Contemporary Communication SystemSolutions Problems Chapter2

Chapter 22.1 Consider the signals displayed in Figure P2.1.Show that each of these signals can be expressed as the sum of rectangular Π (t ) and/or triangular Λ (t ) pulses. Figure P2.1 Solution: a. ⎛t⎞ ⎛t⎞ x1 (t ) = Π ⎜ ⎟ + Π ⎜ ⎟ ⎝2⎠ ⎝4⎠ ⎛t −3⎞ ⎛t −3⎞ b. x2 (t ) = 2Λ ⎜ ⎟− Λ⎜ ⎟ ⎝ 6 ⎠ ⎝ 2 ⎠ c. ⎛ 11 ⎞ ⎛ 7⎞ ⎛ 3⎞ ⎛ 1⎞ ⎛ 5⎞ ⎛ 9⎞ x3 (t ) = ... + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + ... 2⎠ ⎝ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ = ∞ ∑ Π ⎡⎣t − ( 2n + 0.5)⎤⎦ n =−∞ 1 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. ⎛t+4⎞ ⎛t⎞ ⎛t−4⎞ d. x4 (t ) = ... + Λ ⎜ ⎟ + Λ⎜ ⎟ + Λ⎜ ⎟ + ... ⎝ 2 ⎠ ⎝2⎠ ⎝ 2 ⎠ ∞ ⎛ t − 4n ⎞ = ∑ Λ⎜ ⎟ ⎝ 2 ⎠ n =−∞ 2.2 For the signal x2 (t ) in Figure P2.1 (b) plot the following signals x2 (t − 3) a. b. x2 (−t ) c. x2 (2t ) d. x2 (3 − 2t ) Solution: x2 (t − 3) x2 (−t ) 1 1 0 0 1 2 3 4 5 6 7 8 9 −6 −5 −4 −3 −2 −1 t 1 2 3 x2 (3 − 2t ) x2 (2t ) 1 1 0 1 2 3 4 5 6 t 0 −2−1 1 2 3 4 t 2.3 Plot the following signals a. x1 (t ) = 2Π (t / 2) cos(6π t ) ⎡1 1 ⎤ b. x2 (t ) = 2 ⎢ + sgn ( t ) ⎥ ⎣2 2 ⎦ 2 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. x3 (t ) = x2 (−t + 2) c. d. x4 (t ) = sinc ( 2t ) Π ( t / 2 ) Solution: x2 (t ) 2 1.5 2 x1(t) 1 0.5 0 0 t 1 2 3 4 5 6 -0.5 -1 1 -1.5 0.8 -2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0.6 xx4(t) (t) tt 0.4 4 x3 (t ) 0.2 2 0 -0.2 0 t -0.4 -1 1 2 3 4 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 tt 2.4 Determine whether the following signals are periodic. For periodic signals, determine the fundamental period. a. x1 (t ) = sin(π t ) + 5cos(4π t / 5) Solution: sin(π t ) is periodic with period T1 = 2π π = 2 . cos(4π t / 5) is periodic with period T 2π 5 = . x1 (t ) is periodic if the ratio 1 can be written as ratio of 4π / 5 2 T2 integers. In the present case, T2 = T1 2 × 2 4 = = T2 5 5 3 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Therefore, x1 (t ) is periodic with fundamental period To such that To = 5T1 = 4T2 = 10 b. x2 (t ) = e j 3t + e j 9t + cos(12t ) Solution: 2π . Similarly, e j 9t is periodic with period 3 2π 2π π T2 = . cos(12t ) is periodic with period T3 = = . x2 (t ) is periodic with 9 12 6 2π fundamental period To = LCM (T1 , T2 , T3 ) = . 3 e j 3t is periodic with period T1 = c. x3 (t ) = sin(2π t ) + cos(10t ) Solution: 2π = 1 . cos(10t ) is periodic with period 2π T 2π π T2 = = . x3 (t ) is periodic if the ratio 1 can be written as ratio of integers. 10 5 T2 In the present case, sin(2π t ) is periodic with period T1 = T1 1× 5 5 = = T2 π π Since π is an irrational number, the ratio is not rational. Therefore, x3 (t ) is not periodic. π⎞ ⎛ d. x4 (t ) = cos ⎜ 2π t − ⎟ + sin(5π t ) 4⎠ ⎝ Solution: π⎞ 2π ⎛ cos ⎜ 2π t − ⎟ is periodic with period T1 = = 1 . sin(5π t ) is periodic with 4⎠ 2π ⎝ T 2π 2 period T2 = = . x4 (t ) is periodic if the ratio 1 can be written as ratio of T2 5π 5 integers. In the present case, 4 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Not authorized for sale or distribution in any manner.5 Classify the following signals as odd or even or neither. This document may not be copied. b. in whole or part. distributed. So x(t ) is even. −t =e − −t . x(t ) = e −t Solution: e c. x4 (t ) is periodic with fundamental period To such that To = 2T1 = 5T2 = 2 2. So x(t ) is odd. forwarded. x(t ) = −4t Solution: x(−t ) = 4t = −(−4t ) = − x(t ) . 5 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.T1 1× 5 5 = = T2 2 2 Therefore. 2 e. x(t ) = sin(3t − ) 2 Solution: π x(t ) = sin(3t − ) = − cos(3t ) which is even. a. scanned. . π d. u (1) = 1 but u (−1) = 0 ≠ −u (1) . x(t ) = u (t ) Solution: u (t ) is neither even nor odd. For example. or posted on a website. 5cos(3t ) is also even. duplicated. x(t ) = 5cos(3t ) Solution: Since cos(t ) is even. x1 (t ) is a power signal. This document may not be copied. distributed. Not authorized for sale or distribution in any manner. x2 (t ) = 4 cos(2π t ) + 3cos ( 4π t ) Solution: 4 cos(2π t ) is a power signal. duplicated.f. b. T /2 T /2 2 2 1 1 ⎡⎣ 4 cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt = lim ⎡⎣ 4 cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt Px2 = lim ∫ ∫ T →∞ T T →∞ T −T /2 −T /2 T /2 1 ⎡16 cos 2 (2π t ) + 9 cos 2 ( 4π t ) + 24 cos(2π t ) cos ( 4π t ) ⎤⎦dt T →∞ T ∫ ⎣ −T /2 = lim Now T /2 T /2 1 8 8T =8 16 cos 2 (2π t )dt = lim 1 + cos(4π t ) ]dt = lim [ ∫ ∫ T →∞ T T →∞ T T →∞ T −T /2 −T /2 lim T /2 1 lim 9 cos 2 (4π t )dt = 4. x(π / 8) = 2 but x(−π / 8) = 0 ≠ − x(π / 8) . forwarded. or posted on a website. scanned. x1 (t ) = u (t ) Solution: The normalized average power of a signal x1 (t ) is defined as T /2 1 1 2 u (t ) dt = lim ∫ T →∞ T T →∞ T −T /2 Px1 = lim T /2 1T 1 = T →∞ 2 2 ∫ 1 dt = lim T 0 Therefore. 3cos ( 4π t ) is also a power signal. For example. 2. in whole or part. x(t ) = sin(2t ) + cos(2t ) Solution: x(t ) is neither even nor odd. . This is proprietary material solely for authorized instructor use. Since x2 (t ) is sum of two power signals. or neither and calculate the corresponding energy or power in the signal: a.5 T →∞ T ∫ −T /2 6 © 2012 by McGraw-Hill Education.6 Determine whether the following signals are energy or power. it is a power signal. or posted on a website. Px2 = 8 + 4. e. scanned.T /2 T /2 24 24 cos(2π t ) cos ( 4π t )dt = lim ∫ ∫ ⎡⎣cos(6π t ) + cos ( 2π t )⎤⎦dt = 0 T →∞ T T →∞ T −T /2 −T /2 lim Therefore. duplicated.5 = 12.5 c. in whole or part. This document may not be copied. d. distributed. Not authorized for sale or distribution in any manner. This is proprietary material solely for authorized instructor use. . x5 (t ) = Π (t / 3) + Π (t ) Solution: x5 (t ) 2 1 −3 / 2 3/ 2 t 7 © 2012 by McGraw-Hill Education. x3 (t ) = 1 t Solution: ⎛ 1 T /2 ⎞ ⎛ −2 ⎞ −2 t dt lim = ⎜− ⎟⎟ = Tlim ⎜ ⎟=0 T →∞ ∫ T →∞ ∫ T →∞ ⎜ →∞ t T / 2 ⎝ ⎠ T /2 − −T /2 −T /2 ⎝ ⎠ T /2 T /2 T /2 ⎞ 1 1 1⎛ 1 1 ⎛ −2 ⎞ 2 −2 Px3 = lim t dt t dt 1/ lim lim = = − = lim ⎜ ⎜ ⎟ ⎟=0 ⎟ →∞ T →∞ T ∫ T →∞ T ∫ T →∞ T ⎜ T T ⎝T / 2⎠ −T /2 −T /2 ⎝ t −T /2 ⎠ T /2 Ex3 = lim T /2 1/ t dt = lim 2 x3 (t ) is neither an energy nor a power signal. x4 (t ) = e−α t u (t ) Solution: T /2 Ex4 = lim T →∞ ∫ T /2 2 −α t e u (t ) dt = lim T →∞ −T /2 ∫ 0 e −2α t ⎛ e −2α t dt = lim ⎜ − T →∞ ⎜ 2α ⎝ T /2 0 ⎞ 1 1 ⎟= lim ( −e −α T + 1) = ⎟ 2α T →∞ 2α ⎠ Thus x4 (t ) is an energy signal. forwarded. T /2 Ex5 = lim T →∞ ∫ Π (t / 3) + Π (t ) dt = 2 −T /2 −1/2 1/ 2 −3/ 2 −1/2 ∫ 1dt + ∫ −1/ 2 4dt + ∫ 1dt −3/2 = 1+ 4 +1 = 6 Thus x5 (t ) is an energy signal.7 Evaluate the following expressions by using the properties of the delta function: a. f. 1 Px7 = To To / 2 2 ⎡⎣ Λ ( t / 2 ) ⎤⎦ dt = ∫ To −To / 2 2 1 1 ( ) 1 2 ∫0 (1 − t ) dt = 2 ∫0 1 − 2t + t dt 2 1 t3 ⎞ 1⎛ 1⎛ 1⎞ 1 = ⎜ t − t 2 + ⎟ = ⎜1 − 1 + ⎟ = 2⎝ 3 ⎠0 2⎝ 3⎠ 6 2. x7 (t ) = ∞ ∑ Λ ⎡⎣( t − 4n ) / 2⎤⎦ n =−∞ Solution: x7 (t ) is a periodic signal with period To = 4 . distributed. g. This document may not be copied. x6 (t ) = 5e( −2t + j10π t )u (t ) Solution: T /2 Ex6 = lim T →∞ ∫ 2 5e( −2t + j10π t )u (t ) dt = lim T →∞ −T /2 ⎛ e = 25 lim ⎜ − T →∞ ⎜ 4 ⎝ −4 t T /2 0 T /2 ∫ 0 T /2 5e −2t e j10π t ) dt = lim 25 ∫ e−4t dt 2 T →∞ 0 ⎞ 25 25 ⎟= lim ( −e −2T + 1) = →∞ T ⎟ 4 4 ⎠ Thus x6 (t ) is an energy signal. in whole or part. Not authorized for sale or distribution in any manner. This is proprietary material solely for authorized instructor use. scanned. forwarded. duplicated. or posted on a website. x1 (t ) = δ (4t ) sin(2t ) Solution: 1 4 δ (4t ) = δ (t ) 1 1 x1 (t ) = δ (t ) sin(2t ) = δ (t ) sin(0) = 0 4 4 8 © 2012 by McGraw-Hill Education. . x3 (t ) = δ (t )sinc(t + 1) Solution: x3 (t ) = δ (t )sinc(t + 1) = δ (t )sinc(0 + 1) = δ (t )sinc(1) = δ (t ) × 0 = 0 d. or posted on a website. ∫ δ (2t )sinc(t )dt x5 (t ) = −∞ Solution: ∞ ∞ 1 x5 (t ) = ∫ δ (2t )sinc(t )dt = ∫ δ (t )sinc(t ) dt 2 −∞ −∞ ∞ ∞ 1 1 1 δ (t )sinc(0)dt = ∫ δ (t )dt = ∫ 2 −∞ 2 −∞ 2 = ∞ f.5π t ) Solution: x4 (t ) = δ (t − 2)e − t sin(2. in whole or part. Not authorized for sale or distribution in any manner.5π × 2) = δ (t − 2)e −2sin(π ) = 0 ∞ e. This is proprietary material solely for authorized instructor use. duplicated. This document may not be copied. scanned. x4 (t ) = δ (t − 2)e − t sin(2.5π t ) = δ (t − 2)e −2sin(2. distributed. x6 (t ) = ∫ δ (t − 3) cos(t )dt −∞ Solution: x6 (t ) = ∞ ∞ −∞ −∞ ∫ δ (t − 3) cos(t )dt = cos(3) ∫ δ (t − 3)dt = cos(3) 9 © 2012 by McGraw-Hill Education. . x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ 4⎠ ⎝ Solution: π⎞ π⎞ ⎛ ⎛ ⎛π ⎞ x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ = δ (t ) cos ⎜ 0 + ⎟ = cos ⎜ ⎟ δ (t ) 4⎠ 4⎠ ⎝ ⎝ ⎝4⎠ c.π⎞ ⎛ b. forwarded. 8 For each of the following continuous-time systems. (3) memoryless. x8 (t ) = ∫ δ (3t − 4)e −3t dt −∞ Solution: ∞ x8 (t ) = −3t ∫ δ (3t − 4)e dt = −∞ ∞ ∞ 1 ⎛ 4⎞ δ ⎜ t − ⎟ e −3t dt ∫ 3 −∞ ⎝ 3 ⎠ 4 1 e −4 ⎛ 4 ⎞ −3× = ∫ δ ⎜ t − ⎟ e 3 dt = 3 −∞ ⎝ 3 ⎠ 3 i. in whole or part. scanned. distributed. (2) time-invariant. x7 (t ) = 3 dt −∞ Solution: ∞ x7 (t ) = ∫ δ (2 − t ) −∞ ∞ 1 1 dt = ∫ δ (t − 2) dt 3 1− t 1− t3 −∞ ∞ ∞ 1 1 1 dt = − ∫ δ (t − 2)dt = − = ∫ δ (t − 2) 3 1− 2 7 −∞ 7 −∞ ∞ h. To prove linearity.5 ) − δ ( t − 0. The response of the system to x(t ) is 10 © 2012 by McGraw-Hill Education. y (t ) = x(t − 1) Solution: The system is linear. duplicated. time-invariant. forwarded. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. or posted on a website. . and (4) casual. The system has memory because current value of the output depends on the previous value of the input. and has memory. a. determine whether or not the system is (1) linear. let x(t ) = α x1 (t ) + β x2 (t ) . The system is causal because current value of the output does not depend on future inputs. This document may not be copied.5 ) ' 2. ∞ ⎛ e −4 4⎞ ∫ δ ⎜ t − 3 ⎟⎠ dt = 3 −∞ ⎝ x9 (t ) = δ '(t ) ⊗ Π (t ) Solution: ∞ x9 (t ) = d ∫ Π (t − τ )δ '(τ )dτ = (−1) dτ Π (t − τ ) τ −∞ =0 = Π (t ) = δ ( t + 0. causal.∞ 1 ∫ δ (2 − t ) 1 − t g. forwarded. The response of the system to x(t ) is y (t ) = 3x(t ) − 2 = 3 [α x1 (t ) + β x2 (t ) ] − 2 ≠ α y1 (t ) + β y2 (t ) = α [3 x1 (t ) − 2] + β [3 x2 (t ) − 2] To show that the system is time-invariant. The system is causal because current value of the output does not depend on future inputs. y (t ) = 3x(t ) − 2 Solution: The system is nonlinear. To show that the system is time-invariant. or posted on a website. let x1 (t ) = x(t − to ) be the system input. .y (t ) = x(t − 1) = α x1 (t − 1) + β x2 (t − 1) = α y1 (t ) + β y2 (t ) T T where x1 (t ) ⎯⎯ → y1 (t ) and x2 (t ) ⎯⎯ → y2 (t ) . The system is memoryless because current value of the output depends only on the current value of the input. scanned. causal. To prove nonlinearity. y (t ) = x(t ) Solution: The system is nonlinear. in whole or part. The corresponding output is y1 (t ) = x1 (t − 1) = x(t − 1 − to ) = y (t − to ) b. time-invariant. To prove nonlinearity. distributed. let x(t ) = α x1 (t ) + β x2 (t ) . duplicated. time-invariant. and memoryless. causal. The system is causal because current value of the output does not depend on future inputs. The corresponding output is y1 (t ) = 3x1 (t ) − 2 = 3x(t − to ) − 2 = y (t − to ) c. let x(t ) = α x1 (t ) + β x2 (t ) . and memoryless. This document may not be copied. This is proprietary material solely for authorized instructor use. The system is memoryless because current value of the output depends only on the current value of the input. let x1 (t ) = x(t − to ) be the system input. The response of the system to x(t ) is 11 © 2012 by McGraw-Hill Education. Not authorized for sale or distribution in any manner. y (t ) = e x (t ) Solution: The system is nonlinear. y (t ) = [ cos(2t )] x(t ) Solution: The system is linear. The system is memoryless because current value of the output depends only on the current value of the input. scanned. Not authorized for sale or distribution in any manner. forwarded. or posted on a website. in whole or part. let x1 (t ) = x(t − to ) be the system input. The system is causal because current value of the output does not depend on future inputs. time-variant. time-invariant. and memoryless. and memoryless. . To prove nonlinearity. causal. The response of the system to x(t ) is y (t ) = [ cos(2t ) ] x(t ) = [ cos(2t ) ][α x1 (t ) + β x2 (t ) ] = α [ cos(2t ) ] x1 (t ) + β [ cos(2t ) ] x2 (t ) = α y1 (t ) + β y2 (t ) To prove time-variance. duplicated. The corresponding output is y1 (t ) = [ cos(2t ) ] x1 (t ) = [ cos(2t ) ] x ( t − to ) ≠ y (t − to ) = cos ⎡⎣ 2 ( t − to ) ⎤⎦ x ( t − to ) e. This document may not be copied. let x(t ) = α x1 (t ) + β x2 (t ) . 