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

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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

by (42.1k points)
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.

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