Question

The value of inverse Z-transform of log(\(\frac{z}{z+1}\)) is _______________

(a) (-1)^n/n for n = 0; 0 otherwise

(b) (-1)^n/n

(c) 0, for n = 0; (-1)^n/n, otherwise

(d) 0

I have been asked this question in an international level competition.

My doubt is from Inverse Z-Transform topic in division Z-Transform and Digital Filtering of Signals and Systems

Answer 1

The correct answer is (c) 0, for n = 0; (-1)^n/n, otherwise

To explain I would say: Putting z = \(\frac{1}{t}\), U (z) = log \(\left(\frac{\frac{1}{y}}{\frac{1}{y}+1}\right)\)

= – log (1+y) = -y + \(\frac{1}{2}\) y^2 – \(\frac{1}{3}\) y^3 + …..

= -z^-1 + \(\frac{1}{2}\) z^-2 – \(\frac{1}{3}\) z^-3 + …..

Thus, un = 0, for n = 0; (-1)^n/n otherwise.