12 © 2012 by McGraw-Hill Education. causal. To show that the system is time-invariant. The corresponding output is y1 (t ) = x1 (t ) = x(t − to ) = y (t − to ) d. The system is memoryless because current value of the output depends only on the current value of the input. the system is nonlinear. distributed. The system is causal because current value of the output does not depend on future inputs. let x1 (t ) = x(t − to ) be the system input. This is proprietary material solely for authorized instructor use.y (t ) = α x1 (t ) + β x2 (t ) ≤ α x1 (t ) + β x2 (t ) Because α x1 (t ) + β x2 (t ) = α x1 (t ) + β x2 (t ) ≠ α y1 (t ) + β y2 (t ) for all α and β . let x1 (t ) = x(t − to ) be the system input. The system is causal because current value of the output does not depend on future inputs. To prove linearity. forwarded. let x(t ) = α x1 (t ) + β x2 (t ) . The system is causal because current value of the output does not depend on future inputs. The response of the system to x(t ) is y (t ) = eα x1 ( t ) + β x2 ( t ) = eα x1 ( t ) e β x2 (t ) ≠ α y1 (t ) + β y2 (t ) = α e x1 ( t ) + β e x2 ( t ) for all α and β To show that the system is time-invariant. The corresponding output is y1 (t ) = e x1 (t ) = e x (t −to ) = y (t − to ) f. This document may not be copied. let x1 (t ) = x(t − to ) be the system input. The response of the system to x(t ) is 13 © 2012 by McGraw-Hill Education. causal.To prove nonlinearity. or posted on a website. time-variant. causal. The corresponding output is y1 (t ) = tx1 (t ) = tx ( t − to ) ≠ y (t − to ) = ( t − to ) x ( t − to ) t g. y (t ) = tx(t ) Solution: The system is linear. time-invariant. This is proprietary material solely for authorized instructor use. and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. y (t ) = ∫ x(2τ )dτ −∞ Solution: The system is linear. let x(t ) = α x1 (t ) + β x2 (t ) . let x(t ) = α x1 (t ) + β x2 (t ) . distributed. in whole or part. Not authorized for sale or distribution in any manner. To prove nonlinearity. The response of the system to x(t ) is y (t ) = t [α x1 (t ) + β x2 (t ) ] = α tx1 (t ) + β tx2 (t ) = α y1 (t ) + β y2 (t ) To prove time-variance. scanned. . duplicated. and has memory. The system has memory because current value of the output depends only on the past values of the input. x(t ) = e − t u (t ) and h(t ) = e −2t u (t ) 14 © 2012 by McGraw-Hill Education. t For 2 ≤ t < 4 . or posted on a website.t t t −∞ −∞ −∞ ∫ [α x1 (2τ ) + β x2 (2τ )] dτ = α ∫ x1 (2τ )dτ + β ∫ x2 (2τ )dτ = α y1 (t ) + β y2 (t ) y (t ) = To check for time-variance. The corresponding output is t y1 (t ) = t − to t ∫ x (2τ )dτ = ∫ x [ 2(τ − t )] dτ = ∫ x ( 2α ) dα = y ( t − t ) o 1 −∞ o −∞ −∞ 2.9 Calculate the output y (t ) of the LTI system for the following cases: a. ⎪ ⎪⎩ 4 ∫e 2τ dτ =e −2 t 2 e2τ 2 4 =e 2 −2 t e4 − 1) e8 − e 4 −2( t − 2) ( =e 2 2 2≤t ≤4 t>4 otherwise b. y (t ) = ∫ e −2( t −τ ) dτ = 2 t −2 ∫e −2τ 0 e −2τ dτ = −2 t −2 0 1 − e−2(t − 2) = 2 For t ≥ 4 . 2 ⎪ ⎪0. forwarded. there is no overlap and y (t ) = 0. scanned. x(t ) = e −2t u (t ) and h(t ) = u (t − 2) − u (t − 4) Solution: ⎛ t −3⎞ h(t ) = u (t − 2) − u (t − 4) = Π ⎜ ⎟ ⎝ 2 ⎠ y (t ) = ∞ ∞ −∞ −∞ ∫ h(τ ) x(t − τ )dτ = ∫ e −2( t −τ ) ⎛τ −3 ⎞ u (t − τ )Π ⎜ ⎟ dτ ⎝ 2 ⎠ For t < 2 . duplicated. in whole or part. 4 y (t ) = ∫ e −2( t −τ ) dτ = e −2 t 2 ⎧1 − e −2(t − 2) . This is proprietary material solely for authorized instructor use. distributed. let x1 (t ) = x(t − to ) be the system input. ⎪ 2 ⎪ ⎪⎪ −2(t − 2) ( e 4 − 1) y (t ) = ⎨e . This document may not be copied. . Not authorized for sale or distribution in any manner. This is proprietary material solely for authorized instructor use. ∫ 3e −2τ 0 ∞ For t ≥ 0 . x(t ) = u (−t ) and h(t ) = δ (t ) − 3e−2t u (t ) Solution: ∞ y (t ) = ∫ −∞ ∞ h(τ ) x(t − τ )dτ = ∫ ⎡⎣δ (τ ) − 3e −2τ u (τ ) ⎤⎦ u (τ − t )dτ −∞ Now ∞ ∫ δ (τ )u(τ − t )dτ = u (−t ) −∞ ∞ For t < 0 . y (t ) = e −2 t t ∫ e dτ =e τ −2 t eτ = e−2t ( et − 1) = e − t − e −2t . scanned. x(t ) = δ (t − 2) + 3e3t u (−t ) and h(t ) = u (t ) − u (t − 1) Solution: 15 © 2012 by McGraw-Hill Education. duplicated. y (t ) = ⎨ ⎪ 3 e −2t . This document may not be copied.Solution: ∞ y (t ) = ∫e −τ u (τ )e −2( t −τ ) −∞ t u (t − τ )dτ = ∫ e −τ e−2(t −τ ) dτ 0 For t < 0 . Not authorized for sale or distribution in any manner. in whole or part. there is no overlap and y (t ) = 0. . ⎪⎩ 2 t −2τ 3e −2τ dτ = −2 ∞ t 3 = e −2t 2 t<0 t≥0 d. distributed. or posted on a website. ∫ 3e −∞ 3e −2τ dτ = −2 −2τ ∞ = 0 3 2 ∞ u (τ )u (τ − t )dτ = ∫ 3e 5 ⎧3 ⎪⎪ 2 + 1 = 2 . t ≥ 0 t 0 0 c. forwarded. ∞ 1 −∞ 0 1 3( t −τ ) 3( t −τ ) 3t −3τ ∫ 3e u(−t + τ ) [u(τ ) − u(τ − 1)] dτ = ∫ 3e dτ = 3e ∫ e dτ 0 = 3e3t −3τ 1 e −3 0 1⎞ ⎛ = e 3t ⎜ 1 − 3 ⎟ ⎝ e ⎠ For 0 < t ≤ 1 . duplicated.10 The impulse response function of a continuous-time LTI is displayed in Figure P2. forwarded. 16 © 2012 by McGraw-Hill Education. .2(b). This is proprietary material solely for authorized instructor use. Solution: For t < 1 .y (t ) = = ∞ ∞ −∞ −∞ ∫ h(τ ) x(t − τ )dτ = ∫ ⎡⎣δ (t − τ − 2) + 3e ∞ ∞ −∞ −∞ 3( t −τ ) u (−t + τ ) ⎤⎦ [u (τ ) − u (τ − 1)] dτ 3( t −τ ) ∫ δ (t − τ − 2) [u (τ ) − u (τ − 1)] dτ + ∫ 3e u (−t + τ ) [u(τ ) − u (τ − 1)] dτ Now ∞ ∫ δ (t − τ − 2) [u(τ ) − u(τ − 1)] dτ = u(t − 2) − u(t − 2 − 1) = Π ( t − 2. Not authorized for sale or distribution in any manner. or posted on a website. distributed. scanned.5) −∞ For t < 0 . determine the system output waveform y (t ) and sketch it. Assuming the input x(t ) to the system is waveform illustrated in Figure P2.2(a). e ⎠ ⎪ ⎝ ⎪1. ∞ ∫ 3e 3( t −τ ) −∞ 1 u (−t + τ ) [u (τ ) − u (τ − 1) ] dτ = ∫ 3e 3( t −τ ) dτ = 3e t 1 3t ∫e −3τ dτ t 1 e −3τ 1⎞ ⎛ = 3e = e3t ⎜ e −3t − 3 ⎟ −3 t e ⎠ ⎝ 3t ⎧ 3t ⎛ 1⎞ ⎪e ⎜ 1 − e3 ⎟ . there is no overlap and y (t ) = 0. This document may not be copied. t≤0 0 < t ≤1 2≤t ≤3 otherwise 2. ⎠ ⎪ ⎝ ⎪ ⎛ 1⎞ y (t ) = ⎨e3t ⎜ e −3t − 3 ⎟ . in whole or part. ⎪ ⎩0. Figure P2. y (t ) = ∫ dτ = t − 1 for 1 ≤ t < 2 0 1 For 2 ≤ t < 3 . . y (t ) = t −3 t Referring to Figure (c). y (t ) = t −2 7 For 8 ≤ t < 10 .2 x(τ ) (a) 1 1 0 τ 3 2 4 5 6 7 (b) h(t − τ ) 1 1< t < 2 τ 0 1 2 4 3 h(t − τ ) (c) 1 5<t <6 τ 0 1 2 4 3 t -1 As shown in Figure (b). or posted on a website. y (t ) = ∫ dτ = 7 − t + 3 = 10 − t t −3 y(t ) 2 1 1 2 3 4 5 6 7 8 9 t 10 17 © 2012 by McGraw-Hill Education. y (t ) = ∫ dτ = 1 0 1 ∫ dτ =1 − t + 3 = 4 − t For 3 ≤ t < 4 . forwarded. duplicated. Not authorized for sale or distribution in any manner. This document may not be copied. This is proprietary material solely for authorized instructor use. y (t ) = ∫ dτ = t − 5 for 5 ≤ t < 7 5 t ∫ dτ = t − t + 2 = 2 For 7 ≤ t < 8 . distributed. scanned. in whole or part. forwarded.5(t − 2)u (t − 2) .5 C3 = −0.5e − jπ /4 a.5e e − 0. we get the following terms: ⎡ e j 40π t − e − j 40π t ⎤ ⎡ e j 80π t e− jπ /2 + e− j 80π t e jπ /2 ⎤ ⎡ e j160π t e jπ / 4 + e− j160π t e− jπ /4 ⎤ + x(t ) = 5 ⎢ 7 ⎥ ⎢ ⎥−⎢ ⎥ 2j 2 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ j 40π t − j 40π t j 80π t − j 80π t jπ /4 j160π t − jπ /4 − j160π t = − j 2. Is the system casual? Solution: Yes. distributed.2. C−2 = j 3.5 x dx = −0. in whole or part.5e + j 3. C−1 = j 2.5(t + 2)u (t + 2) .5e − 0. Solution: The system is not causal but stable. Is the system stable? Solution: ∞ The system is stable because ∫ ∞ h(t ) dt = −∞ ∫e 0 −0.5e e The Fourier coefficients are C1 = − j 2. Repeat parts (a) and (b) for h(t ) = e −0. This is proprietary material solely for authorized instructor use.12 Write down the exponential Fourier series coefficients of the signal x(t ) = 5sin(40π t ) + 7 cos ( 80π t − π / 2 ) − cos (160π t + π / 4 ) Solution: Applying the Euler’s formula. or posted on a website.11 An LTI system has the impulse response h(t ) = e −0. C−3 = −0. Not authorized for sale or distribution in any manner. what is its period? Solution: 18 © 2012 by McGraw-Hill Education. duplicated.5e jπ /4 . Is x(t) periodic? If so.5 C2 = − j 3. This document may not be copied.5e + j 2.5.5 x e −0. The system is casual because h(t ) = 0 for t < 0 . scanned.5e − j 3.5 ∞ =2 0 c. 2. . b.5. a. in whole or part. This is proprietary material solely for authorized instructor use. . This document may not be copied. To = 3. ωo = 40π ⇒ f o = 20.14 Write down the complex exponential Fourier series for each of the periodic signals shown in Figure P2. forwarded. 