Digital Signal Processing, Homework 2, Spring 2013, Prof. C.D. Chung. n; 0 n N 1, x [n] = N; N n. ) (n N) u [n N], z N 1. x [n] = u [ n 1] + Y (z) =

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1 Digital Signal Processing, Homework, Spring 0, Prof CD Chung (05%) Page 67, Problem Determine the z-transform of the sequence n; 0 n N, x [n] N; N n x [n] n; 0 n N, N; N n nx [n], z d dz X (z) ) nu [n], z d dz z nu [n], ( z ) jzj > nu [n] (n N) u [n N] z jzj > x [n n 0 ], X (z) z n 0 ) (n N) u [n N], z N ( z ) jzj > Therefore, X (z) z z N ( z ) z z N ( z ) (05%) Page 68, Problem 7 The input to a causal LTI system is x [n] u [ n ] + The z-transform of the output of this system is n u [n] Y (z) z z ( + z ) (a) Determine H (z), the z-transform of the system impulse response specify the ROC (b) What is the ROC for Y (z)? (c) Determine y [n] (a) ) X (z) x [n] u [ n ] + n u [n] z + z < jzj < Be sure to

2 Now to nd H (z) we simply use H (z) Y (z) X (z); ie, H (z) Y (z) X (z) H (z) causal ) ROC jzj < z z ( + z ) ( z ) z z ( z ) ( + z ) (b) Since one of the poles of X (z), which limited the ROC of X (z) to be less than, is cancelled by the zero of H (z), the ROC of Y (z) is the region in the z-plane that satis es the remaining two constraints jzj > and jzj > Hence Y (z) converges on jzj > (c) Therefore, Y (z) (05%) Page 70, Problem 6 y [n] z + jzj > + z n u [n] + ( )n u [n] When the input to an LTI system is n x [n] u [n] + () n u [ n ], the corresponding output is y [n] 5 n n u [n] 5 u [n] (a) Find the system function H(z) of the system Plot the pole(s) and zero(s) of H(z) and indicate the ROC (b) Find the impulse response h [n] of the system (c) Write a di erence equation that is satis ed by the given input and output (d) Is the system stable? Is it causal? (a) To determine H(z), we rst nd X (z) and Y (z): X (z) Y (z) z 5 z z ( z ), 5 5 z 5 z z < jzj < z z, jzj >

3 Now H (z) Y (z) X (z) The pole-zero plot of H(z) is plotted below z z jzj > (b) Taking the inverse z-transform of H(z), we get h [n] n n u [n] u [n ] n (u [n] u [n ]) (c) Since we can write Y (z) H (z) Y (z) X (z) z z, z X (z) z, whose inverse z-transform leads to y [n] y [n ] x [n] x [n ] (d) The system is stable becase the ROC includes the unit circle It is also causal since the impulse response h [n] 0 for n < 0 4 (05%) Page 7, Problem An LTI system is characterized by the system function H (z) z z z, jzj > 4

4 (a) Determine the impulse response of the system (b) Determine the di erence equation relating the system input x [n] and the system output y [n] (a) (b) y [n] H (z) 5 (05%) Page 7, Problem h [n] 4 [n] z z z z z 4 + z 8 z + 7 n u [n] z n u [n] 4 4 y [n ] + 8 y [n ] x [n] x [n ] Consider a right-sided sequence x [n] with z-transform X (z) ( az ) ( bz ) z (z a) (z b) In Section, we considered the determination of x [n] by carrying out a partial fraction expansion, with X (z) considered as a ratio of polynomials in z Carrry out a partial fraction expansion of X (z), considered as a ratio of polynomials in z, and determine x [n] from this expansion X (z) Obtain a proper fraction: z (z a) (z b) z z (a + b) z + ab z (a + b) z + abj z z (a + b) z + ab (a + b) z ab (a+b)a ab (a+b)a ab (a + b) z ab X (z) + (z a) (z b) + a b z a + b a z b + a a b z a b a b z b + a 4 b a z az b z bz

5 6 (05%) Page 7, Problem 9 x [n] [n] + a a b an b u [n ] a b bn u [n ] [n] + a n+ b n+ u [n ] a b A causal LTI system has system function H (z) z 0:5z z ( 0:5z ) ( + 0:5z ) (a) Determine the output of the system when the input is x [n] u [n] (b) Determine the input x [n] so that the corresponding output of the above system is y [n] [n] [n ] (c) Determine the output y [n] when the input is x [n] cos (0:5n) for < n < You may leave your answer in any convenient form H (z) z 0:5z z ( 0:5z ) ( + 0:5z ) (a) Given x [n] u [n], we have X (z) z, < jzj Then Y (z) H (z) X (z) z ( 0:5z ) ( + 0:5z ) z ( 0:5z ) ( + 0:5z ) ( 0:5z ) +, ( + 0:5z ) 0:5 < jzj (The ROC for Y (z) includes the intersection of the ROC of H (z) with the ROC of X (z)) Inverse z-transform gives y [n] (0:5)n u [n] + ( 0:5)n u [n] (b) If y [n] [n] [n ], then Y (z) z, 0 < jzj We have Inverse z-transform gives X (z) Y (z) H (z) z z 0:5z 0:5z, 0 < jzj x [n] [n] 0:5 [n ] 5

