Communication Systems Solutions

March 26, 2018 | Author: Naveen Mahesh | Category: Frequency Modulation, Wireless, Signal Processing, Data Transmission, Broadcasting


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ACE AcademySystems 1 Solutions to Communication CHAPTER- 2 Random Signals & Noise 01. From the property of CDF is that Fx () = 1. So, the options ‘c’ and ‘d’ can be eliminated since Fx ( ) is Zero in both of them. if CDF is a Ramp, the corresponding pdf will be given d (Ramp)= Step . But, since the dx pdf is not step, the option ‘b’ also can be eliminated. Hence, the correct option is ‘a’. 1 02. H (f)  1  J 2 π fRC & f 3db  f c  1 2  R C  H (f)  1 1  J  f fc  o p PSD  H (f) o p . i p PSD  2 Noise Power = k 2 1   f fc  k .df  kfc . 2  1   f f c    Ans: ‘c’ 03. Auto correlation is maximum at =0 i.e. R (O)  |R()| Ans :- ‘b’ 04. Power spectral density is always non negative i.e. S(f)  0 Ans:- ‘b’ 05. This corresponds to Binomial distribution. When an experiment is repeated for n times, the probability of getting the success ‘m’ times, independent of order is P(x=m) = n c m . pm . (q)n-m Where p = Prob. of success & q = 1-p In the present problem, success is getting an error. The corresponding probability is given as ‘p’. P(At most one error) = P(no errors) + P(one error) = P(X=0) + P(X=1)  n c . (p) 0 . (1  p) n  n c . (p)1 (1  p) n 1 0 = (1-p)n + np(1-p)n-1 Ans:- ‘c’ 06. The random variable y is taking two values 0 & 1. 1 2 Solutions to Communication Systems Academy P(y=1) = P (-2.5  x  2.5) P(y=0) = P (x  2.5) + P(x  -2.5) 2.5  f ( x ) dx  0.5  P (-2.5  x  2.5) =  2.5 5 P(x  2.5) =  f ( x )dx  0.25 2.5 P(x   2.5 )  2.5  f(x) dx  0.25 5  P(y = 1) = 0.5 ; P(y=0) = 0.25 + 0.25 = 0.5  f (y) = 0.5 (y) + 0.5 (y-1) Ans :- ‘b’ 07. Ans: ‘b’ 08. PSD of PSD of i p process Sxx () = 1 o p process Syy () = | H ()|2 = H()  SYY () 16  SXX () 16  ω 2 4 16  ω 2 16 16  ω 2  H( )  4 4Jω R We have H() for an RL – Low Pass Filter as H() = R  JL  Ans :- (a) 09. R = 4 ; L = 4H Ans :- ‘a’ 10. Noise Power = ( o p ) PSD  B. H () = 2 . exp (-Jtd) | H () | 2 = 4  o p Noise PSD = 4NO  o p Noise Power = 4NO B o p Ans :- ‘b’  k  r for 0  k  4  4 11. P( r )   = 0 elsewhere 4 Since  P( r ).d r  1  k  0 1 2 4 Mean Square Value is r 0 2 . P( r ). d r  8 ACE ACE Academy Engineering Electronics & Communication 3 Ans :- ‘c’ 12. |H(f)|2 = 1 + (0.1  10-3)f for -10 KHz  f  0 = 1  (0.1  10-3)f for 0  f  10 KHz 2 ( o p) PSD = H (f )  i p PSD Power of o p Process = 10103  10103 (o p) PSD. df  1 10 6  Ans:- ‘b’ 13. R ()  FT  PSD  Sxx  ω  Since PSD is sinc – squared function, its inverse Fourier Transform is a Triangular pulse. Ans:- ‘b’ 14. Var [d(n)] = E[d2(n)]  {E[d(n)]}2 E[d(n)] = E[x(n)  x(n1)] = E[x(n)]  E[x(n1)] = 0 Var[d(n)] = E[d2(n)] = E[{x(n)  x(n1)}2] = E[x2(n)] + E[x2(n1)]  2.E[x(n).x(n1)] =  2x +  2x  2.Rxx (1)  2  2x – 2Rxx(1) =  1  10 2 x R xx (k ) at k = 1 = 0.95  2x Ans: ‘a’   x  4 2  exp    18     x  4 2  1 exp =    29  2  9  1 P  X  4 = PX ( x ) x 4 = 3 2 1 3 2 15. PX(x) = Ans: b 16. P(at most one bit error) = P(No error) + P(one error) = n C . (P)0 (1-P)n-0 + n C (P)1 (1-P)n-1 = (1-P)n + n P(1-P)n-1 0 Ans: d 1 4 Solutions to Communication Systems Academy 17. g(t) a . g(t) = g1(t) a  H( ) ACE 2 = a  PSD of g1(t) = a .Sg ()   2 Rg1 ( ) = F 1 a .Sg () = a 2 . Rg    power of Rg1(  ) = a 2 . Rg  0  = a 2 . Pg Ans: a 18. The fourier Transform of a Gaussian Pulse is also Gaussian. Ans: ‘c’ 19. The Auto correlation Function (ACF) of a rectangular Pulse of duration T is a Triangular Pulse of duration 2T Ans: ‘d’ 20. The Prob. density function of the envelope of Narrow band Gaussian noise is Rayleigh Ans: ‘c’ 21. P(x) = K. exp (- x 2 2) , -   x        P ( x ) . dx = 1   k. exp(  x 1 2 We have 1 2 e x2 2  e x2 2 2) dx  1 .dx = 1, since 2  is the Normal density N (m,  2 ) = N (0,1) k  1 2 Ans: ‘a’ 22. F-1  PSD  =Auto correlation Function R( )    sin f     f  R( ) = F-1    2   , which is a triangular pulse. Ans: ‘d’ 23. R( ) =R(- )  Even symmetry Ans : ‘d’ 24. Rayleigh ACE Academy Engineering Electronics & Communication 5 Ans : ‘d’ 25. R2 (TK) R1 (TK) The Noise equivalent circuit is R1  R2 = 4R2KT2B = 4R1KT1B  (R1 + R2) = 4(R1T1+R2T2) KB  = 4RKTB R  RT = R1T1 +R2T2 R 1T1  R 2 T2 R1  R 2  T= 3 26. E(X) =  x. P( x )dx  1 1 3 E(X2) =  x. P( x )dx  7 / 3 1 Var (X) = E(X2) – [E(X)]2 = 7 4 1  3 3 Ans: ‘b’ 27. Half wave rectification is Y = X for x  0 = 0 elsewhere 1 f(y) = δ (y)  2  y 1 e 2π N 2 2N E(Y) = 0 & E(Y2) = N Ans: ‘d’ 28. P(X = at most one error) = P(X = 0) + P(X = 1) 6 Solutions to Communication Systems Academy = 8C 0 . (P)0 (1-P)8 + 8C 1 . (P)1 (1-P)8-1 = (1P)8 + 8P (1P)7 Ans: ‘b’ 29. Var [(kx)] = E[( kx)2]  E(kx) 2 = k2 E (x2)  [ k. E (x)]2 = k2 E (x2)  k2. [E (x)]2 = k2 [E (x2)  E(x) 2 ] = k2 . x2 Ans: ‘d’ CHAPTER – 3 Objective Questions Set – A 01. (B.W)AM = 2 ( Highest of the Baseband frequency available) = 2(20 KHZ) = 40 KHZ 02. Percentage Power saving = PT  PTX 100 % PT 2 = 2  m 2 100 % For m = 1 , Power saving =  03. PT = PC  1   m2 2 2 100 % = 66.66 % 3    For m = 0 ; P T = PC For m = 1 ; PT = 1.5 PC  TX. Power increased by 50% ACE ACE Academy Engineering Electronics & Communication 7 04. mT =  06. m= 07. The given AM signal is of the form [A + m(t)] cos c t, which is an AM-DSB-FC signal. It can be better detected by the simplest detector i.e. Diode Detector 08. MW/Broadcast band is 550 KHz – 1650 KHz. 09. Hence the received 1 MHz signal lies outside the MW band. 10. Q= 12. m2 PT = PC + PC 2 m12  m 22 Vmax  Vmin Vmax  Vmin (0.3) 2  (0.4) 2 = 0.5 1 2 f0 110 6 = 3 =100 BW 10 10 Pc . m 2 2 =  Pc (0.4) 2 2 = 0.08 Pc  PT = 1.08Pc  Increase in Power is 8%. 14. em(t) = 10(1+0.4 cos 10 3 t + 0.3 cos 104 t) cos ( 106t ) This is a multi Tone AM signal with m1=0.4 and m2=0.3  m= 15. m12  m 22 =0.5 Image freq(fi) = fs +2 IF  fs = fi – 2 IF = 2100 – 900 = 1200 KHz. 16. Same as Prob. 2 18. Same as 3 19. m2 PSB = 75 + 75 = 150 = PC 2 and Pc=PT - PSB = 600 – 150 = 450  m2 PC 2 = 450  m 2 2/3 2 =150  m= 20. Pc = 450  22. BW of each AM station = 10 KHZ. No. of stations = 100 10 3 =10 10 10 3 8 Solutions to Communication Systems Academy ACE E m 15  m=25% = E c 60 25. m= 26. (B.W)AM = 2 27. Message B.W = Band limiting freq. of the baseband signal = 10 KHz. 28. B.W = 2(10 KHz) = 20 KHz. 29. The various freq. in o/p are 1000 KHz, (1000  1) KHz & (1000  10) KHz.  1500 = 3 KHz.  The freq. which will not be present in the spectrum is 2 MHz. 30. Highest freq. = USB w.r.t highest baseband freq. available = (1000 + 10) KHz = 1010 KHz CHAPTER – 3 Objective Questions – SET C 5. A freq. tripler makes the freq. deviation, three times the original. f  New Modulation Index = 3. f = 3 mf m 6. Mixer will not change the deviation. Thus, deviation at the o/p of the mixer is  . 20. B.W1 = 2(  f + 10 KHz) B.W2=2(  f + 20 KHz)  B.W increases by 20 KHz. 29. In NBFM, Modulation Index is always less than 1. CHAPTER – 3 Additional objective questions – SET D 1. Amplitude of each sideband = = m Ec 2 0.3  10 3 2 = 150v Ans: ‘b’ ACE Academy Engineering 2 Ec = 1 KV  Electronics & Communication 9 m E c 1000 m = =200 2 2  m = 0.4 Ans: 3. ‘c’ Pc = 1 KW; PSB = PC = 0.5 KW 2  PT = PC + PSB = 1.5 KW. Ans: 4. As per FCC regulations, in AM, (fm)max = 5 KHz Ans: 5. ‘b’ Ec + Em = 130  Em = 130 – 100 = 30 V m= Ans: 6. ‘b’ Em 30 = = 0.3 Ec 100 ‘b’ V(t) = A[1 + m sin  m t ] sin c t By comparing the given with above V(t), the unmodulated carrier peak A = 20  rms value = 20/ 2 Ans : b’ 7. Side band peak = mE c 0.5  20 = =5 2 2 Rms value = 5/ 2 Ans: 8. m = 0.5  50% Modulation Ans: 09. a’ b’ V = A[1+msin  m t ] sin c t  m =6280 Ans: 10. c’ c =6.28  106 10 Solutions to Communication Systems Academy ACE Ans : ‘a’ 11. m  1 results in over Modulation, causing distortion . Ans : ‘d’ 12. Ans: ‘b’ 13. EC + Em = 2Ec  Em = Ec m = Ans: 14. Em = 100% Ec ‘d’ Ec + Em = 110 Ec - Em = 90  Ec = 100V; Em = 10V Ans: 15. Using the above results, m = Ans: 16. Em 10 = = 0.1 Ec 100 ‘a’ using the above results, the sideband amplitude is Ans: 17. ‘c’ m= ‘b’ Em  Em = m.Ec Ec The carrier peak is (100) 2  Em = (0.2)(100) 2 = 20 2  Ec + Em = (120) 2 The corresponding rms value = 120 V Ans: 20. It = I c ‘d’ 1 m2 2 mE c 0.1100 = = 5V 2 2 ACE Academy Engineering Electronics & Communication 11 Ic = 10 Amp; It = 10.4 Amp.  m = 0.4  Ans: b 21. m= (0.3) 2  (0.4) 2 = 0.5  Modulation Index = 50% Ans: 23. Pc = PT - PSB = 1160 – 160 = 1000 Watts Ans: 24. ‘a’ m= ‘a’ I max  I min 6 = = 0.3 I max  I min 20  Percent Modulation = 30% Ans: 27. ‘b’ To implement Envelope detection, Tc < RC < Tm Tc = 1  sec; Tm = 0.5 msec = 500  sec Since Tc < RC < Tm  RC = 20  sec. Ans: 28. As per FCC regulations in FM, (fm)max = 15 KHz Ans: 29. ‘b’ ‘c’ In FM, (  f)  Em  if Em is doubled, δf also gets doubled Ans: 30. ‘a’ If FM, (δf) is independent of Base Band signal frequency. Thus, δf remains unaltered. Ans: ‘d’ 31 Ans: ‘d’ 32. frequency doubler doubles the freq. deviation. Thus at the o/p of the doubler, the modulation index is 2.mf 12 Solutions to Communication Systems Academy Ans: 33. mf = Ans: 38. ‘b’ δf = (fc)max  fc = 210  200 = 10 KHZ Ans: 37. ‘a’ Mixer will not change the freq. deviation. Thus freq. deviation at the o/p of Mixer is δ Ans: 35. ACE ‘b’ δf 5 KHz  10 = fm 500 Hz ‘a’ δf  Em  E m1 δ f1  δf 2 E m2  δf 2  (f1 )(E m 2 ) (E m1 )   5 KHZ10 V   2.5 V   20 KHz 39. f 2 20  10 3   40 m= fm 500 40. δf2 = Ans: 41. E m1  5  20  50 KHz 2 ‘b’ Assuming the signal to be an FM signal, the Power of the Modulated signal is same as that of un Modulated carrier. Ans: 43.  δf1   E m2  ‘a’  FM  t  = A cos (ct + mf . Sin mt)  c = 6.28  108 Ans: 44. ‘a’ m = 628 Hz Ans: f ‘a’ 45. mf = f  f m  4 f m = 25/2 Hz ACE Academy Engineering Ans: 46. Electronics & Communication 13 ‘c’ 3 2 Figure of Merit in FM is  = m f , where mf is the Modulation Index. 