50 c. It is periodic with fundamental period To = 1 = 0. The period is = 0. or posted on a website. Is the signal periodic? If so. what is its period? Solution: Yes. f o = 1/ 3 19 © 2012 by McGraw-Hill Education. Figure P2. Does the signal have energy at DC? Solution: Yes as indicated by the presence of DC term C0 = 10 . Use odd or even symmetry whenever possible. To = 1 1 .Yes.13 A signal has the two-sided spectrum representation shown in Figure P2. scanned.3 a. Solution: a. duplicated. distributed. 2. Solution: π⎞ π⎞ ⎛ ⎛ x(t ) = 14 cos ⎜100π t − ⎟ + 10 + 8cos ⎜ 300π t − ⎟ 3⎠ 2⎠ ⎝ ⎝ b.05sec .4. Write the equation for x(t).3. Not authorized for sale or distribution in any manner.02 . 20 20 2. scanned. f o = 1/ 2 1 1 ⎛ t2 ⎞ ⎤ 2A ⎡ 1 A 1 C0 = − t dt = A ) ⎥ ⎜ t − ⎟ = A ⎜⎛1 − ⎟⎞ = ⎢∫ ( 2 ⎣0 2 ⎠0 ⎝ 2⎠ 2 ⎦ ⎝ 20 © 2012 by McGraw-Hill Education. forwarded.. ... Not authorized for sale or distribution in any manner. This is proprietary material solely for authorized instructor use.2 3 ⎤ 1 1⎡ C0 = ⎢ ∫ 2dt − ∫ 1dt ⎥ = ( 4 − 1) = 1 3 ⎣0 2 ⎦ 3 2 3 2π nt 2π nt −j −j ⎤ 1⎡ Cn = ⎢ ∫ 2e 3 dt − ∫ e 3 dt ⎥ 3 ⎣0 2 ⎦ 2 2π nt 3 ⎤ −j j 3 ⎡ − j 2π3nt ⎢ 2e = −e 3 ⎥ 3 × 2π n ⎢ 0 2⎥ ⎣ ⎦ j ⎡⎣ 2e − j 4π n /3 − 2 − e − j 2π n + e − j 4π n /3 ⎤⎦ = 2π n 3e − j 2π n /3 ⎛ e + j 2π n /3 − e − j 2π n /3 ⎞ j 3 − j 4π n /3 = − = 1 e ( ) πn ⎜ ⎟ 2π n 2j ⎝ ⎠ 3e − j 2π n /3 ⎛ 2π n ⎞ = sin ⎜ ⎟ . in whole or part. πn ⎝ 3 ⎠ b. duplicated. f o = 1/ 2 1 ⎤ 1 t2 1⎡ C0 = ⎢ ∫ tdt ⎥ = 2 ⎣ −1 ⎦ 2 2 Cn = 1 = −1 1 (1 − 1) = 0 4 1 1 ⎤ 1 ⎡ − jπ nt ⎤ 1 ⎡ − jπ nt = − te dt ) ⎢∫ ⎥ ⎢ ∫ td ( e ⎥ 2 ⎣ −1 j 2π n ⎣ −1 ⎦ ⎦ = 1 ⎤ j ⎡ − jπ nt 1 − jπ nt − te e dt ⎢ ⎥ ∫ −1 2π n ⎣ −1 ⎦ = 1 − jπ nt 1 ⎤ j ⎡ − jπ n + jπ n +e + e e ⎢ −1 ⎥ 2π n ⎣ jπ n ⎦ j ⎡ e − jπ n e + jπ n e − jπ n e jπ n ⎤ = ⎢ + + 2 2 − 2 2⎥ 2 ⎣ πn jπ n jπ n ⎦ πn je − jπ n j ( −1) = = . πn πn n c... n = ±1. n = ±1. ±2.. distributed. This document may not be copied. or posted on a website. ±2.... To = 2.. To = 2. Cn = 1. scanned. distributed. . ⎪ = ⎨ 2A n odd ⎪⎩ π 2 n 2 . 1 1 ⎤ An ⎡ 2 A 1 π π 1 cos( ) cos( ) t nt dt A nt dt A =⎢ − = − ( ) ⎥ ∫0 ∫0 t cos(π nt )dt 2 ⎣ 2 ∫0 ⎦ 1 1 ⎤ A A ⎡ 1 π π sin( ) sin( ) sin(π nt )dt ⎥ td nt dt t nt = 0− = − + ( ) ⎢ ∫ ∫ 0 πn 0 πn ⎣ 0 ⎦ 1 A ⎡ cos(π nt ) ⎤ A A 0 = − ⎢ ⎥ = − 2 2 [ cos(π n) − 1] = 2 2 [1 − cos(π n)] π n ⎢⎣ π n 0 ⎥⎦ π n π n n even ⎧0. f o = 1/ 3 1 1 3/ 2 ⎤ 1 ⎡1 1 ⎤ 2A ⎤ 2A ⎡t2 2A ⎡ 3/2 C0 = +t1 ⎥ = ⎢ + ⎥ = ⎢ ⎢ ∫ tdt + ∫ 1dt ⎥ = 3 ⎣0 ⎥⎦ 3 ⎣ 2 2 ⎦ 3 1 ⎦ 3 ⎢⎣ 2 0 Since x4 (t ) is an even function of time.5 3/2 ⎤ 2A ⎡1 ⎤ An ⎡ 2 A =⎢ = + x t nt dt t nt dt ( ) cos(2 / 3) cos(2 / 3) cos(2π nt / 3)dt ⎥ π π ⎥ ⎢∫ 4 ∫ ∫ 2 ⎣ 3 0 1 ⎦ 3 ⎣0 ⎦ Now 1 1 1 ⎤ ⎤ 3 ⎡ 3 ⎡ 1 π π td sin(2 nt / 3) t sin(2 nt / 3) sin(2π nt / 3)dt ⎥ = − ( ) ⎢∫ ⎥ ⎢ ∫ 0 2π n ⎣ 0 0 ⎦ 2π n ⎣ ⎦ 3 ⎡ 3 1⎤ π sin(2 n / 3) cos(2π nt / 3) 0 ⎥ = + ⎢ 2π n ⎣ 2π n ⎦ 3 ⎡ 3 sin(2π n / 3) + = [cos(2π n / 3) − 1]⎤⎥ ⎢ 2π n ⎣ 2π n ⎦ ∫ t cos(2π nt / 3)dt = 0 3/2 3 ∫ cos(2π nt / 3)dt = 2π n sin(2π nt / 3) 1 3/2 1 =− 3 2π n sin(2π n / 3) Substituting yields 21 © 2012 by McGraw-Hill Education. or posted on a website. duplicated. This document may not be copied. To = 3.Since x3 (t ) is an even function of time. Not authorized for sale or distribution in any manner. in whole or part. forwarded. Cn = d. This is proprietary material solely for authorized instructor use. forwarded. distributed. or posted on a website. P ( f ) = 2sinc ( 2 f ) e− j 4π f ⎡⎣1 − e− j 8π f ⎤⎦ + 4sinc ( 4 f ) e− j 4π f ⎡⎣1 − e− j 8π f ⎤⎦ The FS coefficients of a periodic signal x5 (t ) are now obtained as 22 © 2012 by McGraw-Hill Education. in whole or part. scanned. This is proprietary material solely for authorized instructor use. . This document may not be copied. 2A ⎛ 3 ⎞ ⎜ ⎟ 3 ⎝ 2π n ⎠ x5 (t ) = 2 ⎡ ⎛ 2π n ⎞ ⎤ 3A ⎢cos ⎜ 3 ⎟ − 1⎥ = 2π 2 n 2 ⎠ ⎦ ⎣ ⎝ ⎡ ⎛ 2π n ⎞ ⎤ ⎢cos ⎜ 3 ⎟ − 1⎥ ⎠ ⎦ ⎣ ⎝ ∞ ∑ p ( t − 8n ) n =−∞ where p ( t ) = Π ⎡⎣( t − 2 ) / 2 ⎤⎦ + Π ⎡⎣( t − 2 ) / 4 ⎤⎦ − Π ⎡⎣( t − 6 ) / 2 ⎤⎦ − Π ⎡⎣( t − 6 ) / 4⎤⎦ The FT of the pulse shape p ( t ) over [ 0.Cn = e. duplicated. To ] is given by To P ( f ) = ∫ p (t )e − j 2π ft dt 0 The FS coefficients of a periodic signal with basic pulse shape p ( t ) are given by Cn = 1 To To ∫ p(t )e − j 2π nf o t dt 0 Comparing yields Cn = 1 P ( f ) f = nf o To Now ℑ Π ⎡⎣( t − 2 ) / 2 ⎤⎦ ←⎯ → 2sinc ( 2 f ) e − j 4π f ℑ Π ⎡⎣( t − 2 ) / 4 ⎤⎦ ←⎯ → 4sinc ( 4 f ) e − j 4π f ℑ Π ⎡⎣( t − 6 ) / 2 ⎤⎦ ←⎯ → 2sinc ( 2 f ) e − j12π f ℑ Π ⎡⎣( t − 6 ) / 4 ⎤⎦ ←⎯ → 4sinc ( 4 f ) e − j12π f Therefore. Not authorized for sale or distribution in any manner. 5To ) be expressed as y (t ) = ∞ ∑C n =−∞ y n e j 2π nfot where Cny = = 1 1 y (t )e − j 2π nfot dt = ∫ x(t − 0. Not authorized for sale or distribution in any manner.5To ) 1 x(ν )e dν = e x(ν )e − j 2π nfoν dν ∫ ∫ To To To To Cnx = e − jπ nCnx Time Shifting introduces a linear phase shift in the FS coefficients. 