISE e. Use (4.188) with x = y1 = Y (t1 ) , y = y2 = Y (t2 ) , mx = my = 0, and σ 2 = σ 2 = 1/10π x y Also ρ (τ ) = RY (τ ) = e−10π|τ | RY (0) Set τ = 0.03 to get ρ(0.03) = 0.3897. Put these values into (4.188). Problem 5.24 Use the integrals I1 = and I2 = Z ∞ Z ∞ −∞ b0 ds jπb0 = (a0 s + a1 ) (−a0 s + a1 ) a0 a1 −∞ ¢ b0 s2 + b1 ds −b0 + a0 b1 /a2 = jπ 2 + a s + a ) (a s2 − a s + a ) (a0 s a0 a1 1 2 0 1 2 Second Order √ πfc /²2 q√ 1+1/²2 1+1/²2 −1 π 6 fc ¡ to get the following results for BN . Filter Type Chebyshev Butterworth Problem 5.25 The transfer function is H (s) = = ω2 c √ √ ¢¡ √ √ ¢ ¡ s + ω c / 2 + jω c / 2 s + ω c / 2 − jω c / 2 A A∗ √ √ ¢+¡ √ √ ¢ ¡ s + ω c / 2 + jω c / 2 s + ω c / 2 − jω c / 2 First Order πfc 2² π 2 fc where A = jωc /21/2 . This is inverse Fourier transformed, with s = jω, to yield ¶ µ √ √ ωct −ωc t/ 2 u (t) sin √ h (t) = 2ω c e 2 5.1. PROBLEM SOLUTIONS for the impulse response. Now 13 πfc ωc |h (t)|2 dt = √ = √ 4 2 2 2 −∞ R∞ after some effort. Also note that −∞ h (t) dt = H (0) = 1. Therefore, the noise equivalent bandwidth, from (5.109), is 100π πfc BN = √ Hz = √ Hz 2 2 2 1 2 Z ∞ Problem 5.26 (a) Note that Hmax = 2, H (f ) = 2 for −1 ≤ f ≤ 1, H (f ) = 1 for −3 ≤ f ≤ −1 and 1 ≤ f ≤ 3 and is 0 otherwise. Thus ¶ µZ 1 Z 2 Z ∞ 1 1 2 2 2 BN = 2 df + 1 df |H (f )| df = 2 Hmax 0 4 0 1 1 = (4 + 1) = 1.25 Hz 4 (b) For this case Hmax = 2 and the frequency response function is a triangle so that Z ∞ Z 1 1 100 BN = [2 (1 − f /100)]2 df |H (f )|2 df = 2 Hmax 0 4 0 ¯1 Z 1 ¯ 100 2 3¯ (1 − v) ¯ = 33.33 Hz (1 − v) dv = − = 3 0 0 Problem 5.27 By definition BN 1 = 2 H0 Z ∞ 0 1 |H (f )| df = 4 2 Z 400 600 · 2Λ µ f − 500 100 ¶¸2 df = 66.67 Hz Problem 5.28 a. Use the impulse response method. First, use partial fraction expansion to get the impulse response: µ ¶ ¢ 1 1 10 10 ¡ −t e − e−20t u (t) Ha (f ) = − ⇔ ha (t) = 19 j2πf + 1 j2πf + 20 19 Problem 5. The result is Snc (f ) = Sns (f ) = α2 α2 + (2πf )2 .30 a. (b) Choosing f0 = f2 moves −f2 right to f = 0 and f2 ´ to f = 0. Thus. all cases give quadrature components that are uncorrelated at the same instant. Use the result from part (a) and the modulation theorem to get p Rn (τ ) = α −α|τ | e 2 α −α|τ | e cos (2πf0 τ ) 2 c.14 Thus BN = CHAPTER 5. the autocorrelation function is Rn (τ ) = where K = α/2. By inverse Fourier transformation. Again use the impulse response method to get BN = 0. the baseband spectrum can be left ³ f 1 written as SLP (f ) = 2 N0 Λ f2 −f1 . However. b.238 Hz −∞ |h (t)| 2 R∞ b.625 Hz. (d) They are 2 not uncorrelated for any case for an arbitrary delay. Problem 5. RANDOM SIGNALS AND NOISE = = = = ¢2 ¡ 10 ¢2 R ∞ ¡ −t dt e − e−20t dt 19 0 hR i2 = £¡ 10 ¢ R ∞ ¤2 ∞ 2 19 0 (e−t − e−20t ) dt 2 −∞ h (t) dt ¢ R∞¡ 1 0 e−2t − 2e−21t + e−40t dt £¡ ¢∞ ¤2 1 2 −e−t + 20 e−20t 0 ¢∞ ¡ 2 1 1 − 1 e−2t + 21 e−21t − 40 e−40t 0 2 2 (1 − 1/20)2 µ ¶ µ ¶ µ ¶ µ ¶ 361 2 1 1 20 2 1 1 20 2 + − = 2 19 2 21 40 2 19 21 × 40 0.29 (a) Choosing f0 = f1 moves −f1 right to f = 0 and f1 left to f = 0. (c) For this case both triangles (left and right) are ³ ´ f centered around the origin and they add to give SLP (f ) = 1 N0 Π f2 −f1 . 5. ´ ³ The cross-spectral density is Snc ns (f ) = 0.32 The result is 1 1 Sn2 (f ) = Sn (f − f0 ) + Sn (f + f0 ) 2 2 . (b) For this case Snc (f ) = Sns (f ) = N0 Π 2(f2f 1 ) 2 −f and the cross-spectral density is ½ N0 − 2 . 0 ≤ f ≤ (f2 − f1 ) (c) For this case. PROBLEM SOLUTIONS and Snc ns (f ) = 0 15 Problem 5.1. (d) for part (a) Rnc ns (τ ) = 0 For part (b) Rnc ns (τ ) = = = = = # Z f1 −f2 N0 j2πf τ N0 j2πf τ e j df + df − e 2 2 0 −(f2 −f1 ) " ¯0 ¯(f −f ) # N0 −e−j2πf τ ¯ e−j2πf τ ¯ 2 1 ¯ j + j2πτ ¯ ¯ ¯ j2πτ 2 0 −(f2 −f1 ) ³ ´ N0 −1 + ej2π(f2 −f1 )τ + e −j2π(f2 −f1 )τ −1 4πτ N0 {2 − 2 cos [2π (f2 − f1 ) τ ]} − 4πτ N0 − sin2 [π (f2 − f1 ) τ ] πτ h i 0 "Z = − N0 π (f2 − f1 )2 τ sinc2 [(f2 − f1 ) τ ] For part (c). the lowpass equivalent power spectral densities are the same as for part (b) and the cross-spectral density is the negative of that found in (b). − (f2 − f1 ) ≤ f ≤ 0 Snc ns (f ) = N0 2 .31 ³ ´ f (a) The equivalent lowpass power spectral density is given by Snc (f ) = Sns (f ) = N0 Π f2 −f1 . the cross-correlation function is the negative of the above. Problem 5. add the two translated spectra. let x2T (t) = N X k=−N nk δ (t − kTs ) . the cdf of R is ½ i¾ 1 h 1 2 2 exp − 2 (n1 − A) + ns fN1 Ns (n1 . and divide the result by 2. ns ) dn1 dns FR (r) = Pr (R ≤ r) = √ 2 2 N1 +Ns ≤r Change variables in the integrand from rectangular to polar with n1 = α cos β and n2 = α sin β Then FR (r) = = Z r Z0 r 0 Z α − 12 [(α cos β−A)2 +(α sin β)2 ] dβdα e 2σ 2πσ 2 0 µ ¶ α − 12 (α2 +A2 ) αA dα I0 e 2σ σ2 σ2 2π Differentiating with respect to r. ns ) = 2πσ 2 2σ Z Z fN1 Ns (n1 . translate it to the right by f0 and to the left by f0 .34 The suggested approach is to apply Sn (f ) = lim n o E |= [x2T (t)]|2 2T T →∞ where x2T (t) is a truncated version of x (t). RANDOM SIGNALS AND NOISE so take the given spectrum.16 CHAPTER 5. we obtain the pdf of R: µ ¶ 1 2 2 rA r fR (r) = 2 e− 2σ2 (r +A ) I0 . Then R= Note that the joint pdf of N1 and Ns is q 2 2 N1 + Ns Thus. r≥0 σ σ2 Problem 5. Problem 5. Thus.33 Define N1 = A + Nc . Consider a 2nT chunk of the waveform and obtain its Fourier transform as = [X2nT (t)] = n−1 X Aτ 0 sinc (f τ 0 ) e−j2πf (∆tk +τ 0 /2+kT ) k=−n Take the magnitude squared and then the expectation to reduce it to n−1 n−1 n h i o X X E e−j2πf [∆tk −∆tm +(k−m)T ] E |X2nT (f )|2 = A2 τ 2 sinc2 (f τ 0 ) 0 k=−n m=−n . PROBLEM SOLUTIONS The Fourier transform of this truncated waveform is = [x2T (t)] = Thus.35 To make the process stationary. T ]. assume that each sample function is displaced from the origin by a random delay uniform in (0.5.1. N X 17 nk e−j2πkTs k=−N ¯ ¯2  N ¯  ¯ X n o ¯ ¯ E |= [x2T (t)]|2 nk e−j2πkTs ¯ = E ¯ ¯  ¯ k=−N ) ( N N X X nk e−j2πkTs nl ej2πlTs = E k=−N l=−N = k=−N l=−N N X = k=−N l=−N N N X X N X N X E [nk nl ] e−j2π(k−l)Ts Rn (0) δ kl e−j2π(k−l)Ts = But 2T = 2N Ts so that Sn (f ) = = lim Rn (0) = (2N + 1) Rn (0) k=−N T →∞ n o E |= [x2T (t)]|2 2T = lim Rn (0) Ts N →∞ (2N + 1) Rn (0) 2N Ts Problem 5. the expectation inside the double sum is 1.18 CHAPTER 5. because the numerator is finite and the denominator goes to infinity. RANDOM SIGNALS AND NOISE We need the expectation of the inside exponentials which can be separated into the product of the expectations for k 6= m.36 The result is £ ¤ 1 E y2 (t) = 2 T Z t+T t Z t+T t £ 2 ¤ 2 RX (τ ) + RX (λ − β) + RX (λ − β − τ ) + RX (λ − β + τ ) dλdβ . the limit can be shown to have the properties of a delta function as n → ∞. which is 4 E {} = T Z T /4 e−j2πf ∆tk d∆tk = e−j2πf (k−m)T sinc2 (f τ /4) 0 For k = m. After considerable manipulation. For f = mT . m an integer. Thus " # ∞ X A2 τ 2 0 SX (f ) = δ (f − m/T ) sinc2 (f τ 0 ) 1 − sinc (f T /4) + 2 T m=−∞ Problem 5. the psd becomes A2 τ 2 0 sinc2 (f τ 0 ) [1 − sinc (fT /4)] T · ¸ 2A2 τ 2 sinc2 (τ 0 f ) sin2 (nπfT ) 2 0 + cos (πfT ) lim n→∞ n sin2 (πf T ) T SX (f ) = Examine L = lim sin2 (nπf T ) n sin2 (πf T ) n→∞ It is zero for f 6= m/T . 37 19 1. PROBLEM SOLUTIONS Problem 5.67π 3 N0 B 3 (c) No.5. Thus. Hence. The variance for Z = dy is found from its Y dt power spectral density. the random process and its derivative are independent. (a) In the expectation for the cross correlation function. Using the fact that the transfer function of a differentiator is j2πf we get SZ (f ) = (2πf )2 so that σ2 Z = 2π N0 = 2 N0 Π (f/2B) 2 4 2 π N0 B 3 3 Z ¯B f3 ¯ f df = 2π N0 ¯ 3 ¯−B −B B 2 2 dR (τ ) y The two processes are uncorrelated because Ryy_dotp (0) = dτ = 0 because the derivative of the autocorrelation function of y (t) exists at τ = 0 and therefore must be 0. . because the derivative of the autocorrelation function at τ = 0 of such noise does not exist (it is a double-sided exponential). β) = 2πN0 B 2.67π 2 N0 B 3 √ √ fY Z (α. their joint pdf is ¢ ¢ ¡ ¡ exp −α2 /2N0 B exp −β 2 /2. write the derivative as a limit: Ryy_dotp · ¸ dy (t + τ ) (τ ) = E y (t) dt ¸¾ ½ · y (t + τ + ²) − y (t + τ ) = E y (t) lim ²→0 ² 1 = lim {E [y (t) y (t + τ + ²)] − E [y (t) y (t + τ )]} ²→0 ² Ry (τ + ²) − Ry (τ ) = lim ²→0 ² dRy (τ ) = dτ (b) The variance of Y = y (t) is σ 2 = N0 B.1. end EX = mean(X).*X). EX2 = mean(X.:).2 Computer Exercises Computer Exercise 5. t = 0:. AVEX = mean(X’).20 CHAPTER 5.*X)’).\theta)’) if n == 1 title(‘Plot of first five sample functions’) end if n == 5 . RANDOM SIGNALS AND NOISE 5.Y). theta = 2*pi*rand(1.plot(t. 60. 20. & 100)’) disp([AVEX(1) AVEX(20) AVEX(40) AVEX(60) AVEX(80) AVEX(100)]) disp(‘ ’) disp(‘Sample mean squares’) disp(EX2) disp(‘ ’) disp(‘Typical time-average mean squares’) disp([AVEX2(1) AVEX2(20) AVEX2(40) AVEX2(60) AVEX2(80) AVEX2(100)]) disp(‘ ’) for n = 1:5 Y = X(n.ylabel(‘X(t.m: Computes approximations to mean and mean-square ensemble and time % averages of cosine waves with phase uniformly distributed between 0 and 2*pi % clf f0 = 1.100). X = [].1. disp(‘ ’) disp(‘Sample means (across 100 sample functions)’) disp(EX) disp(‘ ’) disp(‘Typical time-average means (sample functions 1. A*cos(2*pi*f0*t+theta(n))]. AVEX2 = mean((X.1 % ce5_1. 80. A = 1.n). for n = 1:100 X = [X. 40.1:1. subplot(5. 2.5387 0.5399 0.0998 -0.1 to theta = (pi/2)*(rand(1.0. they should be zero. .0292 Columns 8 through 11 0.100).0875 -0.4717 0. COMPUTER EXERCISES xlabel(‘t’) end end A typical run follows: >> ce5_1 Sample means (across 100 sample functions) Columns 1 through 7 0.5136 0.4960 0.4960 Typical time-average mean squares 0. but the ensembleaverage means will vary depending on the time at which they are computed.5499 Columns 8 through 11 0.5381 0.2 by printing the covariance matrix.100) .4477 0.4628 Computer Exercise 5. 40.5277 0. The diagonal terms give the variances of the X and Y vectors and the off-diagonal terms give the correlation coefficients.1. 60.5).0909 Typical time-average means (sample functions 1.0998 0.5348 0.0909 -0.5499 0.0815 -0.5.0872 0. Ideally.0292 -0.1179 -0. 80.4960 0.0437 0.1179 0.2 This is a matter of changing the statement theta = 2*pi*rand(1. 21 The time-average means will be the same as in Computer Exercise 5.5348 0.4717 0. 20.0273 Sample mean squares Columns 1 through 7 0. Computer Exercise 5. in the program of Computer Exercise 5. & 100) -0.3 This was done in Computer exercise 4.0881 0.0733 -0.0437 -0.4477 0.0909 0. ’. sigma = input(‘Enter desired value of sigma ’).‘-. n = 1.^(KK/10).‘:’.*exp(-(r. r = 0:. KdB(n) = KK. A(n.^2/(2*sigma^2)+K)).. fR = (r/sigma^2).1: Computer Exercise 5. RANDOM SIGNALS AND NOISE Figure 5. plot(r.m: plot of Ricean pdf for several values of K % clf A = char(‘-’.’.‘—’.22 CHAPTER 5.‘-. KdB = [].1:15*sigma. K = 10. ylabel(’f_R(r)’) grid on . fR.’).4 % ce5_4. sqrt(2*K)*(r/sigma)).*besseli(0. for KK = -10:5:15.‘—.:)) if n == 1 hold on xlabel(’r’). [‘K = ’.[‘K = ’.‘ dB’].‘ dB’]. num2str(KdB(4)).5. num2str(KdB(5)).‘ dB’]. COMPUTER EXERCISES 23 Figure 5.2: end n = n+1.2. num2str(KdB(2)). [‘K = ’. num2str(sigma)]) A plot for several values of K is shown below: . end legend([‘K = ’.‘ dB’]) title([‘Ricean pdf for \sigma = ’.[‘K = ’. num2str(KdB(6)). num2str(KdB(3)).‘ dB’].‘ dB’]. num2str(KdB(1)). [‘K = ’. 116). The received signal is xr (t) = Ac [m (t) cos (!c t + µ) ¨ m (t) sin (!c t + µ)] b 1 . Thus for large n.2 We express n (t) as ¸ · ¸ 1 1 n (t) = nc (t) cos !c t § (2¼W) t + µ + ns (t) sin ! ct § (2¼W ) t + µ 2 2 · PT sin (¼=2n) PT = N0BN ¼=2n N0 W where we use the plus sign for the U SB and the minus sign for the LSB. the SNR at the …lter output is close to PT =N0 W . where BN is the noise-equivalent bandwidth of the …lter. f (n) = Problem 6. From (5. we know that the noise-equivalent bandwidth of an nth order Butterworth …lter with 3 dB bandwidth W is Bn (n) = ¼W=2n sin (¼=2n) Thus.1 Problems Problem 6. The noise power is N0BN . the signal-to-noise ratio at the …lter output is SNR = so that f (n) is given by sin (¼=2n) ¼=2n We can see from the form of f (n) that since sin (x) ¼ 1 for x ¿ 1.1 The signal power at the output of the lowpasss …lter is PT . The plot is shown In Figure 6. f (1) = 1.1.Chapter 6 Noise in Modulation Systems 6. 5 0.3 The received signal and noise are given by xr (t) = Acm (t) cos (! ct + µ) + n (t) At the output of the predetection …lter.9 0. The power spectral densities of nc (t) and ns (t) are illustrated in Figure 6. the signal and noise powers are given by 1 ST = A2m2 2 c The predetection SNR is (SN R)T = A2m2 c 2N0 BT NT = n2 = N0 BT . NOISE IN MODULATION SYSTEMS 1 0.2 CHAPTER 6.6 f(n) 0.3 0.1 0 1 2 3 4 5 n 6 7 8 9 10 Figure 6.7 0.2 0.1: Multiplying xr (t) + n (t) by 2 cos (! ct + µ) and lowpass …ltering yields yD (t) = Acm (t) + nc (t) cos (¼W t) § ns (t) sin (¼W t) From this expression. Problem 6.2.8 0.4 0. we can show that the postdetection signal and noise powers are given by SD = A2 m2 c ST = Ac m2 ND = N0W NT = N0 W This gives a detection gain of one. yD (t) is y D (t) = Ac m (t) + nc (t) The output signal power is A2m2 and the output noise PSD is shown in Figure 6. c Case I: BD > 1 BT 2 For this case.3: If the postdetection …lter passes all of the nc (t) component. all of the noise.1. and the output noise power is Z 1 BT 2 ND = N0df = 2N0 BT This yields the detection gain ¡ 1 BD 2 (SNR)D A2 m2=N0BT = c =2 (SN R)T A2 m2 =2N0BT c . Sns ( f ) N0 f 1 − W 2 1 W 2 Figure 6. is passed by the postdetection …lter. PROBLEMS 3 S nc ( f ) .6. nc (t).2: Snc ( f ) N0 f 1 − BT 2 0 1 B 2 T Figure 6.3. 3 except that the predetection signal power is i 1 h ST = A2 1 + a2m2 c n 2 and the postdetection signal power is SD = A2 a2 m2 c n The noise powers do not change. Case I: BD > 1 BT 2 (SNR)D A2 a2 m2 =N0BT 2a2 m2 n n h c i = = 2 m2 (SN R)T 2 1 + a2m2 =2N B 1+ a n Ac 0 T n Case II: BD < 1 BT 2 (SNR)D A2a2m2 =2N0 BD a2m2 B n i n hc = = = T 2 m2 (SN R)T BD 2 1 + a2m2 =2N B 1+ a n Ac 0 T n Problem 6.5 Since the message signal is sinusoidal mn (t) = cos (8¼t) .4 CHAPTER 6. the postdetection …lter limits the output noise and the output noise power is Z BD ND = N0df = 2N0BD This case gives the detection gain ¡BD (SN R)D A2m2=2N0BD BT = c = (SNR)T BD A2m2=2N0 BT c Problem 6. NOISE IN MODULATION SYSTEMS Case II: BD < 1 BT 2 For this case.4 This problem is identical to Problem 6. 29). PROBLEMS Thus. the output SNR is (SNR)D = 0:2424 = ¡6:15 dB PT =N0W If the modulation index in increased to 0:9. (SN T)D = 0:2424 Relative to baseband.6.33). the detection gain is (SN R)D = 2Eff = 0:4848 = ¡3:14dB (SNR)T and the output SNR is. m2 = 0:5 n The e¢ciency is therefore given by Ef f = (0:5) (0:8)2 = 0:2424 = 24:24% 1 + (0:5) (0:8)2 5 From (6. is (SNR)D = 0:2883 = ¡5:40 dB PT =N0W This represents an improvement of 0:75 dB. the e¢ciency becomes Ef f = This gives a detection gain of (SN R)D = 2Eff = 0:5765 = ¡2:39dB (SNR)T The output SNR is (SNR)D = 0:2883 which. from (6. relative to baseband.6 PT N0W (0:5) (0:9)2 = 0:2883 = 28:83% 1 + (0:5) (0:9)2 PT N0 W . Problem 6.1. First we compute µ ¶ Z 1 1 M ¡y2=2¾2 p P fX > Mg = e dy = Q = 0:005 ¾ 2¼¾ M This gives Thus. we have 2 Ef f = and the detection gain is 0:151 m2 ¾2 = = 0:151 6:605¾2 (2:57¾)2 ¡ 1 ¢2 2 2 M = 2:57 ¾ m (t) 2:57¾ 1 + 0:151 (SNR)D = 2E = 0:0728 (SN R)T Problem 6. since m2 = ¾2 m2 = n Since a = 1 . Thus fRn (rn ) = r ¡r2 =2N e N where N is the predetection noise power.7 The output of the predetection …lter is ¡ 1 ¢2 = 3:64% e (t) = Ac [1 + amn (t)] cos [!c t + µ] + rn (t) cos [! c t + µ + Án (t)] The noise function rn (t) has a Rayleigh pdf.6 CHAPTER 6. M = 2:57¾ and mn (t) = Thus. NOISE IN MODULATION SYSTEMS The …rst step is to determine the value of M. This gives 1 1 N = n2 + n2 = N0B T c 2 2 s From the de…nition of threshold 0:99 = Z Ac 0 r ¡r 2=2N e dr N . (SNR)D. (SNR)D.8 Since m (t) is a sinusoid.6. µ 1 1 2 2a + 1 a2 2 = a2 2 + a2 a2 PT 2N W 2+ a 0 ¶ + 10 log 10 PT N0 W a2 2 + a2 (SNR)D.dB = 10 log 10 For a = 0:9.dB = 10 log 10 For a = 0:7. PROBLEMS which gives Thus. 7 0:99 = 1 ¡ e¡Ac =2N ¡ A2 c = 1n (0:01) 2N A2 = 9:21 N c 2 which gives The predetection signal power is i 1 1 h PT = Ac 1 + a2 m2 ¼ A2 [1 + 1] = A2 n c 2 2 c PT A2 = c = 9:21 ¼ 9:64dB N N which gives Problem 6.1. m2 = 1 .dB = 10 log 10 PT ¡ 11:3033 N0W PT ¡ 9:5424 N0 W PT ¡ 7:0600 N0 W PT ¡ 5:4022 N0 W .dB = 10log10 For a = 0:4.dB = 10log10 For a = 0:5. (SNR)D. and the e¢ciency is n 2 Ef f = and the output SNR is (SN R)D = Eff = In dB we have (SNR)D. The …rst term is µ ¶ 1 2 1 2 1 z1 (t) = Ac 1 + a + A2 a cos !m t + A2 a2 cos 2! mt c 2 2 4 c z2 (t) = Acnc (t) z3(t) = Acanc (t) cos !m t . The modulation index only biases (SNR)D. Problem 6. The received signal (with noise) at the predetection …lter output is represented xr (t) = Ac [1 + amn (t)] cos !c t + nc (t) cos ! c t + ns (t) sin ! ct which can be represented xr (t) = fAc [1 + amn (t)] + nc (t)g cos ! c t ¡ ns (t) sin !c t The output of the square-law device is y (t) = fAc [1 + amn (t)] + nc (t)g2 cos2 ! ct ¡fAc [1 + amn (t)] + nc (t)gns (t) cos ! c (t) +n2 (t) sin2 cos !c (t) s Assuming that the postdetection …lter removes the double frequency terms. yD (t) can be written 1 1 yD (t) = fAc [1 + amn (t)] + nc (t)g2 + n2 (t) 2 2 s Except for a factor of 2. We now examine the power spectral density of each term. NOISE IN MODULATION SYSTEMS The plot of (SNR)dB as a function of PT =N0W in dB is linear having a slope of one. We now let mn (t) = cos ! mt and expand.50). this is (6.9 Let the predetection …lter bandwidth be BT and let the postdetection …lter bandwidth be BD.dB down by an amount given by the last term in the preceding four expressions. This gives yD (t) = 1 2 A [1 + a cos !m t]2 + Ac nc (t) 2 c 1 1 +Acanc (t) cos !m t + n2 (t) + n2 (t) 2 c 2 s We represent yD (t) by y D (t) = z1 (t) + z2 (t) + z3 (t) + z4 (t) + z5 (t) where zi (t) is the ith term in the expression for yD (t).8 CHAPTER 6. we have. from (6.10 Assume sinusoidal modulation since sinusoidal modulation was assumed in the development of the square-law detector. Adding these …ve terms together gives the PSD of yD (t). Problem 6.` This is approximately ¡1:8 dB.S` = 2 2 + a2 Taking the ratio (SN R)D.4. Problem 6. we use the coherent result). For a = 1 and mn (t) = cos (2¼fm t) so that m2 = 0:5. PROBLEMS 1 z4(t) = n2 (t) 2 c and 9 1 z5(t) = n2 (t) 2 s A plot of the PSD of each of these terms is illiustrated in Figure 6.1. n for linear envelope detection (since PT =N0 W À 1.S` = (SNR)D.6.61) with a = 1 a (SNR)D. (SNR)D.11 For the circuit given H (f ) = and R 1 ³ ´ = R + j2¼f L 1 + j 2¼f L R 1+ ³ 1 2¼f L R 2 PT 9 N0 W 1 PT 3 N0 W µ ¶2 = PT 2 PT = N0W 9 N0W 2 = ¡1:76 dB 3 jH (f)j2 = The output noise power is Z 1 N0 N= ¡1 2 ´2 µ ¶ R NR dx = 0 2¼L 4L 1+ df ³ 2¼fL R ´2 = N0 Z 1 0 1 1 + x2 .` = Ef f 1 PT a2 m2 PT PT 1 PT n = = 2 1 = 2m2 N0W N0 W 3 N0W 1 + 2 N0 W 1 +a n For square-law detection we have. Sz5 ( f 1 BT + f m 2 ) 1 2 N0 BT 4 f 0 Figure 6. NOISE IN MODULATION SYSTEMS Sz1 ( f ) 2 1 4 AC 64 1 4 2 AC a 4 1 4 1 2 AC  1 + a  4  2  1 4 2 AC a 4 0 1 4 4 AC a 64 f −2 f m − fm − Sz 2 ( f ) fm − 2 AC N0 2 fm f 1 − BT 2 0 Sz3 ( f ) 1 BT 2 f 1 1 − BT − f m − BT + f m 0 2 2 1 BT − f m 2 Sz4 ( f ) .10 CHAPTER 6.4: . the preceding expression becomes Z 2¼RCW N0 x2 N= dx 2¼RC 0 1 + x2 Since x2 1 =1¡ 1 + x2 1 + x2 .1. Sn (f ) = jf j > W 0. PROBLEMS The output signal power is S= This gives the signal-to-noise ratio S 2A2 RL h i = N N0 R 2 + (2¼fc L)2 Problem 6.13 The transfer function of the RC highpass …lter is H (f ) = so that H (f) = R j2¼f RC = 1 1 + j2¼f RC R + j2¼C (j2¼f RC)2 1 + (j2¼fRC)2 Thus.12 For the RC …lter 11 1 A2R2 2 R2 + (2¼fc L)2 S 2A2 RL h i = N N0 1 + (2¼fc RC)2 Problem 6.6. The PSD at the output of the ideal lowpass …lter is ( 2 N0 (2¼fRC) jf j < W 2 1+(2¼fRC)2 . The noise power at the output of the ideal lowpass …lter is Z W Z W (2¼f RC)2 N= Sn (f ) df = N0 df 1 + (2¼fRC)2 ¡W 0 with x = 2¼f RC. for a given acceptable level of ²2 . Problem 6. The plots follow by simply plotting the two preceding expressions.14 For the case in which the noise in the passband of the postdetection …lter is negligible we have ²2 = ¾2 . S N ! 0.15 From the series expansions for sin Á and cos Á we can write 1 1 cos Á = 1 ¡ Á2 + Á4 2 24 1 3 sin Á = Á ¡ Á 6 . D 4 Á DSB Note that for reasonable values of phase error variance. NOISE IN MODULATION SYSTEMS ½Z Z ¾ 2¼RCW 2¼RCW 0 df ¡ 0 dx 1 + x2 ¢ N0 ¡ 2¼RCW ¡ tan¡1 (2¼RCW ) 2¼RC N0 tan¡1 (2¼RCW ) 2¼RC N = N0 W ¡ 1 A2 (2¼fc RC)2 S = A2 jH (fc )j2 = 2 2 1 + (2¼fcRC)2 Thus. Another way of viewing this is to recognize that. Problem 6.12 we can write N0 N= 2¼RC or N= which is The output signal power is CHAPTER 6. DSB is much less Á sensitive to phase errors in the demodulation carrier than SSB or QDSB. SSB and QDSB Q Á and 3 ²2 = ¾4 . namely ¾2 ¿ 1. the phase error variance for can be much greater for DSB than for SSB or QDSB. the signal-to-noise ratio is S A2 (2¼fcRC)2 2¼RC = N 2N0 1 + (2¼fc RC)2 2¼RCW ¡ tan¡1 (2¼RCW ) Note that as W ! 1. 70) and recognizing that m1 (t). For DSB. we have Á Á "2 = ¾ 2 ¾ 2 + ¾ 2 m Á n which yields (6.73). we let ¾2 1 = ¾2 and ¾2 2 = 0. and discarding all terms Ák for k > 4. PROBLEMS Squaring these. This gives (6.79) m m m 3 "2 = ¾2 ¾4 + ¾2 n 4 m Á 13 which can be expressed . This gives m m m µ ¶ 1 4 2 2 2 = ¾2 " m2 ¾Á ¡ ¾ Á + ¾ n 4 For ¾2 >> ¾4 .6. m2 (t). yields 1 cos2 Á = 1 ¡ Á2 + Á4 3 1 4 sin2 Á = Á2 ¡ Á 3 Using (6. yields "2 = ¾2 1 ¡ 2¾2 1 cos Á + ¾2 1 cos2 Á + ¾2 2 sin2 Á + ¾2 m m m m n Assuming Á (t) to be a zero-mean process and recalling that Á4 = 3¾4 Á gives 1 1 cos Á = 1 + ¾2 + ¾4 2 Á 8 Á cos2 Á = 1 ¡ ¾2 + ¾4 Á Á sin2 Á = ¾2 ¡ ¾4 Á Á This yields "2 = ¾2 1 ¡ 2¾2 1 m m µ ¶ ¡ ¢ ¡ ¢ 1 1 1 + ¾2 + ¾4 + ¾2 1 1 ¡ ¾2 + ¾4 + ¾2 2 ¾2 ¡ ¾4 + ¾2 Á Á m Á Á m Á Á n 2 8 3 "2 = ¾ 2 1 ¾ 4 + ¾ 2 2 ¾ 2 ¡ ¾ 2 2 ¾ 4 + ¾ 2 m Á m Á n 4 m Á For QDSB we let ¾2 1 = ¾2 2 = ¾2 . and Á (t) are independent.1. 