2  Noise Performance increases with increase in freq. deviation. Ans: 47. 1 In FM, Modulation Index  f m Ans: 48. ‘c’ B W = 2nfm = 2(8) (15 KHz) = 240 KHz Ans: 54. ‘d’ B.W = 2 ( f + fm ) = 2 (75 + 15) =180 KHz Ans: 53. ‘a’ In FM, o/p Power is independent of modulation Index. Ans: 52. ‘a’ ‘d’ B. W = 2nfm & n = mf + 1 = 8  2(8) (fm) = 160  103  fm = 10 KHz  f (mf) (fm) = (7) (10) KHz = 70 KHz Ans: 55. ‘c’ B.W = 2nfm The modulation Index mf = δf 10 6   100 fm 10  103  n = 100 + 1 = 101  B.W = 2(101) (10  103) = 2.02 MHz Ans: 56. ‘b’ If Em gets doubled, f also get doubled. 14 Solutions to Communication Systems Academy δf 2 10 6   200 fm 10 103  mf = n = 201 B.W = 2(201) (10  103) = 4.02 MHz Ans: 58. For WBFM, B.W = 2(f + fm). Ans: 59. ‘c’ At the o/p of the mixer, ‘’ remains the same. Ans: 67. ‘d’ Since (f) is independent of carrier freq.  the peak deviations are same. Ans: 66. ‘b’ In WBFM, f  fm  B.W  2 f Ans: 63. ‘d’ For NBFM, B.W = 2 fm Ans: 60. ‘d’ ‘d’ i ( t ) = 50t + sin 5t i = d  i ( t ) = 50 + 5 cos 5t dt  At t = 0, i = 55 rad /sec Ans: 75. ‘c’ IF = 455 KHz; fs = 1200 KHz.  Image freq. = fs + 2 IF = 2110 KHz 76. Ans: Refer Q. No. 26 Set–F 77. fi = fs + 2 IF = 1000 + 2(455) ACE ACE Academy Engineering Electronics & Communication 15 = 1910 KHz Ans: 78. ‘d’ fi = fs + 2 IF = 1500 + 2(455) = 2410 KHz Ans: 82. ‘d’ fi = fs + 2 IF = 500 + 2 (465) = 1430 KHz Ans; ‘b’ Chapter – 3 Additional objective Questions  Set E 01. By comparing with the general AM  DSB  FC signal Ac . cos ct + m(t) . cos ct, it is found that m(t) = 2 cos mt. To demodulate using Envelope detector, Ac  mp, where mp is the Peak of the baseband signal, which is 2.  (Ac)min = 2 Ans: 02. ‘a’  FM (t) = 10 cos 2  105t + 5 sin (2  1500t) + 7.5 sin (2  1000t)] i (t) = 2  105t + 5 sin (2  1500)t + 7.5 sin (2  1000)t] i = d i(t) = 2  105 + 5(2  1500) cos (2  1500t) + 7.5(2  1000) cos (2  dt 1000t)  = 5(2  1500) + 7.5(2  1000) f = 7500 + 7500 = 15000 Hz Fm = 1500 Hz `  Modulation Index = Ans: 03. δf  10 fm ‘b’  (t) = cos ct + 0.5 cos mt . sinct 16 Solutions to Communication Systems Academy Let ACE r(t). cos (t) = 1 r(t). sin (t) = 0.5 cosmt  (t) = r(t). cos ct. cos (t) + r(t). sin (t). sin ct = r(t). cos [ct  (t)] Where r(t) = 1  (0.5 cosω m t) 2 = [1 + 0.25 cos2 mt]1/2 = [1 + 0.25 1  cos2ω m t 2  1/2 = [1.125 + 0.125 cos2mt]1/2  1.125 +  0.125 cos2mt 2  (t) = [1.125 + 0.0625 cos2mt] cos[ct  (t)] Hence it is both FM and AM Ans: 04. To avoid diagonal clipping, Rc  Ans: 05. ‘c’ 1  ‘a’ The LSB  Modulated signal f c1  fm = 990 KHZ Considering this as the Baseband signal, the B. of resulting FM signal is 2(990 103) = 1.98 MHz  2 MHz Ans: ‘b’ 1 06. P(t) = and 0 1 g (t) = t 0 1 2 XAM (t) = 100 [P (t) + 0.5 g(t)] cosct for 0  t  1 By Comparing the above with an AM  DSB  FC signal under arbitrary Modulation ACE Academy Engineering Electronics & Communication 17 i.e. A [ 1 +  . m(t) ] cos ct  = 0.5 & m(t) = g(t) is a ramp over 0  t  1  one set of Possible values of modulating signal and Modulation Index would be t, 0.5 Ans: 07. ‘a’ XAM (t) = 10 [ 1 + 0.5 sin2fmt ] cos2fct The above signal is a Tone Modulated signal. Pc m 2 2 The AM Side band Power =  100  0.5 2  2 2 = 6.25  Ans: 08. ‘c’ Mean Noise Power is the area enclosed by noise PSD Curve, and is equal to N   1  B  0  = N0 B 2   2 4  The ratio of Ave. sideband Power to Mean noise Power = Ans: 10. ‘b’ y(t) = x2 (t) A squaring circuit acts as a frequency doubler  New f = 180 KHZ  B.W of o/p signal = 2(180 + 5) = 370 KHZ Ans: 11. ‘a’ ()PM = Kf Em Wm, Where Kf Em is the Phase deviation. Since, it is given that Phase deviation remains unchanged, ( )PM  m  δ ω1 ω m1  δ ω2 ω m2 6.25 25  N0B 4N 0 B 18 Solutions to Communication Systems Academy  δ f1 f m1  δ f2 f m2  10 KHZ 1 KHZ    f2 = 20 KHZ  2 2 KHZ ACE  B. = 2 ( f2 + fm2) = 2 (20 + 2 ) KHZ = 44 KHZ Ans: 13. ‘d’ Power efficiency  = PSB  100  PT The sidebands are m(t). cos ct  1  1 cosω1 t  sinω 2 t  cosc t =  2  2  1  cos ω c ω1  t  cos  ω c ω1  t  + 1  sin  c 2  t  sin  c  2  t  4 4  1 2 PSB = 4  1 4    1 8  2  1 1  PT = PC + PSB = 2 8 = 18   = 5 8 100 0 0  20 0 0 Ans: 14. ‘c’  C1 = B log  1   Since S  bps N S  1 N C1 = B log S N C2 = B log (2. S N ) = B log 2 + Blog S N = B + C1  C2 = C 1 + B Ans: 15. ‘b’ Tc  RC  Tm  1  sec  RC  500  sec  RC = 20 sec ACE Academy Engineering Ans; 16. Electronics & Communication 19 ‘b’  AM (t) = A cosct + 0.1 cosmt. cosct = A cosct + 0.05 [cos(c+ m)t + cos(c  m)t] NBFM is similar to AM signal, except for a Phase reversal of 1800 for LSB  NBFM (t) = Acosct + 0.05 [cos (c + m)t  cos (c  m)t]   AM (t) +  NBFM (t) = 2A cosct + cos(c + m)t This is SSB with carrier. Ans: ‘b’ 17. Noise Power = 1020  100 106 = 1012  Loss = 40 dB  loss = 104 10 3 Signal Power at the receiver =  