23 © 2012 by McGraw-Hill Education.5To )e − j 2π nfot dt ∫ To To To To 1 − j 2π nf o (ν + 0.5To ) Solution: The FS expansion of a periodic pulse train x(t ) = ∞ ⎡ ( t − nTo ) ⎤ ⎥ of ⎣ τ ⎦ ∑ Π⎢ n =−∞ rectangular pulses is given by x(t ) = ∞ ∑C e n =−∞ x n j 2π nf o t where the exponential FS coefficients are Cnx = τ To sinc ( nf oτ ) Let the FS expansion of a periodic pulse train y (t ) = x(t − 0. This is proprietary material solely for authorized instructor use. in whole or part. or posted on a website. their magnitudes are not changed.23. . distributed.5To ) − j 2π nf o ( 0. forwarded. duplicated.Cn = = { } 1 2sinc ( 2nf o ) e− j 4π nfo ⎡⎣1 − e − j 8π nfo ⎤⎦ + 4sinc ( 4nf o ) e− j 4π nfo ⎡⎣1 − e− j 8π nfo ⎤⎦ To { } 1 sinc ( n / 4 ) e − jπ n /2 ⎡⎣1 − e − jπ n ⎤⎦ + 2sinc ( n / 2 ) e − jπ n /2 ⎡⎣1 − e − jπ n ⎤⎦ 4 2. This document may not be copied. compute the Fourier coefficients of the new periodic signal y(t) given by a.15 For the rectangular pulse train in Figure 2. y (t ) = x(t − 0. scanned. … α α α 2. y (t ) = x(t )e j 2π t / To Solution: Cny = = 1 1 y (t )e − j 2π nfot dt = ∫ x(t )e j 2π tfo e− j 2π nfot dt ∫ To To To To 1 − j 2π tf o ( n −1) x(t )e dt ∫ To To = Cnx−1 c.b. This is proprietary material solely for authorized instructor use. or posted on a website.5 with duty cycle 50%. This document may not be copied. in whole or part. y (t ) = x(α t ) Solution: Cny = 1 1 y (t )e − j 2π nfot dt = ∫ To To To ∫ x (α t )e − j 2π nf o t dt To ⎛f ⎞ − j 2π n ⎜ o ⎟ν 1 ⎝α ⎠ ( ) = x ν e dν α To α∫To = Cnx Time Scaling does not change FS coefficients but the FS itself has changed as the f 2f 3f harmonic components are now at the frequencies ± o . scanned. ± o .16 Draw the one-sided power spectrum for the square wave in Figure P2. Figure P2. duplicated. distributed. forwarded. ± o .5 Solution: The FS expansion for the square wave is given by 24 © 2012 by McGraw-Hill Education. . Not authorized for sale or distribution in any manner. x(t ) = ∞ ∑Ce n =−∞ j 2π nf o t n where 1 Cn = ∫ x(t )e − j 2π nfot dt To To To ⎧⎪ To / 2 − j 2π nf t ⎫⎪ − j 2π nf o t o A e dt − A e dt ⎨ ∫ ⎬ ∫ To / 2 ⎪⎩ 0 ⎪⎭ To / 2 To A = e − j 2π nfot − e− j 2π nfot 0 To /2 To ( − j 2π nf o ) 1 = To { = } jA − jπ n e − 1 − e− j 2π n + e− jπ n ) ( 2π n Therefore. This is proprietary material solely for authorized instructor use. Solution: 25 © 2012 by McGraw-Hill Education. This document may not be copied. or posted on a website. n odd otherwise Average power in the frequency component at f = nf o equals Cn . .4 0. in whole or part. duplicated.2 0. Not authorized for sale or distribution in any manner.8 One-sided Power Spectrum 0.6 0. ⎪− Cn = ⎨ π n ⎪⎩0.7 0. One-sided Power Spectrum of Square wave 0.1 0 0 5 10 15 Frequency (xfo) 20 25 a.9 0. distributed. scanned. ⎧ j2 A .3 0. Figure 2 displays the one-sided power spectrum for the square wave. forwarded.5 0. Calculate the normalized average power. 9775 0. Solution: n 1 3 5 7 9 11 13 15 17 19 21 2.6.17 Total Power including current Fourier coefficient 0. Figure P2.9816 98% Power bandwidth = 21× f o Determine the Fourier transforms of the signals shown in Figure P2. scanned.The normalized average power for a periodic signal x (t) is given by 1 Px = To To /2 ∫ x(t ) dt = 2 −To /2 A2To = A2 To b.9596 0. Determine the 98% power bandwidth of the pulse train.9331 0.9496 0.8106 0.9006 0.9711 0.9747 0. distributed.9663 0.6 26 © 2012 by McGraw-Hill Education. This document may not be copied. Not authorized for sale or distribution in any manner. forwarded. duplicated. This is proprietary material solely for authorized instructor use. or posted on a website. in whole or part.9798 0. . scanned.5 ( f − 1) ⎤⎦ + sinc ⎡⎣0. or posted on a website.5 ) ←⎯ → sinc ( f ) e jπ f ℑ Π ( t − 0.5 ) ←⎯ → sinc ( f ) e − jπ f Adding X 1 ( f ) = A ⎡⎣sinc ( f ) e jπ f − sinc ( f ) e− jπ f ⎤⎦ = Asinc ( f ) ⎡⎣e jπ f − e− jπ f ⎤⎦ = Aj 2sinc ( f ) sin (π f ) = j 2π fA ⎡⎣sinc 2 ( f ) ⎤⎦ Solution (b) The pulse x2 (t ) can be expressed as x2 (t ) = Π ( 2t ) cos ( 2π t ) Now 1 ℑ Π (2t ) ←⎯ → sinc( f / 2) 2 1 ℑ cos(2π t ) ←⎯ → ⎡⎣δ ( f − 1) + δ ( f + 1) ⎤⎦ 2 1 1 X 2 ( f ) = ℑ{Π (2t )} ⊗ ℑ{cos(2π t )} = sinc( f / 2) ⊗ ⎡⎣δ ( f − 1) + δ ( f + 1) ⎤⎦ 2 2 1 = ⎡⎣sinc ⎡⎣0. duplicated.Solution (a) The pulse x1 (t ) can be expressed as x1 (t ) = A ⎡⎣Π ( t + 0. distributed. Not authorized for sale or distribution in any manner. forwarded. we obtain ℑ Π ( t + 0. This is proprietary material solely for authorized instructor use. This document may not be copied.5 ( f + 1) ⎤⎦ ⎤⎦ 4 Solution (c) The pulse x3 (t ) can be expressed as 27 © 2012 by McGraw-Hill Education.5 ) ⎤⎦ Now ℑ Π ( t ) ←⎯ → sinc ( f ) Applying the time-shifting property of the FT. . in whole or part.5 ) − Π ( t − 0. scanned. Not authorized for sale or distribution in any manner. This document may not be copied. ℑ p ( −t ) ←⎯ → P (− f ) = 1 ⎡1 − e −(1− j 2π f ) ⎤ ⎦ 1 − j 2π f ⎣ X3 ( f ) = P ( f ) + P (− f ) = = = 1 1 −(1− j 2π f ) ⎡1 − e−(1+ j 2π f ) ⎤ + ⎡ ⎤ ⎦ 1 − j 2π f ⎣1 − e ⎦ 1 + j 2π f ⎣ {(1 − j 2π f ) ⎡⎣1 − e 1 + ( 2π f ) 1 2 2 1 + ( 2π f ) 2 − (1+ j 2π f ) } ⎤ + (1 + j 2π f ) ⎡1 − e−(1− j 2π f ) ⎤ ⎦ ⎣ ⎦ ⎡⎣1 − e−1 cos ( 2π f ) + 2π fe−1 sin ( 2π f ) ⎤⎦ Solution (d) The pulse x4 (t ) can be expressed as x4 (t ) = Π ( t / 2 ) + Π ( t / 4 ) Now 28 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. . forwarded. or posted on a website. distributed.2.x3 (t ) = p ( t ) + p (−t ) where p ( t ) = e − t ⎡⎣u ( t ) − u ( t − 1) ⎤⎦ From Table 2. we obtain e u ( t − 1) = e e −t −1 − ( t −1) e − j 2π f e −(1+ j 2π f ) = u ( t − 1) ←⎯→ e 1 + j 2π f 1 + j 2π f ℑ −1 Combining 1 e −(1+ j 2π f ) 1 ⎡1 − e−(1+ j 2π f ) ⎤ p(t ) ←⎯→ P ( f ) = − = ⎦ 1 + j 2π f 1 + j 2π f 1 + j 2π f ⎣ ℑ By time-reversal property. in whole or part. duplicated. we have ℑ e− t u (t ) ←⎯ → 1 1 + j 2π f Using the time-shifting property of the FT. 18 Use properties of the Fourier transform to compute the Fourier transform of following signals. in whole or part. This document may not be copied. .ℑ Π ( t / 2 ) ←⎯ → 2sinc ( 2 f ) ℑ Π ( t / 4 ) ←⎯ → 4sinc ( 4 f ) Adding X 4 ( f ) = 2sinc ( 2 f ) + 4sinc ( 4 f ) 2. Π (t / T ) cos(2π f c t ) Solution: ℑ Π ( t / T ) ←⎯ → Tsinc ( fT ) 1 ℑ cos(2π f ct ) ←⎯ → ⎡⎣δ ( f − f c ) + δ ( f + f c ) ⎤⎦ 2 1 X ( f ) = ℑ{Π ( t / T )} ⊗ ℑ{cos(2π f c t )} = Tsinc ( fT ) ⊗ ⎡⎣δ ( f − f c ) + δ ( f + f c ) ⎤⎦ 2 T sin c ⎡⎣T ( f − f c ) ⎤⎦ + sinc ⎡⎣T ( f + f c ) ⎤⎦ = 2 { c. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. or posted on a website. b. we obtain ℑ Wsinc 2 (Wt ) ←⎯ → Λ ( f / 2W ) ℑ → sinc 2 (Wt ) ←⎯ 1 Λ ( f / 2W ) W Thus the Fourier transform of a sinc2 pulse is a triangular function in frequency. forwarded. distributed. duplicated. scanned. sinc 2 (Wt ) Solution: τ ⎛ fτ ⎞ ℑ Λ (t / τ ) ←⎯ → sinc 2 ⎜ ⎟ 2 ⎝ 2 ⎠ Using the duality property. (e −t } cos10π t ) u (t ) Solution: 29 © 2012 by McGraw-Hill Education. a. te− t u (t ) Solution: Applying the differentiation in frequency domain property in (2. forwarded.From Table 2. −1 ⎡ ⎤ j d ⎣(1 + j 2π f ) ⎦ j 1 j 2π ℑ{te u (t )} = =− 2 2π df 2π (1 + j 2π f ) −t = e. e −π t 1 (1 + j 2π f ) 2 2 Solution: 30 © 2012 by McGraw-Hill Education.85).2. we obtain ℑ → te −t u (t ) ←⎯ j d ℑ{e − t u (t )} 2π df That is. in whole or part. . ℑ e− t u (t ) ←⎯ → 1 1 + j 2π f Now { } { } ℑ ( e − t u (t ) ) cos10π t = ℑ ( e −t u (t ) ) ⊗ ℑ{cos10π t} = 1 1 ⊗ ⎡⎣δ ( f − 5 ) + δ ( f + 5 ) ⎤⎦ 1 + j 2π f 2 = ⎤ 1⎡ 1 1 + ⎢ ⎥ 2 ⎣1 + j 2π ( f − 5 ) 1 + j 2π ( f + 5 ) ⎦ = 1 ⎡1 + j 2π ( f + 5 ) + 1 + j 2π ( f − 5 ) ⎤ ⎢ ⎥ 2 2 2 ⎢⎣ (1 + j 2π f ) − ( j 2π 5) ⎥⎦ ⎡ ⎤ 1 + j 2π f =⎢ ⎥ 2 2 ⎢⎣ (1 + j 2π f ) + 100π ⎥⎦ ⎡ ⎤ 1 + j 2π f ⎥ =⎢ 2 2 2 ⎢⎣ (1 + 100π ) + j 4π f − 4π f ⎥⎦ d. Not authorized for sale or distribution in any manner. duplicated. or posted on a website. This document may not be copied. distributed. scanned. This is proprietary material solely for authorized instructor use. 5 ( f − 50 ) ⎤⎦ + Λ ⎡⎣ 0. 