Problem Set Solution 22: The z-Transform

22 The z-Transform Solutions to Recommended Problems S22.1 (a) The z-transform H(z) can be written as H(z) = z z -2 Setting the numerator equal to zero to obtain the zeros, we find a zero at z = 0. Setting the denominator equal to zero to get the poles, we find a pole at z = 1. The pole-zero pattern is shown in Figure S22.1. z plane ROC: IzI > R 2 Figure S22.1 (b) Since H(z) is the eigenvalue of the input z' and the system is linear, the output is given by y[n] = = 1 3) +3 1 (2)" 3()" + 4(2)" S22.2 (a) To see if x[n] is absolutely summable, we form the sum N-I SN N-i x[nhl = n=o 2' = 1 2 n=O Since limN-oSN diverges, x[n] is not absolutely summable. (b) Since x[n] is not absolutely summable, the Fourier transform of x[n] does not converge. N N(N- (C) SN " n=o(2 S22-1 Signals and Systems S22-2 limN-oN is finite for Ir > 2. Therefore, the Fourier transform of r-"x[n] con­ verges for Ir| > 2. 2"z~" (d) X(z) = = E n=O n.O 1 __ = (2z- 1) 2z- for 12z~'I < 1 Therefore, the ROC is IzI > 2. 1 (e) X 1(e") = e -s 1 Therefore, x 1 [n] = (2)"u[n]. S22.3 (a) Since x[n] is right-sided, the ROC is given by Iz I > a. Since the ROC cannot include poles, for this case the ROC is given by Iz I > 2. (b) The statement implies that the ROC includes the unit circle Iz = 1. Since the ROC is a connected region and bounded by poles, the ROC must be 5 < IzI < 2 (c) For this situation there are three possibilities: (i) (ii) (iii) Izi < 3 -1 < 2 < I z| IzI > 2 (d) This statement implies that the ROC is given by |z < S22.4 (a) (i) X, (z) x I~ = (-2 T -n )f n=-oon=O 1 1 ' - with an ROC of -1 (ii) X2 (z) = E < 1, or IzI > 2. 2z (i)z~ Letting n = -m, we have X 2(z) = l()--m' - m= 1 2z GO L (2z) = - m = - m1 1 - z ' with an ROC of 12z| < 1, or Iz < i. 1-2z 1. The z-Transform / Solutions S22-3 (b) (i) Im (z ROC Re lz Figure S22.4-1 (ii) ImZl ROC Re (z) lzl= 1 Figure S22.4-2 (c) (i) X 3(z) = 2z z- 2 = 2 (1 -1z) n=O > 1, as shown .The ROC is Izi -Z-1 in Figure S22.4-3. -1i (ii) X 4(z) = 2"z~ = - - Z 2-"z n= n=Al z/2 z 1 -(z/2) z -2' with an ROC of Iz/21 < 1, or IzI < 2, shown in Figure S22.4-4. Signals and Systems S22-4 Im /j A. Re Iz) Izl=2 Figure S22.4-4 (d) For the Fourier transform to converge, the ROC of the z-transform must include the unit circle. Therefore, for x 1 [n] and x 4[n], the corresponding Fourier trans­ forms converge. S22.5 Consider the pole-zero plot of H(z) given in Figure S22.5-1, where H(a/2) = 1. z plane K 1 to 5zeros ' Figure S22.5-1 (a) When H(z) = z/(z - a), i.e., the number of zeros is 1, we have H(ej) = cos 9 + j sin 0_ (cos 0 - a) + j sin 12 Therefore, |H(e )| 2 = 1 + a - 2a and we can plot IH(ejQ)I as in Figure S22.5-2. cos 0' The z-Transform / Solutions S22-5 IH(en2)I Figure S22.5-2 When H(z) =z/(z - a), i.e., the number of zeros is 2, we have - cos 20 + j sin 22 (cos 0 - a) + j sin 2 Therefore, |H(es")| = 1 + a2 - 2a cos 0 Hence, we see that the magnitude of H(eia) does not change as the number of zeros increases. (b) For one zero at z = 0, we have H(z)= z-a' e H(e") = e- a We can calculate the phase of H(e ") by [Q - 4 (denominator)]. For two zeros at 0, the phase of H(ein) is [29 - 4 (denominator)]. Hence, the phase changes by a linear factor with the number of zeros. (c) The region of the z plane where IH(z) I = 1 is indicated in Figure S22.5-3. A z plane a +- Re