14 CHAPTER 6. NOISE IN MODULATION SYSTEMS Problem 6.16 From (6.73) and (6.79), we have 0:05 = ¾2 + Á 0:05 = ¾2 n ; SSB ¾2 m 3 ¡ 2 ¢2 ¾2 n ¾ + 2 ; DSB 4 Á ¾m SSB DSB Thus we have the following signal-to-noise ratios ¾2 m = ¾2 n ¾2 m = ¾2 n 1 ; 0:05 ¡ ¾2 Á 1 3 4 The SSB characteristic has an asymptote de…ned by ¾2 = 0:05 Á and the DSB result has an asymptote de…ned by ³ ´2 ; 0:05 ¡ ¾2 Á or 3 ¡ 2 ¢2 ¾ = 0:05 4 Á ¾2 = 0:258 Á The curves are shown in Figure 6.5. The appropriate operating regions are above and to the left of the curves. It is clear that DSB has the larger operating region. Problem 6.17 Since we have a DSB system 3 ¾2 n "2 = ¾ 4 + 2 N 4 Á ¾m Let the bandwidth of the predetection …lter be BT and let the bandwidth of the pilot …lter be B Bp = T ® This gives ¾2 NB B NB ® n = 0 2 T = T 02 p = 2 ¾m ¾m Bp ¾m ½ 6.1. PROBLEMS 15 1000 900 800 700 600 500 SNR 400 300 200 100 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Phase error variance SSB DSB Figure 6.5: From (6.85) and (6.86), we have ¾2 = Á so that "2 = N For an SNR of 15 dB ½ = 101:5 = 31:623 Using this value for ½ and with k = 4, we have ¡ ¢ "2 = 7:32 10¡7 + 0:032® N 1 1 k2 2½ 3 1 ® 4½ 2 + ½ 16 k The plot is obviously linear in ® with a slope of 0:032. The bias, 7:32 £ 10 ¡7 , is negligible. Note that for k > 1 and reasonable values of the pilot signal-to-noise ratio, ½, the …rst term (the bias), which arises from pilot phase jitter, is negligible. The performance is limited by the additive noise. Problem 6.18 The mean-square error © ª "2 (A; ¿) = E [y (t) ¡ Ax (t ¡ ¿)]2 16 can be written CHAPTER 6. NOISE IN MODULATION SYSTEMS In terms of autocorrelation and cross correlation functions, the mean-square error is "2 (A; ¿) = Ry (0) ¡ 2ARxy (¿) + A2Rx (0) Letting Py = R y (0) and Px = Rx (0) gives "2 (A; ¿) = Py ¡ 2ARxy (¿) + A2Px In order to minimize "2 we choose ¿ = ¿ m such that Rxy (¿) is maximized. This follows since the crosscorrelation term is negative in the expression for "2 . Therefore, "2 (A; ¿ m) = Py ¡ 2ARxy (¿ m ) + A2 Px The gain Am is determined from d"2 = ¡2ARxy (¿ m ) + A2 Px = 0 dA which yields Am = This gives the mean-square error "2 (Am; ¿ m) = Py ¡ 2 which is "2 (Am; ¿ m ) = Py ¡ The output signal power is © ª SD = E [Amx (t ¡ ¿ m)]2 = A2 Px m SD = R2 (¿ m) R2 (¿ m) xy = xy Px Rx (0) 2 R2 (¿ m ) Rxy (¿ m) xy + Px Px 2 Pxy (¿ m ) Px © ª "2 (A; ¿) = E y 2 (t) ¡ 2Ax(t ¡ ¿)y (t) + A2 x2 (t ¡ ¿) Rxy (¿ m ) Px which is Since ND is the error "2(Am; ¿ m ) we have R2 (¿ m ) SD xy = ND Rx (0) Ry (0) ¡ R2 (¿ m ) xy 6.1. PROBLEMS 17 Note: The gain and delay of a linear system is often de…ned as the magnitude of the transfer and the slope of the phase characteristic, respectively. The de…nition of gain and delay suggested in this problem is useful when the magnitude response is not linear over the frequency range of interest. Problem 6.19 The single-sided spectrum of a stereophonic FM signal and the noise spectrum is shown in Figure 6.6. The two-sided noise spectrum is given by SnF (f ) = 2 KD N0f , A2 C ¡1< f < 1 The predetection noise powers are easily computed. For the L + R channel Z 15;000 2 2 ¡ ¢ KD KD Pn;L+R = 2 N0 f 2 df = 2:25 1012 N0 A2 A2 0 C C For the L ¡ R channel Pn;L¡R = 2 Z 53;000 2 2 ¡ ¢ KD KD N0 f 2df = 91:14 1012 N0 A2 A2 C C 23;000 Thus, the noise power on the L ¡ R channel is over 40 times the noise power in the L + R channel. After demodulation, the di¤erence will be a factor of 20 because of 3 dB detection gain of coherent demodulation. Thus, the main source of noise in a stereophonic system is the L ¡ R channel. Therefore, in high noise environments, monophonic broadcasting is preferred over stereophonic broadcasting. Problem 6.20 The received FDM spectrum is shown in Figure 6.7. The kth channel signal is given by xk (t) = Ak mk (t) cos 2¼kf1t Since the guardbands have spectral width 4W , f k = 6kW and the k th channel occupies the frequency band (6k ¡ 1) W · f · (6k + 1) W Since the noise spectrum is given by SnF = 2 KD N0 f 2 , 2 AC jfj < BT The noise power in the kth channel is Z (6k+1)W i BW 3 h Nk = B f 2df = (6k + 1)3 ¡ (6k ¡ 1)3 3 (6k¡1)W 18 CHAPTER 6. NOISE IN MODULATION SYSTEMS Pilot Noise Spectrum L ( f ) + R( f ) L( f ) + R( f ) S nF ( f ) 0 15 23 53 f ( kHz ) Figure 6.6: where B is a constant. This gives Nk = ¢ BW 3 ¡ 216k2 + 2 » 72BW 3 k2 = 3 The signal power is proportional to A2 . Thus, the signal-to-noise ratio can be expressed as k µ ¶2 ¸A2 Ak k (SNR)D = 2 = ¸ k k where ¸ is a constant of proportionality and is a function of KD,W ,AC ,N0. If all channels are to have equal (SNR)D, Ak =k must be a constant, which means that Ak is a linear function of k. Thus, if A1 is known, Ak = k A1; k = 1; 2; :::; 7. A1 is …xed by setting the SNR of Channel 1 equal to the SNR of the unmodulated channel. The noise power of the unmodulated channel is Z W B N0 = B f 2df = W 3 3 0 yielding the signal-to-noise ratio (SNR)D = where P0 is the signal power. Problem 6.21 P0 B W3 3 6.1. PROBLEMS 19 Noise PSD f 0 f1 f2 f3 f4 f5 f6 f7 Figure 6.7: From (6.132) and (6.119), the required ratio is ³ ´ K2 3 2 AD N0 f3 W ¡ tan¡1 W 2 f3 f3 C R= 2 KD 2 3 3 A2 N0W C or R=3 This is shown in Figure 6.8. For f3 = 2:1kHz and W = 15kHz, the value of R is µ ¶2 µ ¶ 2:1 15 ¡1 15 R=3 ¡ tan = 0:047 15 2:1 2:1 Expressed in dB this is The improvement resulting from the use of preemphasis and deemphasis is therefore 13:3 dB. Neglecting the tan ¡1 (W=f3) gives an improvement of 21 ¡ 8:75 = 12:25 dB. The di¤erence is approximately 1 dB. Problem 6.22 From the plot of the signal spectrum it is clear that k= A W2 R = 10log10 (0:047) = ¡13:3 dB µ f3 W ¶3 µ W W ¡ tan¡1 f3 f3 ¶ the SNR becomes (SNR)2 = 2 A 3 N0 2 AW 3 N0B W This increases the signal-to-noise ratio by a factor of B=W .4 0. This yields the signal-to-noise ratio (SNR)1 = If B is reduced to W.6 R 0. Problem 6.2 0.8 0.1 0 0 1 2 3 4 5 W/f3 6 7 8 9 10 Figure 6.23 From the de…nition of the signal we have x(t) = A cos 2¼fc t dx = ¡2¼fc A sin 2¼f ct dt d2x = ¡(2¼fc )2 A cos 2¼fct dt2 .7 0. NOISE IN MODULATION SYSTEMS 1 0.5 0.8: Thus the signal power is S =2 Z A 2 2 f df = AW 2 W 3 0 The noise power is N0B.20 CHAPTER 6.3 0.9 0. the signal-to-noise ratio at y (t) is 5 A2 (SNR)D = 2 N0W µ fc W ¶4 Z W 21 N0 (2¼f )4 2 16 N ¼ 4W 5 5 0 f 4df = ¡W Problem 6.25 The signal-to-noise ratio at y (t) is 2A (SNR)D = N0 µ f3 W ¶ tan¡1 W f3 . PROBLEMS The signal component of y (t) therefore has power 1 4 SD = A2 (2¼fc )4 = 8A2¼4 fc 2 The noise power spectral density at y (t) is Sn (f ) = so that the noise power is N ND = 0 (2¼)4 2 Thus.6. The output signal is ys (t) = A cos 2¼f ct and the output signal power is 1 SD = A2 2 The noise power is the same as in the previous problem. thus ND = This gives the signal-to-noise ratio (SNR)D = 5 A2 32¼4 N0W 5 16 N0¼4W 5 5 Problem 6. the output signal is equal to the input signal.24 Since the signal is integrated twice and then di¤erentiated twice.1. 9 0. x m j±f j = jxj = Z jxj ¡x2 =2¾2 x dx p e 2¼¾x ¡1 r 1 Recognizing that the integrand is even gives j±fj = Therefore.5 0.1 0 1 2 3 4 5 W/f3 6 7 8 9 10 Figure 6. Thus.2 0.4 0. NOISE IN MODULATION SYSTEMS 1 0.22 CHAPTER 6. de…ned by (SNR)D =(2A?N0).8 0. is illustrated in Figure 6.7 Normalized SNR 0.9: The normalized (SNR)D.9. Problem 6.6 0.26 The instantaneous frequency deviation in Hz is ±f = fdm (t) = x (t) 2 where x (t) is a zero-mean Gaussian process with variance ¾2 = fd ¾2 . j±fj = r 2 f ¾m ¼ d r 2 1 ¼ ¾x Z 1 xe ¡x2 =2¾ 2 x dx = 0 2 ¾x ¼ .3 0. by letting D= and BT = 2 (D + 1) W fd W Problem 6.e. With Pw = nPb.6. · ¸ PT 1 ¼ ¡2 ln 2(1¡2n) = 2(2n ¡ 1) ln(2) + 2 ln(2) N0Bp n The values of PT =N0Bp for n = 4. PROBLEMS Substitution into (6. along with the solution follow. D. the e¤ect of quantizating errors and word errors are equivalent. i. 8.148) gives 3 ·q P m2 N0T n W ¸ q h i A2 ¡A2 2 ¾ C C + 6 ¼ fdWm exp 2N0 BT N0BT 23 (SN R)D = p 1 + 2 3 BT Q W ³ fd W ´2 The preceding expression can be placed in terms of the deviation ratio.) Assume a PCM system in which the the bit error probability Pb is su¢ciently small to justify the approximation that the word error pobability is Pw ¼ nPb. The new problem.175) gives 2¡2n = nPb(1 ¡ 2¡2n) Solving for Pb we have Pn = From (6.1.27 (Note: This problem was changed after the …rst printing of the book. Also assume that the threshold value of the signal-to-noise ratio.1. · ¸ P 1 exp ¡ T = 2Pb ¼ 2(1¡2n) 2N0 Bp n . Compare the results with Figure 6. equating the two terms in the denominator of (6. and 12 are given in Table 6. de…ned by (6. occurs when the two denominator terms are equal..5.22 shows close agreement. 8. Comparison with Figure 6.175).22 derived in Example 6. and 12. Using this assumption derive the threshold value of PT =N0Bp in dB for n = 4.178) 1 2¡2n 1 ¼ 2¡2n n 1 ¡ 2¡2n n Solving for PT =N0Bp gives. at threshold. NOISE IN MODULATION SYSTEMS Table 6.28 The peak value of the signal is 7:5 so that the signal power is SD = 1 (7:5)2 = 28:125 2 If the A/D converter has a wordlength n.1: Threshold value of PT =N0Bp 10:96 dB 13:97 dB 15:66 dB n 4 8 12 Problem 6. the width of each quantization level is SD = This gives ¡ ¢ 15 = 15 2¡n n 2 ¡ ¢ ¡ ¢ 1 2 1 SD = (15)2 2 ¡2n = 18:75 2¡2n 12 12 This results in the signal-to-noise ratio "2 = SNR = ¡ ¢ SD 28:125 ¡ 2n ¢ = 2 = 1:5 22n 18:75 "2 . Since these span the peak-to-peak signal range.24 CHAPTER 6. there are 2n quantizing levels. 078 × 10−7 × 20000 = 0.Chapter 7 Binary Data Communication 7.078 PE = Q Thus A2 = 9. ³q ´ 2 z|required .1 The signal-to-noise ratio is z= A2 T A2 = N0 N0 R ¢ ¡√ ¢ ¡ Trial and error using the asymptotic expression Q (x) ≈ exp −x2 /2 / 2πx shows that ³√ ´ 2z = 10−5 for z = 9.135 V Problem 7.078 N0 R or A = p 9. By trial and error. z = Eb /N0 PE |desired = Q 1 .2 The bandwidth should be equal to the data rate to pass the main lobe of the signal spectrum.58 dB = 9.1 Problem Solutions Problem 7. solving the following equation for z.078N0 R p = 9. we have signal power = A2 = z|required N0 R = 1010.565 W (c) For R = B = 500.08 × 10−3 W 9. R. the result is signal power = A2 = 0. 10−4 .65 W (d) For R = B = 1. 000 10.58 dB (9. 9. choose + A < ε. required R = z|required N0 R For example.3 × 10−2 W 1.0565 W . 10. choose − A = 11. 000 bps.90 ratio). 000 bps. respectively. 000 where z|required = 10. The required signal power is Ps.76 × 10−1 W PE = 10−4 6. 10−5 . 000 bps. giving the following results.9 × 10−2 W 6.53/10 × 10−6 × 5. required = Eb. for PE = 10−3 and R = 1000 bits/second. we get Ps.53 dB from Problem 7.76 × 10−6 × 1000 = 4.3 (a) For R = B = 5.9 × 10−1 W PE = 10−5 9.08 × 10−1 W PE = 10−6 11. required /Tsymbol = Eb.78 dB (4. 000 PE = 10−3 4. 10−6 .2 CHAPTER 7.39 dB (6.08 × 10−2 W 9.13 W Problem 7.3 × 5 × 10−3 = 0.2. 000 bps. we get signal power = A2 = 5. required = z|required N0 R = 4. bps 1.76 × 10−3 W Similar calculations allow the rest of the table to be filled in.4 The decision criterion is V V > ε.53 dB (11. 8. 000.9 × 10−3 W 6.76 × 10−3 W 4. the result is signal power = A2 = 11. (b) For R = B = 50.3 × 10−3 W 11.08 ratio).76 × 10−2 W 4.3 W Problem 7.3 ratio) for PE |desired = 10−3 .76 ratio). 000 100. BINARY DATA COMMUNICATION we find that z|required = 6. 0 The results are given in the table zdB for PE = 10−6 11. z = N0 −AT +ε √ dη = Q  πN0 T s 2A2 T N0 + s 2ε2 N0 T   Use the approximation Z ∞ 2 2 e−η /N0 T e−η /N0 T √ √ = dη = dη πN0 T πN0 T −∞ AT −ε s  s ³√ ´ 2T 2 √ 2A 2ε  = Q =Q 2z − 2²/σ − N0 N0 T e−u /2 √ u 2π 2 Z Q (u) = to evaluate 1 (P1 + P2 ) = 10−6 2 by solving iteratively for z for various values of ε/σ.4 zdB for PE = 10−6 10.98 .7. below. PROBLEM SOLUTIONS where V = ±AT + N N is a Gaussian random variable of mean zero and variance σ 2 = N0 T /2. it follows that 1 1 1 PE = P (−AT + N > ε | − A sent) + P (AT + N < ε | A sent) = (P1 + P2 ) 2 2 2 where P1 = Z ∞ 2 e−η /N0 T 3 Because AT +ε = Q Similarly P2 ´ ³√ √ A2 T 2z + 2²/σ .2 0.6 0.8 1.34 11.01 ε/σ 0. P (A sent) = P (−A sent) = 1/2. ε/σ 0 0. σ = N0 T.54 10.1.67 11.69 11. BINARY DATA COMMUNICATION Problem 7. we obtain the following values of D: (a) for 1/T = R = 10 kbps.46 dB. which occur the other half the time.6 The integrate-and-dump detector integrates over the interval [0. result in the i hp 2Eb /N0 (1 − 2 |∆T | /T ) (sketches of the pulse sequences with the error probability Q integration interval offset from the transition between pulses is helpful here). bps 10−4 10−5 10−6 8.04 dB.3) 57. which occur half´ time.8 dB.1 for the giving timing errors. the average error probability is the given result in the problem statement.4 Problem 7. −A) and (A. D = 0. Thus T z0 =1− z δ and the degradation in z in dB is µ ¶ T D = −10 log10 1 − δ Using δ = 10−6 s and the data rates given.2 along with z|required = A2 /N0 R to get R = A2 / z|required N0 . no degradation the ³p 2Eb /N0 . The error probability results from timing error. D = 0.1 for timing errors of 0 and 0.58 (9. A).05 35. A2 (T − δ) N0 . T − δ] so that the SNR is z0 = instead of z = A2 T /N0 . Substitute N0 = 10−6 V2 /Hz. A = 20 mV to get R = 400/ z|required . PE z|required . dB (ratio) R. A) and (A.22 dB. (c) for R = 100 kbps. This results in the values given in the table below. (b) for R = 50 kbps.4 CHAPTER 7.39 (6. Thus.9) 9.08) 10.5 Use the z|required values found in the solution of Problem 7.7 (a) For the pulse sequences (−A. −A).53 (11. so the error probability is Q for the sequences (−A.15 indicate that the degradation at PE = 10−6 is about 2. Plots are shown in Figure 7. Problem 7.97 43. (b) The plots given in Figure 7. D = 0. PROBLEM SOLUTIONS 5 Figure 7. the signal out at t = T is ³ ´ s0 (T ) = A 1 − e−2πf3 T Sn (f ) = N0 /2 1 + (f/f3 )2 The power spectral density of the noise at the filter output is so that the variance of the noise component at the filter output is Z ∞ £ 2¤ N0 πf3 = N0 BN E N = Sn (f ) df = 2 −∞ . or ³ ´ a (t) = 1 − e−2πf3 t u (t) Thus.1.8 (a) The impulse response of the filter is h (t) = (2πf3 ) e−2πf3 t u (t) and the step response is the integral of the impulse response.7.1: Problem 7. for a step of amplitude A. given these two signaling events.199/T = 0. Solve the equation for f3 .25. are µ ¶ 1 P (error | A sent) = P AT + N < AT 2 µ ¶ 1 = P N < − AT 2 Z −AT /2 −η2 /N0 T e √ = dη πN0 T −∞ s  2T A  = Q 2N0 .6 CHAPTER 7. differentiate with respect to f3 and set the result equal to 0. 0 sent The probabilities of error. Thus ¡ ¢2 2A2 1 − e−2πf3 T s2 (T ) SNR = 0 2 = E [N ] N0 πf3 (b) To maximize the SNR. Thus f3.25 = 0. opt = 1. BINARY DATA COMMUNICATION where BN is the noise equivalent bandwidth of the filter. This results in the equation 4πf3 T e−2πf3 T − 1 + e−2πf3 T = 0 Let α = 2πf3 T to get the equation (2α + 1) e−α = 1 A numerical solution results in α ≈ 1. 1 var [N ] = N0 T 2 The output of the integrator at the sampling time is ½ AT + N.9 As in the derivation in the text for antipodal signaling. A sent V = N.199R 2πT Problem 7. 27) and (7.1. k.10 Substitute (7. PROBLEM SOLUTIONS Similarly. ¶ µ 1 P (error | 0 sent) = P N > AT 2 Z ∞ −η2 /N0 T e √ = dη πN0 T AT /2  s 2T A  = Q 2N0 PE = 1 1 P (error | A sent) + P (error | 0 sent) 2 2  s A2 T  = Q 2N0 7 Thus But. the average signal energy is Eave = P (A sent) × A2 T + P (0 sent) × 0 1 2 A T = 2 Therefore.28) into PE = pP (E | s1 (t)) + (1 − p) P (E | s2 (t)) h h i i Z k exp − (v − s02 )2 /2σ 2 Z ∞ exp − (v − s01 )2 /2σ 2 0 0 p p dv + (1 − p) dv = p 2 2 2πσ 0 2πσ 0 k −∞ # Eave N0 Use Leibnitz’s rule to differentiate with respect to the threshold.7. The differentiation gives h i i¯ h 2 2 exp − (k − s02 )2 /2σ 2 ¯ exp − (k − s01 ) /2σ 0 0 ¯ ¯ p p + (1 − p) =0 −p ¯ 2πσ 2 2πσ 2 ¯ 0 0 k=ko p t . set the result of the differentiation equal to 0. the average probability of error is "r PE = Q Problem 7. and solve for k = kopt . . For t0 = T /2. (c) The realizable matched filter has zero impulse response for t < 0.8 or CHAPTER 7.11 (a) Denote the output due to the signal as s0 (t) and the output due to noise as n0 (t). For t0 = T. we find the following results: For t0 = 0. We wish to maximize ¯R ¯2 ¯ ∞ ¯ 2 (t) ¯ −∞ S (f ) Hm (f ) ej2πf t0 df ¯ s0 2 R∞ £ ¤= ∗ N0 E n2 (t) 0 −∞ Hm (f ) Hm (f ) df (b) The matched filter impulse response is hm (t) = s (t0 − t) by taking the inverse Fourier transform of Hm (f ) employing the time reversal and time delay theorems. we have R∞ R∞ Z ∞ ∗ ∗ s2 (t) 2 −∞ S (f ) S (f ) df −∞ Hm (f ) Hm (f ) df 2 0 R∞ £ 2 ¤≤ = |S (f )|2 df ∗ N0 N0 −∞ Hm (f ) Hm (f ) ej2πf t0 df E n0 (t) −∞ Problem 7. s0 (t0 ) = 0. s0 (t0 ) = A2 T /2. s0 (t0 ) = A2 T . s0 (t0 ) = A2 T. For t0 = 2T. (d) By sketching the components of the integrand of the convolution integral. BINARY DATA COMMUNICATION or h h i i exp − (kopt − s01 )2 /2σ 2 exp − (kopt − s02 )2 /2σ 2 0 0 p p p = (1 − p) 2πσ 2 2πσ 2 0 0 − (kopt − s01 ) + (kopt − s02 ) (s01 − s02 ) kopt + 2 2 s2 − s2 02 01 2 kopt ¶ 1−p = p µ ¶ 1−p 2 = σ 0 ln p ¶ µ 2 s02 − s01 σ0 1−p + = ln s01 − s02 p 2 2σ 2 ln 0 µ where the maximum value on the right-hand side is attained if Hm (f ) = S ∗ (f ) e−j2πf t0 By Schwartz’s inequality. 0 ≤ t0 ≤ = N0 2 2 This result increases linearly with t0 and has its maximum value for for t0 = T /2. which is 5A2 T /N0 . use the maximum value of ζ.1. Problem 7. (d) A2 T /3 (Note that AΛ (t−T /2) T i h 2(t−T /2) ).7. Problem 7. (b) 3A2 T /8. (d) The optimum receiver would have two correlators in parallel.30). assuming the signals are equally likely. we obtain Z ∞ 2 2 ζ = [s2 (t) − s1 (t)]2 dt N0 −∞ · µ ¶¸ T 2 A2 t0 + 4A2 t0 + A2 = − t0 N0 2 · 2 ¸ T 2 A T + 4A2 t0 . one for s1 (t) and one for s2 (t).13 (a) The matched filter impulse response is given by hm (t) = s2 (T − t) − s1 (T − t) (b) Using (7. (c) The outputs for the two cases are So2 (t) = 7B 2 τ Λ [(t − t0 ) /7τ ] So1 (t) = A2 τ Λ [(t − t0 ) /τ ] = 7B 2 τ Λ [(t − t0 ) /τ ] (d) The shorter pulse gives the more accurate time delay measurement. where Es is the signal energy.14 Use ζ 2 = 2Es /N0 for a matched filter. which is given in (b). (e) Peak power is lower for y (t). The required values max h i for signal energy are: (a) A2 T . (c) A2 T /2.12 (a) These would be x (t) and y (t) reversed and then shifted to the right so that they are nonzero for t > 0. (c) The probability of error is · ¸ ζ PE = Q √ 2 2 To minimize it. The outputs for the correlators at t = T are differenced and compared with the threshold k calculated from (7. √ (b) A = 7B. in the problem statement should be AΛ T . PROBLEM SOLUTIONS 9 Problem 7.55) and a sketch of s2 (t) − s1 (t) for an arbitrary t0 . EB = B 2 T /2. EC = 3C 2 T /8 (a) For s1 = sA and s2 = sB .15 The required formulas are CHAPTER 7. the average signal energy is A2 T E= 2 The correlation coefficient is R12 = = = The optimum threshold is kopt The probability of error is PE = Q "s T 2N0 µ ¶# B 2 4AB A2 + − 2 π 1 1 = (EB − EA ) = 2 2 µ ¶ B2 2 −A T 2 ¸ π (t − T /2) AB cos dt T 0 · ¸ Z πt 2ABT 1 T AB sin dt = E 0 T πE 4AB 2 + B 2 /2) π (A 1 E Z T ¶ µ B2 1+ 2A2 · . BINARY DATA COMMUNICATION E = R12 kopt PE 1 (E1 + E2 ) 2 Z 1 T = s1 (t) s2 (t) dt E 0 1 (E2 − E1 ) = 2  s (1 − R12 ) E  = Q N0 The energies for the three given signals are EA = A2 T .10 Problem 7. 1.7. PROBLEM SOLUTIONS (b) s1 = sA and s2 = sC : E = R12 = ¢ 1 1¡ 2 (EA + EC ) = A + 3C 2 /8 T 2 2 AC A2 + 3C 2 /8 µ ¶ 3C 2 2 −A T 8 11 kopt 1 1 = (EC − EA ) = 2 2 "s PE = Q (c) s1 = sB and s2 = sC : E = R12 = T 2N0 µ ¶# 3C 2 A2 + − AC 8 µ ¶ 3C 2 2 B + T 4 1 1 (EB + EC ) = 2 4 16BC 3π (B 2 + 3C 2 /4) kopt 1 1 = (EC − EB ) = 2 4 "s µ ¶ 3C 2 2 −B T 4 PE = Q (d) s1 = sB and s2 = −sB : T 4N0 µ ¶# 3C 2 16BC B2 + − 4 π E = EB = B 2 T /2 R12 = −1 kopt = 0 s  B2T  N0 PE = Q  . 6 dB p For PSK with m = 0.’.2. Curves corresponding to the numbers given in Table 7.’).19 or zdB = 12.47 or zdB = 9. fixed variance clf sigma_psi2 = input(’Enter vector of variances of phase error in radians^2 ’).265 or z = 9.’-. L_sig = length(sigma_psi2).m calls user-defined subprogram pb_phase_pdf. 2 (1 − m2 ) z = 4. ASK hp i 2 (1 − m2 ) z = 10−5 .’—’.16 The appropriate equations to solve are £√ ¤ PE = Q z = 10−5 . a_mod = 0.’:’. PSK. Uses subprogram pb_phase_pdf % bep_ph_error.265 or z = 18. z = 4. BINARY DATA COMMUNICATION E = EC = 3C 2 T /8 R12 = −1 kopt = 0 s  3C 2 T  8N0 PE = Q  Problem 7.2 PE = Q £√ ¤ PE = Q z = 10−5 . A = char(’-. Thus..12 (e) s1 = sC and s2 = −sC : CHAPTER 7. % Program bep_ph_error. Eb_N0_dB_max = input(’Enter maximum Eb/N0 desired ’). we get the following results: √ For ASK and FSK. m = 0. for m = 1:L_sig Trial and error shows that the Q-function has a value of 10−5 for it argument equal to 4.2 are shown in Figure 7.765 dB .m.17 The program is given below.m % Effect of Gaussian phase error in BPSK.2. FSK Problem 7. Eb_N0_dB_min = input(’Enter minimum Eb/N0 desired ’).265. []..A(m. k=1 for snr = Eb_N0_dB_min:.pi.5*erfc(sqrt(Eb_N0)). For the constant phase error φ φ . legend([’BPSK. Eb_N0 PE(k) = quad8(’pb_phase_pdf’.. [’BPSK.xlabel(’E_b/N_0. called by bep_ph_error.a) arg = Eb_N0*(1-a^2)*(cos(psi)-a/sqrt(1-a^2)*sin(psi)). Eb_N0 = []. axis([Eb_N0_dB_min Eb_N0_dB_max 10^(-7) 1]).05 it is about 0. T2 = exp(-psi.num2str(sigma_psi2(1))].5 dB.a_mod) k = k+1. PE = []. \sigma_\theta_e^2 = 0’].1.Eb_N0.01 is about 0.sigma_psi.m. P_ideal).18 From the figure of Problem 7...num2str(sigma_psi2(2))].[]. [’BPSK. if m == 1 hold on grid on end end P_ideal = ... sigma_psi = sqrt(sigma_psi2(m)).7.3) % pb_phase_pdf. end PE semilogy(Eb_N0_dB.6 dB.5*erfc(sqrt(arg)).num2str(sigma_psi2(3))]. \sigma_\theta_e^2 = ’. 13 Problem 7. dB’).^(Eb_N0_dB(k)/10).ylabel(’P_b’). \sigma_\theta_e^2 = ’.1 it is about 9. semilogy(Eb_N0_dB. T1 = .-pi.^2..*T2.m and bep_ph_error2. XX = T1.m % function XX = pb_phase_pdf(psi.^2/(2*sigma_psi^2))/sqrt(2*pi*sigma_psi^2). PROBLEM SOLUTIONS Eb_N0_dB = [].Eb_N0(k).sigma_psi.:)).17. \sigma_\theta_e^2 = ’.. the degradation for σ 2 = 0. [’BPSK. PE.5:Eb_N0_dB_max Eb_N0_dB(k) = snr Eb_N0(k) = 10. for σ 2 = 0. for φ σ 2 = 0.045 dB.. Problem 7. . we add the degradation of part (d) to the SNR found in part (e) to get 12. PE = Q 2z = 10−5 gives z = 9. 0. thus. (c) Binary FSK is the same as ASK.044.6 + 0.441 dB. (b) For BPSK.6 dB. BINARY DATA COMMUNICATION Figure 7. (d) The degradation of BPSK with a phase error of 5 degrees is Dconst = −20 log10 [cos (5o )] = 0. hp i √ √ 2 (1 − 1/2) z = Q [ z] = 10−5 gives z = 18.6 dB.19 or (e) For PSK with m = 1/ 2. 0.033 dB. PE = Q [ £√ =¤10−5 gives z = 18.224.1 or z = 9.59 dB.623 dB. (f) Assuming that the separate effects are additive. The degradations corresponding to these phase errors are 0.14 CHAPTER 7. The constant phase error degradations are much less serious than the Gaussian phase error degradations.1. respectively.033 = 12.033 = 9. the degradation is Dconst = −20 log10 [cos (φconst )] As suggested in the problem statement. we set the constant phase error equal to the standard deviation of the Gaussian phase error for comparison putposes.219.59 + 0.19 √ z] (a) For ASK.316 radians. so the required SNR to give a bit error probability of 10−5 is 9. we consider constant phase errors of φconst = 0. and 0.2: model. PE = Q z = 12. and 0.19 or z = 12.633 dB. For BPSK and ASK. A plot of SNR versus m is given in Fig. an SNR of 10.73). For ASK and FSK it is 3 dB more. the required bandwidth is 2R Hz where R is the data rate in bps. the additional SNR ¢ required due to a carrier component is −10 log10 1 − m2 .1. Problem 7. Hence.7.4. Problem 7.3. 5.53 dB is required to give an error probability of 10−6 . For FSK with minimum tone spacing of 0.4.4. The degradations for phase errors of 3.22 (a) For BPSK.3: Problem 7. A plot of SNR versus m is given in Fig. and 15 degrees are 0. √ (b) Solve PE = Q ( z) = 10−5 to get z = 9.5R Hz. for BPSK and ASK. . From (7.54 dB.21 ¡√ ¢ (a) Solve PE = Q 2z = 10−4 to get z = 8. the required bandwidth is 200 kHz. √ (c) Solve PE = Q ( z) = 10−6 to get z = 10.5R Hz. the required bandwidth is 2. 