ECE 301: Signals and Systems Homework Assignment #6

1 ECE 301: Signals and SystemsHomework Assignment #6 Due on November 30, 2015 Professor:Aly El GamalTA:Xianglun Mao1 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 1 Note: Homework 6 will have only 4 problems since there will be a midterm exam and then Thanksgivingbreak, each problem is assigned 1 Letx1[n] be the discrete-time signal whose Fourier transformX1(ejw) is depicted in Figure 1(a).(a) Consider the signalx2[n] with Fourier transformX2(ejw), as illustrated in Figure 1(b). Expressx2[n]in terms ofx1[n]. (Hint: First expressX2(ejw) in terms ofX1(ejw), and then use properties of theFourier transform.)(b) Repeat part (a) forx3[n] with Fourier transformX3(ejw), as shown in Figure 1(c).(c) Let = n= nx1[n] n= x1[n]This quantity, which is the center of gravity of the signalx1[n], is usually referred to as thedelay timeofx1[n].

2 Find . (You can do this without first determiningx1[n] explicitly.)(d) Consider the signalx4[n] =x1[n] h[n], whereh[n] =sin( n/6) nSketchX4(ejw).Figure 1: The graph of Fourier transforms of signalsX1(ejw),X2(ejw),X3(ejw).Problem 1 continued on next page..2 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 1 (continued)Solution(a) We may expressX2(ejw) asX2(ejw) = Re{X1(ejw)}+ Re{X1(ej(w 2 /3))}+ Re{X1(ej(w+2 /3))}.Therefore,x2[n] = Ev{x1[n]}[1 +ej2 /3+e j2 /3].(b) We may expressX3(ejw) asX3(ejw) = Im{X1(ej(w ))}+ Im{X1(ej(w+ ))}.Therefore,x3[n] = Od{x1[n]}[ejwn+e jwn] = 2( 1)nOd{x1[n]}.(c) We may express as =jdX1(ejw)dw|w=0X1(ejw)|w=0=j( 6j/ )1=6 (d) Using the fact thatH(ejw) is the frequency response of an ideal lowpass filter with cutoff frequency 6, we may drawX4(ejw) as shown in Figure 2: The graph of Fourier transforms of signalsX4(ejw).

3 3 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 2 Problem 2 The signalsx[n] andg[n] are known to have Fourier transformsX(ejw) andG(ejw), respectively. Further-more,X(ejw) andG(ejw) are related as follows:12 X(ej )G(ej(w ))d = 1 +e jw(a) Ifx[n] = ( 1)n, determine a sequenceg[n] such that its Fourier transformG(ejw) satisfies the aboveequation. Are there other possible solutions forg[n]?(b) Repeat the previous part forx[n] = (12)nu[n].SolutionLet12 X(ej )G(ej(w ))d = 1 +e jw=Y(ejw)Taking the inverse Fourier transform of the above equation, we obtaing[n]x[n] = [n] + [n 1] =y[n].(a) Ifx[n] = ( 1)n,g[n] = [n] [n 1](b) Ifx[n] = (1/2)nu[n],g[n] has to be chosen such thatg[n] = 1,n= 02,n= 10,n >1any value,otherwiseTherefore, there are many possible choices forg[n].

4 4 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 3 Problem 3In lecture, we indicated that the continuous-time LTI system with impulse responseh(t) =W sinc(Wt ) =sin(Wt) tplays a very important role in LTI system analysis. The same is true of the discrete-time LTI system withimpulse responseh[n] =W sinc(Wn ) =sin(Wn) n(a) Determine and sketch the frequency response for the system with impulse responseh[n].(b) Consider the signalx[n] =sin( n8) 2cos( n4).Suppose that this signal is the input to LTI Systems with the following impulse responses. Determineand sketch the frequency response of the output in each case.(i)h[n] =sin( n/6) n(ii)h[n] =sin( n/6) n+sin( n/2) n(iii)h[n] =sin( n/6)sin( n/3) 2n2(iv)h[n] =sin( n/6)sin( n/3) n(c) Consider an LTI system with unit sample responseh[n] =sin( n/3) and sketch the frequency response of the output in each case.

5 (i)x[n] = the square wave depicted in Figure 3.(ii)x[n] = k= [n 8k](iii)x[n] = ( 1)ntimes the square wave depicted in Figure 3.(iv)x[n] = [n+ 1] + [n 1]Figure 3: The square wave that constructx[n].Solution(a) The frequency response of the system is as shown in Figure 4.(b) The Fourier transformX(ejw) ofx[n] is as shown in Figure 4.(i) The frequency responseH(ejw) is as shown in Figure 5. Therefore,y[n] =sin( n/8).Problem 3 continued on next page..5 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)(ii) The frequency responseH(ejw) is as shown in Figure 5. Therefore,y[n] = 2sin( n/8) 2cos( n/4).(iii) The frequency responseH(ejw) is as shown in Figure 5. Therefore,y[n] =16sin( n/8) 14cos( n/4).

6 (iv) The frequency responseH(ejw) is as shown in Figure 5. Therefore,y[n] = sin( n/4).(c)(i) The signalx[n] is periodic with period 8. The Fourier series coefficients of the signal areak=187 n=0x[n]e j(2 /8) Fourier transform of this signal isX(ejw) = k= 2 ak (w 2 k/8).The Fourier transformY(ejw) of the output isY(ejw) =X(ejw)H(ejw). Therefore,Y(ejw) = 2 [a0 (w) +a1 (w /4) +a 1 (w+ /4)]in the range 0 |w| . Therefore,y[n] =a0+a1ej n/4+a 1e j n/4=58+ [(1/4) + (1/2)(1/ 2)]cos( n/4).(ii) The signalx[n] is periodic with period 8. The Fourier series coefficients of the signal areak=187 n=0x[n]e j(2 /8) Fourier transform of this signal isX(ejw) = k= 2 ak (w 2 k/8).The Fourier transformY(ejw) of the output isY(ejw) =X(ejw)H(ejw).

7 Therefore,Y(ejw) = 2 [a0 (w) +a1 (w /4) +a 1 (w+ /4)]in the range 0 |w| . Therefore,y[n] =a0+a1ej n/4+a1e j n/4=18+14cos( n/4).(iii) The Fourier transformX(ejw) of the signalx[n] is of the form shown in part (i). Therefore,y[n] =a0+a1ej n/4+a1e j n/4=18+ [(1/4) (1/2)(1/ 2)]cos( n/4).(iv) The output isy[n] =h[n] x[n] =sin[ /3(n 1)] (n 1)+sin[ /3(n+ 1)] (n+ 1).Problem 3 continued on next page..6 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)Figure 4: The resulting frequency 3 continued on next page..7 Aly El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 3 (continued)Figure 5: The resulting frequency El GamalECE 301: Signals and Systems Homework Assignment #6 Problem 4 Problem 4An LTI systemXwith impulse responseh[n] and frequency responseH(ejw) is known to have the propertythat, when w0 ,cos(w0n) w0cos(w0n)(a) DetermineH(ejw).

8 (b) Determineh[n].Solution(a) From the given information, it is clear that when the input to the system is a complex exponentialfrequencyw0, the output is a complex exponential of the same frequency but scaled by the|w0|.Therefore, the frequency response of the system isH(ejw) =|w|,for 0 |w0| .Note thatH(ejw) here should be a well-defined frequency response regardless ofw0.(b) Taking the inverse Fourier transform of the frequency response, we obtainh[n] =12 H(ejw)ejwndw=12 0 wejwndw+12 0wejwndw=1 0wcos(wn)dw=1 [cos(n ) 1n2]9