Question

hello, I need help on this z transform problem please try to show steps by steps so I can understand I attached the way my Prof wants to do, please try to follow that I also give answer with the problem for ur help. thanks

OPINA Real fxpnessim o X CD X2= 1-21 4-2 + X2 t21 4-21+0,52 Amswer: gnuSo ): - 0)xD x= eos(-1-56l R1.666uD

please disregard the last pictures try to solve this problem I also include the solution for the problem, please show steps by steps so I can understand thanks

Answer 1

(I) Given, X Z) = X(Z) = 72 72 7 +0.5 Z XEZ) Z EZ- 27 +0.5 (Z-0.5-j0.5) (2-0.5+ 2015) XIZ) A to Z-0.5-jo.5 -0.5tjo.5 find a A= (2-015-j0.5) x (2) z=0.5+ j0.5 7-015+j05 17=0.5+0.5 - In polar form. A = 141 Le A-12 12 Now, *= 2 1 1

P=0.5+ jo.5 in polar form, p= 1PILO P= . Now, p*= 2 Lt Therefore, the inverse - transform is yon) ="( 422 +*) = 2141(1P1) cos (ne+o) ulim) = 29 (3) cos( m un) 400) = 2 (1990s (in-1) en). Given, X(z) = 147-1 1-740.572 2 17+18) 2² _ z +0.5 X(z) = 7 (2+1) z2z+015 → X1Z) = 24 727 40.5 X(z) = (2+1) -0.5-0.5j) (7-1.570558) Z. A (7-0.500.5j) (+-0.570.5;)

To find A, A = (-0.5-10. 5x (2) Z lz=0.5+10.5 = 3+1 z - 0.5 +10.5 z=0.5+10.5 Now, in polar form, a= |AILY | A = |-1241 A = IAILO = VID L.249. Jo find p in polar form, P= IPLO P=0.5+10.5 = 2* Now, p*= V2 = IPIL 2

Therefore, _ D Z -P Inverse - transform is 3tn) = = ( A + 4* = 21A1(1P1)"cos (10+)uln) - * (T) )" cou ( un 1247) uen) .: x(n)= vio ( com-1-244) uin)