To find the ZIR, set X (a)=0 and use the fact that y[-1=1 回m]=-3(-3)uln]=(-3)+um] Co find the ZSR, set y[-1]=0 and use x[n]=(5)(n] as given in the problem, X(z (1+32-1)(1 6 2=(-37+(2) m]-n-1],y-1=0,xn]=ul Taking the unilateral z-transform of both sides of the difference equation Y(z)-32-Y(x)-5y-1]=X(z) X(a-o[ Solving for Y(z) gives To find the ZIR, set X(a)=0(note that this means x[-1=0), and use the fact that y-1]=0, 0 To find the ZSR, set yl-1=0 and use n]=un as given in the problem, X(a) Usain= unlTo find the ZIR, set X(z) = 0 and use the fact that y[−1] = 1, YZIR (z) = −3 1 + 3z −1 yZIR [n] = −3(−3)nu[n] = (−3)n+1u[n]. To find the ZSR, set y[−1] = 0 and use x[n] = ￾ 1 2
n u[n] as given in the problem, X(z) = 1 1 − 1 2 z −1 YZSR (z) = 1 (1 + 3z −1 ) ￾ 1 − 1 2 z −1
= 6 7 1 + 3z −1 + 1 7 1 − 1 2 z −1 yZSR [n] = 6 7 (−3)nu[n] + 1 7  1 2 n u[n]. (b) y[n] − 1 2 y[n − 1] = x[n] − 1 2 x[n − 1], y[−1] = 0, x[n] = u[n] Taking the unilateral z-transform of both sides of the difference equation, Y (z) − 1 2 z −1Y (z) − 1 2 y[−1] = X(z) − 1 2 z −1X(z) − 1 2 x[−1]. Solving for Y (z) gives, Y (z) = (1 − 1 2 z −1 )X(z) 1 − 1 2 z −1 + − 1 2 x[−1] + 1 2 y[−1] 1 − 1 2 z −1 = X(z) | {z } ZSR + − 1 2 x[−1] + 1 2 y[−1] 1 − 1 2 z −1 | {z } ZIR . To find the ZIR, set X(z) = 0 (note that this means x[−1] = 0), and use the fact that y[−1] = 0, YZIR (z) = 0 =⇒ yZIR [n] = 0. To find the ZSR, set y[−1] = 0 and use x[n] = u[n] as given in the problem, X(z) = 1 1 − z −1 YZSR (z) = 1 1 − z −1 yZSR [n] = u[n]. 6