7. respectively.53 dB. 0.57 dB.0331.20 The degradation in dB is Dconst = −20 log10 [cos (φ)]. For FSK. 7.1330. and 0. 10. A plot of SNR versus m is given in Fig.3011 dB.4 ¡dB. 0. PROBLEM SOLUTIONS 15 Figure 7.0119. or 13. it is 250 kHz. It is plotted in Figure 7. 7. The minimum value at this argument value is about -0.6 dB. or ∆f = 0. (d)1 010 101 010 101. For ASK and FSK it is 12.7/T . BINARY DATA COMMUNICATION Figure 7. Problem 7. (h) 1 100 001 000 010.216. The required bandwidths are now 400 kHz for BPSK and ASK. (b) 1 100 110 011 001. (f) 1 110 000 011 011. are: (a) 1 100 001 000 010. (g) 1 111 010 000 101. The improvement in SNR is −10 log10 (1 − R12 ) = −10 log10 (1 + 0.6 dB. Problem 7. (c) 1 111 111 111 111. Using a calculator.4: (b) For BPSK. the required SNR is 9.4.216) = 0. (e) 1 111 111 010 101.85 dB. assuming a reference 1 to start with. it is 500 kHz for FSK. . we find that the sinc-function takes on its first minimum at an argument value of about 1.23 The correlation coefficient is R12 = Z 1 T s1 (t) s2 (t) dt E 0 Z 1 T 2 = A cos (ω c t) cos [(ω c + ∆ω) t] dt E 0 = sinc (2∆fT ) where ∆f is the frequency separation between the FSK signals in Hz.16 CHAPTER 7.24 The encoded bit streams. The phase of the transmitted carrier is 000ππππ0ππ00π. each time it is less than 0. Therefore.7. PROBLEM SOLUTIONS 17 Problem 7.26 Note that E [n1 ] = E ·Z 0 −T ¸ Z n (t) cos (ωc t) dt = 0 E [n (t)] cos (ωc t) dt = 0 −T which follows because E [n (t)] = 0. Problem 7. n3 . Each time the integrated product is greater than 0.27 The two error probabilities are ¡√ ¢ 1 PE. and n4 . the same demodulated signal is detected as if it weren’t inverted. Consider the variance of each of these random variables: ·Z 0 Z 0 ¸ £ 2¤ n (t) n (λ) cos (ω c t) cos (ω c λ) dtdλ var [n1 ] = E n1 = E = Z 0 −T Z 0 Z 0 −T −T E [n (t) n (λ)] cos (ω c t) cos (ω c λ) dtdλ N0 δ (t − λ) cos (ω c t) cos (ωc λ) dtdλ 2 = = N0 2 −T Z −T Z 0 −T 0 −T cos2 (ωc t) dt = N0 T 4 where the fact that E [n (t) n (λ)] = N0 δ (t − λ) has been used along with the sifting property 2 of the δ-function to reduce the double integral to a single integral. Similarly for n2 . opt = e−z and PE. If the signal is inverted. which is the message signal. delay-and-mult ≈ Q z 2 Comparative values are: . n3 . Problem 7. Similarly for n2 .1. Integrate the product of the received signal and a 1-bit delayed signal over a (0. decide 0. decide 1. and n4 . T ) interval. w3 . The resulting decisions are 110 111 001 010. and w4 . w1 has zero mean and variance 1 1 N0 T var [w1 ] = var [n1 ] + var [n2 ] = 4 4 8 Similarly for w2 .25 The encoded bit stream is 1 110 000 100 110 (1 assumed for the starting reference bit). NCFSK = e−zN C F S K /2 2 Solve for the SNR to get zNCFSK = −2 ln (2PE. DPSK ) The values asked for are given in the table below: PE 10−3 10−5 10−7 zNCFSK .89 zDPSK .1 × 10−4 1. dB 8 9 10 11 PE. DPSK = e−zD P S K 2 Solve for the SNR to get zDPSK = − ln (2PE.5R.5 = 40 kbps BNCFSK = 4R. dB 7. opt 9.93 10.18 z. BINARY DATA COMMUNICATION PE.34 11.35 14.29 Bandwidths are given by BBPSK = BDPSK = 2R. NCFSK ) DPSK has error probability 1 PE.8 × 10−3 2.7 × 10−3 8. delay-and-mult 6. dB 10. RBPSK = RBPSK = BDPSK /2 = 50 kbps BCFSK = 2.28 Noncoherent binary FSK has error probability 1 PE.1 × 10−4 Problem 7. RCFSK = BBPSK /2.3 × 10−5 1.5 × 10−4 2.8 × 10−4 2.7 × 10−6 CHAPTER 7.94 13. RNCFSK = BBPSK /4 = 25 kbps .88 Problem 7. The output of the bandpass filter is u (t) = Ak cos (ω c t + Θ) + n (t) = x (t) cos (ωc t + Θ) − ns (t) sin (ωc t + Θ) = Ak cos (ω c t + Θ) + nc (t) cos (ω c t + Θ) − ns (t) sin (ω c t + Θ) where Ak = A if signal plus noise is present at the receiver input. followed by an envelope detector with a sampler and threshold comparator at its output.7. and approximating the square root as √ 1 1 + ² ≈ 1 + ². PROBLEM SOLUTIONS 19 Problem 7. r ≥ 0. we observe that for large SNR at the receiver input r (t) ≈ A + nc (t) . dropping the squared terms.1. for large SNR conditions. In the last equation x (t) = Ak + nc (t) Now u (t) can be written in envelope-phase form as u (t) = r (t) cos [ω c t + Θ + φ (t)] where r (t) = · ¸ p ns (t) x2 (t) + n2 (t) and φ (t) = tan−1 s x (t) The output of the envelope detector is r (t). and A = 0 if noise alone is present at the input. Thus. noise only R where N is the variance (mean-square value) of the noise at the filter output. |²| << 1 2 Thus. we need the pdf of the envelope for noise alone at the input and then with signal plus noise at the input. Rather than use this complex expression. which is fR (r) = r −r2 /2N e . large SNR. just wide enough to pass the ASK signal. however. For signal plus noise at the receiver input. signal present which follows by expanding the argument of the square root. the pdf of the envelope is Ricean. to compute the error probability.30 Consider the bandpass filter. It has mean A and variance N (recall that the inphase and . For noise alone. it was shown in Chapter 4 that the pdf of r (t) is Rayleigh. the envelope detector output is approximately Gaussian for signal plus noise present. which is sampled each T seconds and compared with the threshold. BINARY DATA COMMUNICATION quadrature lowpass noise components have variance the same as the bandpass process). Problem 7.32 Do the inverse Fourier transform of the given P (f ) to show that p (t) results.31 This is a matter of following the steps as outlined in the problem statement. It is also useful to note that for an even function. probability of error is e−z 1 1 1 + e−z/2 PE = P (E | S + N ) + P (E | 0) ≈ √ 2 2 2 4πz Problem 7. the probability of error is approximately P (E | S + N present) = ≈ Now the average signal-to-noise ratio is z= Z A/2 2 p π/N A A2 1 A2 /2 = 2 N0 BT 4N Therefore. Only a sketch is given here.20 CHAPTER 7. The derivation is quite long and requires a fair amount of trigonometry. Thus. for noise alone present. large SNR fR (r) = 2πN For large SNR. the average −∞ 2 e−A /4N ´ ³p e−(r−A) /2N √ A2 /2N dr = Q 2πN 2 which follows because the signal is present only half the time. the inverse Fourier transform integral reduces to Z ∞ Z ∞ p (t) = P (f ) exp (j2πf t) df = 2 P (f ) cos (2πf t) df 0 0 . the threshold is approximately A/2. the probability of error is exactly Z ∞ r −r2 /2N 2 P (E | 0) = e dr = e−A /8N A/2 R For signal plus noise present. The pdf of the envelope detector output with signal plus noise present is approximately e−(r−A) /2N √ . 124) to get 21 p (t) = 2 Z 1−β 2T T cos (2πft) df + 2 = T = cos (πβt/T ) sinc (t/T ) ¶ ¶ µ ¸ · µ ¸¾ Z 1+β ½ · 2T π π πT πT 2β T 2β t f− t f− sin 1+ + sin 1− df − 2 1−β β T 2β β T 2β 2T +T 2πt 2πt µ ¶ ¸ · µ ¶ ¸¾ Z 1+β ½ · 2T π 2β π 2β π πT π πT T 1+ + cos 1− df t f− + t f− + cos + 2 1−β β T 2β 2 β T 2β 2 2T h ³ ´ i sin 2π 1−β t 2T 0 ½ · µ ¶¸¾ 1−β πT 1 + cos f− cos (2πf t) df 1−β β 2T h 2T³ ´ i sin 2π 1+β t 2T Z 1+β 2T T 2 = cos (πβt/T ) sinc (t/T ) h ³ i ´ h  π cos πT 1 + 2β t 1+β − 2β − cos πT 2T β T β T ´ ³ +  2β πT 2 1+ T t β h ³ i ´ h  π cos πT 1 − 2β t 1+β − 2β − cos πT 2T β T β T ´ ³ +  πT 2 1 − 2β t β T = cos (πβt/T ) sinc (t/T ) h ³ ´  cos πT 1 + 2β t 1+β − 2T β T T ´ ³ +  2β πT 2 1+ T t β h ³ ´  cos πT 1 + 2β t 1−β − 2T β T T ´ ³ −  πT 2 1 + 2β t β T = cos (πβt/T ) sinc (t/T ) h ³ ´  2β πT T cos β 1 + T t f − ´ ³ +  πT 2 1 + 2β t β T π 2β i cos + h πT β ´ ³ 2β 1− T t f − ´ ³ πT 1 − 2β t β T πT β π 2β i  1+β 2T  1−β 2T π 2β i i cos + cos + h h π 2β πT β ³ 1+ ³ 1− ³ ´ 1 − 2β t 1+β − 2T T ´ ³ 2β πT 1− T t β ³ ´ 1 − 2β t 1−β − 2T T ´ ³ πT 1 − 2β t β T 2β T t π 2β i i   π 2β ´ ´ 1−β 2T − π 2β i i   2β T t 1−β 2T − π 2β . Now substitute for P (f ) from (7.7.1. PROBLEM SOLUTIONS because P (f ) is even. 22 CHAPTER 7. 0 ≤ |f¢¤ª 2T 1 + cos 0.25 2T Hz The optimum transmitter and receiver transfer functions are |HT (f )|opt = A |P (f )|1/2 Gn (f ) |HC (f )|1/2 1/4 . BINARY DATA COMMUNICATION Expand the trigonometric function arguments to get p (t) = cos (πβt/T ) sinc (t/T )  £ ¤ ¤ £ πt π πt π T cos (1 + β) T + 2 − cos (1 − β) T − 2  ´ ³ +  πT 2 1 + 2β t β T  ¤ ¤ £ £ T  cos (1 + β) πt − π − cos (1 − β) πt + π  2 2 T T ´ ³ + 2β πT 2 1− t β T Expand the cosines using to get p (t) = cos (πβt/T ) sinc (t/T )  £  £ ¤ £ ¤ ¤ £ ¤ T  sin (1 + β) πt + sin (1 − β) πt  T  sin (1 + β) πt + sin (1 − β) πt  T³ T T³ T ´ ´ + − 2β 2β πT πT 2 2 1+ T t 1− T t β β     cos (πβt/T ) sin (πt/T )   cos (πβt/T ) sin (πt/T )  ´ ³ ³ ´ = cos (πβt/T ) sinc (t/T ) −  + 2πt 2πt T T +1 −1 2βt 2βt T T   1 1 T  ´+ ³ ´  cos (πβt/T ) sin (πt/T ) 1− ³ = T T πt 2 +1 2 −1 2βt 2βt = cos (πβt/T ) sinc (t/T ) 1 − (2βt/T )2 Problem 7. |f | > 2T 1.25.75 .75 ≤ |f | ≤ 2T 2T 0.25 |f | − 0.33 For β = 0.75 0. 0.    T 2 P (f ) = | ≤ 0.75 © £ πT ¡ T. P2 = 0. for n = 1:4 beta = (n-1)*del_beta+beta00. f_3 = R/2. f = [f1 f2 f3]. del_beta = input(’Enter step in beta: starting value + 4*step <= 1 ’).5*(1-beta)*R))).1. R = 1. Gn = 1. HR_opt = []. Plots for the various cases are then given in Figures 7.5*(1+cos(pi/(R*beta)*(f2-0. f3 = (1+beta)*R/2:delf:R. P1 = ones(size(f1)). f1 = 0:delf:(1-beta)*R/2.’—’. clf beta00 = input(’Enter starting value for beta: >=0 and <= 1 ’).7. P3 = zeros(size(f3)). % Problem 7. HT_opt = []. PROBLEM SOLUTIONS and |HR (f )|opt = K |P (f )|1/2 23 Gn (f ) |HC (f )|1/2 1/4 Setting all scale factors equal to 1.^2)./(1+(f/f_3).’:’. f2 = (1-beta)*R/2:delf:(1+beta)*R/2. these become q |P (f )|1/2 1/4 1 + (f /fC )2 q = 1/4 1 + (f /f3 )2 1/2 1/4 |HT (f )|opt |HR (f )|opt = |P (f )| A MATL:AB program for computing the optimum transfer functions is given below.01.7.33: Optimum transmit and receive filters for raised-cosine pulse shaping.7. and first-order noise spectrum % A = char(’-’.5 . delf = 0. beta0(n) = beta.’-.’). f_C = R/2. % first-order Butterworth channel filtering. q q 2 1/4 1 + (f /f3 ) 1 + (f/fC )2 . num2str(beta0(4))]./sqrt(1+(f/f_C).1).num2str(beta0(3))].num2str(f_C).1) (a) f3 = fC = 1/2T (b) fC = 2f3 = 1/T . HR_opt = sqrt(P).A(n.’ bps.1._o_p_t(f)|’) if n == 1 hold on end end legend([’\beta = ’.25). P = [P1 P2 P3].*(Gn.’ Hz’]) if n == 1 hold on end subplot(2. subplot(2.plot(f./((Gn. HT_opt = sqrt(P).[’\beta = ’.2). [’\beta = ’.1.ylabel(’|H_T_._o_p_t(f)||’) title([’R = ’. Hz’).HR_opt.:)).num2str(f_3)./sqrt(HC).25).5: f3 = fC = 1/2T HC = 1.[’\beta = ’. Hz’). channel filter 3-dB frequency = ’.24 CHAPTER 7.num2str(beta0(1))]. noise 3-dB frequency = ’.num2str(R).xlabel(’f.HT_opt.^.ylabel(’|H_R_.plot(f.xlabel(’f.^.^2).’ Hz.*sqrt(HC)). BINARY DATA COMMUNICATION Figure 7.:)).A(n.num2str(beta0(2))]. 35 Follow the steps outlined in the problem statement.34 £¡ ¢ ¤ This a simple matter of sketching Λ (f/B) and Λ f − 1 B /B side by side and showing 2 that they add to a constant in the overlap region.7.6: fC = 2f3 = 1/T (c) f3 = 2fC = 1/T Problem 7.36 (a) The optimum transmitter and receiver transfer functions are |HT (f )|opt = and |HR (f )|opt = K |P (f )|1/2 A |P (f )|1/2 Gn (f ) |HC (f )|1/2 1/4 Gn (f ) |HC (f )|1/2 1/4 where P (f ) is the signal spectrum and A and K are arbitrary constants which we set equal .1. Problem 7. Problem 7. PROBLEM SOLUTIONS 25 Figure 7. 0 ≤ |f | ≤ 4800 Hz =¯ 9600 ¯ · µ ¶¸ π |f | 1 1 + cos . BINARY DATA COMMUNICATION Figure 7. . Gn (f ) = 10−12 1 |HC (f )| = q 1 + (f /4800)2 The signal spectrum is. |HT (f )|opt = |HR (f )|opt ¯ µ ¶¯ q ¯ ¯ ¯cos π |f | ¯ 1 + (f/4800)2 .26 CHAPTER 7. P (f ) = = Thus. From the problem statement. for β = 1. 0 ≤ |f | ≤ 4800 Hz 4800 9600 where all scale factors have been set equal to 1. 0 ≤ |f | ≤ 4800 Hz 2 × 4800 4800 µ ¶ π |f | 1 cos2 .7: f3 = 2fC = 1/T ¡ ¢1/4 1/4 = 10−3 and to 1 for convenience. 3) elseif L_taum == 2 legend([’\tau_m/T = ’. num2str(delta)]) if L_taum == 1 legend([’\tau_m/T = ’.8 shows the probability of error curves.37 % clf A = char(’-’.num2str(taum_T0(1))].num2str(taum_T0(1))]. z0_dB = 0:.’:’.:)) if ll == 1 hold on.[’\tau_m/T = ’. semilogy(z0_dB. xlabel(’z_0.num2str(taum_T0(2))].’-.. 7.5*qfn(sqrt(2*z0)*(1+delta))+0. grid on.A(ll.1:15.[’\tau_m/T = ’. % Plots for Prob.5*qfn(sqrt(2*z0)*((1+delta)-2*delta*taum_T)).1.’-.38 The program given in Problem 7.3) elseif L_taum == 3 legend([’\tau_m/T = ’. dB’).num2str(taum_T0(2))]. 1]).^(z0_dB/10).’—. PROBLEM SOLUTIONS 27 Problem 7.num2str(taum_T0)]. ylabel(’P_E’) end end title([’P_E versus z_0 in dB for \delta = ’.3) elseif L_taum == 4 legend([’\tau_m/T = ’.’). 10^(-6). inf. delta = input(’ Enter value for delta: ’) taum_T0 = input(’ Enter vector of values for tau_m/T: ’) L_taum = length(taum_T0).37 can be adapted for making the asked for plot. axis([0.’.37 The MATLAB program is given below and Figure 7.3) end Problem 7. . z0 = 10.num2str(taum_T0(3))].num2str(taum_T0(4))]. P_E.’. [’\tau_m/T = ’.[’\tau_m/T = ’.7. [’\tau_m/T = ’.[’\tau_m/T = ’. for ll = 1:L_taum ll taum_T = taum_T0(ll) P_E = 0.num2str(taum_T0(2))].num2str(taum_T0(3))].’—’.num2str(taum_T0(1))]. 28 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.8: Problem 7.39 (a) The output of the channel in terms of the input is y (t) = x (t) + βx (t − τ m ) Fourier transform both sides and take the ratio of output to input transforms to obtain the channel transfer function: HC (f ) = Y (f ) = 1 + βe−j2πf τ m X (f ) Using Euler’s theorem to expand the exponential and finding the maginitude gives q |HC (f )| = 1 + 2β cos (2πf τ m ) + β 2 Plots are shown in Fig. 7.9 for β = 0.5 and 1.0. % Plots for Prob. 7.39 % clf A = char(’-’,’—’,’-.’,’:’,’—.’,’-..’); tau_m = input(’ Enter value for tau_m: ’) 7.1. PROBLEM SOLUTIONS beta0 = input(’ Enter vector of values for beta: ’) L_beta = length(beta0) f = -1:.005:1; for ll = 1:L_beta beta = beta0(ll); arg = 1+2*beta*cos(2*pi*tau_m*f)+beta^2; mag_H = sqrt(arg); plot(f, mag_H, A(ll,:)) if ll == 1 hold on; grid on; xlabel(’f’); ylabel(’|H(f)|’) end end title([’|H(f)| versus f for \tau_m = ’, num2str(tau_m)]) if L_beta == 1 legend([’\beta = ’,num2str(beta0)],3) elseif L_beta == 2 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))],3) elseif L_beta == 3 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))], [’\beta = ’,num2str(beta0(3))],3) elseif L_beta == 4 legend([’\beta = ’,num2str(beta0(1))],[’\beta = ’,num2str(beta0(2))], [’\beta = ’,num2str(beta0(3))],[’\beta = ’,num2str(beta0(4))],3) end (b) The output of the tapped delay line filter in terms of the input is z (t) = N X 29 n=0 β n y [t − (n − 1) ∆] Fourier transform this equation and take the ratio of the output to the input transforms to obtain the transfer function of the equalizer as Z (f ) X n −j2π(n−1)∆f = β e Heq (f ) = Y (f ) n=0 ∞ (c) From part (a) and using the geometric expansion, we obtain X 1 1 = = (−1)n β n e−j2πnτ m f −j2πf τ m HC (f ) 1 + βe n=0 ∞ 30 CHAPTER 7. BINARY DATA COMMUNICATION Figure 7.9: Equate this to the result of part (b) with τ m = ∆. Clearly, β n = (−β)n−1 . Problem 7.40 The appropriate equations are: BPSK: (7.73) with m = 0 for nonfading; (7.174) for fading; DPSK: (7.111) for nonfading; (7.176) for fading; CFSK: (7.87) for nonfading; (7.175) for fading; NFSK: (7.119) for nonfading; (7.177) for fading. Degradations for PE = 10−2 are: Modulation type BPSK DPSK Coh. FSK Noncoh. FSK Fading SNR, dB 13.85 16.90 16.86 19.91 Nonfaded SNR, dB 4.32 5.92 7.33 8.93 Fading Margin, dB 9.52 10.98 9.52 10.98 Problem 7.41 See Problem 7.40 for equation reference numbers. The inverse equations, expressing signalto-noise ratio in dB, are: 7.1. PROBLEM SOLUTIONS (a) BPSK zBPSK, NF = 10 log10 zBPSK, F = 10 log10 ½ ¤2 1 £ −1 Q (PE ) 2 ¾ dB ) 31 ( (1 − 2PE )2 1 − (1 − 2PE )2 dB The margin in dB is the difference between these two results for a given PE . (b) DPSK zDPSK, NF = 10 log10 {− ln (2PE )} dB ½ ¾ 1 zDPSK, F = 10 log10 − 1 dB 2PE (c) Coherent FSK zBPSK, NF = 10 log10 zBPSK, F (d) Noncoherent FSK zNCFSK, NF = 10 log10 {−2 ln (2PE )} dB ½ ¾ 1 zNCFSK, F = 10 log10 − 2 dB PE The fading margin for these various modulation cases are given in the table below for PE = 10−3 . Modulation type Fading SNR, dB Nonfaded SNR, dB Fading Margin, dB BPSK 23.97 6.79 17.18 DPSK 26.98 7.93 19.05 Coh. FSK 26.98 9.80 17.18 Noncoh. FSK 29.99 10.94 19.05 Problem 7.42 The expressions with Q-functions can be integrated by parts (BPSK and coherent FSK). The other two cases involve exponential integrals which are easily integrated (DPSK and noncoherent FSK). n£ ¤2 o Q−1 (PE ) dB ( ) 2 (1 − 2PE )2 = 10 log10 dB 1 − (1 − 2PE )2 32 Problem 7.43 We need to invert the matrix CHAPTER 7. BINARY DATA COMMUNICATION Using a calculator or MATLAB, we find the inverse matrix to be  1.02 −0.11 0.02 =  −0.21 1.04 −0.11  0.06 −0.21 1.02   1 0.1 −0.01 1 0.1  [PC ] =  0.2 −0.02 0.2 1  [PC ]−1 The optimum weights are the middle column. (a) The equation giving the equalized output is peq (mT ) = −0.11pC [(m + 1) T ] + 1.04pC [mT ] − 0.21pC [(m − 1) T ] (b) Calculations with the above equation using the given sample values result in peq (−2T ) = −0.02; peq (−T ) = 0; peq (0) = 1; peq (T ) = 0; peq (2T ) = −0.06 Note that equalized values two samples away from the center sample are not zero because the equalizer can only provide three specified values. Problem 7.44 (a) The desired matrix is ¢  ¡ 1 + b2 z + π 2 N0  bz + π e−2π [Ryy ] = 2 T π −4π 2e  π −2π ¡ bz + 22 e ¢ 1+b z+ π 2 bz + π e−2π 2 π −4π 2e π −2π ¡ bz + 22 e ¢ 1+b z+ π 2   (b) The following matrix is needed: Multiply the matrix given equations ¢  ¡ 1 + b2 z + π 2 N0  bz + π e−2π 2 T π −4π 2e in (a) by the column matrix of unknown weights to get the π −2π ¡ bz + 22 e ¢ 1+b z+ π 2 bz + π e−2π 2 π −4π 2e π −2π ¡ bz + 22 e ¢ 1+b z+ π 2    Ryd (−T ) 0 [Ryd ] =  Ryd (0)  =  A  Ryd (T ) bA     a0 0 −1   a0  =  A  0 a0 bA 1 7.1. PROBLEM SOLUTIONS or ¢  ¡ 1 + b2 z + π 2  bz + π e−2π 2 π −4π 2e 33 π −2π ¡ bz + 22 e ¢ 1+b z+ π 2 bz + π e−2π 2 π −4π 2e π −2π ¡ bz + 22 e ¢ 1+b z+ π 2       a−1 0 0   a0  =  A2 T /N0  =  z  a1 bA2 T /N0 bz where the factor N0 /T has been normalized out so that the right hand side is in terms of the signal-to-noise ratio, z (the extra factor of A needed mulitplies both sides of the matrix equation and on the left hand side is lumped into the new coefficients a−1 , a0 , and a1 . Invert the matrix equation above to find the weights. A MATLAB program for doing so is given below. For z = 10 dB and b = 0.1, the weights are a−1 = −0.2781, a0 = 0.7821, and a1 = 0.0773. % Solution for Problem 7.44 % z_dB = input(’Enter the signal-to-noise ratio in dB ’); z = 10^(z_dB/10); b = input(’Enter multipath gain ’); R_yy(1,1)=(1+b^2)*z+pi/2; R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=b*z+(pi/2)*exp(-2*pi); R_yy(1,3)=(pi/2)*exp(-4*pi); R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’R_yy’) disp(R_yy) B = inv(R_yy); disp(’R_yy^-1’) disp(B) R_yd = [0 z b*z]’ A = B*R_yd Typical output of the program: >> pr7_44 Enter the signal-to-noise ratio in dB 10 Enter multipath gain .5 R_yy R_yd = n_sim). zdB = input(’Enter z = A^2*T/N_0 vector in dB: ’).0000 5. PE(k) = ET(k)/length(Y).0338 0. disp(’ ’) disp(’ SNR.5*(1+sign(S)).0338 0.0831 -0.0000 5./z.7821 0.0029 14.0120 -0.1 % ce7_1.0773 7.0120 -0. A = 1. E = xor(V.0708 R_yy^-1 0.n_sim)-. Z = sign(Y).0338 0.2781 0. dB Errors P_E est.5).5*N0(k)*T)*randn(1.0029 0. N0 = (A^2*T).5*erfc(sqrt(z)). W = 0.2 Computer Exercises Computer Exercise 7. T = 1.W). PE = [].0338 0. V = 0. z = 10.0831 CHAPTER 7.34 14. Y = S+N.0708 5.^(zdB/10). P_E theory’) disp(’ ________________________________________’) . ET = [].m: Simulation BEP for antipodal signaling % clf n_sim = input(’Enter number of bits for simulation: ’). Lz = length(z). for k = 1:Lz N = sqrt(0.0029 0.0951 -0.0029 14. BINARY DATA COMMUNICATION 0 10 5 A= -0. end PEth = 0. ET(k) = sum(E).0708 5.5*(1+Z). S = A*T*sign(rand(1. 0000 0.0000 0.’—’). PE. Eb1 = []. for Y = 0:. dB Errors P_E est.0352 0. num2str(n_sim). []. ’ bits per SNR value ’]) legend([’Simulated’].0022 0. P_E theory ______________________________________ 2.0002 35 A plot also appears showing the estimated and theoretical probabilities of error plotted versus signal-to-noise ratio. [].’-. ylabel(’P_E’). grid hold on semilogy(zdB. delT = []. end Eb = fzero(@PE. k = k+1. k = 0. for m = 1:4 PE0 = 10^(-m).0002 0.2 % ce7_2: Compute degradation in SNR due to bit timing error % clf A = char(’-’. COMPUTER EXERCISES disp(’ ’) disp([zdB’ ET’ PE’ PEth’]) disp(’ ’) semilogy(zdB. xlabel(’z in dB’).’:’. deg = [].0375 4.45.0000 4.0000 238. Eb1(k) = fzero(@PE1. ’o’). Computer Exercise 7. delT(k). [’Theoretical’]) A typical run is given below: >> ce7_1 Enter number of bits for simulation: 20000 Enter z = A^2*T/N_0 vector in dB: [2 4 6 8] SNR.’. PE0).0000 0. . delT(k) = Y.0024 8. PE0).2.0125 6.0000 0.0000 44.7. 2.025:.0119 0. PEth) title([’BER simulated with ’.0000 704. 2. 3 A MATLAB program for this computer exercise is given in the solution for Problem 7.. deg. BINARY DATA COMMUNICATION Figure 7. plot(delT. grid end end legend([’P_E = 10^-^1’]. ylabel(’Degradation in dB’). [’ = 10^-^4’].1. [’ = 10^-^2’].. .. Computer Exercise 7.45 0 20]) xlabel(’Fractional timing error.17. 7.Eb.. \DeltaT/T’).36 CHAPTER 7. title([’Degradation in SNR for bit synchronization timing error’]). [’ = 10^-^3’].. if m == 1 hold on axis([0 . .:)).10: deg = Eb1 . A(m. 2) The output plot is shown in Fig. elseif I_BW_R == 2 Rkbps = input(’Enter vector of desired data rates in kbps: ’). find the required % bandwidth and Eb/N0 in dB. COMPUTER EXERCISES Computer Exercise 7. % Rate for baseband elseif I_mod == 1 | I_mod == 2 Rkbps = BkHz/2. % I_mod = input(’Enter desired modulation: 0 = baseband. 3 = CFSK. 4 = NCFSK: ’).5*erfc(sqrt(z)) MATLAB has an inverse erf but not an inverse Q-function .7. % Rate in kbps for CFSK elseif I_mod == 4 Rkbps = BkHz/4. if I_BW_R == 1 if I_mod == 0 Rkpbs = BkHz. B = BkHz*1000. % Transmission BW for BPSK or DPSK in kHz elseif I_mod == 3 BkHz = 3*Rkbps. BPSK/DPSK.4 37 % ce7_4: For a given data rate and error probability. N0 = 1. coherent FSK % and noncoherent FSK modulation may be specified. % Transmission BW for CFSK in kHz elseif I_mod == 4 BkHz = 4*Rkbps. % Rate in kbps for NFSK end elseif I_BW_R == 2 if I_mod == 0 BkHz = Rkbps. 2 if data rate specified: ’). % Transmission BW for baseband in kHz elseif I_mod == 1 | I_mod == 2 BkHz = 2*Rkbps. I_BW_R = input(’Enter 1 if bandwidth specified. 2 = DPSK.2. 