10  7 ω 10 4 S 10 7  10 log = 10 log = 10 log105 N 10 12 = 50 db Ans: ‘a’ 18. Carrier = cos 2 (101  106)t Modulating signal = cos 2 (106)t o/p of BM = 0.5 [cos 2(101  106)t + cos 2 (99  106)t] o/p of HPF = 0.5 cos2(101  106)t o/p of Adder is = 0.5 cos 2(101  106)t + sin 2(100  106)t = 0.5 cos2 [(100 + 1)  106]t + sin 2(100  106)t = 0.5 [cos 2(100  106)t. cos2  106t  sin 2 (100  106)t.sin2106t] + sin2(100  106)t = 0.5 cos 2(100  106)t. cos2  106t  sin 2 (100  106)t [1 0.5 sin(2 106 )t] Let. 0.5 cos(2  106)t = R(t). sin(t) 1 0.5 sin(2  106)t = R(t).cos(t) The envelope R(t) = {[0.5 cos(2106)t]2 + [1 0.5 sin(2106)t]2}1/2 20 Solutions to Communication Systems Academy ACE = [1.25  sin(2  106)t]1/2 12  5   sin (2π 10 6 )t  =   4  Ans: 19. ‘b’ A frequency detector produces a d.c voltage (constant) depending on the difference of the two i/p frequencies. Ans: ‘d’ 20. Ans: ‘c’ 21. o/p of Balanced Modulator is  13  11  10  9 7 0 7 9 10 11 13 f(KHz) o/p of HPF is  13  11  10 11 10 f(KHz) 13 The freq. at the o/p of 2nd BM are 0 2 3 23 24 26 f(KHz)  The +ve frequencies where Y(f) has spectral peaks are 2 KHZ & 24 KHZ Ans: 22. ‘b’ 1 V0 = a0 [Ac .cos(2fc1t) + m(t)] + a1 [Ac = a0 [Ac 1 cos(2fc1t) + m(t)]3 cos(2fc1t) + m(t)] + a1[(Ac1)3 cos3(2fc1t) + m 3(t) + 3 (Ac1)2 cos2 (2fc1t). m (t) + 3 (Ac1). Cos (2fc1t). m2 (t)] The DSB  Sc Components are 2 fc1 ± fm 1 ACE Academy Engineering Electronics & Communication 21 These should be equal to fc ± fm  2fc1 = fc  fc1 = f c 2 = 0.5 MHZ Ans: 23. ‘c’ 2 Pc m Total side band Power 2  carrier power Pc  Ans: 24. m2  1 8 2 ‘d’ fm = 2KHZ; fc = 106 HZ f = 3(2fm) = 12 KHZ Modulation index  =   FM (t) =  A.J n  n δf 6 fm () cos (ω c nω m ) t  =  5. Jn (6) cos {2 [{1000 + n(2)}103] t} n   the coefficient of cos 2 (1008  10 )t is 5. J 3 Ans: 25. P  6 ; Q  3; R  2; S  4 Ans: 26. ‘d’ ‘a’ f0 = fs + IF (f0) max = (fs)max + IF = 1650 + 450 = 2100 (f0) min = (fs)min + IF = 1650  450 = 1200 1  2100 (f0) max = 2π Lc min 1  1200 (f0) min = 2π Lc max 4 (6) 22 Solutions to Communication Systems Academy c max 2100  7 4 c min 1200   ACE c max =3 c min Image freq. = fs + 2 IF = 700 + 2 (450) = 1600 KHZ Ans: 27. ‘c’ Let the i/p signal be cosct. cosm t + n (t) = cosct. cosmt + nc(t) cosct  ns (t). sinc t = [nc(t) + cosmt] cosct  ns (t). sinct When this is multiplied with local carrier, the o/p of the multiplier is [nc (t) + cosmt ] cos2ct  n s (t) . sin2ct 2 n (t)  1 cos2ω c t   s sin2ω c t  2 2   = [nc(t) + cosmt]  The o/p of Base band filter is 1 [nc(t) + cosmt] 2 Thus, the noise at the detector o/p is nc(t) which is the inphase component. Ans: ‘a’ 28. The o/p noise in an Fm detector varies parabolically with frequency. 29. Ans: ‘a’ 30. m(t) 100 sec t ACE Academy Engineering Electronics & Communication 23 1 fm = 100  10 6  10 KHZ Its Fourier series representation is 4  [cos2 (10  103)t  1 1 cos2(30  103)t  cos2 (50  103) t + -----] 3 5 The frequency components present in the o/p are fc ± 10KHZ = (1000 ± 10) KHZ fc ± 30 KHZ = (1000 ± 30) KHZ ------i.e. 970 KHZ , 990KHZ, 1010KHZ, 1030 KHZ -----etc. Hence, among the frequencies given, the frequency that is not present in the modulated signal is 1020 KHZ Ans: 31. ‘c’ S(t) = cos 2 (2  106t + 30 sin 150 t + 40 cos 150t) i (t) = 2 (2  106 t + 30 sin 150t + 40 cos150t)  Phase change = 2 [30 sin150t + 40 cos150t] Let r cos = 30 ; r sin = 40  Phase Change = 2 r cos (150t - ) Where r = (30) 2  (40) 2  50  Phase change = 100 .cos (150t  ).  Max Phase deviation = 100 i = d i (t) = 2 [2  l06 + (30)(150) cos(150t)  (40) (150) sin 150t] dt Frequency change = 2 [(30)(150)cos150t  (40)(150)sin150t] This can be written as (2) (150) r. cos(150 t + ), Frequency change = (2)(150)(50) cos(150t + ) Max frequency deviation  = 2 (150)(50)  f = (150) (50) = 7.5 KHz Ans: 32. ‘d’ LPF can be used as reconstruction filter. Ans: ‘d’ Where r = 50 24 Solutions to Communication Systems Academy 33. The envelope of an AM is the baseband signal. Thus, the o/p of the envelope detector is the base band signal Ans: 34. ACE ‘a’ 2(f + fm) = 106 HZ  f = 495 KHZ For y(t), f = 3(495 KHZ ) = 1485KHZ and fc = 300 MHZ  B. of y(t) = 2 (1485 + 5) KHZ = 2980 KHZ = 2.9 MHZ  3 MHZ adjacent frequency components in FM signal will be separated by fm = 5 KHz. Ans: 35. ‘a’ o/p of multiplier = m(t) cos0t .cos(0t + ) = m(t)  cos(2ω0 t  θ)  cosθ 2 o/p of LPF = m(t) . cosθ 2 Power of o/p = m 2 (t) . cos 2θ 4 Since, m 2 ( t ) = Pm, the Power of output signal is Ans: Pm . cos 2 θ . 4 ‘d’ 36. ‘a’ 37. ‘a’ 38. The frequency components available in S(t) are (fc  15) KHZ, (fc  10) KHZ, (fc + 10) KHZ, (fc + 15) KHZ.  B. = (fc + 15) KHZ  (fc  15) KHZ = 30 KHZ. Ans: ‘d’ ACE Academy Engineering 39. Electronics & Communication 25 Complex envelope or pre envelope is S(t) + J . Sh(t), Where S(t) is the Hilbert Transform of S(t). Let S(t) = eat . cos (c + )t.  