University of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing

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Oswald Baldric Hall
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1 University of Illinois at Urbana-Champaign ECE 0: Digital Signal Processing Chandra Radhakrishnan PROBLEM SET : SOLUTIONS Peter Kairouz Problem. Hz z 7 z +/9, causal ROC z > contains the unit circle BIBO stable. Hz z 7 z +/9, anticausal ROC z < doesn t contain the unit circle NOT BIBO stable. Input example: for x[n] δ[n], output y[n] n cos [n ]π/u[ n] + 7 n cos [n ]π/u[ n + ] is unbounded. z. Hz z 0.7z +z+, h[n] is two-sided ROC 0.7 < z < doesn t contain the unit circle NOT BIBO stable. Xz z z +z+ and Y z Input example: for x[n] sin nπ/u[n], z z 0.7z +z+. Therefore, y[n] contains a term ne j nπ, which is unbounded.. Hz z+ z, causal ROC z > doesn t contain the unit circle NOT BIBO stable. Input example: for x[n] u[n], Y z z +z z. Therefore, y[n] contains a term of the form nu[n], which is unbounded. 5. Hz z z +j, causal ROC z > doesn t contain the unit circle NOT BIBO stable. Input example: for x[n] cos nπ/u[n], Y z has double poles at e jπ/. Therefore, y[n] contains a term of the form ne jπ/n u[n], which is unbounded. Problem a Let the output of the two adders be u[n] and v[n], Uz z zxz + V z 8z + 80Xz Uz z 0 V z 5z Xz + Uz 5z + 80 V z z 0 Xz Y z Uz + V z z + 60 z 0 Xz Y zz 0 Xzz + 60 y[n + ] 0y[n] x[n + ] + 60x[n] b The direct form structure is shown in Fig. Problem Hz Y z z + 60 Xz z 0 z + 60z 0z

2 Using the MATLAB program in c as in the code below: % Part d a Let m a n + Let, m then n +, then xn [ ] yn [ ] z x [ 5 0 ]; h [ 0 0 ]; cconv len lengthx+lengthh ; zp x padarrayx, [0 cconv len lengthx], 0, 'post'; zp h padarrayh, [0 cconv len lengthh], z 0, 'post'; y circular convzp x,zp h; gives {y n } 0 60 n0 {, 7,,, 0,,, 9, 0, 8,,, }. Therefore, both approaches give the same result. Figure : Direct Form II structure for Problem b Problem 0 0 y[m] y[m] y[m y[m ] ] + x[m] + x[m] + x[m + x[m ] ] y[n] y[n] y[n y[n ] ] x[n] x[n] x[n x[n ] ] b Two possible b Two block possible diagrams blockare diagrams given below: are given below: Direct Form I Direct Form II c For the LSI system, the output Figure y[n] : is Filter the impulse structure response for Problem h[n] when b x[n] δ[n]: h[n] h[n ] δ[n] δ[n ] c For the LSI system, the output y[n] is the impulse response h[n] when x[n] δ[n]: Using the zero initial conditions h[ ] h[ ] 0, h[n] h[n ] δ[n] δ[n ] Using the zero initial conditions h[ ] h[ ] 0, n 0: h[0] h[ ] h[0] n : h[] h[ ] h[] n : h[n] h[n ] 0 z 0 h[n] A n + B n h[0] A + B h[] A + B A, B h[n] n u[n] Therefore, y[ ] 0, y[0], y[], y[], y[] d From y[n] y[n ] x[n] x[n ] in a, Y z z Xz z Hz Y z z Xz z

3 e Therefore, Hz z z z Hz z z z + z A z + + B z A z z z B z z + z Hz z z + z where system is causal z h[n] n u[n] h[ ] 0, h[0], h[], h[], h[] which is identical to the samples of the unit pulse response computed above via direct iteration of the difference equation. Problem. y[n] x[n] + x[n 0] Y z + z 0 Xz Hz + z 0 H d ω + e j0ω e j5ω cos 5ω H d ω cos 5ω { 5ω for cos 5ω 0 H d ω 5ω + π for cos 5ω < 0.5 H d ω ω x π rad/sec pi/ H d ω pi/ ω x π rad/sec Figure : Magnitude and Phase for Problem. Note h[n] is real since H d ω H d ω.

