Question

em 2: Given two sequences x[n] = 8 8[n - 8] and h[n] = (0.7)"u[n] Determine the z-transform of the convolution of the two sequences using the convolution property of the Z-transform Y(z) = X(z) H(2) Determine the convolution y[n] = x[n] * h[n] by using the inverse z-transform

Answer 1

Problema Ginen x En] =8 8 [n-8 ] h En] = (0.73 u [n a-transferm g (m) Bytime shifting property 2-transform_k g - kJ 2 " x [n] 23, X(2) x [n] = 88 [n-8] taking z-transform both side. we get X (2) = 8 28 ( By time shifting property hm] = 0.7" un] taking z-trans from both side we get H2) = I ILOR, - Ontz Now Y(z) = x (Z). H(2) |Y(z) = 828 T-0.72'

To find the convolution take inverse z-transform of Y (2) Y GJ = 8278 1 -0.77' Y (23= 828 Y, (2) Y,(z)= I 1-0.72 z" (Y, (2J) = 0.7 um) New from time shifting we have 28 ) K = 8. Now shift the time by 8 we get y [ ] = 0.7*-&cm-8)

PAshge 3 X(2) = 4²' +2 - 2-1 (2-122 Now rewrite X (2) in +28 +25 2.05 z' form. X GJ = Yzt + zu Z(1-3') 2² (1-²) ² + 28 1-0.52 35 . or X (ZJ= 422 + 2b + 28 +25 1-27 -27 (lizija + 2 5+25 Now take the z-inverse trans farm.

423 Š recom) 2 yu(n-2) T-2 using time shifting - Property) 2 num] 2 Now using time shifting property. with time shift k=8. 25 21 2² (1-5) u (n-5) oes n-8 u(n-8 J. Ray (1-0.5 25 2 8 (2-5) Now N-8 x [n] = 4 4 (n=8) (0-2) + (n-5) 46n-5)+0.5 + g(n-s)

Chiron), s X(z) = 2 1 222 +2.72 +2 ar XCZ) = 27 2 +2.72 +22' Now pasital using Partial feaction . salue fraction using "Residue 2" command. Code eng. a=[lo] , b = [2 2.72] [r,a}= residuez (a, b) XCZ) = 0.25 +0.9875 € +0.25 +0.2872 1 -(-0.675-0.73781)2' 1-60.67 +0.73) z' n x[M] = (0.95+0.98758.(-0.675 -0.73781) ucm) + (0.95 -0.28751°). (0.675 +0.7378