Question

What is the inverse z-transform of X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\) if ROC is 0.5<|z|<1?

(a) -2u(-n-1)+(0.5)^nu(n)

(b) -2u(-n-1)-(0.5)^nu(n)

(c) -2u(-n-1)+(0.5)^nu(-n-1)

(d) 2u(n)+(0.5)^nu(-n-1)

I have been asked this question in a job interview.

The question is from Inversion of Z Transform in chapter Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

Right choice is (b) -2u(-n-1)-(0.5)^nu(n)

Easy explanation: The partial fraction expansion of the given X(z) is

\(X(z)= \frac{2z}{z-1}-\frac{z}{z-0.5}\)

In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the signal corresponds to a causal signal and the other corresponds to an anti causal signal. Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get

x(n)= -2u(-n-1)-(0.5)^nu(n).