4sinc (t ) cos (100π t ) 2 Solution: ⎛ f ⎞ ℑ 4sinc 2 ( t ) ←⎯ → 4Λ ⎜ ⎟ ⎝2⎠ 1 ℑ cos (100π t ) ←⎯ → ⎡⎣δ ( f − 50 ) + δ ( f + 50 ) ⎤⎦ 2 ⎛ f ⎞ 1 X ( f ) = ℑ{4sinc 2 ( t )} ⊗ ℑ{cos (100π t )} = 4Λ ⎜ ⎟ ⊗ ⎡⎣δ ( f − 50 ) + δ ( f + 50 ) ⎤⎦ ⎝2⎠ 2 { } = 2 Λ ⎡⎣0. we obtain X(f ) =e −π f 2 ∞ −πν ∫ e dν 2 −∞ 1 f. This is proprietary material solely for authorized instructor use. . ℑ x(t ) ⊗ y (t ) ←⎯→ X ( f )Y ( f ) Now ℑ → sinc(2Wt ) ←⎯ 1 Π ( f / 2W ) 2W 31 © 2012 by McGraw-Hill Education. or posted on a website. sinc(Wt ) ⊗ sinc(2Wt ) Solution: We use the convolution property of Fourier transform in (2.∞ X(f ) = ∫e −π t 2 e − j 2π ft ∞ dt = −∞ ∫e ( −π t 2 + j 2 ft ) dt −∞ Multiplying the right hand side by e −π f eπ f yields 2 X(f ) =e −π f 2 ∞ ( ∫e −π t 2 + j 2 ft − f 2 ) dt = e −π f −∞ 2 ∞ 2 −π ( t + jf ) dt ∫e 2 −∞ Substituting t + jf = ν . forwarded. distributed. Not authorized for sale or distribution in any manner. duplicated.79).5 ( f − 50 ) ⎤⎦ 2. scanned. in whole or part. This document may not be copied.19 Find the following convolutions: a. distributed. This document may not be copied. forwarded. duplicated. in whole or part. sinc 2 (Wt ) ⊗ sinc(2Wt ) Solution: Again using the convolution property of Fourier transform 1 1 Λ ( f / 2W ) × Π ( f / 2W ) 2W W 1 = Λ ( f / 2W ) 2W 2 ℑ → sinc 2 (Wt ) ⊗ sinc(2Wt ) ←⎯ 2. scanned. or posted on a website. . Not authorized for sale or distribution in any manner. 1 1 Π( f / W ) × Π ( f / 2W ) 2W W 1 = Π( f / W ) 2W 2 ℑ → sinc(Wt ) ⊗ sinc(2Wt ) ←⎯ b. v(t ) = x(5t − 1) Solution: ⎡ ⎛ 1 ⎞⎤ v(t ) = x(5t − 1) = x ⎢5 ⎜ t − ⎟ ⎥ ⎣ ⎝ 5 ⎠⎦ Let 1 ⎛ f ⎞ 1 1 X⎜ ⎟= 5 ⎝ 5 ⎠ 5 ( j 2π f / 5 ) + 5 1 = j 2π f + 25 ℑ y (t ) = x(5t ) ←⎯ →Y ( f ) = − j 2π f − j 2π f 1 ⎛ 1⎞ ℑ 5 = v(t ) = y ⎜ t − ⎟ ←⎯→ V ( f ) = Y ( f )e e 5 j 2π f + 25 ⎝ 5⎠ 32 © 2012 by McGraw-Hill Education.Therefore. This is proprietary material solely for authorized instructor use.20 The FT of a signal x(t ) is described by X(f ) = 1 5 + j 2π f Determine the FT V ( f ) of the following signals: a. Not authorized for sale or distribution in any manner. This document may not be copied. duplicated. distributed.b. This is proprietary material solely for authorized instructor use. forwarded. or posted on a website. in whole or part. . we can write 33 © 2012 by McGraw-Hill Education. v(t ) = x(t ) ⊗ u (t ) Solution: ℑ Using the convolution property of FT x(t ) ⊗ y (t ) ←⎯→ X ( f )Y ( f ) . we get V( f ) = j 2π f + 5 j 20π f + (π 2104 + 25 − 4π 2 f 2 ) c. scanned. v(t ) = x(t ) cos(100π t ) Solution: ⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ v(t ) = x(t ) cos(100π t ) = 2 Now ⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ ℑ 1 ←⎯→V ( f ) = ⎡⎣ X ( f − 50 ) + X ( f + 50 ) ⎤⎦ 2 2 1 1 = + j 2π ( f − 50 ) + 5 j 2π ( f + 50 ) + 5 After simplification. v(t ) = dt Solution: Using the differentiation property of FT V ( f ) = j 2π fX ( f ) = d ℑ x(t ) ←⎯→ j 2πfX ( f ) . we obtain dt j 2π f 5 + j 2π f e. v(t ) = x(t )e j10t Solution: 5⎞ 1 ⎛ ℑ →V ( f ) = X ⎜ f − ⎟ = v(t ) = x(t )e j10t ←⎯ 5⎞ π⎠ ⎛ ⎝ j 2π ⎜ f − ⎟ + 5 π⎠ ⎝ dx(t ) d. the impulse response of the delay element is given from (2. distributed. The frequency response function of the system is given by H( f ) = 2 2 + j 2π f Figure P2.⎛1 1 1 ⎞ ⎜ δ( f )+ ⎟ j 2π f ⎠ 5 + j 2π f ⎝ 2 1 1 1 = δ(f ) + 2 5 + j 2π f j 2π f ( 5 + j 2π f ) V ( f ) = X ( f )U ( f ) = = 1 1 δ(f )+ 10 j 2π f ( 5 + j 2π f ) 2. forwarded. What is the impulse response h(t)? Solution: Since y (t ) = h(t ) ⊗ x(t ) = x(t − 3) . What is the magnitude and phase response function of the system? Solution: H ( f ) = ℑ{δ (t − 3)} = e − j 6π f H( f ) =1 H ( f ) = −6π f 2. a. This is proprietary material solely for authorized instructor use. This document may not be copied.22 The periodic input x(t ) to an LTI system is displayed in Figure P2. or posted on a website. scanned.7 34 © 2012 by McGraw-Hill Education. Not authorized for sale or distribution in any manner.7. .21 Consider the delay element y (t ) = x(t − 3) .16) as h(t ) = δ (t − 3) b. in whole or part. duplicated. The exponential FS coefficients x(t ) are Cn = 0. That is. we showed that time shifting introduces a linear phase shift in the FS coefficients.5sinc ( 0.5n ) ⎤⎦e jπ nt n =−∞ b. This document may not be copied.5 ∑ sinc ( 0.a.24 with To = 2 ( f o = 0.5 ∑ ⎡⎣e − jπ n /2sinc ( 0. This is proprietary material solely for authorized instructor use. in whole or part. distributed.24 shifted by To / 4 and duty cycle τ = 0.5 is given by ∞ gTo (t ) = 0. To x(t ) = gTo ( t − To / 4 ) = ∞ ⎡ ( t − nTo − To / 4 ) ⎤ ⎥ 0. Solution: The input x(t ) is rectangular pulse train in Example 2.5 ) and τ To = 0. duplicated. The phase shift is equal to e − j 2π nfou for a time shift of u. Plot the magnitude and phase response functions for H ( f ) . Write the complex exponential FS of input x(t ) . forwarded.5n ) e − j 2π nf o ( 0. their magnitudes are not changed. or posted on a website.5sinc ( 0. scanned.5 .15(a). .5To ⎣ ⎦ ∑ Π⎢ n =−∞ The complex exponential FS of gTo (t ) from Example 2. Not authorized for sale or distribution in any manner.5n ) ∞ x(t ) = 0.25To ) = e − jπ n /2 0.5n )e jπ nt n =−∞ In Exercise 2. Solution: H( f ) = 2 1 = 2 + j 2π f 1 + jπ f H ( nf o ) = 1 1 = 1 + jπ f 1 + jπ nf o 35 © 2012 by McGraw-Hill Education. forwarded. in whole or part.0 0 -10 -5 Phase Response(degrees) Magnitude Response(dB) -20 -10 -15 -20 -30 -40 -50 -60 -70 -25 -80 -30 0 2 4 6 Frequency (f) 8 -90 0 2 4 6 Frequency (f) 8 c. This document may not be copied.0025π f .5n)e π j nt n =−∞ n ⎡ ⎤ ⎢ 0. This is proprietary material solely for authorized instructor use. Solution: Using (2. duplicated.23 The frequency response of an ideal LP filter is given by ⎧⎪5e − j 0. H( f ) = ⎨ ⎪⎩0. f < 1000 Hz f > 1000 Hz Determine the output signal in each of the following cases: a. scanned.114).5e − jπ n / 2 n =−∞ sinc ( 0. Not authorized for sale or distribution in any manner. . x(t ) = 5sin ( 400π t ) + 2 cos (1200π t − π / 2 ) − cos ( 2200π t + π / 4 ) 36 © 2012 by McGraw-Hill Education. distributed.5e − jπ n /2 ⎥ sinc ( 0. the output of an LTI system to the input x(t ) = ∞ ∑ ⎡⎣0. Compute the complex exponential FS of the output y (t ) . or posted on a website.5n ) ⎤⎦e jπ nt Cn is given by y (t ) = ∞ ∑ C H (0.5n ) ⎥ e jπ nt = ∑⎢ n =−∞ ⎢1 + j 05π n ⎥ FS coefficient of y ( t ) ⎣⎢ ⎦⎥ ∞ 2. 