Question

What is the partial fraction expansion of X(z)=\(\frac{1}{(1+z^{-1})(1-z^{-1})^2}\)?

(a) \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z+1)^2}\)

(b) \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} – \frac{z}{2(z+1)^2}\)

(c) \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}\)

(d) \(\frac{z}{4(z+1)} + \frac{z}{4(z-1)} + \frac{z}{2(z+1)^2}\)

I had been asked this question by my college director while I was bunking the class.

Question is taken from Inversion of Z Transform in division Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

Correct option is (c) \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}\)

Easiest explanation: First we express X(z) in terms of positive powers of z, in the form X(z)=\(\frac{z^3}{(z+1)(z-1)^2}\)

X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is

\(\frac{X(z)}{z} = \frac{z^2}{(z+1)(z-1)^2} = \frac{A}{z+1} + \frac{B}{z-1} + \frac{C}{(z-1)^2}\)

On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.

Therefore, we get \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}\).