4 a x[n] cos 0 n + sin 0 n + π y[n] H d 0 cos 0 n + H d 0 y[n] sin n + π 0 + π π y[n] sin n + π 0 + H d sin n + π 0 + H d b x[n] cos 5 n + π y[n] 0H d H d 5 cos 5 n + π + H d 5 y[n] cos 5 n + π Problem 5 n u[n ] h [n] 7 n u[ n] n u[ n] + n u[n ] 7 7 n u[ n ] + 7 δ[n] + n u[n] 7 7 δ[n] 7 n u[ n ] + n u[n] 6δ[n] 7 H z 6 z + z 7 z 7 z, < z < 7 zz 7 zz 6z z z z, z z z, < z < h [n] δ[n] n u[ n] δ[n] n u[ n] δ[n] n u[ n ] δ[n] δ[n] + n u[ n ] H z + z, z < z z z, z < z z H zh z z z z, < z < z z z, z > < z <

5 H z is not causal, BIBO stable H z is not causal, BIBO stable H zh z is causal, BIBO stable. Problem 6 The inputs correspond to outputs Therefore, x [n], x [n] cos n + 0 sin n + 00, x [n] sin n + 5 y [n] 9, y [n] sin n + 0, y [n] 0 H d 0 H d e j90 H d 0 x[n] 5 + sin n cos π n sin n cos n 5 ỹ[n] 5H d 0 + H d sin n H d + 0 H d cos n 5 + H d 5 + sin n 75 Problem 7 Note h[n] is NOT real since H d ω H d ω.. x[n] 5 + 0e j n+5 + j n 5 + 0e j n+ π + e j π n y[n] 5H d 0 + 0e j n+ π H d + ej π n H d 0 + 0e j n+ π π e j + e j π n π ej 5π ej n+ π + + π ej n+. x[n] cos n j n 5 + 5e j n+ π + 5e j n+ π + e j π n y[n] 5H d 0 + 5e j n+ π H d + 5e j n+ π H d π + ej π n H d 0 + 5e j n+ π π e j 5e j n+ π π 5jπ sin π n + π + + π ej n+ e j + e j π n π ej Problem 8 5

6 a Y z z, z > z z z z, z > Xz z, z > z 6z 6z, z > z z Hz z 6z, subject to z > 6z z, subject to z > One-sided, causal solution ROC: z > 6: h[n] n + 6n u[n] Two-sided solution ROC: < z < 6: h[n] n u[n] 6n u[ n ] b The solution is not unique. There are two possible answers as shown above. The system is unstable for both solutions. Therefore, even if the system is known to be unstable, there are two possible solutions. If the system is given to be causal, then the ROC is z > 6 and there is a unique, one-sided solution for h[n]. Problem 9 a The poles are at re ±jω0 and zeros at e ±jω0. The sketch is shown in Fig. 0 0 Figure : Pole-zero plot for Problem 9a 6

7 b For ω ω 0, H d ω 0 0. For values of ω 0, the poles and zeros cancel resulting in H d ω. c where ω 0 π. Then Hω G e jω0 e jω e jω0 e jω re jω0 e jω re jω0 e jω G [ cosω ω0 + r rcosω ω 0 d Hω dω 0 ω π Hπ G + r + r G + r + r ] [ cosω + ω0 + r rcosω + ω 0 ] d The direct form II structure is shown in Fig. 5. Figure 5: Pole-zero plot for Problem 9d Problem 0 All three systems are causal. a H z Y z 7 z Y z + z Y z Xz + z Xz zz + z 7 z + zz + z z Hence ROC : z >. The system is not BIBO stable. Similarly, H z z z zz + 7

8 ROC: z >, BIBO stable. H z zz z + 5 z ROC: z >. BIBO stable. b Since H z and H z are BIBO stable we can connect them in series in any order to create a BIBO stable system. H zh z z z z z + z + 5 z Note: We can also have cascade of H z and H z. In this case H z must be the first system in the cascade. H zh z 8 z z + z z + z ROC: z > c Y zz + 5 z z Xz8 z z + Y z z 9 0 z + 8 Xz 0 z + z y[n] 9 0 y[n ] + 8 y[n ] 0 X[n] X[n ] Problem a By looking at the plots, the only system that is BIBO stable is system C, whose poles all lie within the unit circle. The other systems have poles have outside the unit circle. b z z + H A z z 0.5z e j π z e j π z z + z 0.5 H B z z e j π z e j π z e j π z e j π z e j π z e j π z e j π z e j π H C z z 0.5e j π z 0.5e j π z 0.5e j π z 0.5e j π z 0.5z H D z z z + c H BC z H B zh C z z z + z 0.5 z 0.5e j π z 0.5e j π z 0.5e j π z 0.5e j π Since the poles are inside the unit circle the system is BIBO stable. 8

9 Now compute the inverse z-transform, 9 H BC z j j z 0.5e j π z 0.5e j π + + j z 0.5e j + j π z 0.5e j π 9 n h BC [n] j 0.5e j π 9 u[n ] + + j 0.5e + + j 0.5e j π j π n u[n ] n u[n ] + j 0.5e j π n u[n ] d H AD z H A zh D z Since the poles are outside the unit circle the system is NOT BIBO stable. The inverse z-transform is, z z e j π z e j π H AD z j z e j + + j π z e j π e h AD [n] j j π n u[n ] + + j e j π n u[n ] 9

Signals and Systems. Problem Set: The z-transform and DT Fourier Transform

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