1 = BPSK. if I_BW_R == 1 BkHz = input(’Enter vector of desired bandwidths in kHz: ’). % PE = 0. % Rate in kbps for BPSK or DPSK elseif I_mod == 3 Rkbps = BkHz/3. end PE = input(’Enter desired BEP: ’). % Transmission BW for NFSK in kHz end end R = Rkbps*1000. baseband. z = sqrt_z^2. BINARY DATA COMMUNICATION sqrt_z_over_2 = erfinv(1-2*PE). end Eb_N0_dB = 10*log10(z). Hz A. kbps BW. DPSK. elseif I_mod == 3 CHAPTER 7. elseif I_mod == 2 z = -log(2*PE). z = 2*sqrt_z_over_2^2. CFSK. elseif I_mod == 4 z = -2*log(2*PE). % z = A^2/(R*N0) for baseband % This is the required Eb/N0 in dB elseif I_mod == 1 | I_mod == 2 | I_mod == 3 | I_mod == 4 A = sqrt(2*R*N0*z). NFSK .38 if I_mod == 0 | I_mod == 1 sqrt_z = erfinv(1-2*PE). if I_mod == 0 A = sqrt(R*N0*z). volts’) disp(’ ___________________________’) disp(’ ’) disp([Rkbps’ BkHz’ A’]) disp(’ ’) % z = A^2/2*R*N0 for BPSK. end disp(’ ’) disp(’Desired P_E and required E_b/N_0 in dB: ’) format long disp([PE Eb_N0_dB]) format short disp(’ ’) disp(’ R. 0000 150.0000 60. clf % Clear any plots left over from previous runs k = 0. % Compute filter bandwidth from BW*T_bit [num.2*BW/samp_bit).0465 50. samp_bit = input(’Enter number of samples per bit used in simulation: ’). . N_bits_block = input(’Enter number of bits/sim.0232 10. % Simulation is broken up into contiguous blocks for memory management. COMPUTER EXERCISES A typical run is given below: >> ce7_4 Enter desired modulation: 0 = baseband. % Eb_N0_dB_max = input(’Enter maximum Eb/N0 in dB: ’).79982256904398 R.7. T_bit = 1. kbps BW. block: ’). N_bits = input(’Enter total number of bits in simulation: ’). 2 = DPSK.2. % Arbitrarily take bit time as 1 second BW = BWT_bit/T_bit. 2 if data rate specified: 2 Enter vector of desired data rates in kbps: [5 10 20 50] Enter desired BEP: 1e-3 Desired P_E and required E_b/N_0 in dB: 0. % Obtain filter num/den coefficients e_tot = [].0248 20. N_blocks = floor(N_bits/N_bits_block).0000 30. BWT_bit = input(’Enter filter bandwidth normalized by bit rate: ’). volts ___________________________ 5.00100000000000 9.0000 437. n_order = input(’Enter order of Butterworth detection filter: ’). 4 = NCFSK: 3 Enter 1 if bandwidth specified.0000 309.den] = butter(n_order.5 39 % ce7_5.0000 977.m: Simulation of suboptimum bandpass filter/delay-and-multiply % demodulator with integrate-and-dump detection for DPSK. Hz A.0000 618. % Ensure that plotting arrays are empty Perror = []. N_bits_sim = []. 3 = CFSK. 1 = BPSK. input bandpass % filter is Butterworth with selectable bandwidth-bit period product.0000 15.2173 Computer Exercise 7. Eb_N0_dB = []. Eb_N0_dB_min = input(’Enter minimum Eb/N0 in dB: ’). % Compare detected bits with input if nn == 1 error_array(1:10) = 0. signal is 1-bit delayed S + Ne y_mult = y. % Build array wth columns samp_bit long sig = sig(:).N_bits_block).samp_bit. of errors nn = nn + 1. % Ref. . % Differentially encode data for DPSK s = 2*data_diff_enc-1. = 1 N0 = Eb/Eb_N0. % Filter S plus N with chosen filter zi = zf. noise samples [y. % Compute std deviations of noise samples e_sum_tot = 0. N_bits_sim(k) = (nn-1)*N_bits_block-10-(nn-2). % Don’t include 1st 5 bits due to transients elseif nn ~= 1 error_array(1:1) = 0. % Logical data is sequence of 1s and 0s data_diff_enc = diff_enc(data).5). Perror(k) = e_sum_tot/N_bits_sim(k). sig = s(ones(samp_bit. for each Eb/N0 k = k+1. % Generate sequence of random +-1s data = 0. % Compute noise PSD from Eb/N0 del_t = T_bit/samp_bit. % Multiply rec’d S + N by 1 bit delay bits_out = int_and_dump(y_mult. % Sum to get total number of errors/blk e_sum_tot = e_sum_tot + e_sum.zi). % Convert matrix of bit samples to vector bits_out = [].40 CHAPTER 7. Eb_N0 = 10^(Eb_N0_dB(k)/10).sig+noise.samp_bit). if ampl. noise = sigma_n*randn(size(sig)). % Save final values for future init. cond’ns y_ref = delay1(y.den. det. error_array = abs(bits_out-data).zf] = filter(num. end e_tot(k) = e_sum_tot. nn = 1. % Running sum of total no.5*(ss+1). % I-&-D det. Eb_N0_dB(k) = kk. while nn <= N_blocks & e_sum_tot < 200 ss = sign(rand(1. % Generate bipolar data for modulation sig = []. % Bit energy is T_bit. % Form sequence of Gauss.*y_ref.:). % Delete 1st bit due to y_ref = 0 end e_sum = sum(error_array).N_bits_block)-. % Compute sampling interval sigma_n = sqrt(N0/(2*del_t)). % Convert desired Eb/N0 from dB to ratio Eb = T_bit. % Make sure various arrays are empty y_det = []. zi = []. BINARY DATA COMMUNICATION for kk = Eb_N0_dB_min:Eb_N0_dB_max % Loop for simul.1). N_bits). ’-order Butterworth filter.^(Eb_N0_dB/10))) % Plot approximate theoretical result for suboptimum detector semilogy(Eb_N0_dB. samp_bit. 0. integrate = sum(samp_array). BT = ’.^(Eb_N0_dB/10))). end end % int_and_dump(input.’.1)). delay/multiply detector’. xlabel(’E_b/N_0. ’. num2str(N_blocks). else output(k) = not(bitxor(input(k). ’. optimum differential detector’.5*exp(-10. num2str(n_order). 3) %diff_enc(input). num2str(BWT_bit)]) legend([’Simulation. ’ blocks’]. qfn(sqrt(10. for k = 1:L_in if k == 1 output(k) = not(bitxor(input(k). hold % Plot theoretical bit error probability for optimum DPSK detector semilogy(Eb_N0_dB.’ bits. . Perror.N_bits). % Integrate (sum) each bit (column) bits_out = (sign(integrate) + 1)/2.’) Title([’Comparison of optimum & delay-and-multiply detectors for DPSK. output = []. grid. COMPUTER EXERCISES 41 end % End of Eb/N0 loop disp(’ ’) disp(’ Eb/N0.samp_bit. BT = ’. of error disp(’ __________________________’) disp(’ ’) disp([Eb_N0_dB’ Perror’ e_tot’]) disp(’ ’) % Plot simulated bit error probabilities versus Eb/N0 semilogy(Eb_N0_dB.7. dB. num2str(N_bits). ’.’Theory.N_bits) % Reshape input vector with each bit occupying a column samp_array = reshape(input. num2str(BWT_bit).samp_bit. % function bits_out = int_and_dump(input. function to differentially encode a bit stream vector % function output = diff_enc(input) L_in = length(input).2.’-. dB’). ’Theory.output(k-1))). PE errors’) % Display computed prob. ylabel(’P_E’).’—’). PE.0000 4.0101 152.0216 216.5 Enter total number of bits in simulation: 15000 Enter number of bits/sim. for delta = -0. PE errors __________________________ 3. T2 = 0.0000 5. z0 = 10.0000 Computer Exercise 7. block: 5000 Eb/N0.6. dB.0000 7.^(z0dB/10).0000 6.5*qfn(sqrt(2*z0)*(1+delta)). PE = T1 + T2.0000 0.6 % ce7_6: program to plot Figure 7-28 % clf z0dB = 0:1:30. if taum_over_T == 0 semilogy(z0dB.9:0. BINARY DATA COMMUNICATION A typical run is given below. end for taum_over_T = 0:1:1 T1 = 0. ’-’) elseif taum_over_T == 1 semilogy(z0dB. A plot is also made comparing probabilities of error for the optimum detector. PE. simulated suboptimum detector.42 CHAPTER 7.0000 0. >> ce7_5 Enter maximum Eb/N0 in dB: 7 Enter minimum Eb/N0 in dB: 3 Enter number of samples per bit used in simulation: 5 Enter order of Butterworth detection filter: 2 Enter filter bandwidth normalized by bit rate: 1.5*qfn(sqrt(2*z0)*(1+delta)-2*delta*taum_over_T).0000 0.0000 0. A minimum of 200 errors are counted unless the total number of specified bits to be simulated is exceeded.0485 242. and approximate theoretical result for the suboptimum detector.1040 519. if delta > -eps & delta < eps delta = 0.0655 327. ’—’) end .0000 0.3:0. and (7. num2str(delta)]) elseif delta > 0. Computer Exercise 7. (7.3 text(z0dB(5).11: if delta <= -. (7.87).119). PE(8).3 & delta <= 0. PE(10).111).7 Use Equations (7.9 hold on ylabel(’P_E’).73) (with m = 0).(7. PE(5).174) . num2str(delta)]) elseif delta > -0.[’\tau_m/T = 1’]) The output plot is shown in Fig. num2str(delta)]) end if delta == -0.2. Solve them for the SNR in terms of the error probability.2. Express the SNR in dB and subtract .7. [’\delta = ’.3 text(z0dB(8).177).3 text(z0dB(10). [’\delta = ’. COMPUTER EXERCISES 43 Figure 7. (7. 7. [’\delta = ’. xlabel(’z_0 in dB’) axis([0 30 1E-5 1]) grid end end end legend([’\tau_m/T = 0’]. disp(’ ’) if mod_type == 1 disp(’ BPSK’) Z_bar = (0./(PE .44 CHAPTER 7.PE. DPSK.^2).2*PE)). 3 = DPSK. z = (erfinv(1 . The MATLAB program below does this for the modulation cases of BPSK.z_dB. elseif mod_type == 4 disp(’ NFSK’) Z_bar = 1.2*PE)). 4 = NFSK: ’). 3 = DPSK.^2. z = -2*log(2*PE). z = -log(2*PE). elseif mod_type == 3 disp(’ DPSK’) Z_bar = 1. and noncoherent (NFSK). % ce7_7.2*PE). BINARY DATA COMMUNICATION the nonfading result from the fading result. PE = input(’Enter vector of desired probabilities of error: ’). % MATLAB has an inv. 2 = CFSK.^2).5*(1 . error function elseif mod_type == 2 disp(’ CFSK’) Z_bar = (0. 4 = NFSK: 4 .^2).PE. disp(’ ’) disp(’ P_E Degradation.25*(1 . z_dB = 10*log10(z).^2)./(2*PE) -1.2*PE). coherent FSK (CFSK). 2 = CFSK. dB’) disp(’_________________________’) disp(’ ’) disp([PE’ deg_dB’]) disp(’ ’) A typical run is given below: >> ce7_7 Enter type of modulation: 1 = BPSK. z = 2*(erfinv(1 . deg_dB = Z_bar_dB .m: Program to evaluate degradation due to flat % Rayleigh fading for various modulation schemes % mod_type = input(’Enter type of modulation: 1 = BPSK.^2./PE -2./(PE . end Z_bar_dB = 10*log10(Z_bar). disp(’ ’) disp(’Inverse of Pc:’) disp(’ ’) disp(Pc_inv) . COMPUTER EXERCISES Enter vector of desired probabilities of error: [1e-2 5e-3 1e-3 5e-4 1e-4] NFSK P_E Degradation. one for the zero-forcing case and one for the MMSE case.0100 10. for m = 1:2*N+1 row = [].0005 21.7.m) = row’.2.0001 27. % ce7_8a. end disp(’ ’) disp(’Pc:’) disp(’ ’) disp(Pc) Pc_inv = inv(Pc). Pc = [].3239 0.0010 19.0050 13.8 45 The solution to this exercise consists of two MATLAB programs. mid_samp = fix(L_pc/2)+1.6859 Computer Exercise 7. for n = 2-m:2*N+2-m row = [row pc(mid_samp-1+n)].9779 0.0469 0. dB _________________________ 0. end Pc(:. disp((L_pc-1)/4) N = input(’Enter number of zeros each side of decision sample desired: ’).m: Plots unequalized and equalized sample values for ZF equalizer % clf pc = input(’Enter sample values for channel pulse response (odd #): ’). L_pc = length(pc). disp(’Maximum number of zeros each side of decision sample: ’).6023 0. num2str(N).1.2500 -0.2500 -0.1500 0..stem(p_eq_p).1500 -0..0000 Inverse of Pc: . disp(’ ’) disp(’Equalized pulse train’) disp(’ ’) disp(p_eq_p) disp(’ ’) t=-5:.0000 -0.2 .3 1 -.3000 0.xlabel(’n’). axis([1 L_pc -.05 -.25 -. title([’Output equalized with ’. for m = mid_samp-floor(steps/2):mid_samp+floor(steps/2) p_eq(m) = sum(coef’.2). output pulse’).05 .2500 -0.2000 1.2000 1.1500 0.5000 Enter number of zeros each side of decision sample desired: 2 Pc: 1. end p_eq_p = p_eq(N+1:mid_samp+floor(steps/2)).0000 -0..5]).3000 0..stem(pc).15 -.1).1 -.3000 0. title([’Ch..1000 0.01:5.1500 0..1.5 1.2000 1. disp(’ ’) disp(’Equalizer coefficients:’) disp(’ ’) disp(coef) disp(’ ’) steps = (L_pc-length(coef))+1.N+1).1500 0.*fliplr(pc(m-N:m+N))).46 CHAPTER 7.ylabel(’Ch. axis([1 L_pc -..1000 -0.0500 0.5]). y=zeros(size(t)).3000 0..2000 1..’ zero samples either side’]) A typical run is given below: >> ce7_8a Enter sample values for channel pulse response (odd #): [-.num2str(pc)..15 . p_eq = []. BINARY DATA COMMUNICATION coef = Pc_inv(:.1000 0.1 ..ylabel(’Equal.02] Maximum number of zeros each side of decision sample: 2.0000 -0. output pulse’). pulse samp.’]’]) subplot(2. = [’.5 1.0000 -0.. subplot(2. 0350 -0.12: 1. COMPUTER EXERCISES 47 Figure 7.2.1317 0.7.1095 0.0956 0.0000 0.2572 1.2842 -0.1686 -0.0554 The MATLAB program for the MMSE case is given below: % Solution for ce 7. z = 10^(z_dB/10).0588 -0.1095 Equalizer coefficients: -0.0588 -0.0366 -0.1777 0.1777 0.0000 0 -0.1403 -0.8(b) % z_dB = input(’Enter the signal-to-noise ratio in dB: ’).0107 0.2572 -0. .1686 1. b = input(’Enter multipath gain: ’).1777 0.1731 0.1403 -0.1403 1.0093 0.0877 1.2572 1.0000 0.1476 0.0366 -0.2842 1.0000 -0.1476 0.1317 0.0877 0.1095 Equalized pulse train -0. 2)=R_yy(1. R_yy(2.3).48 CHAPTER 7. disp(’ ’) disp(’Optimum coefficients:’) disp(A) disp(’ ’) A typical run is given below: >> ce7_8b Enter the signal-to-noise ratio in dB: 7 Enter multipath gain: .2).1).3869 3.0101 8.1)=(1+b^2)*z+pi/2.2)=b*z+(pi/2)*exp(-2*pi). R_yy(1. BINARY DATA COMMUNICATION R_yy(1. disp(’ ’) disp(’R_yy:’) disp(’ ’) disp(R_yy) B = inv(R_yy).1).2)=R_yy(1. disp(’ ’) disp(’R_yy^-1:’) disp(’ ’) disp(B) R_yd = [0 z b*z]’. R_yy(1.3)=(pi/2)*exp(-4*pi).1)=R_yy(1.0576 0.1399 -0.3)=R_yy(1. R_yy(3.0000 3.3)=R_yy(1.3869 3.0101 0.2).0000 3.0101 8.0576 0. R_yy(2.6 R_yy: 8. R_yy(3. R_yy(3.0101 0.3869 R_yy^-1: 0.0207 -0.1606 -0.0576 . R_yy(2.1)=R_yy(1.2). disp(’ ’) disp(’R_yd:’) disp(’ ’) disp(R_yd) A = B*R_yd. 2.6316 0. COMPUTER EXERCISES 0.0207 -0.0119 3.0576 R_yd: 0 5.2267 0.1399 49 .1319 0.7.0071 Optimum coefficients: -0. · · ·. Rb = 12. Then d2 (t) is staggered. 1. 000 bps.1 Problem Solutions Problem 8. −1. −1. In this case. For M = 4. 000 bps. 5π/4. so θ2 (t) = 7π/4. After Ts = 2Tb seconds d1 (t) takes on the second 1-value. so θ1 (t) = π/4.000 symbols per second. 7π/4. Rb = 2 × 2. 1. 7π/4. At 3Tb seconds. 1. So the first phase shift is for d1 (t) = 1 and d2 (t) = 1 or θ2 (t) = π/4. 7π/4. −1. so θ2 (t) = 7π/4. and for M = 64. 000 bps. At 4Tb seconds.Chapter 8 Advanced Data Communications Topics 8. θi (t) = tan d1 (t) Alternating every other bit given in the problem statement between d1 (t) and d2 (t) gives d1 (t) = 1. Problem 8. for M = 8. 1.1 Use the relationship Rb = (log2 M ) Rs bps where Rb is the data rate. (c) Now the switching instants are each Tb seconds. −1. d1 (t) changes to −1. 7π/4. but d2 (t) is still 1. Start with d1 (t) = 1. −1 which results in θi (t) = π/4. Rs = 2000 symbols per second. −1 and d2 (t) = 1. 000 = 4. etc. −1.2 (a) 5. and M is the number of possible signals per signaling interval. by 1 bit time. −1. 1 . Rs is the symbol rate. or offset. For QPSK the symbol switching points occur each Ts seconds. (b) Recall that · ¸ √ −1 d2 (t) y (t) = A [d1 (t) cos (ωc t) + d2 (t) sin (ωc t)] = 2A cos [ω c t − θi (t)] . d2 (t) changes to −1. Rb = 6. 1. 3π/4. (ii) θi (t) = 45o . (c) 0.4 × 10−5 R. then the amplitude of the √ ¡√ ¢2 2A T / (2N0 ) = envelope-phase-modulated form of the carrier is 2A. (iv) θi (t) = −45o . CHAPTER 8. Hence. . symbol = 2Q Trial and error using the asymptotic approximation for the Q-function gives [Es /N0 ]req’d ≈ 12. (d) 0. ADVANCED DATA COMMUNICATIONS TOPICS PE. If the quadrature-carrier amplitudes are A.0442 V. A −11 ) (19. (iii) If θi (t) = −45o . d1 (t) = −1 and d2 (t) = 1. (b) Error in detecting d1 (t): (i) θi (t) = 135o . d1 (t) = 1 and d2 (t) = −1. d1 (t) = 1 and d2 (t) = 1.2 Problem 8. Problem 8. (b) 0.0099 V.5 (a) Use θi (t) = tan −1 · d2 (t) d1 (t) ¸ (i) If θi (t) = 45o . (iv) If θi (t) = −135o . d1 (t) = −1 and d2 (t) = −1. (iii) θi (t) = −135o . (e) 0.5.5) R = 1.0044 V.014 V. (f) 0.9 dB = 19.0014 V. The N0 R [Es /N0 ]req’d = (10 A 0 req’d = answers are as follows: (a) 0.0031 V. (ii) If θi (t) = 135o .4 Take the expectation of the product after noting that the average values are 0 because E[n (t)] = 0: ·Z Ts ¸ Z Ts E [N1 N2 ] = E n (t) cos (ω c t) dt n (t) sin (ω c t) dt = = = = Z Ts ´ ³p Es /N0 = 10−5 Z 0 Ts Z Z 0 0 Ts E [n (t) n (λ)] cos (ω c t) sin (ω c λ) dtdλ Ts 0 Z N0 Ts cos (ω c t) sin (ω c t) dt 2 0 Z N0 Ts sin (2ω c t) dt = 0 4 0 0 0 N0 δ (t − λ) cos (ωc t) sin (ωc λ) dtdλ 2 Problem 8.3 For QPSK. and Es /N0 = q p √ 2 /N R. 8 For the data stream 11100 10111 00100 00011. the error ³q ´ Es term becomes negligible compared with PE1 = Q N0 as Es /N0 becomes larger. (iv) θi (t) = 135o . the one corresponding to d2 (t) is delayed by Ts /2 = Tb seconds with respect to d1 (t). and type II MSK corresponds to d1 (t) and d2 (t) multiplied by absolute-value cosine and sine waveforms with periods of Ts . respectively (one half cosine or sine per Ts pulse). other factors might favor BPSK. (i) θi (t) = −45o . . (b) QPSK is 3 dB worse in terms of symbol error probability for equal transmission bandwidths. these pulse sequences are aligned and for OQPSK. Waveforms for QPSK.8.3 for a random (coin toss) serial bit sequence. respectively. For QPSK.1 . the quadrature data streams are d1 (t) = 11-1-11-11-1-11 d2 (t) = 1-1111-1-1-1-11 Each 1 above means a positive rectangular pulse Ts seconds in duration and each -1 above means a negative rectangular pulse Ts seconds in duration. and Type II MSK generated by a MATLAB program are shown in Figures 8. (iii) θi (t) = 45o .8. PROBLEM SOLUTIONS 3 (c) Error in detecting d2 (t). since the Q-function is monotonically decreasing with increasing Es /N0 . Problem 8. (c) Choose QPSK over BPSK in terms of performance. Problem 8.1. however. (ii) θi (t) = −135o . Problem 8.7 The exact result is Psymbol = 1 − (1 − PE1 )2 2 = 2PE1 − PE1 The approximation is Psymbol ≈ 2PE1 .6 (a) Both are equivalent in terms of symbol error probabilities. so the error term is ²= 2 PE1 = Q " Ãr Es N0 !#2 Clearly. such as simpler implementation. but it handles twice as many bits per second. Type I MSK corresponds to the OQPSK waveforms d1 (t) and d2 (t) multiplied by cosine and sine waveforms with periods of 2Ts . OQPSK. 2: . ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.4 CHAPTER 8.1: Figure 8. For MSK. For OQPSK. The maximum phase change is ±π/2 radians. Use appropriate trigonometric identities to reduce the product of the sinc-functions to the given form. the instantaneous frequency is 1/4Tb Hz above the carrier (with the definition of θi (t) given by (8. −3π/4. π/4. −π/4. −3π/4. −π/4.10 Write the sinc-functions as sin(x) /x functions. π/4. Problem 8. PROBLEM SOLUTIONS 5 Figure 8. If it is a sequence of . the phase can change each Ts /2 = Tb seconds because d2 (t) is delayed by Ts /2 seconds with respect to d1 (t). the excess phase corresponds to a stepwise function that may change values each Ts seconds. The phase deviation is given by · ¸ −1 d2 (t) θi (t) = tan d1 (t) so the sequence of phases for QPSK computed from d1 (t) and d2 (t) given in Problem 8.18)).1.8 is π/4. Problem 8. −3π/4. the excess phase trajectories are straight lines of slopes ±π/2Tb radians/second.8. π/4 radians. 3π/4.17) and (8.9 For QPSK.11 If d (t) is a sequence of alternating 1s and 0s.3: Problem 8. symbol log2 (M ) and the energy per bit-to-noise spectral density ratio is Es 1 Eb = N0 log2 (M ) N0 Thus. the bit error probability is given in terms of symbol error probability as PE. 000. M = 32   3 × 10−5 . 000/4 = 975. Problem 8.5 × 10−5 . M = 64 "r # Eb 2 log2 (M ) sin (π/M ) = 10−5 N0 Q 2 log2 (M ) . ADVANCED DATA COMMUNICATIONS TOPICS 1s or 0s. Problem 8. Assuming Gray encoding. M = 8   2 × 10−5 .5 × 10−5 .5 degrees (360/16). 025. 000 − 100. the actual error probability is very close to the upper bound. The decision regions are pie wedges centered over each signal point.12 √ The signal points lie on a circle of radius Es centered at the origin equally spaced at angular intervals of 22.6 CHAPTER 8. (a) The instantaneous frequency is 5. 000/4 = 1. 000. the instantaneous frequency is 1/4Tb Hz below the carrier. (b) The instantaneous frequency is 5.13 The bounds on symbol error probability are given by P1 ≤ PE. 000 + 100. we solve 2 Q log2 (M ) or "r Eb 10−5 sin (π/M ) = N0 2 #   1. symbol ≤ 2P1 where P1 = Q "r # 2Es sin (π/M ) N0 For moderate signal-to-noise ratios and M ≥ 8. M = 16 × log2 (M ) =  2. 000 Hz. bit ≈ PE. 000 Hz. 06.06 N0 r Eb 10 × 0.17 N r 0 Eb 6 × 0.098 ¶2 # = 22.3827 For M = 16.1951 = 4.1. we have r Eb 2 log2 (8) sin (π/8) = 4. for M = 8. approximately.17 N0 Eb N0 For M = 32. we have r Eb 2 log2 (16) sin (π/16) = 4.1.17 2 1 = 17.57 dB = 10 log10 8 0. M = 16 Q-function argument = 4. M = 64 Thus. M = 32    4. by trial and error (either using MATLAB and a program for the Q-function or a calculator and the asymptotic expansion of the Q-function):   4. PROBLEM SOLUTIONS 7 The argument of the Q-function to give these various probabilities are found.35 dB .098 = 4.17 N0 Eb N0 " µ ¶ # 1 4.1 N0 r Eb 8 × 0.06 N0 Eb N0 = 10 log10 " µ ¶ # 4.3827 = 4. we have r Eb 2 log2 (32) sin (π/32) = 4.06 0.8.17 2 = 12.02. M = 8   4.17.1951 " 1 10 µ 4.96 dB = 10 log10 6 0. given signal labeled 1111 was sent. 1100 1101 1110 1111 Gray 1010 1011 1001 1000 Problem 8. P (C | I) = P (0 ≤ X ≤ 2a) P (0 ≤ Y ≤ 2a) Z 2a −(x−a)2 /N0 Z 2a −(y−a)2 /N0 e e √ √ = dx dy πN0 πN0 0 0 Let √ √ 2 (x − a) 2 (y − a) u= and u = N0 N0 . No.02 N0 r Eb 12 × 0. No. Y ) = (a + N1 . 0000 0000 0100 0001 0001 0101 0010 0011 0110 0011 0010 0111 for the decimal digits 0 − 15 and their Gray code equivGray 0110 0111 0101 0100 Bin. No.02 N0 Eb N0 = 10 log10 " 1 12 µ 4. Gray Bin. be (X. a + N2 ) where N1 and N2 are zero-mean.049 ¶2 # = 27.49 dB Problem 8. uncorrelated Gaussian random variables with variances N0 /2.15 Let the coordinates of the received data vector. Since they are uncorrelated. 1000 1001 1010 1011 Gray 1100 1101 1111 1110 Bin.02 0.8 For M = 64. Thus.049 = 4. No. ADVANCED DATA COMMUNICATIONS TOPICS Eb 2 log2 (64) sin (π/64) = 4. we have r CHAPTER 8.14 The binary number representation alents are given below: Bin. they are also independent. A = P (C | III) = [1 − Q (A)]2 Thus PE = 1 − ¾ 1 1 1 [1 − 2Q (A)]2 + [1 − 2Q (A)] [1 − Q (A)] + [1 − Q (A)]2 4 2 4 ½ ¾ ¤ 1£ ¤ 1£ ¤ 1£ 2 2 2 1 − 4Q (A) + 4Q (A) + 1 − 3Q (A) + 2Q (A) + 1 − 2Q (A) + Q (A) = 1− 4 2 4 ½ s 2a2 N0 · P (C | II) = [1 − 2Q (A)] [1 − Q (A)] ≈ 1 − {1 + 3Q (A)} .16 The symbol error probability expression is ¸ 1 1 1 Ps = 1 − P (C | I) + P (C | II) + P (C | III) 4 2 4 where P (C | I) = [1 − 2Q (A)]2 .1. neglecting all terms in Q2 (A) s  2a2  = 3Q  N0 . PROBLEM SOLUTIONS This results in Z r 2a2 N0 2 e−u /2 9 P (C | I) = Z 2 e−u /2 √ du = 4 2π 0 0  s 2 2a2  = 1 − 2Q  N0 Z 2a2 N0 r 2 − 2a N0 r √ du 2π Z r 2 − 2a N0 r r 2a2 N0 e−v /2 √ dv 2π e−v /2 √ dv 2π 2 2 2a2 N0 Similar derivations can be carried out for the type II and III regions. Problem 8.8. 19 Use the same relations as in Problem 8.3 dB.5 dB. . symbol and = 2 (M − 1) N0 log2 (M ) N0 A tight bound for the symbol error probability for coherent M -ary FSK is Ãr ! Es PE. PE. the given expression for PE follows. and ³√ ´2 M − 2 type I regions. For M = 32. To accomplish this. A sketch will help in determining this. The expression for a follows by computing the average symbol energy in terms of a. bit = Es Eb M 1 PE. II. The following results are obtained: For M = 2.3 dB. bit = 10−4 for Eb /N0 = 5.18 Use PE.5 dB. Problem 8. symbol = M−1 µ X k=1 M −1 k ¶ µ ¶ k Es (−1)k+1 exp − k+1 k + 1 N0 Use MATLAB to perform the sum.10 CHAPTER 8. The approximate probability of error expression follows in a manner similar to the derivation for M = 16 outlined in Problem 8. except now PE.18 for relating signal-to-noise ratio per bit and symbol and for relating bit to symbol error probability. Problem 8. bit = 10−4 for Eb /N0 = 12. PE. the sums m X i=1 m X m (m + 1) m (m + 1) (2m + 1) i= i2 = and 2 6 i=1 are convenient.9 dB. ³√ four of which are type III regions (the corner regions). There are M´× M total regions. 4 M − 2 type II regions. PE.16. bit = 10−4 for Eb /N0 = 5. and III regions. For M = 16. ADVANCED DATA COMMUNICATIONS TOPICS Problem 8. PE.17 √ √ This follows by counting type I. bit = 10−4 for Eb /N0 = 6. Thus. symbol ≤ M Q N0 Use the asymptotic expression for the Q-function and iteration on a calculator to get the following results: For M = 8. PE. For M = 64. bit = 10−4 for Eb /N0 = 7. The decision regions are formed by the positive coordinate axes and the 45-degree bisector of the first quadrant angle.8/Tb = 0. 11 Problem 8.   