MAS160: Signals, Systems & Information for Media Technology Problem Set 7 DUE: November 19, 2003 Instructors: V. Michael Bove, Jr. and Rosalind Picard T.A. Jim McBride Problem 1: z-Transforms, Poles, and Zeros Determine the z-transforms of the following signals. Sketch the corresponding pole-zero patterns. (a) x[n] = δ[n − 5] (b) x[n] = nu[n] 1 n (c) x[n] = − 3 u[n] (d) x[n] = (an + a−n)u[n]; a real n (e) x[n] = (na cos !0n)u[n]; a real 1 n (f) x[n] = 2 (u[n − 1] − u[n − 10]) SOLUTION : PS 7-1 Signals, Systems & Information : Problem Set 7 Solutions PS 7-2 (a) X1 δ[n − 5] −!Z δ[n − 5]z−n n=−∞ = δ[5 − 5]z−5 = z−5 ROC : all z, except z = 0. Pole-zero plot corresponding to x[n] = δ[n-5] 1 5 zeros at ∞ 0.5 5 0 Imaginary part -0.5 -1 -1 -0.5 0 0.5 1 Real part Figure 1: z-plane plot for (a) x[n] = δ[n − 5] PS 7-2 Signals, Systems & Information : Problem Set 7 Solutions PS 7-3 (b) X1 nu[n] −!Z nu[n]z−n = X(z) n=−∞ X(z) = z−1 + 2z−2 + 3z−3 + 4z−4 + ::: −z−1X(z) = −z−2 − 2z−3 − 3z−4 − ::: (1 − z−1)X(z) = −1 + 1 + z−1 + z−2 + z−3 + z−4 + ::: 1 = −1 + 1 − z−1 z−1 = 1 − z−1 z−1 X(z) = (1 − z−1)2 ROC : jz−1j < 1 ) jzj > 1 1 One zero at z=∞ 0.5 2 0 Imaginary part -0.5 -1 -1 -0.5 0 0.5 1 Real part Figure 2: z-plane plot for (b) x[n] = nu[n] PS 7-3 Signals, Systems & Information : Problem Set 7 Solutions PS 7-4 (c) 1 n X1 1 n − u[n] −!Z − u[n]z−n 3 3 n=−∞ X1 1 n = − z−n 3 n=0 X1 1 n = − z−1 3 n=0 1 = 1 −1 1 + 3 z 1 −1 1 ROC : j 3 z j < 1 ) jzj > 3 1 0.5 0 Imaginary part -0.5 -1 -1 -0.5 0 0.5 1 Real part 1 n Figure 3: z-plane for (c) x[n] = − 3 u[n] PS 7-4 Signals, Systems & Information : Problem Set 7 Solutions PS 7-5 (d) X1 (an + a−n)u[n] −!Z (an + a−n)u[n]z−n n=−∞ X1 = (an + a−n)z−n n=0 X1 X1 = (az−1)n + (a−1z−1)n n=0 n=0 1 1 = + 1 − az−1 1 − a−1z−1 1 − a−1z−1 + 1 − az−1 = (1 − az−1)(1 − a−1z−1) a2+1 −1 2 − a z = (1 − az−1)(1 − a−1z−1) 1 ROC : jzj > max fa; a g. For example, at a = 2: 1 0.5 0 Imaginary Part −0.5 −1 −1 −0.5 0 0.5 1 1.5 2 Real Part Figure 4: z-plane for (d) x[n] = (an + a−n)u[n]; a real PS 7-5 Signals, Systems & Information : Problem Set 7 Solutions PS 7-6 Z d n (e) Since nx[n] −! −z dz X(z), first find the z-Transform of x[n] = a cos(!0n)u[n]: X1 n Z n −n (a cos(!0n))u[n] −! (a cos(!0n))u[n]z n=−∞ X1 n −n = a cos(!0n)z n=0 X1 1 = an ej!0n + e−j!0n z−n 2 n=0" # 1 X1 X1 = (aej!0 z−1)n + (ae−j!0 z−1)n 2 n=0 n=0 1 1 1 = + 2 1 − aej!0 z−1 1 − ae−j!0 z−1 1 1 − aej!0 z−1 + 1 − ae−j!0 z−1 = 2 1 − aej!0 z−1 − ae−j!0 z−1 + a2z−2 1 2 − az−1(ej!0 + e−j!0 ) = 2 1 − az−1(ej!0 + e−j!0 ) + a2z−2 −1 1 − az cos !0 = −1 2 −2 1 − 2az cos !0 + a z 2 z − az cos !0 = 2 2 z − 2az cos !0 + a Now we take the derivative and multiply by −z: 2 d z − az cos !0 X(z) = −z 2 2 dz z − 2az cos !0 + a 2 2 2 (z − 2az cos !0 + a )(2z − a cos !0) − (z − az cos !0)(2z − 2a cos !0) = −z 2 2 2 (z − 2az cos !0 + a ) 2 2 3 z a cos !0 − 2za + a cos !0 = z 2 2 2 (z − 2az cos !0 + a ) PS 7-6 Signals, Systems & Information : Problem Set 7 Solutions PS 7-7 We can solve for the poles and zeros using the quadratic formula: p 2a cos ! ± 4a2 cos2 ! − 4a2 poles: z = 0 0 p 2 2 = a(cos !0 ± cos !0 − 1) = a(cos !0 ± j sin !0) zeros: z = 0 p 2a2 ± 4a4 − 4a4 cos2 ! z = 0 2a cos !0 a ± a sin ! = 0 cos !0 a(1 ± sin ! ) = 0 cos !0 1 π ROC : jzj > jaj. For example, at a = 2 and !0 = 4 : 1 0.8 0.6 0.4 2 0.2 2 0 Imaginary Part −0.2 2 −0.4 −0.6 −0.8 −1 −1 −0.5 0 0.5 1 Real Part n Figure 5: z-plane for (e) x[n] = (na cos !0n)u[n]; a real PS 7-7 Signals, Systems & Information : Problem Set 7 Solutions PS 7-8 (f) 1 n X1 1 n (u[n − 1] − u[n − 10]) −!Z (u[n − 1] − u[n − 10])z−n 2 2 n=−∞ X9 1 n = z−n 2 n=1 1 z−1 − 1 z−1 10 = 2 2 1 −1 1 − 2 z 1 −1 1 −1 9 2 z 1 − 2 z = 1 −1 1 − 2 z 1 ROC : The pole and zero at z = 2 cancel ! all z, except z = 0. 