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<math>X(z) = \log \left( 1+z \right), \quad |z|<1 </math>. | <math>X(z) = \log \left( 1+z \right), \quad |z|<1 </math>. | ||
| − | + | Hint: expand the function into a power series using either the Taylor series formula or a [[PowerSeriesFormulas|table of power series formulas]]. | |
---- | ---- | ||
Post Your answer/questions below. | Post Your answer/questions below. | ||
| − | + | ||
| + | The power series expansion of the given function is: | ||
| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | Substitute n = -k | ||
| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | <math>\begin{align} | ||
| + | x[n] &= \frac{(-1)^{-n}}{n} u[-n-1] \\ | ||
| + | &= \frac{(-1)^{n}}{n} u[-n-1] | ||
| + | \end{align}</math> | ||
| + | |||
| + | since it doesn't matter if the (-1) is in the num or denom. | ||
| + | ---- | ||
*Answer/question | *Answer/question | ||
*Answer/question | *Answer/question | ||
Revision as of 09:09, 29 November 2010
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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