0.23 Use Figure 8. MSK = 1. M = 32 16-QAM: R B FSK: Problem 8. M = 16 . QPSK.0 bps/Hz.24 The baseband power spectrum is G (f ) = 2A2 Tb [log2 (M )] sinc2 [log2 (M ) Tb f ] = 2.67 kHz.5 bps/Hz.0 kHz   30. M = 8 R PSK: B = 2.1. Problem 8. (b) B99. M = 16.8/Tb = 0. M = 32 . BPSK = 1.33 bps/Hz. M -PSK and M -QAM R = log2 (M) B M+1 . QPSK.0 bps/Hz. PE.0 kHz. MSK = 0. (a) B90.23/Tb = 1.8Rb .5 log2 (M ) .21 The bandwidth efficiencies are given by ( 0. PROBLEM SOLUTIONS For M = 4. M = 8 = 0. M = 16 . bit = 10−4 for Eb /N0 = 9.  0.15 bps/Hz. BPSK = 14/Tb = 14Rb .2 dB.6 kHz. Problem 8. Problem 8. (a) B99. PE. M = 32 R B   6. M = 8 B= 5.6Rb .8.0 kHz.19 noting that the abscissa is normalized baseband bandwidth. (c) B90. M = 16  4. M = 16  66. The 90% power containment bandwidth is defined by the ordinate = -10 dB. The 99% power containment bandwidth is defined by the ordinate = -20 dB. RF bandwidthis twice as large.22 Use Figure 8.8Rb . RF bandwidth is twice as great.  2. (c) B99.6 dB. bit = 10−4 for Eb /N0 = 8. M = 32 B = 5.5/Tb = 3. M = 8 B= 42. (b) B90. OQPSK = 3. OQPSK = 0.19 noting that the abscissa is normalized baseband bandwidth.5 bps/Hz.20 This is a normal cartesian coordinate system with the signal points located on the φ1 and √ φ2 axes at Es . For M = 8.6/Tb = 1.0 kHz.23Rb . coherent M -FSK This gives the results given in the table below:   1.24 bps/Hz.5Rb .0 kHz. 8/Tb = 0.8Rb / log2 (M ) = 0. m 6= 0 4 4 4 1 4 Sr (f ) = Tb sinc2 (f Tb ) SNRZ.267Rb Hz.12 where CHAPTER 8. M = 8 Problem 8. ADVANCED DATA COMMUNICATIONS TOPICS A2 = Es cos2 θi = Es sin2 θi · · ¸ ¸ M M 1 X 2 2π (i − 1) 1 X 2 2π (i − 1) = = cos sin M M M M i=1 i=1 The 10% power containment bandwidth is defined by (see Problem 8.16Rb Hz. m 6= 0 µ f Tb 2 ¶ f Tb 2 ¶ Tb sinc2 Sr (f ) = 2 A2 Tb Spolar RZ (f ) = sinc2 4 µ . MPSK   0. M = 16  0. MPSK = 0. mark (f ) = A2 Tb sinc2 (f Tb ) (b) For polar RZ: Rm = ½ A2 . m = 0 2 2 (A) (A) + (−A) (A) 1 + (A) (−A) 1 + (−A) (−A) 1 = 0.22) log2 (M ) B90.8Rb B90.25Rb Hz.25 (a) For NRZ mark: Rm = ½ (A) (A) 1 + (−A) (−A) 1 = A2 . M = 8 = 0. m = 0 0. 1.26 To show: cos µ πt 2Tb ¶ t Π 2Tb µ ¶ ←→ 4Tb cos (2πTb f ) π 1 − (4Tb f )2 Use the modulation theorem together with the Fourier transform of a rectangular pulse to get ¶ µ ¶ µ t πt Π ←→ Tb sinc [2Tb (f − 1/4Tb )] + Tb sinc [2Tb (f + 1/4Tb )] cos 2Tb 2Tb Rewrite the RHS as RHS = Tb Note that sin (2πTb f ± π/2) = ± cos (2πTb f ) Then RHS = = · ¸ 1 1 2Tb + cos (2πTb f ) π 1 − 4Tb f 1 + 4Tb f 4Tb cos (2πTb f ) π 1 − (4Tb f )2 · sin (2πTb f − π/2) sin (2πTb f + π/2) + 2πTb f − π/2 2πTb f + π/2 ¸ .8. m = 0  2 ¡ 1 ¢2 ¡ 1 ¢2 2 ¡ 1 ¢2 2 ¡ ¢2 ¡ ¢¡ ¢ Rm = (A) (A) ¡4 ¢ ¡ ¢ (−A) 4 ¡ +¡ ¢ (A) 4 + (−A) (−A) 1 ¡ + ¡ ¢ (0) 1 1 + (A) (−A) (A) 4 2   + (0) (A) 1 1 + (−A) (0) 1 ¢ 1 + (0) (−A) ¡ 1 ¢ ¡ 1 ¢ + (0) (0) 1 ¢ 1 = 0. PROBLEM SOLUTIONS 13 (c) Bipolar RZ:  2  (A) (A) 1 + (0) (0) 1 = A . 4m 6= 0 2 4 4 2 2 4 2 2 Tb Sr (f ) = sinc2 2 µ f Tb 2 ¶ f Tb 2 ¶ A2 Tb sinc2 Sbipolar RZ (f ) = 8 µ Problem 8. 28).28 This is a matter of MATLAB programming and plotting.30 The states are (read from top to bottom. column by column): 1111 0100 1011 0111 0010 0101 0011 1001 1010 0001 1100 1101 1000 0110 1110 The output sequence is the last digit in each 4-element word.27 Apply (5.14 CHAPTER 8. .29 A plot is given in Figure 8. Problem 8. Problem 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. Problem 8. The autocorrelation function is a triangular pulse centered at the origin two units wide repeated every 15 units and a horzontal line between the triangles at −1/15 units.4.4: Problem 8. -2. 0. (a) -1. 0. 0. column by column): 11111 00101 10100 01101 01111 10010 01010 00110 00111 01001 11010 00011 10011 00100 11101 10001 11001 00010 01110 11000 01100 00001 01110 11100 10110 10000 10111 11110 01011 01000 11011 The output sequence is the last digit in each 5-element word. (the maximum absolute value correlation is 4 . The result can be scaled by the length of the sequence if desired. Problem 8. 0. -1.31 The states are (read from top to bottom. 1. . -4. The expectation can be taken inside. The result for the variance (same as the mean square) is then found to be N0 Tb . -1. . 15.nonzero delay).34 The random variable to be considered is NI = Z Tb 0 AI c (t) cos (∆ωt + θ − φ) dt . 0. The integral of the resulting cosine squared is Tb /2.32 To find the aperiodic correlation function. -1. -1. 3. 7. 0.33 Note that the expectation of Ng is zero because n (t) has zero mean. 0. 0. -1 (the maximum absolute value correlation is 1 nonzero delay).8. . -1. we slide the sequence past itself and compute the number of alike bits minus the number of unalike bits for the overlap (the sequence is not periodically repeated). 0. PROBLEM SOLUTIONS 15 Problem 8. Problem 8. The autocorrelation function is a triangular pulse centered at the origin two units wide repeated every 31 units and a horzontal line between the triangles at −1/31 units. 0. Problem 8. Write the expectation of the square of Ng as an iterated integral.1. 0. -1. -1. -1. (b) -1. Use the fact that E [n (t) n (λ)] = N0 δ (t − λ) 2 to reduce the double integral to a single integral. 2π).09 × 10−5 for a JSR of 10 dB.37 ³√ ´ In the limit at SNR Eb /N0 → ∞. PE = 1.34 × 10−4 for a JSR of 10 dB. Rc = 100 megachips per second. (b) BRF = 2Rc = 200 MHz.16 CHAPTER 8. PE = 1. PE = 0. Problem 8.34 for a JSR of 30 dB.03 × 10−5 for a JSR of 15 dB.72 × 10−6 for a JSR of 5 dB. (c) Use (8.36 The results are PE = 5. Problem 8.35 (a) The processing gain is Gp = Tb Rc = or Rc = Gp Rb Tc Rb Z 0 Tb Z Tb 0 A2 c (t) c (λ) cos (∆ωt I ¾ + θ − φ) cos (∆ωλ + θ − φ) dtdλ For Gp = 100 and a data rate of 1 Mbps.0895 for a JSR of 30 dB.03 × 10−5 for a JSR of 5 dB. PE = 0. Clearly. which is given by var (NI ) = E = Z ½Z Tb Tb A2 E [c (t) c (λ)] cos (∆ωt + θ − φ) cos (∆ωλ + θ − φ) dtdλ I 0 0 Z Tb Z Tb 1 1 = A2 Tc Λ [(t − λ) /Tc ] cos [∆ω (t − λ)] dtdλ I Tc 2 0 0 Z Tb 2T T A c b 1 = A2 Tc dt = I I 2 2 0 Problem 8. ADVANCED DATA COMMUNICATIONS TOPICS where θ − φ is assumed uniform in[0. The variance is the same as the mean square.99) to get the following: PE = 5. the argument of the Q-function for PE = Q SNR becomes r √ 3N ≈ 3. PE = 5. PE = 8.81 to give PE = 10−4 SNR = K −1 . the expectation of NI is zero due to this random phase.42 × 10−2 for a JSR of 15 dB. 7 17 Round this down to K = 53 users since the number of users must be an integer. PROBLEM SOLUTIONS Thus.9 Problem 8. the uplink and downlink probabilities of error are related to the uplink and downlink SNRs by Ãsµ ¶ !# " "s µ ¶ # Eb Eb 1 1 − erf = 2 pu = Q N0 u 2 N0 u Ãsµ ¶ !# " "s µ ¶ # Eb Eb 1 1 − erf = 2 pd = Q N0 d 2 N0 d where these probabilities have been put in terms of the error function because MATLAB includes an inverse error function.81)2 = 53.8. When these relations are inverted. See the text for the bent-pipe case.121) becomes pd = 10−5 − pu 1 − 2pu For BPSK.39 (a) Only the OBP case will be done. .1. K = 3N +1 (3. Equation (8. these become µ ¶ £ ¤2 Eb = erf −1 (1 − 2pu ) N0 µ ¶u £ ¤2 Eb = erf −1 (1 − 2pd ) N0 d Typical values are given in the table below.38 The result is Tacq ¢ ¡ 2 (1000) 10−3 2 − 0.22 seconds = = 2 (1 − PFA ) PH 2 (1 − 10−3 ) 0. Problem 8. Part (b) is just a matter of inputting the desired p_overall in the program. The MATLAB program for computing them is also given below along with a plot.9 nTc Te 2 − PH = 1. dB 9.993 × 10−6 9..999 × 10−6 9.2*pd)).39 % p_overall = input(’Enter desired end-to-end probability of error ’) pu = logspace(log10(p_overall)-4. xlabel(’(E_b/N_0)_d.2 × 10−7 (Eb /N0 )u . plot(Eb_N0_d.5639 10.2*pu)).997 × 10−6 9. pd = (p_overall .6497 pd 9..0839 11.. log10(p_overall)+1e-20). grid.5880 9./(1 .6217 % Solution for Problem 8. dB 12. Eb_N0_d = 20*log10(erfinv(1 . dB’). dB’).5495 12.3 × 10−8 7.280 × 10−6 (Eb /N0 )d .5879 9.957 × 10−6 9.5898 9. Eb_N0_u).3228 12.2*pu). ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. axis square .18 CHAPTER 8.5: pu 10−9 3 × 10−9 7 × 10−9 4. Eb_N0_u = 20*log10(erfinv(1 .5882 9.pu). ylabel(’(E_b/N_0)_d. .40 (a) This is similar to the previous problem except that the error probability expresseion for noncoherent FSK is µ ¶ Eb 1 PE = exp − 2 2N0 Solving for Eb N0 . the equations describing the link are 10−6 − pu pd = 1 − 2pu µ ¶ Eb = − ln (2pu ) N0 u µ ¶ Eb = − ln (2pd ) N0 d A MATLAB program for computing the performance curves for both parts (a) and (b) is given below.000001*p_overall)).8. we have Eb = −2 ln (2PE ) N0 Thus.40 % mod_type = input(’Enter 1 for NFSK. 10−6 − pu pd = 1 − 2pu µ ¶ Eb = −2 ln (2pu ) N0 u µ ¶ Eb = −2 ln (2pd ) N0 d (b) For DPSK. log10(1.1. 2 for DPSK ’) p_overall = input(’Enter desired end-to-end probability of error ’) pu = logspace(log10(p_overall)-6. Typical performance curves are also shown. the probability of error is PE = ¶ µ Eb 1 exp − 2 N0 Therefore. PROBLEM SOLUTIONS 19 Problem 8. % Solution for Problem 8. . elseif mod_type == 2 Eb_N0_u = 10*log10(-log(2*pu)). Eb_N0_d = 10*log10(-2*log(2*pd)). Eb_N0_u). ylabel(’(E_b/N_0)_d. grid.. num2str(p_overall). Eb_N0_d = 10*log10(-log(2*pd)). xlabel(’(E_b/N_0)_d./(1 . dB’). dB’). ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. NFSK modulation’]) elseif mod_type == 2 title([’OBP link performance for overall P_E = ’. ’. DPSK modulation’]) end .pu). end plot(Eb_N0_d. ’. axis square if mod_type == 1 title([’OBP link performance for overall P_E = ’..6: pd = (p_overall . if mod_type == 1 Eb_N0_u = 10*log10(-2*log(2*pu)).2*pu).20 CHAPTER 8. num2str(p_overall). 54). PROBLEM SOLUTIONS 21 Figure 8. the equations for the OBP link are pd = pu ≈ µ Eb N0 ¶ = 10−6 − pu 1 − 2pu Ãs µ ¶ ! Eb M Q = log2 (M ) 2 N0 u µ ¶¸ · 4pu 2 2 −1 1− erf log2 (M ) M Ãs µ ¶ ! Eb M Q = log2 (M ) 2 N0 d µ ¶¸ · 4pd 2 2 −1 1− erf log2 (M ) M " Ãs µ ¶ !# log2 (M ) Eb M 1 − erf 4 2 N0 u " Ãs µ ¶ !# log2 (M ) Eb M 1 − erf 4 2 N0 d u pd ≈ µ Eb N0 ¶ = d .51) and (8.41 For noncoherent FSK. the bit error probability.1.8.7: Problem 8. is Ãr Ãr ! ! M M2 Eb Eb Q Q Pb = log2 (M ) log2 (M ) ≈ 2 (M − 1) N0 2 N0 So. using (8. ’-FSK modulation’]) . num2str(M). % Solution for Problem 8. grid. 8.22 CHAPTER 8.8: A MATLAB program is given below and a plot for 16-FSK is also given in Fig.pu). Eb_N0_u).. pd = (p_overall .41 % p_overall = input(’Enter desired end-to-end probability of error ’) M = input(’Enter the desired M ’) pu = logspace(log10(p_overall)-4..(4/M)*pd)). plot(Eb_N0_d. num2str(p_overall). ylabel(’(E_b/N_0)_d. dB’). ’. log10(p_overall)+1e-20). coherent ’.8. xlabel(’(E_b/N_0)_d. Eb_N0_u = 20*log10((2/log2(M))*erfinv(1 .(4/M)*pu)). Eb_N0_d = 20*log10((2/log2(M))*erfinv(1 ./(1 .. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. dB’).2*pu). axis square title([’OBP link performance for overall P_E = ’. 43 = 3N or N = d29. we have µ ¶ Dco 20 = 10(3) log10 − 1 − 7.8.7815 dA or µ ¶ Dco 20 + 7.7815 = 0.32e = 19 (i = 2.7815 dA µ ¶ Dco 14 + 7.67e = 31 (i = 1.794 + 1 dA √ = 9.7815 = 0. j = 3) The efficiency is ηv = = ¥ 120 ¦ 19 6 MHz 6 = 1 voice circuits per base station per MHz 6 . PROBLEM SOLUTIONS 23 Problem 8. we have µ ¶ Dco 14 = 10(3) log10 − 1 − 7. For 20 dB minimum signal-to-interference ratio and a propagation power law of α = 3.726 log10 −1 = dA 30 Dco = 100.1.32 = 3N or N = d13.726 + 1 dA √ = 6.42 (a) The number of users per reuse pattern remain the same at 120.794 −1 = log10 dA 30 Dco = 100. j = 5) The efficiency is ηv = ¥ 120 ¦ 31 6 MHz 3 1 = = voice circuits per base station per MHz 6 2 (b) For 14 dB minimum signal-to-interference ratio and a propagation power law of α = 3. . if I_mod == 1 z_dB = 3:.24 CHAPTER 8.1 % ce8_1.^(z_dB/10). kk=log2(M)/2. NCMFSK = 3: ’). for j = 1:6 M=2^j.2 Computer Exercises Computer Exercise 8. elseif I_mod == 2 | I_mod == 3 z_dB = 3:. kk=(sin(pi/M))^2*log2(M). . elseif I_mod == 3 A=M/(2*(M-1)).ylabel(’P_b’)..1:15. CMFSK = 2.xlabel(’E_b/N_0. Pb = Pb+A*B*exp(-alpha*z). dB’). if I_mod == 1 A=2/log2(M).1:30.Pb). alpha=k*log2(M)/(k+1). end if j == 1 & I_mod == 1 Pb = . kk = 1. noncoherent FSK (exact) % clf I_mod = input(’Enter type of mod: MPSK = 1.5*erfc(sqrt(z)). elseif I_mod == 2 A=M/2. % coherent FSK (uses union bound)..m: BEPs for M-ary PSK in AWGN (uses upper bound for M > 4). elseif I_mod == 3 Pb = zeros(size(z)). elseif (j >= 2 & I_mod == 1) | (j >= 1 & I_mod == 2) Pb = (A/2)*erfc(sqrt(kk*z)). end end semilogy(z_dB. for k = 1:M-1 B = (prod(1:(M-1))/(prod(1:k)*prod(1:(M-1-k))))*(-1)^(k+1)/(k+1). ADVANCED DATA COMMUNICATIONS TOPICS 8. end z = 10. ’:’. with the resulting plot shown in Fig. COMPUTER EXERCISES axis([min(z_dB) max(z_dB) 10^(-6) 1]). [’M = ’. Pb(25). Pb(35).8.025:7. [’M = ’.’—. if I_mod == 1 title(’BEP for MPSK’) if j <= 2 text(z_dB(35)+.. num2str(M)]) end elseif I_mod == 3 title(’BEP for noncoherent MFSK’) if j <= 3 text(z_dB(50)+.2. QPSK (OQPSK).’.’. Pb(50).2..2.’). num2str(M)]) end end if j==1 hold on grid end end A typical run follows. MFSK = 3: 1 Computer Exercise 8.m: Out-of-band power for M-ary PSK.2. 8.’-. Pb(50).’-. [’M = ’. MFSK = 2. num2str(M)]) end elseif I_mod == 2 title(’BEP for coherent MFSK’) if j <= 3 text(z_dB(50)+.9: >> ce8_1 Enter type of modulation: MPSK = 1.. and MSK % clf A = char(’-’. coh.. Tbf = 0:. Pb(25). [’M = ’. noncoh. num2str(M)]) elseif j > 3 text(z_dB(25)+. 25 .’—’. ’M = 2 & 4’) elseif j >= 3 text(z_dB(105)+. num2str(M)]) elseif j > 3 text(z_dB(25)+.2 % ce8_2.2. [’M = ’.2. Pb(105).2. Tbf(k).:)) plot(Tbf. . A(1. OBP_MSK. A_QPSK = [].9: LTbf = length(Tbf). A_QPSK(k) = quadl(’G_QPSK’. 7). for k = 1:LTbf A_BPSK(k) = quadl(’G_BPSK’.:)).grid. 7)..10. end OBP_BPSK = 10*log10(A_BPSK/A_BPSK(1)). OBP_MSK = 10*log10(A_MSK/A_MSK(1)). dB’). xlabel(’T_bf’). 8. A(2. A_MSK(k) = quadl(’G_MSK’. OBP_QPSK. hold on plot(Tbf. A(3. Tbf(k). OBP_BPSK. A_MSK = []. plot(Tbf.:)) legend(’BPSK’. Tbf(k)...26 CHAPTER 8. OBP_QPSK = 10*log10(A_QPSK/A_QPSK(1)). axis([0 5 -40 0]).. 14).. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. ylabel(’Relative out-of-band power. ’QPSK/OQPSK’. A_BPSK = [].. ’MSK’) A typical plot generated by the program is shown in Fig. Tbf0 = input(’Enter f_0 times T_b ’) delTbf0 = . COMPUTER EXERCISES 27 Figure 8.’:’.8. A(k. L_delTbf = length(delTbf0) Tbf = 0:.025:10.’).’.10: Computer Exercise 8..’—.:)) if k == 1 hold on grid xlabel(’T_bf’).m: Computes spectrum of continuous phase FSK % clf A = char(’-’.5. ylabel(’Spectral amplitude’) end .’—’. for k = 1:L_delTbf delTbf = delTbf0(k). G_FSK = (S_CFSK(Tbf-Tbf0) + S_CFSK(Tbf-Tbf0-delTbf)). area = sum(G_FSK)*0. G_FSKN.^2.025.5:. plot(Tbf. G_FSKN = G_FSK/area.3 % ce8_3.25:1.’.’-.’-.2. o’).’-. 8.11.11: end legend([’\DeltafT_b = ’.’—.28 CHAPTER 8. % For a desired P_E and given JSR and Gp.num2str(delTbf0(5))]) title([’Spectrum for continuous phase FSK.’-.4 % ce8_4. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8. .num2str(delTbf0(2))]. Gp_dB = input(’Enter desired processing gain in dB ’).num2str(delTbf0(4))].’-.m: Computes bit error probability curves for jamming in DSSS.num2str(delTbf0(3))]. [’\DeltafT_b = ’. Computer Exercise 8. num2str(Tbf0)]) % Function to approximate CFSK spectrum % function y = S_CFSK(Tbf) y = sinc(Tbf).’—’. computes the required Eb/N0 % clf A = char(’-’.’.’:’.x’.’. T_bf_0 = ’. [’\DeltafT_b = ’. [’\DeltafT_b = ’.[’\DeltafT_b = ’. A typical plot is shown in Fig.num2str(delTbf0(1))]. num2str(JSR0(4))].:)) if k == 1 hold on axis([0 30 1E-15 1]) grid on xlabel(’E_b/N_0. COMPUTER EXERCISES Gp = 10^(Gp_dB/10). Gp_dB_0 = input(’Enter given value of processing gain in dB ’)./(1+z*JSR/Gp). JSR_dB_0 = input(’Enter given value of JSR in dB ’). num2str(JSR0(2))]. [’JSR = ’. k = 1. num2str(JSR0(1))]. text(z_dB(30)-5. num2str(JSR0(5))]. ’No jamming’) legend([’JSR = ’. PE = 0. z_dB = 0:. z = 10. semilogy(z_dB. num2str(JSR0(3))]. JSR0 = 10^(JSR_dB_0/10).5*erfc(sqrt(arg)). semilogy(z_dB. PE.5*erfc(sqrt(Gp0/JSR0)). num2str(Gp_dB). Gp0 = 10^(Gp_dB_0/10). PEG(30).’ dB’]) end k = k+1.2. dB’). 29 .^(z_dB/10). & desired P_E. end PEG = 0. JSR0 = []. for JSR_dB = 5:5:25. A(k+1. pause PE0 = input(’Enter desired value of P_E ’).5:30. JSR0(k) = JSR_dB. PELIM = 0.5*qfn(sqrt(2*z)). arg = z. [’JSR = ’. JSR = 10^(JSR_dB/10).8. [’JSR = ’. JSR. a solution is not possible ’). ylabel(’P_E’) title([’BEP for DS BPSK in jamming for proc. 3) disp(’Strike ENTER to continue’). gain = ’. PEG). [’JSR = ’. disp(’ ’) disp(’Infinite Eb/N0 BEP limit’) disp(PELIM) if PELIM >= PE0 disp(’For given choices of G_p. SNR0_dB = 10*log10(SNR0). Gp0_over_JSR0_dB = 10*log10(Gp0_over_JSR0). % h = input(’Enter satellite altitude in km: ’).30 CHAPTER 8. 8. determines circular antenna aperture diameter and % maximum antenna gain. SNR0 = 1/(1/arg0 . ADVANCED DATA COMMUNICATIONS TOPICS Gp0_over_JSR0 = (erfinv(1-2*PE0))^2. end disp(’ ’) disp(’To give BEP of:’) disp(PE0) disp(’Requires GP. JSR.JSR0/Gp0).6085 Computer Exercise 8. JSR. f0 = input(’Enter carrier frequency in GHz: ’).0010 Requires GP. and Eb/N0 in dB of:’) disp([Gp_dB JSR_dB SNR0_dB]) A typical run follows. .0000 9.12 and a specific output generated: >> ce8_4 Enter desired processing gain in dB: 30 Strike ENTER to continue Enter desired value of P_E: 1e-3 Enter given value of JSR in dB: 20 Enter given value of processing gain in dB: 30 Infinite Eb/N0 BEP limit 3.5 % ce8_5. with both a plot given in Fig.m: Given a satellite altitude and desired spot diameter for % illumination. d_spot = input(’Enter desired illuminated spot diameter at subsatellite point in km: ’). and Eb/N0 in dB of: 30. rho = input(’Enter desired antenna efficiency: 0 < \rho <= 1: ’).8721e-006 To give BEP of: 0.0000 25. disp(’The minimum required value of G_p over JSR in dB is:’) disp(Gp0_over_JSR0_dB) else arg0 = (erfinv(1-2*PE0))^2. 3) disp(’Wavelength in meters:’) disp(lambda) disp(’Antenna diameter in meters:’) disp(d) disp(’Antenna gain in dB:’) disp(G0_dB) disp(’ ’) A typical run follows: . phi3dB = theta.12: theta = d_spot/h. disp(’ ’) disp(’3-dB beamwidth in degrees: ’) disp(phi3dB*57.2. % G0 = rho*(pi*d/lambda)^2 G0 = (pi/phi3dB)^2. COMPUTER EXERCISES 31 Figure 8. d = lambda/(phi3dB*sqrt(rho)). lambda = 3E8/(f0*1E9).8. G0_dB = 10*log10(G0). 000001:. EbN0d = 1.m: Performance curves for bent-pipe relay and mod/demod relay % clf A = char(’—’. EbN0d_dB = 10. FSK./EbN0u. elseif I_mod == 2 EbN0r = 2*(erfinv(1 .6 % ce8_6. elseif I_mod == 4 EbN0r = -2*log(2*PE0). . end EbN0r_dB = 10*log10(EbN0r) EbN0u_dB = EbN0r_dB+. 3 = DPSK.’./den.01:35. elseif I_mod == 3 EbN0r = -log(2*PE0).0375 Antenna diameter in meters: 0. 2 = coh.^(EbN0u_dB/10). I_mod = input(’Enter type of modulation: 1 = BPSK. PE0 = input(’Enter desired value of P_E ’).7300 Wavelength in meters: 0.32 CHAPTER 8.2*PE0))^2.4482 Antenna gain in dB: 29.9430 Computer Exercise 8. 4 = noncoh.2*PE0))^2. EbN0u = 10.’-. FSK ’). ADVANCED DATA COMMUNICATIONS TOPICS >> ce8_5 Enter satellite altitude in km: 200 Enter desired illuminated spot diameter at subsatellite point in km: 20 Enter carrier frequency in GHz: 8 Enter desired antenna efficiency: 0 < rho <= 1: .’:’).7 3-dB beamwidth in degrees: 5. den = 1/EbN0r-1.’-’. for I_type = 1:2 if I_type == 1 if I_mod == 1 EbN0r = (erfinv(1 .*log10(EbN0d). both in dB. end EbN0u_dB = 10*log10(EbN0u). EbN0d = -log(2*pd).^2. FSK modulation’]) elseif I_mod == 3 title([’Uplink E_b/N_0 versus downlink E_b/N_0.num2str(PE0).^2. grid on.num2str(PE0).2*pd)).2*pu)). xlabel(’(E_b/N_0)_d. ylabel(’(E_b/N_0)_u. BPSK modulation’]) elseif I_mod == 2 title([’Uplink E_b/N_0 versus downlink E_b/N_0.:)). to give P_E = ’. EbN0d = (erfinv(1 . axis([5 30 5 30])..pu).2*pd)). d2. both in dB. if I_type == 1 axis square. elseif I_mod == 4 EbN0u = -2*log(2*pu). pu = logspace(d1.’.00000001*PE0).1) if I_mod == 1 title([’Uplink E_b/N_0 versus downlink E_b/N_0.2*pu)). end plot(EbN0d_dB.9999999*PE0)..[’Demod/remod’]. pd = (PE0 . elseif I_mod == 2 EbN0u = 2*(erfinv(1 .’. to give P_E = ’.’.8. both in dB.2./(1 . if I_mod == 1 EbN0u = (erfinv(1 . both in dB. EbN0d = -2*log(2*pd). dB’).^2. d2 = log10(0. coh. to give P_E = ’.A(I_type. EbN0u_dB. COMPUTER EXERCISES elseif I_type == 2 d1 = log10(0. 5000).^2.2*pu). 33 . EbN0d_dB = 10*log10(EbN0d).. elseif I_mod == 3 EbN0u = -log(2*pu). EbN0d = 2*(erfinv(1 . dB’) hold on end end legend([’Bent pipe’]. DPSK modulation’]) elseif I_mod == 4 title([’Uplink E_b/N_0 versus downlink E_b/N_0.num2str(PE0). ’—’.’. 0 for normal MSK: ’).7 % ce8_7. N_bits = input(’Enter total number of bits: ’). end . FSK modulation’]) end A typical run follows: >> ce8_6 Enter type of modulation: 1 = BPSK. N_samp = 50.’). FSK.34 CHAPTER 8.m: plots waveforms or spectra for gmsk and msk % A = char(’-’. samp_bit = input(’Enter number of samples per bit used in simulation: ’). if GMSK == 1 BT_bit = input(’Enter vector of bandwidth X bit periods: ’).’-.num2str(PE0). 4 = noncoh.’.7998 Computer Exercise 8. noncoh.’:. FSK 2 Enter desired value of P_E 1e-3 EbN0r_dB = 9. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.13: to give P_E = ’. 2 = coh. GMSK = input(’Enter 1 for GMSK. 3 = DPSK. L_t = length(t). L = length(s). COMPUTER EXERCISES 35 I_plot = input(’Enter 1 to plot waveforms. end clf kp = pi/2. t_max = max(tp). elseif GMSK == 0 freq = kp*s_t. L_tp=length(tp).1. for k = 1:LB B = BT_bit(k)/T_bit.:). if I_plot == 1 f0 = 1. alpha = 1. T_bit = 1.2. end . % Note that B is two-sided bandwidth del_t = T_bit/samp_bit. data = 0. end phase = del_t*cumsum(freq). if GMSK == 1 hG = (sqrt(pi)/alpha)*exp(-pi^2*(tp-t_max/2). 2 to plot spectra and out-of-band power ’). L_hG = length(hG). s = 2*data .8.5)+1). tp=0:del_t:N_samp*T_bit-del_t. freq1 = kp*del_t*conv(hG. % Build array whose columns are samp_bit long s_t = s_t(:)’.1774/B.s_t). s_t = s(ones(samp_bit. fs = 1/del_t. elseif I_plot == 2 f0 = 0. LB = length(BT_bit).5*(sign(rand(1. % Convert matrix where bit samples occupy columns to vector L_s_t = length(s_t). elseif GMSK == 1 y_mod = exp(j*2*pi*f0*t+j*phase). t=0:del_t:L*T_bit-del_t. freq = freq1(L_hG/2:length(freq1)-L_hG/2).1).N_bits)-. if GMSK == 0 y_mod = exp(j*2*pi*f0*t+j*phase).^2/alpha^2). plot(t.A(k.1).2). axis([0 3 10^(-8) 1]) if k == 1 hold on grid ylabel(’Fraction of out-of-band power’).1). J/Hz’). ylabel(’Modulated signal’) elseif I_plot == 2 Z=PSD(real(y_mod).-nn*(pi/2)*ones(size(t)).semilogy(FR. 1.1.36 CHAPTER 8.2). ylabel(’Excess phase.1) elseif LB == 2 .1.semilogy(FR.3). LZ = length(Z).axis([min(t).ZN.2. OBP = 1-cumsum(ZN)/ET.real(y_mod)). num2str(BT_bit(k))]) end subplot(3. ADVANCED DATA COMMUNICATIONS TOPICS if I_plot == 1 subplot(3.:)).nn*(pi/2)*ones(size(t)).phase). ET = sum(ZN).2).plot(t.A(k. axis([0 3 10^(-8) 1]) if k == 1 hold on grid ylabel(’PSD.2048. del_FR = fs/2/LZ.:)).fs).plot(t.1. xlabel(’t’).plot(t. xlabel(’fT_b_i_t’) end if LB == 1 legend([’BT_b_i_t = ’.2.’m’) end subplot(3. max(t). subplot(1.num2str(BT_bit(1))].’m’) subplot(3.1. ylabel(’Data’) if GMSK == 0 title(’MSK waveforms’) elseif GMSK == 1 title([’GMSK for BT_b = ’.2]).OBP. -1.s_t).’) for nn = 1:11 if nn == 1 hold on end subplot(3. xlabel(’fT_b_i_t’) title(’Power spectra and out-of-band power for GMSK’) end subplot(1. FR = [0:del_FR:fs/2-del_FR].plot(t. % MATLAB supplied function ZN = Z/max(Z).2. rad.1.2). num2str(BT_bit(1))].2. [’BT_b_i_t = ’.5 1 1.5] Enter 1 to plot waveforms.num2str(BT_bit(2))]. 