0025π f Π⎜ Π⎜ ⎟ 5e ⎟ = 5Π ⎜ ⎟e ⎝ 2200 ⎠ ⎝ 2000 ⎠ ⎝ 2000 ⎠ 37 © 2012 by McGraw-Hill Education. the output of the system for an input 2 cos (1200π t − π / 2 ) can now be expressed using (2. Next H (600) = 5 and H (600) = −3π / 2 = π / 2 . H( f ) = ⎨ ⎪⎩0. The output of the LP filter is therefore given by y (t ) = 25sin ( 400π t − π / 2 ) + 10 cos (1200π t ) b. the output of LP filter to sin ( 2200π t ) is πt ⎛ f ⎞ − j 0.0025π f . x(t ) = 2sin(400π t ) + sin ( 2200π t ) πt Solution: Since H (200) = 5 and H (200) = −π / 2 .0025π f ⎛ f ⎞ ⎛ f ⎞ − j 0. .Solution: ⎧⎪5. the output of the system for an input 2sin(400πt) is 10sin(400πt − π/2). So the LP filter doesn’t pass cos ( 2200π t + π / 4 ) . distributed. This document may not be copied. duplicated. or posted on a website.117) as 25sin(400πt − π/2). Not authorized for sale or distribution in any manner. H( f ) = ⎨ ⎪⎩0. in whole or part. we note that πt sin ( 2200π t ) ⎛ f ⎞ ℑ = 2200 × sinc ( 2200t ) ←⎯ →Π⎜ ⎟ πt ⎝ 2200 ⎠ In frequency domain. To calculate the response to sin ( 2200π t ) .117) as 10 cos (1200π t − π / 2 + π / 2 ) = 10 cos (1200π t ) . This is proprietary material solely for authorized instructor use. forwarded. ⎧− ⎪ 0. the output of the system for an input 5sin(400πt) can now be expressed using (2. scanned. Now H (1100) = 0. Since H (200) = 5 and f < 1000 Hz f > 1000 Hz f < 1000 Hz f > 1000 Hz H (200) = −π / 2 . the output of the system for an input cos(400π t ) is 5cos ( 400π t − π / 2 ) . Not authorized for sale or distribution in any manner.00125 ) ⎤⎦ ⎟e ⎝ 1000 ⎠ The output of the LP filter is therefore given by y (t ) = 5cos ( 400π t − π / 2 ) + 5000sinc ⎡⎣1000 ( t − 0. scanned.Now ⎛ f ⎞ − j 0. x(t ) = cos(400π t ) + sin (1000π t ) πt Solution: Since H (200) = 5 and H (200) = −π / 2 .00125 ) ⎤⎦ c.00125 ) ⎤⎦ d.0025π f Π ⎜ ⎟ = 5Π ⎜ ⎟e ⎝ 1000 ⎠ ⎝ 1000 ⎠ Now ⎛ f ⎞ − j 0.0025π f ℑ Π⎜ ←⎯→1000sinc ⎡⎣1000 ( t − 0. To calculate the response to sin (1000π t ) . x(t ) = 5cos(800π t ) + 2δ (t ) Solution: 38 © 2012 by McGraw-Hill Education. duplicated. forwarded. in whole or part. . This is proprietary material solely for authorized instructor use.0025π f 5e− j 0. we note that πt sin (1000π t ) ⎛ f ⎞ ℑ = 1000 × sinc (1000t ) ←⎯ →Π ⎜ ⎟ πt ⎝ 1000 ⎠ In frequency domain.0025π f ℑ Π⎜ ←⎯→ 2000sinc ⎡⎣ 2000 ( t − 0. distributed. This document may not be copied. the output of LP filter to sin (1000π t ) is πt ⎛ f ⎞ ⎛ f ⎞ − j 0.00125 ) ⎤⎦ ⎟e ⎝ 2000 ⎠ The output of the LP filter is therefore given by y (t ) = 10sin ( 400π t − π / 2 ) + 10000sinc ⎡⎣ 2000 ( t − 0. or posted on a website. This document may not be copied. x(t ) = 2 cos (1850π t − π / 2 ) − cos (1900π t + π / 4 ) Solution: H (925) = 2 and H (950) = 2 .00125 ) ⎤⎦ . 900< f < 1000 Hz. scanned. 39 © 2012 by McGraw-Hill Education. duplicated. H( f ) = ⎨ ⎪⎩0. This is proprietary material solely for authorized instructor use. distributed. forwarded.Since H (400) = 5 and H (400) = −π .0005π f .25 The frequency response of an ideal BP filter is given by ⎧⎪2e− j 0. f < 20 Hz Determine the output signal y (t ) for the input a. the output of the system for an input 5cos ( 800π t ) is 25cos ( 800π t − π ) . H( f ) = ⎨ otherwise ⎪⎩0. Not authorized for sale or distribution in any manner. The impulse response of the ideal LP filter is 10000sinc ⎡⎣ 2000 ( t − 0. The response of the system to δ (t ) is h(t ) . Combining y (t ) = 25cos ( 800π t − π ) + 2h(t ) = 25cos ( 800π t − π ) + 20 × 103 sinc ⎡⎣ 2000 ( t − 0.24 The frequency response of an ideal HP filter is given by ⎧⎪4.00125 ) ⎤⎦ 2. in whole or part. x(t ) = 5 + 2 cos ( 50π t − π / 2 ) − cos ( 75π t + π / 4 ) Solution: y (t ) = 8cos ( 50π t − π / 2 ) − 4 cos ( 75π t + π / 4 ) b. or posted on a website. . f > 20 Hz. Determine the output signal y (t ) for the input a. x(t ) = cos ( 20π t − 3π / 4 ) + 3cos (100π t + π / 4 ) Solution: y (t ) = 12 cos (100π t + π / 4 ) 2. 00025) ⎤⎦ = 2sinc ⎡⎣60 ( t − 0.462π H (950) = −0. in whole or part.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠ Therefore.0005π f = −0. Not authorized for sale or distribution in any manner.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣ 60 ( t − 0.00025 ) ⎤⎦ ⎡⎣e j1900π ( t −0. This document may not be copied. Therefore.0005π f 1 ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.H ( f ) = −0. y (t ) = sinc ⎡⎣60 ( t − 0. H (925) = −0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣60 ( t − 0. .0005π f 1 ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0. distributed.0005π × 950 = −0.00025) + e − j1900π ( t −0.0005π f .0005π f × 1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ × ⎢Π ⎜ ⎟+Π⎜ ⎟⎥ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Now 1 ⎛ f − 950 ⎞ ℑ j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e − j 0.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠ Similarly 1 ⎛ f + 950 ⎞ ℑ − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e− j 0.0005π f = −0.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.0005π × 925 = −0. This is proprietary material solely for authorized instructor use. forwarded.475π y (t ) = 4 cos (1850π t − π / 2 − 0.462π ) − 2 cos (1900π t + π / 4 − 0. duplicated.475π ) b.00025 ) ⎤⎦ 40 © 2012 by McGraw-Hill Education. x(t ) = sinc ( 60t ) cos(1900π t ) Solution: X(f ) = = 1 ⎛ f ⎞ 1 Π ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 60 ⎝ 60 ⎠ 2 1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Π⎜ ⎟+Π⎜ ⎟⎥ ⎢ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Y( f )= H( f )X ( f ) = 2e − j 0. scanned. or posted on a website. 0005π f 1 ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0.00025 ) ⎤⎦ ⎡⎣ e j1900π ( t −0. Solution: 41 © 2012 by McGraw-Hill Education. in whole or part. Find the energy density spectrum of the output of the filter.00025) ⎤⎦ = 2sinc 2 ⎡⎣30 ( t − 0.00025) 2 ×Λ⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0. Calculate the energy of the input signal and the output signal. distributed. Not authorized for sale or distribution in any manner.0005π f × 1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟+ Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Now 1 ⎛ f − 950 ⎞ ℑ 2 j1900π t ×Λ⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.00025 ) 2 ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0. . scanned.