6 (c) Now x [n] cos (0:5n), < n < At! 0:5 we have H e j0:5 e j0:5 0:5e j :e j 4 Then y [n] : cos 0: (05%) Page 74, Problem Using any method, determine the inverse z-transform for each of the following: (a) X (z) (+ z ) ( z )( z ), (x [n] is a stable sequence) (b) X (z) e z (c) X (z) z (a) z Therefore, z X (z), (x [n] is a left-sided sequence) + z ( z ) ( z ) 5 + z z 5 x [n] n 5 (n + ) u [n + ] + 58 (5) (5) ()n u [ n ] < jzj < 700 ( z ) + 5 ( z ) n u [n] 700 (5) ()n u [ n ] (b) (c) Therefore, x [n] n! u [n] Therefore, X (z) e z + z + z! + z! + z 4 4! + : : : X (z) z z z z + z + jzj < z x [n] [n + ] + [n + ] () n u [ n ] 8 (05%) Page 74, Problem 5 For each of the following sequences, determine the z-transform and ROC, and sketch the pole-zero diagram: 6

7 (a) x [n] a n u [n] + b n u [n] + c n u [ n ] ; jaj < jbj < jcj (b) x [n] n a n u [n] (c) x [n] e n4 cos (a) (b) n u [n] e n4 cos n u [n ] x [n] a n u [n] + b n u [n] + c n u [ n ] jaj < jbj < jcj X (z) az + jbj < jzj < jcj bz cz X (z) cz + (bc + ac ab) z ( az ) ( bz ) ( cz ) jbj < jzj < jcj Poles: a, b, c, Zeros: z, z, where z and z are roots of numerator quadratic x [n] n a n u [n] x [n] a n u [n], X (z) az jzj > a x [n] nx [n] na n u [n], X (z) z d dz X az (z) ( az ) jzj > a x [n] nx [n] n a n u [n], z d dz X (z) z d az dz ( az ) jzj > a X (z) az ( + az ) ( az ) jzj > a (c) x [n] e n4 cos n u [n] e n4 cos n u [n ] Therefore, X (z) for all jzj e n4 cos n (u [n] u [n ]) [n] 7

8 9 (05%) Page 67, Problem 45 Consider the system of Figure 40, with the discrete-time system an ideal lowpass lter with cuto frequency 8 radians/s (a) If x c (t) is bandlimited to 5kHz, what is the maximum value of T that will avoid aliasing in the C/D converter? (b) If T 0kHz, what will the cuto frequency of the e ective continuous-time lter be? (c) Repeat part (b) for T 0kHz A plot of H (e j! ) appears below (a) (b) x c (t) 0, jj 5000 The Nyquist rate is times the highest frequency ) T sec This avoids 0;000 all aliasing in the C/D converter T 0kHz! T 8 0; 000 c c 65rad/sec f c 65Hz 8

9 (c) T 0kHz! T 8 0; 000 c c 50rad/sec f c 50Hz 0 (05%) Page 7, Problem 4 Consider a continuous-time signal x c (t) with Fourier transform X c (j) shown in Figure P4- (a) A continuous-time signal x r (t) is obtained through the process shown in Figure P4- First, x c (t) is multiplied by an impulse train of period T to produce the waveform x s (t), ie, x s (t) +X n x [n] (t nt ) Next, x s (t) is passed through a low pass lter with frequency response H r (j) H r (j) is shown in Figure P4-9

10 Determine the range of values for T for which x r (t) x c (t) (b) Consider the system in Figure P4-4 The system in this case is the same as the one in part (a), except that the sampling period is now T The system H s (j) is some continuous-time ideal LTI lter We want x o (t) to be equal to x c (t) for all t, ie, x o (t) x c (t) for some choice of H s (j) Find all values of T for which x o (t) x c (t) is possible For the largest T you determined that would still allow recovery of x c (t), choose H s (j) so that x o (t) x c (t) Sketch H s (j) (a) The impulse-train signal x s (t) has spectrum X s (j) given by X s (j) X X j k T T An example is shown below k 0

11 We will have x r (t) x c (t) provided T 0 (b) We will have x 0 (t) x c (t) under any of the following circumstances: As illustrated above, T 0 As illustrated below, :5 0 T 0 As illustrated below, T 0 The frequency response of the lter that is needed to recover x c (t) is shown below