2 to plot spectra and out-of-band power: 1 >> ce8_7 Enter number of samples per bit used in simulation: 10 Enter total number of bits: 2000 Enter 1 for GMSK.1) elseif LB == 4 legend([’BT_b_i_t = ’.1) end end end Two typical runs follow .8. >> ce8_7 Enter number of samples per bit used in simulation: 20 Enter total number of bits: 50 Enter 1 for GMSK.1) elseif LB == 3 legend([’BT_b_i_t = ’.num2str(BT_bit(4))].one for plotting waveforms and one for plotting spectra. [’BT_b_i_t = ’. [’BT_b_i_t = ’.num2str(BT_bit(1))]. COMPUTER EXERCISES legend([’BT_b_i_t = ’. [’BT_b_i_t = ’.num2str(BT_bit(3))].num2str(BT_bit(2))].num2str(BT_bit(3))]. 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: . 2 to plot spectra and out-of-band power: 2 37 .num2str(BT_bit(1))]. [’BT_b_i_t = ’.num2str(BT_bit(2))]. 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: [. [’BT_b_i_t = ’.5 Enter 1 to plot waveforms. 38 CHAPTER 8. ADVANCED DATA COMMUNICATIONS TOPICS Figure 8.14: . 2.8. COMPUTER EXERCISES 39 Figure 8.15: . The likelihood ratio is Λ (Z) = ¡ ¢ fZ (Z|H2 ) = 0. λ > 0 0 which integrates to the result given in the problem statement.Chapter 9 Optimum Receivers and Signal Space Concepts 9.1 a. The threshold is . Given H1 . so fZ (z|H1 ) = fN (n) |z=n = 10e−10z u (z) Given H2 . b. Z > 0 fZ (Z|H1 ) η= 1 P0 (c21 − c11 ) = P1 (c12 − c22 ) 3 1 c. Thus. the resulting pdf under H2 is the convolution of the separate pdfs of S and N : fZ (z|H2 ) = Z ∞ −∞ fs (z − λ) fN (λ) dλ = 20 Z z e−2z e−8λ dλ. where S and N are independent. Z = S +N . Z = N .1 Problems Problem 9.25 e8Z − 1 . 106 Z < 8 H1 e.346 0. g. Then PF = e−10η ⇒ η ≥ 0. PF = e−10η and PD = 1.2 CHAPTER 9. Also.25e−2η − 0. from part (e). PD = 1. the risk is Risk = Z ∞ 10e−10z dz = 0. Consider the threshold set at η.106 The probability of false alarm is PF = Therefore. PD is approximately equal to the Þrst term of the above expression.714 4 4 4 f.925 PD = 0.25e−10η For values of η ≥ 0.925) − (5) (1 − 0.921 for PF ≤ 10−4 .921.921 to give PF ≤ 10−4 .106 3 1 1 (5) + (5) (1 − 0.25e−10η . From part (f). The resulting PD is 0.198. For this range of ηPD is strictly decreasing with η .346) = 0. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS d.5 e−2z − e−10z dz = 0. The likelihood ratio test becomes H2 ¢ > 1 ¡ 8Z 0.25e−2η − 0.25 e − 1 < 3 H1 This can be reduced to H2 > ln (7/3) = 0. so the best value of PD occurs for η = 0. The probability of detection is Z ∞ ¡ ¢ 2. The properties listed under ”Structure of Signal Space” in the text then become: . b2 .2 for η = 1. αa3 ) and vector addition by (a1 .2 a. It is shown in Figure 9. b3 ) = (a1 + b1 . DeÞne scalar multiplication by α(a1 . PROBLEMS 3 Figure 9. Problem 9. The decision regions R1 and R2 are indicated in Figure 9. b2 . b3 ). a3 ) +(b1 .9. The conditional pdfs under each hypothesis are plotted below. a2 + b2 . a2. a2. a3 + b3 ).1. αa2. a3 ) and B = (b1 .1. a3 ) = (αa1 . The likelihood ratio is Λ (Z) = 1 |Z| 2e e√ 2π − 1 Z2 2 = r π − 1 Z 2 −|Z| e 2 2 b. the likelihood ratio test is H2 > ln (4η + 1) =γ Z < 8 H1 A plot of PD versus PF as a function of η or γ constitutes the operating characteristic of the test.1: From part (d). a2.3 DeÞne A = (a1 . Problem 9. a3 + b3 ) = (αa1 + αb1. 0. α1 (α2 A) = (α1 α2 A) ² R3 (follows by writing out in component form). α1 a3 + α2 b3 ) ²R3 . The unique element 0 is (0. − a3 ) so that A + (−A) = 0. 6. 2.2: 1. 4.4 Consider (x. αa2. αa3 + αb3 ) = (αa1 . −A = (−a1 . Problem 9. 3.4 CHAPTER 9. αb2 . y)∗ ∗ T (αx. αa3 ) + (αb1. −a2. αa2 + αb2 . αb3 ) = αA + αB ² R3 . y) = lim T →∞ −T Z T αx (t) y (t) dt = α lim ∗ T →∞ −T Z x (t) y∗ (t) dt = α (x. 0) so that A + 0 = A. α (A + B) = α (a1 + b1. y) = lim Also Z T T →∞ −T Z T x (t) y ∗ (t) dt Z T T →∞ −T y (t) x (t) dt = ∗ · T →∞ −T lim ¸∗ x (t) y (t) dt = (x. y) = lim Then (x. 5. y) . α1 A + α2 B = (α1 a1 + α2 b1 . a2 + b2 . 1 • A = A (follows by writing out in component form). α1 a2 + α2 b2 . OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9. x2 ) = lim e−(5+j)t dt = T →∞ 0 5+j c. x2 ) = lim Z T T →∞ −T x (t) x∗ (t) dt = lim T →∞ −T Z T |x (t)|2 dt ≥ 0 T →∞ 0 2e−6t dt = 1 3 b. x2 ) = lim T →∞ 2T 0 Problem 9. This scalar product is (x1 . z) = = lim 5 = (x. PROBLEMS and (x + y. Use the power signal scalar product deÞnition: Z T 1 cos (2πt) dt = 0 (x1 . Use the energy signal scalar product deÞnition as in (a): Z T 1 (x1 . z) + (y.5 a. Use the power signal product deÞnition: Z T 1 cos (2πt) sin2 (2πt) dt = 0 (x1 . x) = lim Z T T →∞ −T Z T T →∞ −T Z T [x (t) + y (t)] z (t)∗ (t) dt x (t) z ∗ (t) dt + lim Z T lim T →∞ −T y (t) z ∗ (t) dt The scalar product for power signals is considered in the same manner. Z T 2 [x1 (t) + x2 (t)]2 dt kx1 + x2 k = lim = = T →∞ −T Z T T →∞ −T kx1 k2 + 2 (x1 . x2 ) lim x2 (t) dt + 2 1 T −→∞ −T + kx2 k2 lim Z T x1 (t) x2 (t) dt + lim T −→∞ −T Z T x2 dt 2 .9. x2 ) = lim T →∞ 2T −T d. Problem 9.1. z) Finally (x.6 Since the signals are assumed real. 6 CHAPTER 9. Problem 9. x2 ) = 0 and (x3 . x2 ) = 0. kx1 + x2 k2 = kx1 k2 + kx2 k2 Problem 9. Clearly. x2 ) = 0. Choose them as basis functions (not normalized). Problem 9. A set of normalized basis functions is 1 φ1 (t) = √ s1 (t) 2 r · ¸ 2 1 φ2 (t) = s2 (t) − s1 (t) 3 2 · ¸ √ 2 2 3 s3 (t) − s1 (t) − s2 (t) φ3 (t) = 3 3 . they are orthogonal.7 By straight forward integration. x1 ) = 2 Since (x1 . x3 is their vector sum. it follows that kx1 k2 = 2 kx2 k2 = kx3 k2 = Also or or or 2 3 8 3 kx1 k = √ 2 r 2 kx2 k = 3 r 8 kx3 k = 3 (x1 .9 a. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS From this it is seen that if and only if (x1 .8 It follows that kx1 k2 = Also n=1 N X |an |2 and kx2 k2 = N X n=1 N X |bn |2 (x1 . x2 ) = The inequality can be reduced to XX m n an b∗ n n=1 |an bm − am bn |2 ≥ 0 which is true because each term in the sum is nonnegative. φ1 ). and (s3 . PROBLEMS b.9. For example.1. unit-width. φ1 )φ1 (t) where (x2 . Problem 9.11 First we set v1 (t) = x1 (t) = exp(−t)u(t) By deÞnition kv1 k = kx1 k = Thus 2 2 r Z ∞ exp(−2t)dt = 0 1 2 1 kv1 k = √ 2 √ 2 exp(−t)u(t) This leads to the Þrst basis function φ1 (t) = For the second basis function we compute <1 v2 (t) = x2 (t) − (x2 . s2 = √ φ1 + φ2 . φ1 ) = Thus Z ∞√ 0 √ Z 2 exp(−t) exp(−2t)dt = 2 ∞ 0 √ 2 exp(−3t)dt = 3 √ · ¸ 2√ 2 2 exp(−t)u(t) = exp(−2t) − exp(−t) u(t) v2 (t) = exp(−2t)u(t) − 3 3 . Problem 9. φ1 ). nonoverlapping pulses spanning the interval could have been used. s3 = 2φ1 + (φ + φ3 ) 2 3 2 2 7 Note that a basis function set could have been obtained by inspection. three unit-height.10 A basis set is φ1 (t) = Ks1 (t) and φ2 (t) = Ks2 (t) where K= fc N A2 A signal point lies on each coordinate axis at ±1/K. (s2 . Evaluate (s1 . φ1 ) to get r r √ √ 3 2 1 s1 = 2φ1 . φ1 ) = so that Z ∞ 2 Z 0 ∞µ ¶ 4 4 1 exp(−4t) − exp(−3t) + exp(−2t) dt = 3 9 36 4 6 exp(−2t) − exp(−t) 5 5 √ 2 4 √ exp(−3t) 2 exp(−t)dt = 1 exp(−t) 2 0 (x3 . φ2 )φ2 (t) − (x3 .12 First we set r 3 6 exp(−2t) + exp(−t) 5 10 229 600 · ¸ 600 6 3 exp(−3t) − exp(−2t) + exp(−t) u(t) 229 5 10 v1 (t) = x1 (t) = t By deÞnition kv1 k = kx1 k = 2 2 Z 1 t2 dt = −1 2 3 . OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Next we compute kv2 k = This gives φ2 (t) = [6 exp(−2t) − 4 exp(−t)] u(t) We next consider v3 (t) = x3 (t) − (x3 . φ2 )φ2 (t) = Also (x3 . φ2 ) = 5 0 so that (x3 . φ1 )φ1 (t) = For t ≥ 0 we then have v3 (t) = exp(−3t) − This leads to kv3 k2 = and φ3 (t) = Problem 9. φ1 )φ1 (t) Using the same procedure as previously used Z ∞ 1 exp(−3t) [6 exp(−2t) − 4 exp(−t)] dt = (x3 .8 CHAPTER 9. 1. φ2 ) = and (x3 . φ1 ) = so that v3 (t) = t − This gives 2 3 Z 1 t3 −1 Ãr ! 5 2 t dt = 0 2 Z 1 t3 r −1 Ãr 32 25 ! r 3 32 t dt = 2 25 3 3 t = t3 − t 2 5 r · ¸ 8 6 4 9 2 2 kv3 k = t − t + t dt = 5 25 175 −1 Z 1 and v3 (t) = φ3 (t) = kv3 k r · ¸ 175 3 3 t − t 8 5 . φ1 ) = Z 1 9 r 3 t. φ1 )φ1 (t) where (x2 . Thus Z 1 2 kv2 k2 = t4 dt = 5 −1 from which t2 = φ2 (t) = kv2 k r 5 2 t 2 For the third basis function we write v3 (t) = x3 (t) − (x3 . 2 −1 < t < 1 t 2 −1 Ãr ! 3 t dt = 0 2 since the intergrand is odd and the limits are even. φ2 )φ2 (t) − (x3 . PROBLEMS Thus the Þrst basis function is φ1 (t) = For the next basis function we set v2 (t) = x2 (t) − (x2 . φ1 )φ1 (t) where (x3 .9. A suitable basis is v4 (t) = φ4 (t) = kv4 k r · ¸ 441 4 5 2 t − t 8 7 φ1 (t) = 2 cos (2πfc t) and φ2 (t) T r 2 = sin (2πfc t) for 0 ≤ t ≤ T T r The coordinates of the signal vectors are µ ¶ Z T √ √ 2πi xi = si (t) φ1 (t) dt = E cos = E cos ψi M 0 and yi = where E = A2 T /2 Z T 0 and ψ i = 2πi . φ3 ) and (x4 . 2. φ1 ) are zero. φ1 )φ1 (t) or v4 (t) = x4 (t) − (x4 . 3. φ2 )φ2 (t) − (x4 . OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS For the next basis function we write v4 (t) = x4 (t) − (x4 . By deÞnition r Z 1 Ãr ! 5 2 52 4 (x4 . 4 . φ3 )φ3 (t) − (x4 . φ2 )φ2 (t) since (x4 . µ ¶ √ √ 2πi si (t) φ2 (t) dt = − E sin = − E sin ψi M i = 1. φ2 ) = t t dt = 2 27 −1 so that v4 (t) = t − This gives 2 4 r 52 27 r 5 2 5 t = t4 − t2 2 7 · ¸ 8 10 6 25 4 8 kv4 k = t − t + t dt = 7 49 441 −1 Z 1 and Problem 9.10 CHAPTER 9. Thus M √ √ si (t) = E cos(ψ i )φ1 (t) − E sin(ψi )φ2 (t) .13 a. 1. d.|si (t)] Z Z 1 = 1− R πN0 · i µ³ ´2 ³ ´2 ¶¸ √ √ 1 x − E cos ψi + y + E sin(ψi ) dxdy · exp − N0 i = 1. Since the error probability is independent of the signal chosen. DeÞning zi (t) = si (t) + n(t) gives √ √ zi (t) = ( E cos(ψ i ) + N1 )φ1 (t) + (− E sin(ψi ) + N2 )φ2 (t) . 3. The angle tan−1 (x/y) is formed to determine which pie-wedge decision region the received data vector is in. Thus. c. 2. Therefore Z Z í h ³ p 1 π/M ∞ drdθ r exp − r2 − 2r E/N0 cos(θ) + E/N0 P [error] = 1 − π −π/M 0 Z Z 1 r P [error |si (t)] = 1 − π Ri · ´¸ p 1 ³¡ 2 ¢ · exp − r N0 − 2r EN0 cos(θ + ψi ) + E drdθ N0 .9. Call these samples x and y. The probability of error can then be written in the form P [error |si (t)] = 1 − Pr[correct recept. let i = M so that ψ i = 2π and cos(θ+ψ i ) = cos θ. Since all decision regions have equal area and the noise has circular symmetry. PROBLEMS 11 b. (Note that the noise variance is N0 /2). The correlator outputs are sampled at the end of each signaling interval. The optimum partitioning of the signal space is a pie-wedge centered on each signal point with the origin as the vertex and having angle 2π/M. The optimum receiver consists of two parallel branches which correlate the incoming signal plus noise with φ1 (t) and φ2 (t). the conditional symbol error probability is equal to the unconditional error probability for the case in which all signals are transmitted with equal probability. 4 where Ri is the region for a correct decision given that si (t) is transmitted. θ plane. the error probability is independent of the transmitted signal. The expression for the probability of error can be changed to polar coordinates by letting p p x = r N0 cos θ and y = r N0 sin θ With this transformation This gives dxdy → N0 rdrdθ where Ri now represents the decision region for a correct decision in the R. These are the most probable errors. Consider S1 . The result then reduces to ´ ³p Pc = 1 − Q E/N0 i h which gives Pc = Q (E/N0 )1/2 . ³ p ´ ³ p ´ ± E/3 . The space is three-dimensional with signal points at the eight points ± E/3 . and (y3 + 1)2 < (y3 − 1)2 . 1. y2 . Given the partitioning discussed in part (a). This was done in Chapter 8. −1. we make a correct decision only if kY − S1 k2 < kY − S2 k2 and kY − S1 k2 < kY − S4 k2 and kY − S1 k2 < kY − S8 k2 where S2 . −1). 1).15 ´ ³ p a. ± E/3 . OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS e. This gives Pc = Z ∞ −∞ e−y √ π 2 "Z y+ √ E/N0 −∞ # 2 e−x √ dx dy π Z √E/N0 −∞ e−z √ π 2/2 Z ∞ −∞ e−2(y+z/2) √ dydz π 2 Complete the square in the exponent of the inside integral and use a table of deÞnite integrals to show that is evaluates to (1/2)1/2 . 1). ). S4 (−1. The optimum partitions are planes determined by the coordinate axes taken two at a time (three planes). the above conditions become (y1 + 1)2 < (y1 − 1)2 .12 CHAPTER 9. and S8 are the nearest-neighbor signal points to S1 . 1. 1. Substituting Y = (y1 . Problem 9. S4 . (y2 + 1)2 < (y2 − 1)2 . the desired result. 1. b.14 Write Pc = and let z = x − y. Thus the optimum decision regions are the eight quadrants of the signal space. y3 ) and S1 = (1. S2 (1. Problem 9. and S8 = (1. We need 1/2T Hz per signal. exp − fR2 (r2 |H1 ) = N0 N0 . The symbol error probability is 1 − Pc . Thus · ¸3 h ³ √ Z ∞ 2´ í3 ³p 1 1 − N y− E/3 0 e dy = 1 − Q 2E/3N0 P [correct dec. PROBLEMS by canceling like terms on each side. these reduce to y1 > 0. y3 > 0) = [Pr (yi > 0)]3 because the noises along each coordinate axis are independent. and y3 > 0 13 to deÞne the decision region for S1 . this is the average probability of correct detection. The noise variances.17 In the equation P (E|H1 ) = substitute fR1 (r1 |H1 ) = and Z ∞ ·Z ∞ r1 0 fR2 (r2 |H1 ) dr2 fR1 (r1 |H1 ) dr1 ¸ ¶ µ 2 2r1 r1 . The symbol error probability is plotted in Figure 9. Problem 9.9.16 a. For vertices-of-a-hpercube signaling. Note that Eb = E/n since there are n bits per dimension.3. Therefore. y2 > 0. we have ín h ³p 2E/nN0 Pe = 1 − 1 − Q where n = log2 M.|s1 (t)] = √ πN0 0 Since this is independent of the signal chosen. M = 2n where n is the number of dimensions.1. all i. Hence M = 22W T . the probability of correct decision is P [correct dec. Note that E [yi ] = (E/3)1/2 . For S1 . b. Generalizing to n dimensions. M = W/ (1/2T ) = 2W T . y2 > 0. c. and therefore the variances of the yi ’s are all N0 . Therefore.|s1 (t)] = Pr (y1 > 0. exp − 2Eσ2 + N0 2Eσ2 + N0 r2 > 0 r1 ≥ 0 µ 2 ¶ 2r2 r2 . Problem 9. 18 a. it can be reduced to µ 2¶ Z ∞ 2r1 r exp − 1 dr1 P (E|H1 ) = 2Eσ 2 + N0 K 0 ¡ ¢ N0 2Eσ2 + N0 K= 2Eσ2 + N0 The remaining integral is then easily integrated since its integrand is a perfect differential if written as µ 2¶ Z ∞ 2r1 r K exp − 1 dr1 P (E|H1 ) = 2+N 2Eσ K K 0 0 Problem 9.14 CHAPTER 9.3: The inside integral becomes µ 2 ¶ r1 I = exp − N0 When substituted in the Þrst equation. all j where . Decide hypothesis i is true if fZ|Hi (z|Hi ) ≥ fZ|Hj (z|Hj ). OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9. The probability of correct decision can be reduced to Pc = M −1 µ X i=0 ¶ m−1 N0 = 1 − Pe 2 + N ) (M − 1 − i) i N0 + (Eσ 0 Problem 9.37.j6=i This constitutes the maximum likelihood decision rule since the hypotheses are equally probable. then the optimum receiver is seen to be an M -signal replica of the one shown in Fig. 9. If the signals have equal energy.8a. 2N σN Γ (N ) 11 N−1 y2 e−y2 /N0 2 y1 ≥ 0 2N (N0 /2)N Γ (N ) .9. Y1 is the sum of squares of 2N independent Gaussian random variables each having mean zero and variance σ2 = 2 11 E 2 N0 σ + N 2 and Y2 is the sum of squares of 2N independent Gaussian random variables each having zero mean and variance N0 σ2 = 21 2 Therefore. from Problem 4.19 a. the pdfs under hypothesis H1 are fYi (y1 |H1 ) = and fY2 (y2 |H1 ) = N−1 y1 e−y1 /2σ11 . we can take the natural log of both sides and use that as a test. PROBLEMS where fZ|Hi (z|Hi ) = ´ ³ 2 z 2 +zsi exp − Eici 2 +N0 σ (Ei σ2 exp −  1 N0  M X ¡ ¢ 2 2 zcj + zsj  15 πM M + N0 ) N0 j=1.1. It can be shown under hypothesis that H1 . y2 ≥ 0 . b. This reduces to computing ¡ 2 ¢ Ei σ2 2 zci + zsi Ei σ 2 + N0 and choosing the signal corresponding to the largest. Alternatively. Z ∞ £ jΛω ¤ β m −βλ m−1 e ejλω λ dλ Φ (jω) = E e = Γ (m) 0 . Plots are given in Figure 9. By deÞnition. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Figure 9.20 a.4: b. Problem 9. 2. and 4 (dash-dot curve). The probability of error is Pe = Pr (E|H1 ) = Pr (E|H2 ) = Pr (Y2 > Y1 |H1 ) ¸ Z ∞ ·Z ∞ = fY2 (y2 |H1 ) dy2 fY1 (y1 |H1 ) dy1 0 yi Using integration by parts. it can be shown that I (x.4 for N = 1 (solid curve). c. The characteristic function is found and used to obtain the moments. 3. a) = Z ∞ z n e−az dz = e−ax x n X i=0 xi n! i!an+1−i This can be used to get the formula given in the problem statement.16 CHAPTER 9. 1. which is obtained as follows: Z ∞ fZ|Λ (z|λ) fΛ (λ) dλ fZ (z) = 0 Z ∞ β m −βλ m−1 λe−λz λ dλ = e Γ (m) 0 Z ∞ (β + z)m+1 m mβ m λ dλ e(β+z)λ = Γ (m + 1) (β + z)m+1 0 mβ m = (β + z)m+1 where the last integral is evaluated by noting that the integrand is a pdf and therefore integrates to unity. Use Bayes’ rule. we Þnd that fΛ|Z (λ|z) = λm (β + z)m+1 e−(β+z)λ Γ (m + 1) . This gives Z ∞ ejλω (−1)m−1 λm−1 e−βλ dλ = 0 Multiply both sides by β m and cancel like terms to get Z ∞ β m −βλ m−1 βm e ejλω λ dλ = Γ (m) (β − jω)m 0 Thus Φ (jω) = βm (β − jω)m £ ¤ m (m + 1) E Λ2 = j 2 Φ00 (0) = β Differentiating this with respect to ω.9. Using Bayes’ rule. note that Z 17 ∞ ejλω e−βλ dλ = 0 1 β − jω (−1)m−1 (m − 1)! (β − jω)m Differentiate both sides m − 1 times with respect to β. PROBLEMS To integrate. But Þrst we need fZ (z). we get E [Λ] = jΦ0 (0) − The variance is m β and ¡ ¢ m V ar (Λ) = E Λ2 − E 2 (Λ) = 2 β b. . z2 ) = λm+1 e−λ(β+z1 +z2 ) (β + z1 + z2 )m+2 Γ (m + 2) m (m + 1) β m (β + z1 + z2 )m+2 Since this is of the same form as in parts (a) and (b).zK (λ|z1 . They are E [Λ|Z] = m+1 β+Z and V ar [Λ|Z] = m+1 (β + Z)2 c. zK ) = Γ (m + 2) and m+2 β + Z1 + Z2 + . . z2 |λ) = λ2 e−λ(z1 +z2 ) . Then fz1 z2 |Λ (z1. . we Þnd that fΛ|z1 z2 (λ|z1 . . we can extend the results there for the moments to E [Λ|Z1 . z1 . Therefore.. we conclude that λm+1+K e−λ(β+z1 +z2 +. . Z2 . ZK ] = (β + Z1 + Z2 + . . . Assume the observations are independent. + zK )m+K fΛ|z1 z2 . .21 The conditional mean and Bayes’ estimates are the same if: . . z2 ≥ 0 The integration to Þnd the joint pdf of Z1 and Z2 is similar to the procedure used to Þnd the pdf of Z above with the result that fz1 z2 (z1. ZK ] = Problem 9. . z2 ) = Again using Bayes’ rule.18 CHAPTER 9. we can infer the moments from part (a) as well. . . (a). . + ZK )2 E [Λ|Z1 .+zK ) (β + z1 + z2 + . . . By examining the pattern developed in parts. (b). + ZK m+K V ar [Λ|Z1 . Z2 ] = m+2 m+2 and V ar [Λ|Z1 . . and (c). OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Note that this is the same pdf as in part (a) except that β has been replaced by β + z and m has been replaced by m + 1. Z2 ] = β + Z1 + Z2 (β + Z1 + Z2 )2 d. . z2... Z2 . . PROBLEMS 1. The variance of the estimate is ¡ ¢ 2σ4 V ar σ2 = n bn K is a sufficient statistic. b. Same . The conditional pdf of the noise samples is ¡ ¢ ¡ ¢−K/2 − P K z2 /2σ2 n i=1 i fZ z1 . . Same . PK 2 i=1 Zi . we Þnd the maximum likelihood estimate to be n σ2 bn b. we Þnd the following to be true: a. fA|Z is symmetric about the mean.all conditions are satisÞed. Condition 3 is not satisÞed. 2. Condition 2 is not satisÞed. K 1 X 2 = Zi K i=1 c.9.22 a. . z2 . C (x) is convex upward. C (x) is symmetric.1. d. . It can be found n by differentiating the pdf with respect to σ2 and setting the result equal to zero. d. With this. zK |σ2 = 2πσ 2 e n n The maximum likelihood estimate for σ 2 maximizes this expression. c. n Solving for σ2 . 3. .all conditions are satisÞed. Yes. 19 Problem 9. 20 CHAPTER 9. The result is · ¸ 1 2 2 2 1 − N (Z1 +Z2 + T A2 ) 2 cosh fZ|Θ (Z1 . but instead estimate m0 as zero (its mean). OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Problem 9. Note that the MAP estimate reduces to the ML estimate when σ2 → ∞. which is m0. The sample value M0 is Gaussian with mean m0 and variance σ2 . where Z0 = M0 + N where N is a Gaussian random variable of mean zero and variance N0 /2.2 ) = 1 1 − N [(z1 ∓A)2 +(z2 ±A)2 ] e 0 πN0 where the top sign in the exponent goes with H1 and the bottom sign goes with H2 . z2 |θ. we can take the natural log of both sides and differentiate it with respect to Θ.24 a. The results are fZ (z1 . and have a condition that the maximum likelihood estimate must satisfy. b. Note also that for N0 → ∞ (i. Thus. m small signal-to-noise ratio) we don’t use the observed value of Z0 .23 We base our estimate on the observation Z0 . The result is #r "r T 2A T 2A (Z1 cos θ − Z2 sin θ) (−Z1 sin θ − Z2 cos θ) = 0 tanh 2 N0 2 N0 .e. m0 = E (m0 |Z0 = z0 ) = ρz0 b where It is of interest to compare this with the ML estimate.ML = Z0 b σm E (M0 Z0 ) p =q ρ= p V ar (M0 ) V ar (Z0 ) σ2 + m N0 2 which results if we have no a priori knowledge of the parameter. The Þgure of merit is mean-squared error. we use a MAP m estimate. set the result to zero. That is. H1.. Problem 9. It can be shown that the estimate for m0 is the mean value of M0 conditioned on Z0 because the random variables involved are Gaussian. Z2 |Θ) = e 0 (A1 Z1 − A2 Z2 ) πN0 N0 Since the log function is monotonic. ¢1/2 £ ¤ ¡ sin cos−1 m = 1 − m2 The development of the log-likelihood function follows similarly to the development of the one for Problem 9. c.9. which are r r Z T Z T 2 2 Z1 = y (t) y (t) cos ω c t dt and Z2 = sin ω c t dt T T 0 0 this can be put into the form · ¸ Z T Z T 4z 4z tanh y (t) cos (ωc t + θ) dt −y (t) sin (ωc t + θ) dt = 0 AT 0 AT 0 21 where z = A2 T /2N0 .1. The block diagram of the phase estimator has another arm that adds into the feedback to the VCO that acts as a phase-lock loop for the carrier component. Note that for x small tanh (x) ' x and the phase estimator becomes a Costas loop if the integrators are viewed as lowpass Þlters. The variance of the phase estimate is bounded by ´ ³ 1 V ar bM L ≥ θ 4z (z + 1) Problem 9. For (a) note that ¤ £ cos cos−1 m = m. PROBLEMS Using deÞnitions for Z1 and Z2 . .24 This is similar to the previous problem. This implies a Costas loop type structure with integrators in place of the low pass Þlters and a tanh-function in the leg with the cosine multiplication.23. 1477 bits The maximum entropy is I (x) = log2 5 = 2.7004 bits (b) I (x) = − log2 [(52)(52)] = 11.3 The entropy is I (x) = −0.Chapter 10 Information Theory and Coding 10.4 1 .05 log2 (0.0969 Hartleys Problem 10.8) = 0.1 Problem Solutions Problem 10.8) = 0.1 The information in the message is I (x) = − log2 (0.4009 bits (c) I (x) = − log2 [(52)(51)] = 11.3 log2 (0.1) −0.25 log2 (0.25) −0.3) − 0.25 log2 (0.05) − 0.2 (a) I (x) = − log2 (52) = 5.25) − 0.3729 bits Problem 10.8) = 0.1 log2 (0.3219 bits I (x) = − loge (0.2231 nats I (x) = − log10 (0.05 log2 (0.05) = 2.3219 bits Problem 10. 1) = 3 1 (0.5 + 0.2 + 0.2) = 3 1 (0.6000  1 (0.4667 1.1: I (x) = 2.2 + 0.3 0.3667 30 11 = 0.1.4000  −0.6 y1 x2 0.6 + 0.5 (a) The channel diagram is shown is illustrated in Figure 10.2 + 0.1333 −0.3 + 0.471 bits Problem 10.2000 −0.1333 −0.2667 30 11 = 0. INFORMATION THEORY AND CODING x1 0.3667 30 .4000 [P (X)] = [0.2 0. (b) The output probabilities are p (y1 ) = p (y2 ) = p (y3) = (c) Since [P (Y )] = [P (X)] [P (Y |X)] we can write [P (X)] = [P (Y )] [P (Y |X)]−1 which is  2.2 0.2 0.333]  −0.5 0.5333 −1.333 0.8000 2.1 y2 0.2 CHAPTER 10.333 0.2 0.7 y3 x3 Figure 10.7) = 3 8 = 0. 