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠ Similarly 1 ⎛ f + 950 ⎞ ℑ 2 − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠ Therefore. duplicated.00025 ) ⎤⎦ 2. y (t ) = sinc 2 ⎡⎣30 ( t − 0. x(t ) = sinc 2 ( 30t ) cos(1900π t ) Solution: X(f ) = = 1 ⎛ f ⎞ 1 Λ ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 30 ⎝ 60 ⎠ 2 1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟ + Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Y( f )= H( f )X ( f ) = 2e − j 0.c. This is proprietary material solely for authorized instructor use.0005π f 1 ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.26 The signal 2e −2t u (t ) is input to an ideal LP filter with passband edge frequency equal to 5 Hz.00025) + e− j1900π ( t −0. This document may not be copied. or posted on a website. forwarded. . forwarded. in whole or part.∞ Ex = ∫ −∞ ∞ e −4t 2e u (t ) dt = ∫ 4e dt = 4 −4 0 2 −2 t ℑ x(t ) = 2e −2t u (t ) ←⎯ → ∞ −4 t =1 0 2 2 + j 2π f 2 The energy density spectrum of the output y(t). This document may not be copied. is related to the energy density spectrum of the input x(t) 2 2 ⎛ f ⎞ Y( f ) = H( f ) X( f ) = Π⎜ ⎟ ⎝ 10 ⎠ 2 + j 2π f 2 2 ∞ Ey = ∫ ∞ y (t ) dt = 2 −∞ ∫ Y( f ) ∞ 2 2 ∫ df = −∞ 5 2 2 −∞ 2 2 2 ⎛ f ⎞ df Π⎜ ⎟ ⎝ 10 ⎠ 2 + j 2π f 5 1 1 df = 2∫ df =∫ 1 + jπ f 1+ π 2 f 2 0 −5 Making change of variables π f = u ⇒ df = Ey = 2 π 5π 1 ∫ 1+ u 2 du = 0 2 π tan −1 ( u ) 5π 0 = 2 π 1 π du . 2. a. Not authorized for sale or distribution in any manner. we obtain (1. distributed. scanned. Y ( f ) . This is proprietary material solely for authorized instructor use.27 Calculate and sketch the power spectral density of the following signals. x(t ) = 2 cos (1000π t − π / 2 ) − cos (1850π t + π / 4 ) x1 ( t ) x2 ( t ) Solution: T /2 1 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) − Rx1x2 (τ ) − Rx2 x1 (τ ) T →∞ T ∫ −T /2 Now 42 © 2012 by McGraw-Hill Education. or posted on a website. Calculate the normalized average power of the signal in each case.507 ) = 0. duplicated.9594 Thus output of the LP filter contains 96% of the input signal energy. This is proprietary material solely for authorized instructor use. Therefore. This document may not be copied. scanned. Similarly. or posted on a website.T /2 1 Rx1 (τ ) = lim x1 (t ) x1 (t − τ )dt T →∞ T ∫ −T /2 1 π⎞ π⎤ ⎛ ⎡ 2 cos ⎜1000π t − ⎟ 2 cos ⎢1000π ( t − τ ) − ⎥ dt T →∞ T ∫ 2⎠ 2⎦ ⎝ ⎣ −T /2 T /2 = lim T /2 { } 1 cos (1000πτ ) + cos ⎡⎣1000π ( 2t − τ ) − π ⎤⎦ dt T →∞ T ∫ −T /2 = 2 lim = 2 cos (1000πτ ) The second term is zero because it integrates a sinusoidal function over a period. in whole or part. duplicated. T /2 1 1 x2 (t ) x2 (t − τ )dt = cos (1850πτ ) ∫ T →∞ T 2 −T / 2 Rx2 (τ ) = lim The cross-correlation term T /2 T /2 1 1 x1 (t ) x2 (t − τ ) = lim 2 cos (1000π t − π / 2 ) cos ⎡⎣1850π ( t − τ ) + π / 4 ⎤⎦dt ∫ T →∞ T T →∞ T ∫ −T / 2 −T / 2 Rx1 x2 (τ ) = lim T /2 1 = lim {cos (850π t − 1850πτ + 3π / 4 ) + sin ( 2850π t − 1850πτ − π / 4 )}dt T →∞ T ∫ −T / 2 is zero because it integrates a sinusoidal function over a period in each case. it can be shown that all other cross-correlation terms are zero. forwarded. distributed. Similarly. 1 Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) = 2 cos (1000πτ ) + cos (1850πτ ) 2 1 ⎧ ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨2 cos (1000πτ ) + cos (1850πτ ) ⎬ 2 ⎩ ⎭ 1 = [δ ( f − 500) + δ ( f + 500) ] + [δ ( f − 925) + δ ( f + 925) ] 4 The normalized average power is obtained by using (2. Not authorized for sale or distribution in any manner. .172) as 43 © 2012 by McGraw-Hill Education. ∞ ∞ −∞ −∞ 1 ∞ ∫ [δ ( f − 500) + δ ( f + 500)] df + 4 ∫ [δ ( f − 925) + δ ( f + 925)] df ∫ Gx ( f )df = Px = −∞ 1 1 5 = 1+1+ + = 4 4 2 Power Spectral Density 1 0. duplicated.2 0 0. This is proprietary material solely for authorized instructor use.4 -0. x(t ) = [1 + sin(200π t ) ] cos(2000π t ) Solution: x(t ) = [1 + sin(200π t ) ] cos(2000π t ) = cos(2000π t ) + sin(200π t ) cos(2000π t ) 1 1 = cos(2000π t ) + sin ( 2200π t ) − sin (1800π t ) 2 2 x (t ) 1 x2 ( t ) x3 ( t ) Now T /2 1 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) T →∞ T ∫ −T /2 where 1 cos ( 2000πτ ) 2 1 Rx2 (τ ) = cos ( 2200πτ ) 8 1 Rx3 (τ ) = cos (1800πτ ) 8 Rx1 (τ ) = 44 © 2012 by McGraw-Hill Education.2 0 -1 -0.2 Frequency (kHz) 0.6 0. . This document may not be copied.8 -0.4 0. in whole or part.6 0.6 -0. scanned.8 0. forwarded.4 0.8 1 b. or posted on a website. distributed. Not authorized for sale or distribution in any manner. 172) as ∞ ∞ ∞ 1 1 Px = ∫ Gx ( f )df = ∫ [δ ( f − 1000) + δ ( f + 1000) ] df + ∫ [δ ( f − 1100) + δ ( f + 1100)] df 4 −∞ 16 −∞ −∞ ∞ + 1 1 1 1 3 [δ ( f − 900) + δ ( f + 900)] df = + + = ∫ 16 −∞ 2 8 8 4 Power Spectral Density 0. duplicated. scanned. Not authorized for sale or distribution in any manner.Therefore.4 -0.25 0. -1 -0. 1 1 1 Rx (τ ) = cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) 2 8 8 1 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) ⎬ 8 8 ⎩2 ⎭ 1 1 = [δ ( f − 1000) + δ ( f + 1000) ] + [δ ( f − 1100) + δ ( f + 1100) ] 4 16 1 + [δ ( f − 900) + δ ( f + 900) ] 16 The normalized average power is obtained by using (2.2 Frequency (kHz) 0. distributed.1 0. .05 0 c. in whole or part.2 0.8 -0. This document may not be copied. forwarded.6 -0.8 1 x(t ) = cos 2 ( 200π t ) sin (1800π t ) Solution: 45 © 2012 by McGraw-Hill Education. or posted on a website.6 0.2 0 0. This is proprietary material solely for authorized instructor use.15 0.4 0. duplicated. . or posted on a website. in whole or part. 1 1 1 Rx (τ ) = cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) 8 32 32 1 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) ⎬ 32 32 ⎩8 ⎭ 1 1 = [δ ( f − 900) + δ ( f + 900) ] + [δ ( f − 700) + δ ( f + 700) ] 16 64 1 + [δ ( f − 1100) + δ ( f + 1100) ] 64 The normalized average power is obtained by using (2. Not authorized for sale or distribution in any manner. This document may not be copied.1 1 1 ⎡⎣1 + cos ( 400π t ) ⎤⎦ sin (1800π t ) = sin (1800π t ) + sin (1800π t ) cos ( 400π t ) 2 2 2 1 1 1 = sin (1800π t ) + sin (1400π t ) + sin ( 2200π t ) 2 4 4 x(t ) = Now Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) where 1 Rx1 (τ ) = cos (1800πτ ) 8 Rx2 (τ ) = 1 cos (1400πτ ) 32 Rx3 (τ ) = 1 cos ( 2200πτ ) 32 Therefore. This is proprietary material solely for authorized instructor use. distributed. forwarded. scanned.172) as ∞ ∞ ∞ 1 1 Px = ∫ Gx ( f )df = [δ ( f − 900) + δ ( f + 900)]df + ∫ [δ ( f − 700) + δ ( f + 700)] df ∫ 16 −∞ 64 −∞ −∞ ∞ + 1 1 1 1 3 [δ ( f − 1100) + δ ( f + 1100)] df = + + = ∫ 64 −∞ 8 32 32 16 46 © 2012 by McGraw-Hill Education. scanned. distributed. or posted on a website. This document may not be copied. in whole or part.07 0.8 -0.6 -0.6 0.0. duplicated.4 0.01 0 -1 -0.06 0. Not authorized for sale or distribution in any manner.8 1 47 © 2012 by McGraw-Hill Education.08 0.09 Power Spectral Density 0. forwarded.2 0 0.4 -0.1 0.05 0.04 0. This is proprietary material solely for authorized instructor use.03 0.02 0. .2 0.