1 One zero at z=∞ 0.5 9 0 Imaginary part -0.5 -1 -1 -0.5 0 0.5 1 Real part 1 n Figure 6: z-plane for (f) x[n] = 2 (u[n − 1] − u[n − 10]) PS 7-8 Signals, Systems & Information : Problem Set 7 Solutions PS 7-9 Problem 2: z-Transform Properties Given x[n] below, use the properties of the z-transform to derive the transform of the following signals. z−1 x[n] ! X(z) = (1 − z−1)2 (a) x[n − 3] (b) x[n] ∗ δ[n − 3] (c) x[n] − x[n − 1] (d) x[n] ∗ (δ[n] − δ[n − 1]) 1 n (e) 5x[n − 1] + 4 − 3 u[n] SOLUTION : (a) x[n − 3] −!Z X(z)z−3 z−4 = (1 − z−1)2 (b) x[n] ∗ δ[n − 3] −!Z X(z)z−3 z−4 = (1 − z−1)2 (c) x[n] − x[n − 1] −!Z X(z) − X(z)z−1 = X(z)(1 − z−1) z−1 = 1 − z−1 (d) x[n] ∗ (δ[n] − δ[n − 1]) −!Z X(z)(1 − z−1) z−1 = 1 − z−1 (e) ! 1 n 1 5x[n − 1] + 4 − u[n] −!Z 5X(z)z−1 + 4 from (1c) above 1 −1 3 1 + 3 z 5z−2 4 = + −1 2 1 −1 (1 − z ) 1 + 3 z PS 7-9 Signals, Systems & Information : Problem Set 7 Solutions PS 7-10 Problem 3: Relating pole-zero plots to frequency- and impulse- response (a) DSP First 8.16 (b) DSP First 8.17 SOLUTION : (a) (a) From the magnitude frequency response, we see that A is a high-pass filter, with six zeros along the frequency axis (i.e. the unit circle). Only two of the pole- zero plots have six zeros on the unit circle (PZ1 and PZ2). From the pole-zero plots, we see PZ1 is a low-pass filter (the zeros are concentrated towards higher frequency), while the zeros of PZ2 are concentrated towards lower frequencies (making it a high-pass filter). Therefore frequency response A corresponds to pole-zero plot PZ2. (b) B is a high-pass filter, with a sharp peak near maximum frequency (π) and a zero at zero frequency. This means that there is a pole near the unit circle at ! = π and a zero on the unit circle at ! = 0 (z = 1). Therefore, the corresponding pole-zero plot is PZ5. (c) C is a low-pass filter, with six zeros along the frequency axis, corresponding to pole-zero plot PZ1. (d) D is a very sharp band-bass filter, indicating poles close to the unit circle at π ! = ± 2 . This is consistent with pole-zero plot PZ6. (e) E is a somewhat complex response, with sharp peaks (indicating poles close to the unit circle) at a low frequency and somewhat smoother peaks at higher frequencies (indicating poles a little bit further from the unit circle). This pattern indicates pole-zero plot PZ3. (b) (a) The first thing to notice about J is that it is an infinite impulse response, and therefore has a pole somewhere other than at zero or 1. It's shape (exponential decay), is consistent with a form h[n] = anu[n], which is a single-pole system with a pole at z = a. And we know from the impulse response that in this case, a is positive, which also indicates a low-pass response. Therefore, the corresponding pole-zero plot is PZ4. (b) K is FIR, with a length of 7 (N = 6) and therefore has six zeros and poles only at zero or 1.