001.2000 02.001.2667 0.08.001 0. 0 0 ··· p(xn ) Problem 10.1.0400  0. .001 0.999 which corresponds to an error probability of 0. The joint probability matrix is a square matrix with the input probabilities on the main diagonal and zeros elsewhere.   .2 0.333]  0 0 0 0.333 0.  .2000 0 [P (X. . . Consider the MATLAB program a = [0.3 0.0267 0.10. a1=a^n.6 For a noiseless channel.0400 0.2   0. Y )] =  0. .2 0.5333 0 0 0. while a1(1. .667] (d) The joint probability matrix is    0. end % channel matrix % initial value % save a .999 0.1067 [P (X. 0.5 0. the transition probability matrix is a square matrix with 1s on the main diagonal and zeros elsewhere.7 This problem may be solved by raising the channel matrix · ¸ 0.2  0.999].2)<0.1200 0.5333 0. . Problem 10. Y )] = [0. In other words   ··· 0 p (x1 ) 0   0 p (x2 ) · · · 0   [P (X. . a1 = a.08 n=n+1.1 0.1867  3 Note that the column sum gives the output possibilities [P (Y )]. n = 1.333 0. PROBLEM SOLUTIONS This gives[P (X)] = [0.001 A= 0. .6 0.1600 0.2667 0.. to increasing powers n and seeing where the error probability reaches the critical value of 0.999 0.0533 0. Y )] =  .7 which gives  0. and the row sum gives the input probabilities [P (X)] . This gives H (Y |X) − H (Y ) = − which is H (Y |X) − H (Y ) = − or XX i j p (xi . yj ) i j .9201 0. There are a number of other practical examples. However cascading 88 channels yields PE > 0. However. yj ) [log2 p (yj |xi ) − log2 p (yi )] or ¸ · p (xi . yj ) 1 XX p (xi . yj ) H (Y |X) − H (Y ) ≤ ln 2 p (xi . the channel A may represent a lengthly cable with a large number of repeaters.0799 À a^88 ans = 0. yj ) i j Since ln x ≤ x − 1. INFORMATION THEORY AND CODING % display result Executing the program yields n − 1 = 87.9192 Thus 87 cascaded channels meets the speciÞcation for PE < 0.0808 0.0799 0. yj ) ln H (Y |X) − H (Y ) = − ln 2 p (xi )p(xi ) i j · ¸ 1 XX p (xi )(yj ) H (Y |X) − H (Y ) = p (xi . the preceding expression can be written ¸ · p (xi )p(yj ) 1 XX −1 p (xi .) Problem 10.8 The Þrst step is to write H (Y |X) − H (Y ). yj ) ln ln 2 p (xi . Thus we compute (MATLAB code is given) À a^87 ans = 0.08. (Note: This may appear to be an impractical result since such a large number of cascaded channels are speciÞed.9201 0. yj ) log2 p (yj |xi ) − X j p (yj ) log2 p (yj ) XX i j p (xi .0808 0.4 n-1 CHAPTER 10.08 and the speciÞcation is not satisÞed.9192 0. Note that p (y1 |x1 ) = 1 p (y1 |x2 ) = 1 p (y1 |x3 ) = 0 p (y2 |x1 ) = 0 p (y2 |x2 ) = 0 p (y2 |x3 ) = 1 .10. H (Y |X) − H (Y ) ≤ 0 which gives H (Y ) ≥ H (Y |X) Problem 10. Thus. yj ) H (Y |X) − H (Y ) ≤  ln 2  i j i j The term in braces is zero since both double sums evaluate to one.10 For this problem. It therefore follows that C = 1 bit/symbol and that the capacity is achieved for p (x1 ) = p (x2 ) = 1 2 Problem 10. the channel diagram is illustrated in Figure 10.1. PROBLEM SOLUTIONS 5 x1 y1 x2 y2 x3 Figure 10.9 For this problem note that H (Y |X) = 0 so that I (Y |X) = H (X).2.2: or    X X XX 1 p (xi )p(yj ) − p (xi . Y ) = H (Y ) − H (Y |X) The conditional entropy H (Y |X) can be written X XX p (xi ) p (yj |xi ) log2 p (yj |xi ) = p (xi ) H (Y |xi ) H (Y |X) = − i j i where H (Y |xi ) = − X j p (yj |xi ) log2 p (yj |xi ) .11 For the erasure channel C=p bits/symbol and capacity is achieved for equally likely inputs.12 The capacity of the channel described by  p  q   0 0 is easily computed by maximizing the transition probability matrix  q 0 0 p 0 0   0 p q  0 q p I (X. INFORMATION THEORY AND CODING XX i j H (Y |X) = − p (xi .6 Thus CHAPTER 10. Problem 10. Problem 10. Y ) = H (Y ) The capacity is therefore C=1 bit/symbol and is achieved for source probabilities satisfying p (x1 ) + p (x2 ) = 1 2 p (x3 ) = 1 2 Note that capacity is not achieved for equally likely inputs. Thus I (X. yj ) log2 p (yj |xi ) = 0 since each term in the summand is zero. 4. This gives a capacity of log2 2 = 1 bit/symbol. and the channel capacity is C = 2 + p log2 p + (1 − p) log2 (1 − p) or C = 2 − H (p) This is shown in Figure 10. . x1 and x2 can be viewed as a single input and x3 and x4 can be viewed as a single input as shown in Figure 10. 4 all j Since p (yj ) = for all j. each input uniquely determines the output and the channel reduces to a noiseless channel with four inputs and four outputs. For such a channel X p (xi ) K = K H (Y |X) = i and I (X. The capacity of such channels are easily computed. This channel is modeled as a pair of binary symmetric channels as shown. PROBLEM SOLUTIONS For the given channel H (Y |xi ) = −p log2 p − q log2 q = K 7 so that H (Y |xi ) is a constant independent of i. See Gallager (1968). H (Y ) = 2.1. Y ) = H (Y ) − K We see that capacity is achieved when each channel output occurs with equal probability. This results since each row of the matrix contains the same set of probabilities although the terms are not in the same order.3. then each of the two subchannels (Channel 1 and Channel 2) have zero 2 capacity. If p = 1 or p = 0. Note that this channel is an example of a general symmetric channel. pages 91-94 for a discussion of these channels. We now show that equally likely outputs result if the inputs are equally likely. Then 4 p (yj ) = For each j p (yj ) = so that 1 4 X i p (xi ) p (yj |xi ) 1 1 1X 1 p (yj |xi ) = (p + q) = [p + (1 − p)] = 4 4 4 4 i 1 p (yj ) = . However. Assume that p (xi ) = 1 for all i. If p = q = 1 . This gives a capacity of log2 4 = 2 bits/symbol.10. The result is equivalent to a noiseless channel with two inputs and two outputs. 3: p x1 Channel 1 q y1 q p p q q p y2 y3 x2 x3 Channel 2 x4 y4 Figure 10. INFORMATION THEORY AND CODING C 2 1 0 1/2 1 p Figure 10.8 CHAPTER 10.4: . yj ) = log2 p (xi |yj ) + log2 p (yj ) 9 Evaluating the resulting sum yields (10.31) except that p (xi . yj ) in (10.2877/a a 3 Z x2 In a similar manner x1 ae−ax dx = e−ax1 − e−ax2 = e−ax2 = 1 2 3 1 − e−ax2 = 4 4 which gives or x2 = Also Z 1 ln (2) = 0. PROBLEM SOLUTIONS Problem 10.7585σ x2 = 1. yj ) is written p (yj |xi ) p (xi ) .28) as p (xi |yj ) p (yj ) and recall that log2 p (xi .6931/a a x3 ae−ax dx = x2 1 4 gives x3 = 1.14 The entropy is maximized when each quantizing region occurs with probability 0.25.1774σ x3 = 1. Thus Z x1 1 ae−ax dx = 1 − eax1 = 4 0 which gives e−ax1 = or 3 4 µ ¶ 1 4 x1 = ln = 0.30). The same method is used to show (10.13 To show (10.6651σ . Problem 10.3863/a Problem 10. write p (xi .30). The results are x1 = 0.15 The thresholds are now parameterized in terms of σ.10.1. 17 Five thresholds are needed to deÞne the six quantization regions.3413 p (m3 ) = √ 2 2πσ 0 Z 2σ 1 2 2 e−y /2σ dy = Q (1) − Q (2) = 0.9674 . m4 .3413 the entropy at the quantizer output is H (X) = −2 [0.9674σ −0.3413 log2 0. The Þve thresholds are −k2 .4303σ −0. −k1 .09 bits/symbol Since the symbol (sample) rate is 500 samples/s. and m5 . we only need compute three states.0228 p (m5 ) = √ 2πσ 2σ By symmetry p (m0 ) = p (m5 ) = 0.4303σ < xi < 0 0 < xi < 0. The thresholds are selected so that the six quantization regions are equally likely.4303σ < xi < 0.4303σ 0. Using symmetry. the information rate is r = 500 H (X) = 1045 symbols/s Problem 10. k1 .0228 log2 0. 0.9674σ < xi < −0. and k2 .1359 p (m2 ) = p (m3 ) = 0. Z σ 1 1 2 2 e−y /2σ dy = − Q (1) = 0. INFORMATION THEORY AND CODING Problem 10.0228] = 2.4303 This gives the quantizing rule Quantizer Input −∞ < xi < −0. m3 .1359 log2 0. Finding the inverse Q-function yields k1 = 0.9674σ < xi < ∞ Quantizer Output m0 m1 m2 m3 m4 m5 k2 = 0.16 First we compute the probabilities of the six messages.0228 p (m1 ) = p (m4 ) = 0.3413 + 0.1359 p (m4 ) = √ 2πσ σ Z ∞ 1 2 2 e−y /2σ dy = Q (2) = 0.9674σ 0.10 CHAPTER 10.1359 + 0. 75 log2 0. the error probability is Q 2z .7 0. PROBLEM SOLUTIONS 11 Problem 10.75 0.7 Therefore.25 log2 0.3113) = 3.005954 Using the techniques of the previous problem.1 log2 0.1 0.9 0.006143 Note that for all practical purposes the performance of the overall system is established by the downlink performance.25 0.3 p11 p12 = = p21 p22 0.3 0.24). the capacity of the overall channel is C = 1 + 0. or equivalently.9 0.1.18 The transition probabilities for the cascade of Channel 1 and Channel 2 are ¸ · ¸· ¸ · ¸ · 0.75 + 0.1887 bits/symbol It can be seen that the capacity of the cascade channel is less than the capacity of either channel taken alone.9 + 0.10. Problem 10.000191 = 0. Equation (10.2451 .8113 4 4 4 4 and the entropy of the fourth-order extension is ¡ ¢ H X 4 = 4H (X) = 4 (0.25 0.7 + 0.3 log2 0. From the data given Uplink error probability Downlink error probability = 0.20 The source entropy is 3 1 1 3 H (X) = − log2 − log2 = 0.75 0.1 = 0. yields the overall error probability PE = 0.19 ¡√ ¢ For BPSK. Problem 10.7 log2 0.1 0.25 = 0.1187 bits/symbol The capacity of Channel 1 is C = 1 + 0.5310 bits/symbol and the capacity of Channel 2 is C = 1 + 0.3 = 0.9 log2 0. 6561 0.0009 0.0081 0.0001 Codeword 0 100 101 110 1110 111100 1111010 1111011 1111100 1111101 1111110 111111100 111111101 111111110 1111111110 1111111111 .0729 0.0009 0.21 For the fourth-order extension we have Source Symbol AAAA BAAA ABAA AABA AAAB AABB ABAB BAAB ABBA BABA BBAA ABBB BABB BBAB BBBA BBBB P (·) 0.0009 0.0081 0.0081 0. is to determine the probability distribution of the extended source.0729 0.2451 bits/symbol Problem 10. We have 1 symbol with probability 4 6 4 1 Thus 81 256 27 symbols with probability 256 9 symbols with probability 256 3 symbols with probability 256 1 symbol with probability 256 µ ¶ µ ¶ ¡ 4¢ 81 81 27 9 27 9 = − log2 −4 log2 −6 log2 H X 256 256 256 256 256 256 µ ¶ 3 3 1 1 −4 log2 − log2 256 256 256 256 = 3.0081 0. INFORMATION THEORY AND CODING The second way to determine the entropy of the fourth-order extension.12 CHAPTER 10.0729 0.0009 0.0081 0.0729 0.0081 0. 22% Problem 10.598 1.4690 0.4 P (·) 0.2 0. This diagram yields the codewords m1 01 m2 10 m3 11 m4 000 m5 001 The average wordlength is L = 0.2 (2 + 2 + 2 + 3 + 3) = 2.2 Codeword 00 01 10 110 111 .22 For the Shannon-Fano code we have Source Symbol m1 m2 m3 m4 m5 The entropy is H (X) = −5 (0.2 log2 0.5.2 0. PROBLEM SOLUTIONS 13 The average wordlength: L = 1.9702/4 = 0.75% 2.4 The efficiency is therefore given by H (X) 2.9702 H (X n ) 0.4070 1.319 = = 0.90% 72.2) = 2.9702 which is L/4 = 1.9380 1.8760 Efficiency 46.10.4 L The diagram for determining the Huffman code is illustrated in Figure 10.2 0.2 0.3219 The average wordlength is I = (0.4926.29 1.05% 95.9675 = 96.1.2) (2 + 2 + 2 + 3 + 3) = 2.71% 88. The efficiencies are given in the following table: n 1 2 3 4 L 1 1. 40 00 1 0.2 m4 0.2 m3 0.2 0 1 Figure 10.16 10 001 m3 0. . INFORMATION THEORY AND CODING 0. the efficiency is also the same.4 m1 0.14 CHAPTER 10.2 m5 0.22 (Shannon-Fano code) (Huffman code) Note that the Huffman code gives the shorter average wordlength and therefore the higher efficiency.2 m2 0.25 L = 2.4 0.2 0 1 0. Problem 10.23 The codewords for the Shannon-Fano and Huffman codes are summarized in the following table: Codewords Source Symbol P (·) Shannon-Fano Huffman m1 0.10 111 011 m5 The average wordlengths are: L = 2.15 110 010 m4 0.5: Since the average wordlength is the same as for the Shannon-Fano code.19 01 000 m2 0.6 0 1 0.4 0 1 0. there are seven codewords of length three and two codewords of length four.85 log2 0.1. for n large rH (X) ≤ 300 or r≤ which gives r ≤ 491.10. The average wordlength is L= 1 29 [7 (3) + 2 (4)] = = 3.2222 9 9 .85 − 0.) Source Symbol m1 m2 m3 m4 m5 m6 m7 m8 m9 Codewords Shannon-Fano Huffman 000 001 001 010 010 011 011 100 100 101 101 110 110 111 1110 0000 1111 0001 300 300 = H (X) 0.932 symbols/s Problem 10.24 The entropy is H (X) = −0. Thus.25 The Shannon-Fano and Huffman codes are as follows. PROBLEM SOLUTIONS Problem 10.15 = 0. both codes have the same average wordlengths and therefore have the same efficiencies.15 log2 0.60984 bits/symbol We know that n→∞ 15 lim r L ≤ 300 n and that L = H (X) n→∞ n lim Therefore.60984 In both cases. (Note that the codes are not unique. 38% L Problem 10.5850 This gives the efficiency E= H (X) = 0.9777 = 97. INFORMATION THEORY AND CODING H (X) = log2 9 = 3.77% L 1 44 [4 (3) + 8 (4)] = = 3.16 Since the source entropy is CHAPTER 10.6667 12 12 P (·) 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 Codeword 000 0010 0011 010 0110 0111 100 1010 1011 110 1110 1111 The diagram for determining the Huffman code is illustrated in Figure 10.26 For the Shannon-Fano code we have Source Symbol m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 The average wordlength is L= and the entropy of the source is H (X) = log2 12 = 3.6.1699 the efficiency is E= H (X) = 98. This yields . 10. we have four codewords of length three and eight codewords of length four.6: the codewords: Source Symbol m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 Codeword 100 101 110 111 0000 0001 0010 0011 0100 0101 0110 0111 As in the case of the Shannon-Fano code. PROBLEM SOLUTIONS 2/3 0 17 m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 0 1 0 1 0 1 0 1 0 1 0 1 1/6 0 1/3 1/3 1 1/6 1 1/6 0 1/3 0 1/6 1 1/6 0 1/3 1 1/6 1 Figure 10.1. The average wordlength is therefore 3.6667 and the efficiency is . 15 0.1 log2 0.01 Codeword 11 000 010 011 100 101 0011 00100 001010 001011 Problem 10.4 − 0.3 − 0.1000 binary symbols/source symbol A source symbol rate of 250 symbols (samples) per second yields a binary symbol rate of rL = 250(3.1 = 1.0488) = 75.17 0.05 0.07 0.27 The probability of message mk is P (mk ) = Z 0.2 log2 0.0488 bits/source symbol and the average wordlength is L = 3. INFORMATION THEORY AND CODING Problem 10.18 97.19 0.3 log2 0. CHAPTER 10.2 − 0.1000) = 775 binary symbols/second and a source information rate of Rs = rH (X) = 250(3.77%.28 The source entropy is H (X) = −0.1k 0.84644 bits/source symbol .4 log2 0.11 0.09 0.03 0.13 0.2 bits/second P (·) 0.1(k−1) 2x dx = 0.01 (2k − 1) A Huffman code yields the codewords in the following table: Source Symbol m10 m9 m8 m7 m6 m5 m4 m3 m2 m1 The source entropy is given by H (X) = 3. 09 + 0.04 0.04 0.08) +3 (0.02 + 0.03 0.12 + 0. 3.06 + 0.12 0. 5.06 0.08 0.38) log2 3 Problem 10.03 + 0. 1.02) +4 (0.01 19 The second-order extension source symbols are shown in the following table.04 0.02 0.06 + 0. The set of code symbols is (0. 3. Codeword 00 01 02 10 11 12 200 201 202 210 211 212 220 221 2220 2221 The average wordlength of the extended source is L = 2 (0.12 + 0.08 0. 3.9% (2.06 0.04 + 0.16 0.02 0.3 is {1. 5}.10.04 + 0.69288 Source Symbol m1 m1 m1 m2 m2 m1 m2 m2 m1 m3 m3 m1 m2 m3 m3 m2 m3 m3 m1 m4 m4 m1 m2 m4 m4 m2 m3 m4 m4 m3 m4 m4 P (·) 0.29 The set of wordlengths from Table 10.12 0. 2).1. PROBLEM SOLUTIONS and the entropy of the second-order extension of the source is ¡ ¢ H X 2 = 2H (X) = 3. 5.03 + 0.38 Thus the efficiency is E= ¡ ¢ H X2 L log2 D = 3.01) = 2.09 0.69288 = 97. The Kraft inequality gives 8 X i=1 ¡ ¢ ¡ ¢ 1 3 4 =1 2li = 2−1 + 3 2−3 + 4 2−5 = + + 2 8 32 .03 0. 5.16 + 0.08 + 0. For a D = 3 alphabet we partition into sets of 3. INFORMATION THEORY AND CODING 6 5 4 Capacity . It can be seen that as B → ∞ the capacity approaches a Þnite constant.7. the Shannon-Hartley law with S = 40 and N = 10−4 B gives Ã ¡ ¢! µ ¶ 40 104 S = B log2 1 + Cc = B log2 1 + N B This is illustrated in Figure 10.31 From the previous problem Cc = B log2 Ã ¡ ¢! 40 104 1+ B .bits/s 3 2 1 0 2 10 10 3 10 4 10 Frequency 5 10 6 10 7 10 8 Figure 10. Problem 10.7: Thus the Kraft inequality is satisÞed.82).20 x 10 5 CHAPTER 10. Problem 10.30 From (10. 94) with PE = 3q 2 (1 − q) + q 3 The results.^4)+(q.^2). ¡ ¢ ¡ ¢ 40 104 Cc = = 5. plot(q. Note that these results are not 5 typically compariable since the values of p and q change as the code rate changes. we can write that. The top curve (poorest performance) is for the rate 1 code 3 and the bottom curve is for the rate 1 repetition code.^2)+(q.^3)+5*(1-q).10. p3 = 3*(1-q).^5).*(q. For a rate q = 1 − p) 1 3 code we have the corresopnding result (See (10.*(q. as B → ∞.p5) xlabel(’Error probability .p3. 11) Hamming code is 0  0 [H] =   1 1  0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0  0 0   0  1 .*(q.32 For a rate 1 repetition code we have. p5 = 10*((1-q). PROBLEM SOLUTIONS Since ln (1 + u) ≈ u for ¿ 1.33 The parity-check matrix for a (15.q.^3). generated by the MATLAB code. assuming an error probability of (1 − p) on a single 5 transmission. Problem 10.7708 105 bits/s ln (2) 21 Problem 10.1. q = 0:0.01:0. µ ¶ µ ¶ µ ¶ 5 2 5 5 3 4 p (1 − p) + p (1 − p) + (1 − p)5 PE = 3 4 5 which is PE = 10p2 (1 − p)3 + 5p (1 − p)4 + (1 − p)5 or PE = 10 (1 − q)2 q 3 + 5 (1 − q) q 4 + q5 where q = 1 − p.1.8.(1-p)’) ylabel(’Word error probability’) are illustrated in Figure 10. 22 CHAPTER 10. INFORMATION THEORY AND CODING 0.03 0.025 W ord error probability 0.02 0.015 0.01 0.005 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Error probability - q=(1-p) 0.08 0.09 0.1 Figure 10.8: Since the columns of [H] are distinct and all nonzero vectors of length three are included in [H], all single errors can be corrected. Thus, e= or dm = 3 Note that it is not necessary to Þnd the set of codewords. Problem 10.34 The parity-check matrix will have 15 − 11 = 4 rows and 15 columns. Since rows may be 1 (dm − 1) = 1 2 10.1. PROBLEM SOLUTIONS permuted, the parity-check  0  0 [H] =   1 1 matrix is not unique. One possible selection is  0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 1 0 1 0 0   0 1 1 0 1 1 0 0 1 1 0 0 1 0  1 0 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1  23 The corresponding generator matrix is  1 0 0 0 0  0 1 0 0 0   0 0 1 0 0   0 0 0 1 0   0 0 0 0 1   0 0 0 0 0    0 0 0 0 0  [G] =  0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 1   0 1 1 1 0   1 0 1 1 0 1 1 0 1 1 where I11 is the 11×11 identity matrix and H11 represents the Þrst 11 columns of the paritycheck matrix. From the structure of the parity-check matrix, we see that each parity symbol is the sum of 7 information symbols. Since 7 is odd, the all-ones information sequence yields the parity sequence 1111. This gives the codeword [T ]t = [111111111111111] A single error in the third position yields a syndrome which is the third column of the parity-check matrix. Thus, for the assumed parity-check matrix, the syndrome is   0  1  [S] =    1  0 Problem 10.35            · ¸  I11  = H11             24 CHAPTER 10. INFORMATION THEORY AND CODING For the given parity-check matrix, the generator matrix is     [G] =      0 1 0 0 1 1 1 0 1 0 1 0 1 0  1 0 0 0 1 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1           1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 This gives the 16 codewords (read vertically) a1 a2 a3 a4 c1 c2 c3 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 1 0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 1 0 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 Problem 10.36 We Þrst determine [Ti ] = [G] [Ai ] for i = 1, 2. The generator matrix for the previous problem is used. For [A1 ] we have   0    1    0    1   1    =  0  = [T1 ] [G] [A1 ] = [G]     1  1    0  0  0 1 1 1 0 0 1 0 and for [A2 ] we have    1   1   [G] [A2 ] = [G]   =   1     0          = [T2 ]     10.1. PROBLEM SOLUTIONS We note that  1  0  [A1 ] ⊕ [A2 ] =    0  0   25 and which is [T1 ] ⊕ [T2 ].    1   0    = [G] {[A1 ] ⊕ [A2 ]} = [G]    0   0   1 0 0 0 1 1 0           Problem 10.37 For a rate 1 repetition code 3 and for a rate 1 5 repetition code  1 [G] =  1  1   [G] =     1 1 1 1 1        The generator matrix for a rate are unity. 1 n repetition code has one column and n rows. All elements Problem 10.38 For a Hamming code with r = n − k parity check symbols, the wordlength n is n = 2r − 1 and k = 2r − r − 1 The rate is R= k 2r − r − 1 = n 2r − 1 26 CHAPTER 10. INFORMATION THEORY AND CODING If n is large, r is large. As r → ∞ both k and n approach 2r . Thus n→∞ lim R = lim R = r→∞ 2r =1 2r Problem 10.39 The codeword is of the form (a1 a2 a3 a4 c1 c2 c3 ) where c1 = a1 ⊕ a2 ⊕ a3 c2 = a2 ⊕ a3 ⊕ a4 c3 = a1 ⊕ a2 ⊕ a4 This gives the generator matrix     [G] =       1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 0 1 1           Using the generator matrix, we can determine the codewords. The codewords are (read vertically) a1 a2 a3 a4 c1 c2 c3 0 0 0 0 0 0 0 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 1 1 1 1 1 1 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 Let αi denote the i-th code word and let αi → αj indicate that a cyclic shift of αi yields αj . The cyclic shifts are illustrated in Figure 10.9. Problem 10.40 The parity-check matrix for the encoder  1 [H] =  0 1 in the previous problem is  1 1 0 1 0 0 1 1 1 0 1 0  1 0 1 0 0 1 10.1. PROBLEM SOLUTIONS 27 α1 α2 α5 α9 α15 α14 α3 α12 α6 α4 α8 α7 α11 α13 α16 α10 Figure 10.9: For the received sequence [R]t = [1101001] the syndrome is  0 [S] = [H] [R] =  0  0 [R]t = [1101011] the syndrome is  This indicates no errors, which is in agreement with Figure 10.17. For the received sequence  0 [S] = [H] [R] =  1  0  This syndrome indicates an error in the 6th position. Thus the assumed transmitted codeword is [T ]t = [1101001] which is also in agreement with Figure 10.17. Problem 10.41 The 16 codewords are (read vertically) a1 a2 a3 a4 c1 c2 c3 0 0 0 0 0 0 0 1 0 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 1 1 1 0 1 0 1 1 1 0 1 0 1 0 0 0 1 .28 CHAPTER 10. the codewords can be constructed. INFORMATION THEORY AND CODING For the modiÞed encoder we can write Step 1 2 3 4 5 6 7 S1 a1 a1 ⊕ a2 a1 ⊕ a2 ⊕ a3 a2 ⊕ a3 ⊕ a4 0 0 0 S2 0 a1 a1 ⊕ a2 a1 ⊕ a2 ⊕ a3 a2 ⊕ a3 ⊕ a4 0 0 S3 0 0 a1 a1 ⊕ a2 a1 ⊕ a2 ⊕ a3 a2 ⊕ a3 ⊕ a4 0 Thus. the output for each step is Step 1 2 3 4 5 6 7 This gives the generator matrix     [G] =       1 0 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 1 1 0 1           Output a1 a2 a3 a4 a1 ⊕ a3 ⊕ a4 a1 ⊕ a2 ⊕ a3 a2 ⊕ a3 ⊕ a4 Using the generator matrix. the uncoded symbol error probability is given by hp i 2z/k qu = Q where z = STw /N0 and k = 11.39.11. PROBLEM SOLUTIONS 29 α1 α2 α13 α16 α10 α5 α12 α7 α6 α3 α4 α8 α9 α15 α14 α11 Figure 10. let αi denote the i-th code word and let αi → αj indicate that a cyclic shift of αi yields αj . the symbol error probability is hp i qc = Q 2z/15 and the coded word error probability is Pec 15 X µ15¶ q i (1 − qc )15−i = i c i=2 The performance curves are generated using the MATLAB M-Þles given in Computer Exercise 10.10: As in Problem 10.42 For a (15.2 in Section 10.1.10. This gives the uncoded word error probability Peu = 1 − (1 − qu )11 For the coded case.2 (Computer Exercises) of this solutions manual. 