Sh(t) = eat . sin (c + )t  pre envelope = eat. [cos (c + )t + J sin (c + )t] = eat . exp [J(c + )t] Ans: 40. ‘a’ To Provide better Image frequency rejection for a superheterodyne receiver, image frequency should be prevented from reaching the mixer, by providing more tuning circuits in between Antenna and the mixer, and increasing their selectivity against image frequency. There circuits are preselector and RF amplifier. Ans: ‘d’ 41. Ans: ‘a’ 42. Ans: ‘b 43. New deviation is 3 times the signal. So, Modulation Index of the output signal is 3(9) = 27 Ans: ‘d’ 44. Ans: ‘b’ 45. Ans: ‘c’ 46. a2;b1;c5 47. a2; b1;c5 48.  (t) = 5 [cos ( = 5 cos 106(t)  106  t)  sin (103 t) sin 106t] 5 [2sin 103t. sin 106t ] 2 = 5 cos 106 t  5 [cos(106  103)t  cos(106 +103)t 2 = 5.cos 106 t + 5 5 cos (106 +103)t  cos (106  103)t. 2 2 It is a narrow band FM signal, where the phase of LSB is 1800 out of phase with that of AM. Ans: d 26 Solutions to Communication Systems Academy ACE 49. B. = 2 (50 + 0.5) KHZ = 101 KHZ 50. a3;b1; c2 51. The given signal is AM  DSB  FC, which will be demodulated by envelope detector. Ans: 52. ‘a’ Image frequency = fs + 2 IF = 1200 KHZ + 2(455) = 2110 KHZ 53. Power efficiency = Puseful  100 % PT = m2  100% 2  m2 For m = 1, the Power efficiency is max. and is 33.3 % 54. Picture  AM  VSB Speech  FM Ans: 55. ‘c’ For the generated DSB  Sc signal, Lower frequency Limit fL = (4000  2) MHZ = 3998 MHZ and Upper frequency Limit fH = (4000 + 2) MHZ = 4002 MHZ. (fs)min = 2 fH = 8.004 GHZ Ans: ‘d’ 56. Ans: ‘a’ 57. mf = δf K E where f = f m fm 2π  f = 10 10 3  2 10 10 3  2π π ACE Academy Engineering Electronics & Communication 27 m = 104    fm =  mf = Ans: 2 10 4 2 π ‘d’ 58. Te = 210 K Loss = 3 db g1 = 13 db Noise fig. of amp. F1 = 1 + = 1+ T0 = 3000 K Te T0 21 300 = 1.07 For a Lossy Network, Boise Figure is same as its loss.  f2 = 3 db  f2 = 1.995  Overall Noise figure f = f1 + g1 = 13db  g1 = 19.95  f = 1.07 + 1.995  1 = 1.1198 19.95  f = 0.49 db Te of cable = (f  1) T0 = (1.995  1) 300 = 298.50 K Overall Te = Te 1 + = 21 + Te 2 g1 298.5 19.95 = 35.960 K Ans: ‘c’ f 2 1 g1 28 Solutions to Communication Systems Academy 60. 61. ACE A preamplifier is of very large gain. This will improve the noise figure (i.e. reduces its numerical value) of the receiver, if placed on the antenna side Ans: ‘a’ Ans: ‘a’ Chapter  4 01. A source transmitting ‘n’ messages will have its maximum entropy, if all the messages are equiprobable and the maximum entropy is logn bits/message. Thus, Entropy increases as logn. Ans: ‘a’ 02. This corresponds to Binomial distribution. Let the success be that the transmitted bit will be received in error. P(X = error) = P(getting zero no. of ones) + P(getting one of ones) = P(X = 0) + P(X = 1) 0 3 2 = 3c 0 (1  p) p  3c1 (1  p) p = p3 + 3p2(1 – p) Ans: ‘a’ 03. 04. Most efficient source encoding is Huffman encoding. 0.5 0 0.5 0 0.25 10 0.5 1 0.25 11 L = 1  0.5 + 2  0.25 + 2  0.25 = 1.5 bits/symbol Ave. bit rate = 1.5  3000 = 4500 bits/sec Ans: ‘b’ Considering all the intensity levels are equiprobable, entropy of each pixel = log2 64 = 6 bits/pixel There are 625  400  400 = 100  106 pixels/sec  Data rate = 6  100  106 bps = 600 Mbps Ans: 05. ‘c’ Source coding is a way of transmitting information with less number of bits without information loss. This results in conservation of transmitted power. Ans. ‘c’ ACE Academy Engineering Electronics & Communication 29 06. Entropy of the given source is H(x) = - 0.8 log 0.8 – 0.2 log 0.2 = 0.722 bits/symbol 4th order extension of the source will have an entropy of 4.H(x) = 2.888 bits/4 symbol As per shanon’s Theoram, H(x)  L  H(x) + 1 i.e., 2.888  L  3.888 bits/4 messges 07. 12  512  log 82 = 18432 bits 08. Code efficiency =  = L min H  100%   100% L L L = 2 bits/symbol and the entropy of the source is 1 1 1 1 2 1 log  log  log 2 2 4 4 8 8 14 = bits/symbol 8 14  100% = 87.5% = 16 H=  Ans : ‘b’ 09. H(X) =  1 1 1 1 2 1 log  log  log 2 2 4 4 8 8 = 1.75 bits/symbol 10.  S  Channel Capacity C = B log 2  1   B    S B S = 30 db   B = 1000  C = 3  103 log2 (1 + 1000) = 29904.6 bits/sec For errorless transmission, information rate of source R < C. Since, 32 symbols are there the number of bits required for encoding each = log2 32 = 5 bits  29904.6 bits/sec constitute 5980 symbols/sec. So, Maximum amount of information should be transmitted through the channel, satisfying the constraint R < C  R = 5000 symbols/sec Ans: ‘c’ 11. Not included in the syllabus 12. H(x) = log2 16 = 4 bits Ans: ‘d’ 30 Solutions to Communication Systems Academy 13. P(0/1) = 0.5 P(1/0) = 0.5 P(Y/X) =   ACE P(0/0) = 0.5 P(1/1) = 0.5 1 1  2 2 1 1   2 2 A channel with such noise matrix is called the channel with independent input and o/p. Such a channel conveys no information.  