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Practice Question 3, ECE438 Fall 2010, Prof. Boutin

On computing the inverse z-transform of a discrete-time signal.

Compute the inverse z-transform of

$ X(z) = \log \left( 1+z \right), \quad |z|<1 $.

Hint: expand the function into a power series using either the Taylor series formula or a table of power series formulas.

Post Your answer/questions below.

The power series expansion of the given function is:

$ \begin{align} X(z) &= \sum_{n=1}^{\infty} (-1)^{n+1} \frac{z^n}{n}, \ -1 < z \le 1 \\ &= \sum_{n=-\infty}^{\infty} (-1)^{n+1} u[n-1] \frac{z^n}{n} \end{align} $

Substitute n = -k

$ \begin{align} X(z) &= \sum_{k=-\infty}^{\infty} (-1)^{-k+1} u[-k-1] \frac{z^{-k}}{-k} \\ &= \sum_{k=-\infty}^{\infty} \frac{(-1)^{-k+1}}{-k} u[-k-1]z^{-k} \\ &= \sum_{k=-\infty}^{\infty} \frac{(-1)^{-k}(-1)}{-k} u[-k-1] z^{-k} \\ &= \sum_{k=-\infty}^{\infty}\frac{(-1)^{-k}}{k} u[-k-1]z^{-k}, \text{ and by comparison with } X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} \end{align} $

$ \begin{align} x[n] &= \frac{(-1)^{-n}}{n} u[-n-1] \\ &= \frac{(-1)^{n}}{n} u[-n-1] \end{align} $

since it doesn't matter if the (-1) is in the num or denom.

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