Question

Q4. a) Use Z-transform method to determine the output time sequence y(n) for the following impulse response and excitation, h(n) = {2,-1,1,3}, x(n) = (-1)" u(n). b) For the DSP system shown, if x(t)=10 sin(300 ft), find x(n) and y(n). x(t) y(t) A/D h(n) =(n + 1)a"u[n] D/A f=1KHz

Answer 1

asion Qu) a) heroine b[n] = {2,-1, 1,3) hon) = 28 (n-1)-16) + 1 &(n+1) + 38 (nt HIZ) 22-1- + 2 + 3 22 2 - 2 +2? +3 23 Z Z - x(y) tonuln) X(2) = 2+1 Y(Z) = X(2) H(2) Z(9-Z+22+373) 12+1 Z Y(2) = 3 z3 + 2² z²_2+2 Z+) Y(z) = 3 32²/2 1 + 2/2 iz + 12 12+1 2+1 zti Zt, yin) = 360) u[n+2)+ El)?*! U(n+1) – 4)?u(n) +8 60.*'ulo-1) (27) D2

) Ts = fal= 10-3 fs=1kHZ fs 103 x(+) = lo sin (3001t) t=n Tg = 110-30) X () = lo sio (300 JT X10-30) X60) = lo Sin L0.371 n) Put Sinwo z sinwo (2-12 Z - 2 ㄹ 22- 2zco subtl Xin)= lo Sin (0.3172) X[Z) = 10_z Sin (0.31) X2) = lo zsin 10:37) 22-2290 5 (0:31) +) 22-22 Cos (0.317)+1 Xt2) 0.164 0.164 ²= azt hin) = (n+1) anuln) namun) + anunl Hz) o t. d+2(2-0) (2-42 zod ( 2-0)2 H(2) z2-data (2-0) 2 YLZ) = X(2) H(z) = (0-164) . (22d2+d) (2-1)² (2-0) Y(2) = 0.164 (2²-d27d) = A + B + c ER + (2-1)2 zod (2-2)² Z Z2-1²fz-2)2 0.16412²d272) 0.169 -12-47-8)? /2=6 of C:= Ou164 (22-d2+d) 0.164 Z(z-d)2 2=1 (1-0)2 00164 (22-02+0) 0.164 d z (2-1² 27 - E = al = 0.164 a (0-12 11-0² 2=d

D=0082 o 2 2 a B = 0.082 x Y(2) = polby 2 polby I too 6841_+ 0.164 1 + 0-0824 +0:164 a (24) (152) 2 (2) 2 Zad a² (2-2) Y(z) = 0.164 + 0.082 2 + 0.164 Z 1-2) 12-12 221222 y (n) = 0.164 (n) +0:082 U[n) + 00164(n+)u(n+1) +0.082,"ol) + (1-2)2 Oolby ntah anti) ť 0.16Y Z X a ZT 400082 Z d z-a AN 이 L d