its capacity = 0 Ans: ‘d’ 14. A ternary source will have a maximum entropy of log2 3 = 1.58 bits/message. The entropy is maximum if all the messages are equiprobable i.e. 1/3 Ans: ‘a’ 15. Ans: ‘b’ 16. Entropy coding – McMillan’s rule Channel capacity – Shanon’s Law Minimum length code – Shanon Fano Equivocation – Redundancy 17. Ans: ‘c’ Since S << 1 N C  B log 1  0  C is nearly o bps Ans: ‘d’ 18. Ans: ‘b’ 19. Ave. information = log2 26 = 4.7 bits/symbol Ans: ‘d’ 20. Ans: ‘d’ 21. Ans: ‘b’ ACE Academy Engineering Electronics & Communication 31 22. Ans: ‘b’ 23. H1 = log2 4 = 2 bits/symbol H2 = log2 6 = 2.5 bits/symbol H1 < H2 24. Ans: ‘a’ The maximum entropy of binary source is 1 bit/message. The maximum entropy of a quaternary source is 2 bits/message. The maximum entropy of an octal source is 3 bits/message. Since the existing entropy is 2.7 b/symbol the given source can be an octal source Ans: ‘c’ Chapter – 5A 01. Set A (fs)min = 4 KHz  (Ts)max = Ans: 1 1   250  sec (f s ) min 4 KHz ‘c’ Set B 05. In PCM, (B.W)min =  fs Hz 2 If Q = 4   = 2  (B.W)min = fs Hz. If Q = 64   = 6 (B.W)min = 3fs Ans: ‘a’ 18. (f s )min = 8 KHz;  = log2 128 = 7  fs  28 KHz B.W = 2 Ans: ‘d’ Set – C 01. Maximum slope = S fs = 02. d d m( t )  (at )  a dt dt 75 10 3 = 50 V/sec 1.5 10 3 Ans: ‘a’ Rate of rise of the modulator = .fs = /Ts  Slope over loading will occur if  fs < a  T  a   < a Ts s Ans: ‘c’ 32 Solutions to Communication Systems Academy ACE 03. Ans: 04. Since with increasing ‘n’ (increased number of Q levels), Nq reduces, S/Nq increases. For every 1 bit increase in ‘n’. Nq S/Nq improves by a factor of 4. 05. ‘c’ Ans: ‘d’ o/p bit rate =  fs, where  = log2 258 = 8   fs = 64 kbps Ans: ‘c’ 06. 07. Ans: 08. (Q. E)max = S/2 = = ‘c’ VH  VL 2Q 1 of the total peak to peak range 264 Ans: ‘c’ 09. Ans: ‘b’ 10. For every one bit increase in the data word length, S/Nq improves by a 6 db.  The total increase is 21 db Ans: ‘b’ 11. Number of samples from the multiplexing system = 4  2  4 KHz = 32 KHz Each sample is encoded into log2 256 = 8 bits So, the bit transmission rate = 32  8 kbps = 256 kbps Ans: ‘c’ 12. fs = 10 KHz;  = log2 64 = 6 Transmission Rate = 60 kbps Ans: ‘a’ 13. VP-P = 2 V;  = 8  Q = 256 Q S/Nq = (1.76 + 20 log 10 ) db = 49.9 db Ans: ‘b’ 14. (fs)Multiplexed system = 200 + 400 + 800 + 200 = 1600 Hz Ans: ‘a’ ACE Academy Engineering Electronics & Communication 33 15. Each sample is represented by 7 + 1 = 8 bits. Total bit rate = 8  20  8000 = 1280 kbps Ans: ‘b’ 16. ‘a’ (Question number 5 in set B) Set – D 01. The power spectrum of Bipolar pulses is PSD fb = 1/Tb 2/Tb f (B.W)min required = fb Here  = 8; fs = 8 KHz  Bit rate = 64 kbps  (B.W)min = 64 KHz Ans: ‘a’ 5 02. Signal power = x 2 f ( x ).dx 5 f(x) = 1 10 -5x5 =0 elsewhere  Signal Power = 25/3 watts. Quantization Noise Power Step size =  Nq = Nq = s2 12 VP  P 10 10  8   0.039 V Q 2 256 (Step size) 2 = 0.126 mW 12 S 10 log N = 48 db q Ans: ‘c’ 03. For every one bit increase in the data word length, Nq reduces by a factor of H. Given  = 8  Required  = 9  Number of Q  levels = 29 = 512 34 Solutions to Communication Systems Academy ACE Ans: ‘b’ 04. Ans: ‘d’ 05. Since, entropy of the o/p of the quantizer is to be maximized, it implies that all the decision boundaries are equiprobable. 1   f(x). dx  13 5 1   b . dx  1 3  b  112 5 1 Similarly  a. dx  1 1 Ans: 06. 3  a 1 6 ‘a’ Reconstruction levels are  3V, 0V and 3V. Step size = 3V  Nq = 9  3 4 12 1  1  2 1  dx   x   dx 1  12   6 1 2 Signal Power = 2.  x  5 1  6   x3       3   1 5  x3     3   1  1  1  124 2 126 21      6 3 3 18 3 21 4 28    SN  q 3 3 9 = 07. g(t) is Periodic with period of 104 sec i.e. 0 0.5104 2(0.5104) 3(0.5104) …. t In its Fourier series representation, a0 = 0. The remaining frequency components will be fs = 10 KHZ; 2fs = 20 KHZ; 3fs = 30 KHz ….etc.  The frequency components in the sampled signal are 10 KHz  500 Hz; 20 KHz  500 Hz ….etc. ACE Academy Engineering Electronics & Communication 35 When the sampled signal is passed through an ideal LPF with Band width of 1 KHz, The o/p of the LPF will be zero. Ans: 08. ‘c’ x(t) = x1(t) + x2(t) Since sin at   F.T  . G 2 a () πt sin 2π1000t πt   2 (1000)   F.T  2(1000)  5  sin 2π 1000t   πt    x1(t) = 5  3  F .T   6(1000) 6(1000)  7  sin 2π 1000t  x2(t) = 7   πt   2  F . T  Thus, x1(t) + x2(t)  F .T   4(1000) 6(1000) 4(1000) 4(1000) 4(1000) 6(1000)  m = 6(1000)  fm = 3 KHz  (fs)min = 6 KHz Ans: 09. ‘c’ 125 x(t) = 0 1 2 To Track the signal, rate of rise of Delta Modulator and of the signal should be same, i.e.  Sfs = 125 125 S = 32 10 3  0.0039 V = 2-8 V Ans: ‘b’   36 Solutions to Communication Systems Academy 10. In the process of Quantization, the quantizer is able to avoid the effect of all channel noise Magnitudes less than or equal to S 2. If the channel noise Magnitude exceeds S / 2 , there may be an error in the output of the quantizer. On the given Problem for y1(t) + c to be different from y2(t), the minimum value of c to be added is half of the step size, i.e. Ans: Δ 2 ‘b’ a 11. ACE  P(x) dx  a 1  3 a 1 1  4 . dx  3 a a = 3 Ans: ‘b’ 2 12. 