ECE 301: Signals and Systems Homework Assignment #5

Transcription

1 ECE 30: Signals and Systems Homework Assignment #5 Due on November, 205 Professor: Aly El Gamal TA: Xianglun Mao

2 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem Problem Compute the Fourier transform of each of the following signals (a) (b) (c) (d) x(t) { + cos(πt), t 0, t > a k δ(t kt ), a < k0 x(t) [te 2t sin(4t)]u(t) { t 2, 0 < t < 0, otherwise (e) x(t) as show in Figure. Figure : The graph of signal x(t) in (e). Solution (a) 2sin(w) w x(t)e jwt dt ( + cos(πt))e jwt dt + sin(w) π w sin(w) π + w (b) x(t)e jwt dt ( a k e jwkt δ(t kt ))dt k0 a k e jwkt k0 αe jwt Problem continued on next page... 2

3 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem (continued) (c) We have Therefore, x(t) (/2j)te 2t e j4t u(t) (/2j)e 2t e j4t u(t). /2j (2 j4 + jw) 2 /2j (2 + j4 jw) 2. (d) 0 x(t)e jwt dt ( t 2 )e jwt dt jw + 2e jw w 2 2e jw 2 jw 2 (e) If then Therefore, x (t) δ(t 2k), x(t) x (t) + x (t ) X (jw)[2 + e w ] π δ(w kπ)[2 + ( ) k ] k 3

4 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 2 Problem 2 Consider the signal x 0 (t) { e t, 0 t 0, elsewhere Determine the Fourier transform of each of the signals shown in Figure 2. You should be able to do this by explicitly evaluating only the transform of x 0 (t) and then using properties of the Fourier transform. Solution Figure 2: The graph of signals x (t), x 2 (t), x 3 (t), x 4 (t). For the given signal x 0 (t), we use the Fourier transform analysis to evaluate the corresponding Fourier transform X 0 (jw) e (+jw) + jw. (a) We know that x (t) x 0 (t) + x 0 ( t). Using the linearity and time reversal properties of the Fourier transform we have (b) We know that X (jw) X 0 (jw) + X 0 ( jw) 2 2e cos(w) 2we sin(w) + w 2 x 2 (t) x 0 (t) x 0 ( t). Using the linearity and time reversal properties of the Fourier transform we have X 2 (jw) X 0 (jw) X 0 ( jw) j[ 2w + 2e sin(w) + 2we cos(w) + w 2 ] Problem 2 continued on next page... 4

5 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 2 (continued) (c) We know that x 3 (t) x 0 (t) + x 0 (t + ). Using the linearity and time reversal properties of the Fourier transform we have (d) We know that X 3 (jw) X 0 (jw) + e jw X 0 ( jw) + ejw e ( + e jw ) + jw x 4 (t) tx 0 (t). Using the linearity and time reversal properties of the Fourier transform we have X 4 (jw) j d dw X 0(jw) 2e e jw jwe e jw ( + jw) 2 5

6 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 3 Problem 3 Determine which, if any, of the real signals depicted in Figure 3 have Fourier tranforms that satisfy each of the following conditions: (a) Re{X(jw)} 0 (b) Im{X(jw)} 0 (c) There exists a real α such that e jαw X(jw) is real (d) X(jw)dw 0 (e) wx(jw)dw 0 (f) X(jw) is periodic Figure 3: The graph of real signals (a), (b), (c), (d), (e), (f). Problem 3 continued on next page... 6

7 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 3 (continued) Solution (a) For Re{X(jw)} 0, the signal x(t) must be real and odd. Therefore, signals in figures (a) and (d) have this property. (b) For Im{X(jw)} 0, the signal x(t) must be real and even. Therefore, signals in figures (e) and (f) have this property. (c) For there to exist a real α such that e jαw X(jw) is real, we require that x(t + α) be a real and even signal. Therefore, signals in figures (a), (b), (e), and (f) have this property. (d) For this condition to be true, x(0) 0. Therefore, signals in figures (a), (b), (c), (d), and (f) have this property. (e) For this condition to be true the derivative of x(t) has to be zero at t 0. Therefore, signals in figures (b), (c), (e), and (f) have this property. (f) The signal in figure (a) and (b) have this property. Figure (a) has this property because it s the integration of shifted square waves. The Fourier transform of the signal that shown in Figure (b) is 2e jw, which is periodic. 7

8 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 4 Problem 4 The input and the output of a stable and causal LTI system are related by the differential equation d 2 y(t) dt 2 (a) Find the impulse response of this system. + 6 dy(t) dt (b) What is the response of this system if x(t) te 2t u(t)? + 8y(t) 2x(t) (c) Repeat part (a) for the stable and causal LTI system described by the equation Solution d 2 y(t) dt dy(t) dt + y(t) 2 d2 x(t) dt 2 2x(t) (a) Taking the Fourier transform of both sides of the given differential equation, we obtain Therefore, we have Using partial fraction expansion, we obtain Taking the inverse Fourier transform, w 2 Y (jw) + 6jwY (jw) + 8Y (jw) 2X(jw) H(jw) Y (jw) 2 w 2 + 6jw + 8 H(jw) (b) For the given signal x(t) te 2t u(t), we have Therefore, Using partial fraction expansion, we obtain Taking the inverse Fourier transform, jw + 2 jw + 4 h(t) e 2t u(t) e 4t u(t) (2 + jw) 2. 2 Y (jw) X(jw)H(jw) ( w 2 + 6jw + 8) (2 + jw) 2. Y (jw) /4 jw + 2 /2 (jw + 2) 2 + (jw + 2) 3 /4 jw + 4 y(t) 4 e 2t u(t) 2 te 2t u(t) + t 2 e 2t u(t) 4 e 4t u(t) (c) Taking the Fourier transform of both sides of the given differential equation, we obtain w 2 Y (jw) + 2jwY (jw) + Y (jw) 2w 2 X(jw) 2X(jw) Problem 4 continued on next page... 8

9 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 4 (continued) Therefore, we have Using partial fraction expansion, we obtain Taking the inverse Fourier transform, H(jw) Y (jw) 2( w 2 ) w 2 + 2jw + H(jw) j jw 2+j j jw 2 j 2 2 h(t) 2δ(t) 2( + 2j)e (+j)t/ 2 u(t) 2( 2j)e ( j)t/ 2 u(t) 9

10 Aly El Gamal ECE 30: Signals and Systems Homework Assignment #5 Problem 5 Problem 5 A causal and stable LTI system S has the frequency response H(jw) jw w 2 + 5jw (a) Determine a differential equation relating the input x(t) and output y(t) of S. (b) Determine the impulse response h(t) of S. (c) What is the output of S when the input is Solution x(t) e 4t u(t) te 4t u(t)? (a) We have Y (jw) jw w 2 + 5jw. Cross-multiplying and taking the inverse Fourier transform, we obtain (b) We have (c) For d 2 y(t) dt dy(t) dt Taking the inverse Fourier transform we obtain, we have Therefore, + 6y(t) dx(t) dt jw + 4 H(jw) 6 w 2 + 5jw jw 3 + jw h(t) 2e 2t u(t) e 3t u(t) x(t) e 4t u(t) te 4t u(t), Y (jw) X(jw)H(jw) 4 + jw (4 + jw) x(t) (4 + jw)(2 + jw) /2 2 + jw /2 4 + jw Finding the partial fraction expansion of Y (jw) and taking the inverse Fourier transform, y(t) 2 e 2t u(t) 2 e 4t u(t), 0