z = Figure S22.5-3 Signals and Systems S22-6 S22.6 (a) (3)"[n] n=O = (3z)n=O i 1 z Z z e Therefore, there is a zero at z 0 and a pole at z = or -<1 IzI> = 3, 1, and the ROC is , as shown in Figure S22.6-1. z plane WZ / 7 A, i - - - - ­ 37 I (b) b[n +1] L Figure S22.6-1 bS[n + 1]z-" = z, with the ROC comprising the entire z-plane, as shown in Figure S22.6-2. Im z plane Re Figure S22.6-2 S22.7 (a) Using long division, we have X(z) = 1 -!Z- + !z - 3 + z­ We recognize that x[n] = (-i)"u[n] The z-Transform / Solutions S22-7 (b) Proceeding as we did in part (a), we have x[n] (c) X(z) = (-I"u[n] -a -az = S_ a2) a (1 1-az 1 1 a a2)z­ a2 1-a-IZ-1 Therefore, x[n) = - a-[n] - aa Solutions to Optional Problems S22.8 (a) 1{r-"x[n]} = 5{r-u[n]} r~"e -jan = n=O = Z (re+yu-n n=O For the sum to converge, we must have < 1 Thus, Ir| > 1. (b) (i) Im r =1z plane Re Figure S22.8-1 I u[n ­ Signals and Systems S22-8 (ii) Im r= z plane Re Figure S22.8-2 (iii) z plane r= 3 Figure S22.8-3 (c) Figure S22.8-4 S22.9 (a) The inverse transform of 1 1 -i - The z-Transform / Solutions S22-9 is (i)"u[n]. But from the given relation, we have 2-"x[n] x[n] 3 + 2z- 1 = (L)"u[n], (I)"u[n] = 3 + 2z-' (2 + z-1)(1 + z-1) 1 1 2+z1 + z-' (b) 2 + 3z-' + z-2 1 2 1+ iz-' 1 + z- 1 -1)u[n] + (-1)"u[n] = S22.10 Az a) - with A a constant (a) H(z) = ( (1 - azTe) Therefore, (e' ) H A(e -j' 1 - I A 2(e = |H~es")2 - a) ae j ' a)(e" - (1 - ae-3)(1 - a) a) = A2 aej") and thus, |H(e'")| = Al (b) (i) (ii) |v 112 = 1 + |v 2 12 = 1 + = - 1 1 = -| a2 - 2a cos 9 1 2 - - cos a a -2 (a 2 + 1 - 9 2a cos Q) vil2 S22.11 In all the parts of this problem, draw the vectors from the poles or zeros to the unit circle. Then estimate the frequency response from the magnitudes of these vectors, as was done in the lecture. The following rough association can be made: (a) (b) (c) (d) (e) (i) (ii) (iv) (iii) (iv) Signals and Systems S22-10 S22.12 (a) x[n] = (I)"[u[n] - u[n - 101]. Therefore, 9 X(z) = E(i)"Z--" n=o (2z)" = n=O 10 1 = 0 -1(zY 1 (2z)- 1 (2z)-1 11 = zl z9(z |z| (1)10I">0 - 2) > 0, shown in Figure S22.12-1. The Fourier transform exists. (b) x[n] = (.)In' = (I)"u[n] + (.) -u[-n - 1]. But (l\n z u[n] z 1 IzI> , , and (1)-n 2 u[-n - ]- ~z z < 2 - 2' Summing the two z-transforms, we have -2z X(z) = 2 (z 2)D(z 1 2 - 2) < |zI < 2 (See Figure S22.12-2.) The Fourier transform exists. (c) x[n] = 7()" cos + u[n] Therefore, 76 X(z) = cos 7 00 n - l 2n=0 j(2r/6)n+(,r/4)] + e j(2w/6)n+(-/4)] ei(2/6) e 4 -1 -jr/ e 1 -j(2r/6)-1)] e (2v/6) 1 e -j4 (2 / 6) z (e Z ire- 7z n 6__ jw/4 _ le(2r/6)Z- 7z Z 3 [ej/4 7 + 4z j(21/6) 7 ,r) 2 cos (2,r ,r 2z cosir 4 3 6 4 j(2/6) z- I (2r/6>)(z - where |z| > 1 The pole-zero plot and ROC are shown in Figure S22.12-3. Clearly, the Fourier transform exists. The z-Transform / Solutions S22-11 9th order pole pole -zero cancel Figure S22.12-1 z plane Figure S22.12-2 Figure S22.12-3 Signals and Systems S22-12 (d) X(z) = z.=, 1 - z' 0 - 1 1 - = z z10 -1 ­ 1 z"(z - Z' The ROC is all z except z = 0, shown in Figure S22.12-4. The Fourier transform exists. 9th-order pole .1.0 xPC 1 -1 1­ pole -zero cancel Figure S22.12-4 S22.13 (a) From log(1 - w)= IwI < 1, i=1 we find log(1 - (2z)' 2z) = 12z| < 1 z", |z| < , 2-n x[n] = s n >- 0 0, (b) We solve this similarly to the way we solved part (a). log 1 (Oz-')'f 12 - 1-1 1 n 2 01 x[n] Z11<1 z-< 2 n > 0, =t 0, n s 0 1