11) Hamming code. The results are shown in Figure 10. The solid curves represent the symbol error probability with the . Problem 10.10. These cyclic shifts are illustrated in Figure 10. The dashed curves represent the word error probability with the coded case (better performance) on bottom. INFORMATION THEORY AND CODING 10 0 10 -1 10 -2 P robability 10 -3 10 -4 10 -5 10 -6 10 -7 10 12 14 16 STw/No in dB 18 20 22 Figure 10. Problem 10.11: coded case on top (worse performance) and the uncoded case on bottom.30 CHAPTER 10.43 The probability of two errors in a 7 symbol codeword is µ ¶ 7 2 2 P {2 errors} = (1 − qc )5 qc = 21qc (1 − qc )5 2 and the probability of three errors in a 7 symbol codeword is µ ¶ 7 3 3 P {3 errors} = (1 − qc )4 qc = 35qc (1 − qc )4 3 The ratio of P {3 errors} to P {2 errors} is R= 3 Pr {3 errors} 5 qc 35qc (1 − qc )4 = = 2 (1 − q )5 Pr {2 errors} 3 1 − qc 21qc c . 0062 31 It is clear that as the SNR. Moving the parity checks to positions 5. 6. Problem 10.0455 0.1642 0.0141 0.0290 0.1. the value of R becomes negligible.0037 R = Pr{3 errors} Pr{2 errors} 0. 4) Hamming code is  0 0 0 1 1 1 1 [H] =  0 1 1 0 0 1 1  1 0 1 0 1 0 1  The parity checks are in positions 1.0278 0. Since BPSK modulation is used Ãr ! 2z qc = Q 7 where z is STw /N0 . PROBLEM SOLUTIONS We now determine the coded error probability qc .0660 0.44 The parity-check matrix for a (7. increases. and 4.1177 0. and 7 yields a systematic code with the generator matrix     [G] =      which has the partity-check matrix  0 1 1 1 1 0 0 [H] =  1 0 1 1 0 1 0  1 1 0 1 0 0 1   1 0 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1 1 1 1           . z.0167 0.2.0897 0.0497 0.0794 0. This gives us the following table z 8dB 9dB 10dB 11dB 12dB 13dB 14dB qc 0.0085 0.10. 45 For the encoder shown. the distance three property is preserved. there are 4! = 24 different (7. The trellis diagram is illustrated in Figure 10. Thus. If the code is not required to be systematic. Problem 10. The output is v1 v2 where v1 = s1 ⊕ s2 ⊕ s3 ⊕ s4 and v2 = s1 ⊕ s2 ⊕ s4 . we have v1 = s1 ⊕ s2 ⊕ s3 and v2 = s1 ⊕ s2 . .13. We Þrst need to compute the states for the encoder. Previous s1 s2 s2 000 001 010 011 100 101 110 111 Current State A E A E B F B F C G C G D H D H State A B C D E F G H Input 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 s1 s2 s3 s4 0000 1000 0001 1001 0010 1010 0011 1011 0100 1100 0101 1101 0110 1110 0111 1111 Output 00 11 11 00 10 01 01 10 11 00 00 11 01 10 10 01 Problem 10. the constraint span is k = 4. This gives the following table. all of which will have equivalent performance in an AWGN channel. 4) systematic codes.12. INFORMATION THEORY AND CODING Note that all columns of [H] are still distinct and therefore all single errors are corrected. and the output corresponding to each transition appears in the following table. there are 7! = 5040 different codes. Since the Þrst four columns of [H] can be in any order. We now compute the state transitions and the output using the state diagram shown in Figure 10.46 For the encoder shown.32 CHAPTER 10. The state transitions. PROBLEM SOLUTIONS (0 0) 33 A (1 1) (1 0) B (1 1) C (0 1) (0 0) (1 1) E (0 1) (0 0) D (1 0) F (1 0) (0 0) (0 1) (1 1) G (1 0) H (0 1) Figure 10.12: State A B C D s1 s2 00 01 10 11 Input 0 1 0 1 0 1 0 1 s1 s2 s3 000 100 001 101 010 110 011 111 State A C A C B D B D Output 00 11 11 00 10 01 01 10 Problem 10.1.10.47 . Let the table contain l = 525 codewords.9 seconds 3500 symbols/second If the burst duration is Tb . 4) block code. INFORMATION THEORY AND CODING (00) (00) (00) (00) (11) (11) (11) (11) (11) (11) (10) (00) (10) (10) (01) (01) (01) (01) (10) Steady-State Transitions Termination Figure 10. Thus 3675 − 525 = 3150 must be transmitted without error.15) = 525 4 rc Tb = rs Tb symbols affected by the burst.34 (00) CHAPTER 10. The channel symbol rate is µ ¶ n 7 rc = rs = 2000 = 3500 k 4 n k channel symbols will be affected. The transmission time for 3150 symbols is 3150 symbols = 0. then . Assuming a (7.13: The source rate is rs = 2000 symbols/second. we have µ ¶ 7 rc Tb = (2000) (0. This corresponds to (525) (7) = 3675 symbols in the table. 10.001:0. The MATLAB program follows a = 0. 10.2 Computer Exercises Computer Exercise 10.48 The required plot is shown in Figure 10.999.1 For two messages the problem is easy since there is only a single independent probability. .001:0.14: Problem 10.14. COMPUTER EXERCISES 35 Figure 10.2. % number of points in vector % increment for zero .9 0. plot(a.15.8 0.15: b =1-a. INFORMATION THEORY AND CODING 1 0.1 0 0 0. lna = length(a).8 0.4 0. Two of them will be speciÞed and the remaining probability can then be determined.2 0. H = zeros(1.5 a 0.H).5 0. The resulting MATLAB code follows: n = 50. xlabel(’a’) ylabel(’Entropy’) The plot is illustrated in Figure 10.*log(a)+b. nn = n+1.7 0. For three source messages we have two independent probabilities.7 0.6 E ntropy 0.3 0.lna).*log(b))/log(2).1 0.2 0.6 0. H=-(a.36 CHAPTER 10.3 0.9 1 Figure 10.4 0. H. One should experiment with various viewing locations.10.17.26) Hamming code are given in Figure 10. % compute entropy end end end colormap([0 0 0]) % b&w colormap for mesh mesh(z. the solid curves represent the symbol error probability with the coded case on top (worse performance) and the uncoded case on bottom. Computer Exercise 10. 10. % calculate third probability if c>0 xx = [z(i) z(j) c].2 The performance curves for a (15. As in the case of the (7.z. % initialize entropy array z = (0:n)/n.16. COMPUTER EXERCISES H = zeros(nn. The dashed curves represent the word error probability with the coded case (better performance) on bottom.30). The result is illustrated in Figure 10. indicating that the maximum entropy occurs when all three source messages occur with equal probability. % set view angle xlabel(’Probability of message a’) ylabel(’Probability of message b’) zlabel(’Entropy’) figure % new figure contour(z. Next we draw a countour map with 10 countour lines. The equivalent results for a (31.42.11) Hamming code.z.33.4) Hamming code. % Es/No in dB . 0. % plot entropy view(-45. The computer code for the (31. on the basis of word energy and word error probability were derived in Problem. zdB = 0:10. H(i. Note the peak at 0. % prevent problems with log(0) for i=1:nn for j=1:nn c = 1-z(i)-z(j).33.nn).18.j)= -xx*log(xx)’/log(2).2.H).10) % plot contour map xlabel(’Probability of message a’) ylabel(’Probability of message b’) 37 The resulting plot is illustrated in Figure 10. k = 26.26) code is as follows: n = 31. % probability vector z(1) = eps. zz. % Q-function argument for coded PSK psc = 0.zz.pwc.m and nkchoose.’k--’.’k-’.’k--’) xlabel(’STw/No in dB’) ylabel(’Probability’) The results for the (7.^(zdB/10).38 CHAPTER 10. The main program calls two functions werrorc.^k.’k-’. % coded word error probability zz = zdB+10*log10(k).4) Hamming code are generated by changing the Þrst line of the program to n = 7.psc).0))). % PSK symbol error probability pwu = 1-(1-psu). % coded PSK symbol error probability pwc = werrorc(n.0))).16: z = 10.zz. % scale axis for plot (STw/No) % semilogy(zz. k = 4.pwu.m.5*(1-erf(arg2/sqrt(2. INFORMATION THEORY AND CODING Figure 10.k. % Es/No in linear units arg1 = sqrt(2*z).1.psu. . % uncoded word error probability arg2 = sqrt(2*k*z/n).psc. % Q-function argument for PSK psu = 0.5*(1-erf(arg1/sqrt(2. function [y] = werrorc(n.7 0.k.8 0. sum = sum+exp(term). end y(i) = sum.j).7 Probability of message a 0.1 0. COMPUTER EXERCISES 39 1 0.6 0.4 0.psc) lx = length(psc).17: These follow.t. for j=(t+1):n nkc = nkchoose(n.8 P robability of m essage b 0.1 0.2.10. end .2 0.9 0. term = log(nkc)+j*log(psc(i))+(n-j)*log(1-psc(i)).2 0. % initialize y for i=1:lx % do for each element of psc sum = 0.5 0.3 0.9 1 Figure 10.5 0. y = zeros(1.3 0.lx).4 0.6 0. k) % Computes n!/k!/(n-k)! a = sum(log(1:n)).3 The MATLAB code for implementing the given equation is . out = round(exp(a-b-c)). % End of function file.18: function out=nkchoose(n. INFORMATION THEORY AND CODING 10 0 10 -1 10 -2 P robability 10 -3 10 -4 10 -5 10 -6 10 -7 14 16 18 20 STw/No in dB 22 24 26 Figure 10. We prefer to compute the binomial coefficient using logorithms in order to obtain the large dynamic range needed for some very long codes. b = sum(log(1:k)). c = sum(log(1:(n-k))).40 CHAPTER 10. % % % % ln of n! ln of k! ln of (n-k)! result Note: The preceding function is called nkchoose rather than nchoosek in order to avoid conßict with the function nchoosek in the communications toolbox. Computer Exercise 10. 1992. % Based on (1. sum1 = sum1+term.beru.vector of channel symbol error rates % ber .J. Torrieri (2nd Edition).4) Hamming code BER for Hamming code plot results . % Parameters: % q .alphabet size (2 for binary) % n . sum2 = sum2+term.block length % d . end % second loop evaluates second sum for i=(d+1):n term = i*nkchoose(n. we use the following MATLAB program: zdB = 0:0.2.7.^(zdB/10).vector of bit error rates % lnps = length(ps). end 41 In order to demonstrate the preceding function with a (7. berham = cer2ber(2.i)*(cerî)*((1-cer))^(n-i).1:10.zdB. % initialize sums % first loop evaluates first sum for i=(t+1):d term = nkchoose(n.1.d. z = 10.berham) % % % % % % set Eb/No axis in dB convert to linear scale PSK result CSER for (7.ps) % Converts channel symbol error rate to decoded BER.3. cerham = qfn(sqrt(4*z*2/7)). Artech House.10. end % comput BER (output) ber(k) = (q/(2*(q-1)))*((d/n)*sum1+(1/n)*sum2).i)*(cerî)*((1-cer)^(n-i)). COMPUTER EXERCISES function [ber] = cer2ber(q.cerham). beru = qfn(sqrt(2*z)).n. % channel symbol error rate sum1 = 0.correctable errors per block (usually t=(d-1)/2) % ps .minimum distance of code % t .lnps). % initialize output vector for k=1:lnps % iterate error vector cer = ps(k). semilogy(zdB.30) in ’’Principles of Secure Communication Systems’’ % by D. % length of error vector ber = zeros(1.7.4) Hamming code.t. sum2 = 0. 1:10.5*erfc(x/sqrt(2)). Computer Exercise 10.4 The MATLAB program is zdB = 0:0. It can be seen that the (7.4) Hamming code yields only a moderate performance improvement at high SNR.19.19: xlabel(’E_b/N_o in dB’) ylabel(’Bit Error Probability’) % label x axis % label y axis The MATLAB function for the Gaussian Q-function follows: function out=qfn(x) % Gaussian Q function out =0.42 CHAPTER 10. Executing the code yields the result illustrated in Figure 10. % set Eb/No axis in dB . INFORMATION THEORY AND CODING 10 0 10 -1 B it E rror P robability 10 -2 10 -3 10 -4 10 -5 10 -6 0 1 2 3 4 5 6 E /N in dB b o 7 8 9 10 Figure 10. 18 in the text.5*z/7).^3).^(zdB/10). +7*(1-qa7).*(qr3. qr7 = 0.4) Hamming code and the bottom curve is for the (23.*(qr7.*(qa7. % CSER for (15.5*exp(-0.pa7) axis([0 30 0.^3).^6)+(qa7.zdB.dB’) ylabel(’Probability ’) Executing the code yields the results illustrated in Figure 10.zdB.12) Golay code. pa7 = 35*((1-qa7). % BER for Hamming code semilogy(zdB.5. The curves can be identiÞed from Figure 10. qa = 0.*(qa7. semilogy(zdB.0001 1]) xlabel(’Signal-to-Noise Ratio.3.. % PSK result ber2 = qfn(sqrt(12*2*z/23)).23.^7).2.*(qr7..5 The MATLAB code for generating the performance curves illustrated in Figure 10. The order of the curves. z = 10.1.18 in the text follows.pa3. The results are illustrated in Figure 10.^2)+(qa3.21.^2).5*exp(-0.zdB./(1+z/2).qr.zdB.zdB. are in order of the error correcting capability./(1+z/(2*7)). for high signal-tonoise ratio.3.^3). z .^5). % CSER for (23. qa7 = 0.berhamming) xlabel(’E_b/N_o in dB’) ylabel(’Bit Error Probability’) 43 The funcrions Q and cer2ber are given in the previous computer exercise. COMPUTER EXERCISES z = 10.12) Golay code ber3 = qfn(sqrt(4*z*2/7)).10.5*z).^2).^(zdB/10). zdB = 0:30.^4)+21*((1-qr7).11) Hamming code bergolay = cer2ber(2. .bergolay.^2)+(qr3. % BER for Golay code berhamming = cer2ber(2. qr = 0.5.zdB.5.*(qa7.7.*(qr7. The middle curve is for the (7.^7). Computer Exercise 10.ber1.^5).20. qr3 = 0.^6)+(qr7..5*exp(-0.5*z/3).pr7.qa.ber3).*(qa3.zdB. % convert to linear scale ber1 = qfn(sqrt(2*z)).^4)+21*((1-qa7).ber2).7./(1+z/(2*3)). pr3 = 3*(1-qr3). The top curve (worst performance) is the uncoded case. pa3 = 3*(1-qa3).pr3.^3). qa3 = 0. pr7 = 35*((1-qr7).. +7*(1-qr7). 31). % Calculations for diversity system pd7 = zeros(1. N = 7. The MATLAB code follows: zdB z = qr1 qr7 pr7 = 0:30. 10. ./(1+z/(2*7)).^(zdB/10). = 35*((1-qr7).17.6 For this Computer Exercise we use the uncoded Rayleigh fading result from the previous Computer Exercise and also use the result for a Rayleigh fading channel with a rate 1/7 repetition code.. The new piece of information needed for this Computer Exercise is the result from Problem 9.*(qr7.*(qr7./(1+z/2)..^5).^4)+21*((1-qr7). = 0.44 CHAPTER 10.5. INFORMATION THEORY AND CODING 10 0 10 -2 B it E rror P robability 10 -4 10 -6 10 -8 10 -10 0 1 2 3 4 5 6 E /N in dB b o 7 8 9 10 Figure 10.^7).^3). = 0.5.20: Computer Exercise 10.^6)+(qr7.^2).*(qr7. +7*(1-qr7). dB 25 30 Figure 10.qr1.pr7.zdB. sum = sum+term. for j=0:(N-1) term = nkchoose(N+j-1. a = 0.5/(1+z(i)/(2*N)). z .dB’) ylabel(’Probability’) . z .0001 1]) xlabel(’Signal-to-Noise Ratio.zdB.10.2. COMPUTER EXERCISES 45 10 0 10 -1 Probability 10 -2 10 -3 10 -4 0 5 10 15 20 Signal-to-Noise Ratio. end % semilogy(zdB. end pd7(i) = (a^N)*sum.j)*((1-a)^j).pd7) axis([0 30 0.21: for i=1:31 sum = 0. out = round(exp(a-b-c)).22.k) % Computes n!/k!/(n-k)! a = sum(log(1:n)).20 inthe text. c = sum(log(1:(n-k))). z . b = sum(log(1:k)). % % % % ln of n! ln of k! ln of (n-k)! result Executing the MATLAB program yields the result illustrated in Figure 10. . The curves can be identiÞed by comparison with Figure 10.22: The function nkchoose follows: function out=nkchoose(n.46 CHAPTER 10.dB 25 30 Figure 10. % End of function file. INFORMATION THEORY AND CODING 10 0 10 -1 P robability 10 -2 10 -3 10 -4 0 5 10 15 20 Signal-to-Noise Ratio. 93 × 10−5 V rms = 69.1 All parts of the problem are solved using the relation √ Vrms = 4kT RB where k = 1. respectively.3 µV rms (b) Vrms is smaller than the result in part (a) by a factor of Vrms = 21. √ 100 = 10. Thus √ 10 = 3.93 µV rms (d) Each answer becomes smaller by factors of 2.9 µV rms (c) Vrms is smaller than the result in part (a) by a factor of Vrms = 6. and 10. Thus .16.38 × 10−23 J/K B = 30 MHz = 3 × 107 Hz (a) For R = 10. 1 √ 10 = 3. 000 ohms and T = T0 = 290 K p Vrms = 4 (1.38 × 10−23 ) (290) (104 ) (3 × 107 ) = 6.16.Appendix A Physical Noise Sources and Noise Calculations Problem A. 1. and for I > 20Is . kT e (a) At T = 290 K. Problem A.0761) 1.6 × 10−19 1. we need exp(400V ) > 21 or V > ln (21) = 7.5 × 10−5 exp (40 × 0.6 × 10−19 1.61 × 10−3 volts 400 Thus i2 rms B ¢¡ ¢ ¡ ¢ ¡ = 2 1.5 × 10−5 exp 400 × 7.2 APPENDIX A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.0075 × 10−22 A2 /Hz as before.0761 volts 40 µ ¶ eV 2eIB ' 2eBIs exp kT µ ¶ eV 2eIs exp kT ¢¡ ¢ ¡ 2 1. then e kT = 400.3 (a) Use Nyquist’s formula to get the equivalent circuit of Req in parallel with R3 . where Req is given by RL (R1 + R2 ) Req = R1 + R2 + RL The noise equivalent circuit is then as shown in Figure A. kT ' 40. The equivalent noise voltages are p 4kT Req B p = 4kT R3 B V1 = V2 .0075 × 10−22 A2 /Hz · µ ¶ ¸ eV I = Is exp −1 kT i2 rms ' or i2 rms B = = = (b) If T = 29 K. so we have exp (40V ) > 21 giving V > ln (21) = 0.2 Use ¡ ¢ We want I > 20Is or exp eV − 1 > 20.61 × 10−3 = 1. 23 × 10−18 V2 /Hz 300 (2000 + 300) = 265.38 × 10−23 (290) (500)2 = 500 + 265. we have Req = and V02 B = 2 (4kT ) R3 (Req + R3 ) ¢ ¡ 4 1.1: Adding noise powers to get V0 we obtain µ ¶2 ¡ 2 ¢ R3 2 V1 + V22 V0 = Req + R3 µ ¶2 R3 = (4kT B) (Req + R3 ) Req + R3 (4kT B) R2 3 = (Req + R3 ) (b) With R1 = 2000 Ω.3 Req R3 V0 V1 V2 Figure A.4 Ω 300 + 2000 + 300 .4 = 5. R2 = RL = 300 Ω. and R3 = 500 Ω. R3 combination and set RL equal to this to get R3 (R1 + R2 ) RL = R1 + R2 + R3 Problem A. we Þnd the equivalent resistance looking backing into the terminals with Vrms across them. we Þrst determine the noise power due to a source at the output. 333 Ω Thus 2 Vrms = 4kT Req B ¡ ¢ ¡ ¢ = 4 1.4 Find the equivalent resistance for the R1 .68 × 10−10 V2 which gives Vrms = 19. only = S RS + R1 + R2 k RL V02 ¯due to R1 and R2 ¯ ¶2 R2 k RL = (4kT R1 B) RS + R1 + R2 k RL µ ¶2 RL k (R1 + RS ) + (4kT R2 B) R2 + (R1 + RS ) k RL µ .6 To Þnd the noise Þgure. R2 .4 APPENDIX A.5 Using Nyquist’s formula. The results are ¶2 µ ¯ R2 k RL 2¯ (4kT RS B) V0 due to R . Initally assume unmatched conditions. and then due to the source and the network. It is Req = 50k k 20 k k (5 k + 10 k + 5 k) = 50k k 20 k k 20 k = 50k k 10 k (50k) (10 k) = 50k + 10 k = 8.18 µV rms Problem A. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.38 × 10−23 (400) (8333) 2 × 106 = 3. these expressions are substituted back into the expression for F . After some simpliÞcation. Matching at the input requires that RS = Rin = R1 + R2 k RL = R1 + and matching at the output requires that RL = Rout = R2 k (R1 + RS ) = R2 (R1 + RS ) R1 + R2 + RS R2 RL R2 + RL Next. this gives ¶2 µ 2R2 RS (R1 + R2 + RS ) / (RS − R1 ) R2 R1 L + F =1+ 2 (R + R + R ) + R2 (R + R + R ) RS RS R2 1 1 2 S L S L Note that if R1 >> R2 we then have matched conditions of RL = R2 and RS = R1 . we want the input matched to the source and the output matched to the load. µ RL k (R1 + RS ) R2 + (R1 + RS ) k RL Ra k Rb = Ra Rb Ra + Rb Note that the noise due to RL has been excluded because it belongs to the next stage. The equivalent noise temperature is found by using Te = T0 (F − 1) · ¸ R1 = T0 1 + 16 R2 . Then. R1 and R2 = µ ¶2 R2 k RL [4kT (RS + R1 ) B] RS + R1 + R2 k RL µ ¶2 RL k (R1 + RS ) + (4kT R2 B) R2 + (R1 + RS ) k RL ¶2 µ RS + R1 + R2 k RL R2 k RL ¶2 R2 RS R1 + F =1+ RS In the above. Since this is a matching circuit. the noise Þgure simpliÞes to R1 F = 2 + 16 R2 Note that the simple resistive pad matching circuit is very poor from the standpoint of noise.5 ¯ V02 ¯due to R The noise Þgure is S. 6 APPENDIX A. No.9 K Gai 10 dB = 10 30 dB = 1000 30 dB = 1000 Te2 Te3 + Ga1 Ga1 Ga2 Hence. F0 = 1 + Te0 T0 = 2. .5 + 300 3360.14 K This gives a noise Þgure of F0 = 1 + 865.9 864.68 dB (b) With ampliÞers 1 and 2 interchanged Te0 = 864.5 + 10 (10) (1000) = 386.14 290 = 3.8 K F not needed 6 dB 11 dB Tei 300 K 864.98 = 6 dB (c) See part (a) for the noise temperatures.9 + 10 (10) (1000) = 865.7 The important relationships are Fl = 1 + Tel T0 Tel = T0 (Fl − 1) Te0 = Te1 + Completion of the table gives Ampl. 1 2 3 Therefore.33 = 3. Te0 = 300 + 3360.5 K 3360. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A. out Psa. for N stages we have Foverall ≈ N F L . we have ¡ ¢ ¡ ¢ Pna.8 (a) The noise Þgure of the cascade is Foverall = F1 + F2 − 1 F −1 =L+ = LF Ga1 (1/L) (b) For two identicalattenuator-ampliÞer stages Foverall = L + F −1 L−1 F −1 + + = 2LF − 1 ≈ 2LF. in = Psa. out = 1. out = 104 = Pna.57 × 10−5 watt For part (b).38 × 10−23 (1000 + 386.29 × 10−8 watt Once again. out = 9.8) 5 × 104 = 9. out 9. and an overall gain of Ga = 107 . for the conÞguration of part (a) Pna. out 1.29 × 10−8 Psa. we have.7 (d) For B = 50 kHz.57 × 10−9 Psa. we desire Psa. out = 107 1. TS = 1000 K.57 × 10−9 watt We desire Psa.14) 5 × 104 = 1. out = 104 = Pna. out = 1.29 × 10−4 watt and Psa.29 × 10−11 watt Ga which gives which gives Problem A. L >> 1 (1/L) (1/L) L (1/L) L (1/L) (c) Generalizing.38 × 10−23 (1000 + 865. out Psa. out = Ga k (T0 + Te0 ) B ¡ ¢ ¡ ¢ = 107 1. 8 = 2.95 K Now use G1 as found in part (a): F Tes Toverall = 3.33 = 6.8 APPENDIX A.58 + Then Tes = T0 (F − 1) = 290 (1.16 ≥ 1.41) (31.62 (1.95 K and Toverall = Tes + Ta = 230.62) = 1.5 dB = 1.41) = 4. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Problem A.31 − (2.42G1 The overall noise Þgure is F = F1 + which gives 5 dB = 3.16 = 290 (3.4 = 926.31 − 1 3.54 dB = 1.31 − 1 + 31. Then F 3.58 8 dB = 6.95 + 300 = 530.16 − 1 6.16 − 1 + G1 1.16 − 1) = 626.37 dB 1.58 + or G1 ≥ 5.62.4 K = 300 + 626.8 − 1) = 230.16/1.41 30 dB = 1000 F2 − 1 F3 − 1 + G1 G1 G2 6.58 (b) First.9 The data for this problem is Stage 1 (preamp) 2 (mixer) 3 (ampliÞer) Fi 2 dB = 1.16 Gi G1 1.4 K . assume that G1 = 15 dB = 31.31 5 dB = 3. 37 dB = 4.9 (c) For G1 = 15 dB = 31.16 − 1 + + 0.33 (1.38 × 10−23 (926.9) 109 = 3. for G1 = 6.631 (0.631 (0.62 (1.58 − 1 6. (d) A transmission line with loss L = 2 dB connects the antenna to the preamp.46 × 104 1.62.58 Assume two cases for G1 : 15 dB and 6. for G1 = 15 dB = 31. Thus ¡ ¢¡ ¢ ¡ ¢ Pna.38 × 10−23 (530.6) (0.37 dB.81 × 10−10 watts Note that for the second case.16 − 1 1.33) (0.41) F1 − 1 F2 − 1 F3 − 1 + + .58 − 1 + + 0.33) (1.41) (1000) = 6. which more than compensates for the larger input noise power.631) (31.631) (31.4) 107 = 7. First. We Þrst Þnd TS for the transmission line/preamp/mixer/amp chain: FS = FTL + where GTL = 1/L = 10−2/10 = 0.84 − 1) = 534 K and Toverall = 534 + 300 = 834 K Now. GTL GTL G1 GTL G1 G2 FS = 1.6) (1.58 + = 5 This gives TS = 290 (5 − 1) = 1160 K 6. we have FS = 1.31 − 1 3.41) . Ga = 31.11 × 103 1.41) (1000) = 4.31 − 1 3.33.84 This gives TS = 290 (2. Thus ¢¡ ¢ ¡ ¢ ¡ Pna.11 × 103 .631) (4. we have 1. we get less noise power out even wth a larger Toverall .58 + = 2.33.631 and FTL = L = 102/10 = 1. out = 4.27 × 10−9 watts For G1 = 6. Ga = 4.6.46 × 104 . This is due to the lower gain of stage 1.631) (4.37 dB = 4. out = 6. 8 K 1. in = Psa.413) (300) = 426.45 × 10−6 = 0. out = Ga kTS B with the given values yields Pna.259 − 1 600 − (1.745 × 10−8 watts Ga = −51. PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS Toverall = 1160 + 300 = 1460 K Problem A. Y = 1. out = 105 Pna.11 (a) For ∆A = 1 dB.4 K 1.28 dBm Problem A.413 and the effective noise temperature is Te = For ∆A = 2 dB.413 − 1 600 − (1. out = 105 7.259 and the effective noise temperature is Te = 600 − (1.5 dB.45 × 10−6 watts (b) We want Psa.10 (a) Using Pna. out = 0. out = 7.585 − 1 Te T0 For ∆A = 1. Y = 1.585 and the effective noise temperature is Te = (b) These values can be converted to noise Þgure using F =1+ .10 and APPENDIX A.259) (300) = 858.745 watts Psa. out or This gives ¡ ¢¡ ¢ Psa.585) (300) = 212.3 K 1. Y = 1. 2 dB = 74.4 + 74.4 dB = 39.8 ratio µ Te T0 ¶ dB (d) Assuming the SNR = z = Eb /N0 = 13.4 dB = 101.6 dBW (c) µ PS Pn ¶ = −127.2 − (−138.5 dB. we get the results for various digital signaling techniques given in the table below: Modulation type BPSK DPSK Noncoh.2 dBW This gives PS = −202. F = 3.2 dBW (b) Using Pn = kTe B for the noise power.12 (a) Using the data given. we get · µ ¶ ¸ Te B Pn = 10 log10 kT0 T0 = 10 log10 [kT0 ] + 10 log10 + 10 log10 (B) µ ¶ ¡ ¢ 1000 + 10 log10 106 = −174 + 10 log10 290 = −108.938 dB.039 m = −202.8.6) = 11.4 × 10−8 1 −z = 5. (3) For ∆A = 2 dB.39 dB.14 = 13. F = 2. (2) For ∆A = 1.11 With T0 = 290 K.2 + 6 − 5 = −127.06 × 10−7 2e 1 −z/2 = 5. we can determine the following: µ λ ¶ λ 2 4πd GT PT GT = 0. FSK QPSK Error probability ¡√ ¢ Q 2z = 7. Problem A. we get the following values: (1) For ∆A = 1 dB.03 × 10−4 2e Same as BPSK .98 dB.6 dBm = −138. F = 5.

Principles of Communication 5Ed R. E. Ziemer, William H. Tranter Solutions Manual

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