Question

3. (10 points) Two linear time invariant (LTI) systems with impulse response hi(k) and h2(k) are connected in cascade as shown in Figure 1. Let x(k) be the input, yı(k) be the output of the first LTI, and y2(k) be the output of the second LTI. Let hi(k) = k(0.7)k u(k), h2(k) = ku(k), and x(k) = (0.3)k u(k). Use z-transform to (a) find yı(k). (b) find y2(k). x(k) yi(k) y2(k) hi(k) h2(k)

Answer 1

(3 x (3) +(3) → Y2) = x.(z). Hi(2) Given x(K) = (0.3jk ucky X(2) = Z-Transform of a(k) x 3) = Z Z-0.3 Z hi (ky = k(0.7* u(k) (0.7) * u(k) a ZT 7 k k (0-7)" ulks Z-0.7 27 Z d dz (zon) (2000-7 (2.0.2² 0.72 k(en) ulky ~ (2-0-772 1,2)=(s) (2 3.) Z (2-07 Y, (2) Z 0.7z (2-0:3) ( Zr 0.7) using partid Fraction method A + + Z-0.3 (2-0.7) 2 Y, (z, = 드 Z Z-0.7

0.72 = A (2-0.2}+B (2-0.3) (2-0.2) +C (2-0.3) O=A+B 2 0.7 = -14 A - BC o=A 0.49 +0.21B-0.36 3 from 0 → A=-B 0.75 -1.4 AT ATC 0.7 = -0.4 ATC 4 from 3 O = 0-49 A-021A = 0.3G 0 = 0.28/A - 0.49=-0.28 A 107c -0.3c from 0.49 = 0.4C 2 = 1.225 Oo of C 0.241.21 -0.4 = 1.3125 B = -1.3125 4,12) 1.3125 13125 + 1.2.2-5 Z Z-s3 Z-o-> Y, (2) = 1.3125 Z-003 -1.3128 2007 (Z-0.72

Apply Inverse z-trantim k K к Y, (K) = 13125 (0-3 jºu(k) — 1.3125 (0.7j u(k) +1.75 k(0.7ulks H₂ (2) HY₂ (²) = Y(2) H₂ (2) C z Z Given hacks kulks H₂(Z) = (2-1)2 Y₂(3) = 0.72 (2-0.3) (2-0.88 (2-1) Y2 (2) (2-03) (20.2} (21) Apply partial frackan 0.7 z Y₂ (2) A E + pwr Z=013 Except (Z-0.3) Tom geo A Oo7 x 3 (0.3-0.7% (0.3-13 A=0.803

hulwering soling simidarlly by 66.092 -67.72 D = B= c=9.527 E = 11:11 Y₂ (2) 0.803 67.72 + 11.11 + 66.92- 9.527 t 2-0.7 '(2-0-0) Z Z-03 2-li (2-1) Y₂ (2) = 0.8032 9.5202 67.722 t 66.922 Z-o.) Z-013 t 11.12 (2-0) Z-1 (2-12 Apply Inverse zarathim (K) = k 0.803 (0:3,"ucks +66-93-(003 > ic 13.6 KC-3 414 ulkst k 67.72 tiks to Hill k ucks