1  x3   x f(x). dx  a 4  3  a 2 2 1  28 = 12  27  Ans: 3 2 3 = 4 81 ‘a’ 5 13. signal Power = x 2 . f(x) dx 5 f(x) = 1 for  5  x   5 10 = 0 elsewhere 25 volts2 3 S S  43.5db   22387.2 Nq Nq  signal power =  Nq = 3.722  10-4 =  stepsize  2 12  step size = 0.0668 V Ans: 14. Total Nq =  0.05 2 12 S  N  40db q Ans: 15. ‘c’   0.1 2 12 1.041 10 3 ‘d’ for every one bit increase in data word length, S N q improves by a factor of 4.Hence, for two bits increase, the improvement factor is 16. ACE Academy Engineering Ans: 16. ‘c’ Between two adjacent sampling instances, if the base band signal changes by an amount less than the step size, i.e. if the variations are very less magnitude, the o/p of the Delta Modulator consists of a sequence of alternate +ve and –ve Pulses. Ans: 17. Electronics & Communication 37 ‘a’ f(x) = 1 for 0  x  1 = 0 elsewhere M.S. value of Quantization Noise 0.3 = x 1 2 . f(x).dx  0  (x  0.7) 2 f(x) dx 0.3 = 0.039 volts2  rms value = 0.198 Volts  Capture effect  Slope overload  Matched filter  Law ‘c’ 18. FM DM PSK PCM Ans: 19. PP Step size = no.of Q levels  128  0.012 V V Nq = Ans: 20. S2  12  10 -6 Volts 2 12 ‘c’ slope overload occurs if S fs  2 fm . Em S fs = 25120  2 (4  103) (1.5) = 37699.11 Ans: 21. 1.536 ‘b’ R =  fs = 8  8 KHz = 64 Kbps S  1.76 + 20 log Q (db) = 49.8 db Nq Ans: 22. ‘b’  δ t  nT  & T s S  100 μ sec Let S(t) = 5  10-6 n The Fourier series representation of S(t) is 1 2  S(t) = 5  10-6 [ T  T s s   cos 2 nf s t ] n   = 10-4 sec 38 Solutions to Communication Systems Academy   cos 2π (n 10 10  = 5  10 + 10  y(t) = S(t). x(t) = S(t). 10 cos 2 (4  103)t -2 -1 n    3 )t   = 5  10 cos 2 (4  10 )t +  cos 2(n 104)t.cos2(4  103)t -1 3 n   The o/p of ideal LPF = 5  10-1 cos (8  103)t Ans: ‘c’ 23. x(t) = 100 cos 2 (12  103)t Ts = 50 sec  fs = 20 KHz The frequency components available in the sampled signal are 12 KHZ, (20  12) KHZ, (40  12) KHZ …..etc. The o/p of the ideal LPF are 8 KHZ and 12 KHZ. Ans: ‘d’ 24. x(t) = sinc (700t) + sinc (500t) sin (700t) sin (500t)  = 700t 500t π  sin (700 t )  π  sin (500 t )  = +    700  t 500  t  The band limiting frequency of above x(t) is m = 700  fm = 350/ 25.  (fs)min = 700 Hz π  (Ts)max =  700 sec x(t) = 6  10-4 sinc2 (400t) + 106 sinc3(100t) Sinc2 (400t)  F .T   800 800 Sinc3 (100t)  F .T  The convolution extends from  =  1100 to  = +1100.  300 300 1100 m = 1100  fm = = 175 Hz (fs)min = 350 Hz 26. step size = 2 28 2 = 0.0078 Volts S2 = 5.08  volts2 12  0.5 2 = 0.125 Volts2 Signal Power = 2 Nq = m m ACE ACE Academy Engineering Electronics & Communication 39 S 10 log N  44db q 27. For every one bit increase in the data word length, quantization noise power reduces by a factor of 4. Ans: 28. Flat Top sampling is observing be baseband signal through a finite time aperture. This results in Aperture effect distortion. Ans: 29. ‘d’ 10 log 4 = 6 db Ans: 33. ‘b’ Irrespective of the value of , for every one bit increase in Data word length, S N q improves by a factor of 4. Ans: 32. ‘a’ Most of the signal strength will be available in the Major lobe. Hence, (fs)min = 2(1 KHZ) = 2 KHZ Ans: 31. ‘a’ In compression the baseband signal is subjected to a non linear Transformation, whose slope reduces at higher amplitude levels of the baseband signal. Ans: 30. ‘c’ ‘b’ The frequency components available in the sampled signal are 1 KHz, (1.8  1) KHz, (3.6  1) KHz etc. The o/p of the filter are 800 Hz and 1000 Hz. Ans: ‘c’ 34. Ans: ‘c’ 35. Ans: a – 2, b – 1, c – 5. 36. Ans: a – 2, b – 1, c – 4. 37. If pulse width increases, the spectrum of the sampled signal becomes zero even before fm. Ans: ‘a’  fs (B.)min = 2 38. 40 Solutions to Communication Systems Academy ACE Q=4  =2 Q = 64   = 6  B. increases by a factor of 3. 39. (B.)min = (3 +  + 2 + 3 + 2) Hz = 11  Hz 40. The given signal is a band pass signal. N= (fs)min = 2 fH f , where N = H N 3 1800 1.8  10 3 = = 1.2 1500 1500 N=1  (fs)min = 2 fH = 3600 Hz 41. 42. LSB = (4000 – 2) MHz = 3998 MHz USB = (4000 + 2) MHZ = 4002 MHz f 4002 N= H = = 1000.5 4 B  N = 1000 2 fH 2  4002 (fs)min = = MHz = 8.004 MHz 1000 N 1  Es  . cos 2  Pe = erfc  2    1/ 2  The factor is cos2 20 Ans: 43. ‘b’ Nq depends on step size, which inturn depends on No. of Q-levels. Ans: ‘c’ 44. (fs)min to reconstruct 3 KHz part = 6 KHz (fs)min to reconstruct 6 KHz part = 12 KHz The frequencies available in sampled signal are 3 KHz, 6 KHz, (8  3) KHz, (8  6) KHz, (16  3) KHz, (16  6) KHz etc. The o/p of LPF are 3 KHZ, 6 KHz, 5 KHz and 2 KHz. Ans: ‘d’ 45. Ans: ‘c’ Chapter – 5 B & C ACE Academy Engineering 01. Electronics & Communication 41 Required Probability = P (No bit is 1 i.e. zero No. of 1’s) + P (one bit is 1) = 3C0 . (P)3 . (1 - P)3-3 + 3C1 . P2 (1 - P)3-2 = P3 + 3P2 (1 - P) Ans: ‘a’ 02. The given raised cosine pulse will be defined only for 0  | f |  2. Thus, at t = 1/4, i.e. f = 4, P(t) = 0. Ans: b 03. Required probability = P(X = 0) + P(X = 1)  n C0 (P) 0 (1 P) n  0  n C1 P(1 P) n 1  (1  P) n  n P (1  P) n 1 Ans: c 04. Constellation – 1: S1(t) = 0; S2(t) = 2 a 1 + 2 a.2 S3(t) = 22a.1; S4(t) = 2 a 1  2 a 2 Energy of S1(t) = E1 = 0 Energy of S2(t) = E2 = 4a2 Energy of S3(t) = E3 = 8a2 Energy of S4(t) = E4 = 4a2 Avg. Energy of Constellation 1 E  E 2  E3  E 4 E C1  1  4a 2 4 Constellation – 2: S1(t) = a 1  E1 = a2 S2(t) = a 2  E2 = a2 S3(t) = a 1  E3 = a2 S4(t) = a 2  E4 = a2 E C2  a 2 E C1 E C2 Ans:  4 b 05. Constellation – 1 Distance d S1 S2  2 a ; d S1 S3  2 2 a ; d S1 S4  2 a ; d S2 S3  2 a ; d S2 S4  2 2 a ; d S3 S4  2 a  (d min ) C1  2a Constellation – 2 d S1 S2  d S3 S4  2 a ; d S1 S3  2 a ; d S1 S4  2 a; 2 a ; d S2 S3  2 a ; d S2 S4  2 a ; 42 Solutions to Communication Systems Academy (d min ) C2  2a Since (d min )C2  (d min )C1 , Probability of symbol error in Constellation – 2 (C2) is more than that of constellation – 1 (C1). Ans: a 06. P(t) = 1 0 1 t g(t) = 2 0 S(t) = g(t)  (t  2) * g(t) We have (t – 2)  g(t) = g(t  2) S(t) = g(t)  g(t  2) S(t) = 0 2 4 The impulse response of corresponding Matched filter is h(t) = S(t + 4) = S(t) = 0 2 Ans: c 07. Since P(t) = 1 for 0  t  1, and g(t) = t for 0  t  1, the given xAM(t) = 100[1 + 0.5t] cosct Ans: a 4 ACE ACE Academy Engineering Electronics & Communication 43 08. Output of the matched filter is the convolution of its impulse response and its input. The given input S(t) = 1 0 t 2 The corresponding impulse response is h(t) = 1 0 t 2 The response should extend from t = 0 to t = 4. Response    s( τ )  Let t = 1 S() h( + 1) = h(t  τ) d τ 0 1 1  The response at t = 1 is 1 Ans: ‘c’ 09. Let z be the received signal. P(z/0) = 1 0.5 for 0.25  z  0.25 = 0 0.25 P(1/0) =  0 .2 elsewhere  1   dz  0 .5   = 0.1 P(z/1) = 1 for 0  z  1 = 0 elsewhere 0.2 P(0/1) =  dz  0.2 0 Ave. bit error prob. = Ans: ‘a’ 10. Ans: ‘c’ 11. (B.W)BPSK = 2fb = 20 KHz (B.W)QPSK = fb = 10 KHz Ans: ‘c’ 0.1  0.2 = 0.15 2 44 Solutions to Communication Systems Academy ACE S0 2E b 2  106 = = = 20 N0 N0 105 10 log 20 = 13 db Ans: ‘d’ 12. 13. B.W efficiency = data rate ( B.W ) min For BPSK, (B.W)min required is same as data rate.  B.W efficiency for BPSK = 1 Since, coherent detection is used for BPSK, Carrier synchronization is required. Ans: ‘b’ 14. (Pe)PSK =  A2 T  1 erfc   2  2  1 2 2 1   erfc  0.6 A T  (Pe)FSK = 2 2    1 2 10 log 0.6 = -2.2 db = -2 db Ans: ‘c’ 15. fH = nfb & fL = mfb, where n and m are integers such that n>m. Ans: ‘d’ 16. Ans: ‘d’ 17. fH = 25 KHz & fL = 10 KHz  fc + fc -     = 25 KHz 2  = 10 KHz 2 = 15 KHz   = 15 103  For FSK signals to be orthogonal, 2  Tb = n   2(15    10 3 ) Tb = n   30  103  Tb should be an integer. This is satisfied for Tb = 280  sec Ans: 18. Ans: ‘d’ ‘c’ 19. In PSK, the signaling format is NRZ and in ASK, it is ON-OFF signaling. Both representations are having same PSD plot. Ans: ‘c’ 20. Ans: ‘d’ 21. Ans: ‘b’ ACE Academy Engineering Electronics & Communication 45 22. Ans: a – 3; b – 1; c – 2 23. b(t) b1(t) 1 b (t – T6) D b(t) b1(t) 1 Phase Ans: 0 1 1 0 0 0 0   0 0 1 1  0 ‘c’ 24. a 25. c 26. QPSK 27. a b(t) 28. b1(t) Tb b(t) b1(t) 1 1 1 1 1 0 0 0 1 1 1 1 1 since the phase of the first two message bits is ,  , the received is 0 0 1 0 1 1 1 1 0 0 1 0 1 1 ______________________________________________ 0 0 0 0 1 1     0 0 Ans : d 29. P(at most one error) = P(X=0) + P(X=1) 7 = 8C 0 .(1-P)8 . P0 + 8C 1 . 1  P  P = (1 – P)8 + 8P (1 – P)7 46 Solutions to Communication Systems Academy ACE Ans: b Chapter – 6 (Objective Questions) 01. (B.W)min = w+w+2w+3w = 7w Ans: ‘d’ 02. The total No.of channels in 5 MHz B.W is 5  106  8 = 200 2  105 With a five cell repeat pattern, the no. of simultaneous channels is Ans : B 03. RC = 1.2288  106 Rc GP =  100 Rb Rc   Rb 100  1.2288  104  Rb  Rb  12.288  103 bps Ans: a 04. Bit rate = 12 ( 2400 + 1200+1200) = 57.6 kbps Ans: c 05. Sample rate = 200+ 200 + 400 +800 = 1600 Hz Ans : a 06. d 07. 12  5 KHz + 1 KHz = 61 KHz 08. b 09. d 1 (data rate) 2 1 = (4  2  5 KHz) 2 10 . Theoritical (B.W)min = = 20 KHz 11. c 12. a 200 = 40 5 ACE Academy Engineering Electronics & Communication 47 13. The path loss is due to a) Reflection : Due to surface of earth, buildings and walls b) Diffraction : This is due to the surfaces between Tx. and Rx. that has sharp irregularities (edges) c) Scatterings: Due to foliage, street signs, lamp posts, i.e. scattering is due to rough surfaces, small objects or by other irregularities in a mobile communication systems. 14. 1333 Hz. 15. Min. Tx. Bit rate = (2  4000 + 2 8000 + 2 8000 + 24000)8 = 384 kbps Ans: ‘d’ 16. 12  8 KHz Ans : c 17. a 18. c 19. b 20. c 21. b
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