Question

What is the inverse z-transform of X(z)=log(1+az^-1) |z|>|a|?

(a) x(n)=(-1)n+1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

(b) x(n)=(-1)n-1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

(c) x(n)=(-1)n+1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

(d) None of the mentioned

This question was addressed to me in a national level competition.

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

Answer 1

Correct choice is (c) x(n)=(-1)n+1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

Explanation: Using the power series expansion for log(1+x), with |x|<1, we have

X(z)=\(\sum_{n=1}^∞ \frac{(-1)^{n+1} a^n z^{-n}}{n}\)

Thus

x(n)=(-1)^n+1 \(\frac{a^n}{n}\), n≥1

=0, n≤0.