Revision as of 12:08, 12 September 2010

Question about computing the inverse z-transform

Computation of inverse z-transform by a student (with question about how to obtain ROC)

Take $ x[n] = a^n\left( u[n-2]+u[n]\right) $. We then have

$ \begin{align} X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)},\\ &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, \\ &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). \\ \text{Now let }k=-n, \\ \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^n + \sum_{k=0}^\infty (a/z)^n ,\\ &=\sum_{k=0}^\infty \left( (a/z)^n + 2)\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ & = \left(\frac{1}{1-a/z}+2\right) + \left(\frac{1}{1-a/z}\right), \\ & = \frac{z}{z-a}+2 + \frac{z}{z-a}, \\ & = \frac{z}{z-a}+2\frac{z-a}{z-a} + \frac{z}{z-a} , \\ & = \frac{4z-2a}{z-a}, \\ & = \frac{4-2a/z}{1-a/z}, \text{ for } |z|<a ??? \end{align} $

So if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a?

~ksoong

Comments/corrections from Prof. Mimi

Take $ x[n] = a^n(u[n-2]+u[n]) $. We then have $ \begin{align} X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n} \text{ (by definition of the z-transform)}, {\color{OliveGreen}\surd}\\ &= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, {\color{OliveGreen}\surd} \\ &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). {\color{OliveGreen}\surd} \\ \text{Now let }k=-n,& {\color{red}\text{This change of variable is not useful, unfortunately.}} \\ \Rightarrow X(z) &= \sum_{k=-2}^{\color{red}-\infty} (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{The terms inside the summation contain n, but the summation is over k.}} \\ &=\sum_{k=0}^{\color{red}-\infty} \left( (a/z)^n {\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \sum_{k=0}^\infty \left( \frac{a}{z}\right)^n \\ & = \left(\frac{1}{1-a/z}+{\color{red} -(a/z)^{-2}-(a/z)^{-1}} )\right) + \left(\frac{1}{1-a/z}\right), {\color{red}\text{For this last step, you need to assume } \left| \frac{a}{z}\right|<1, \text{ else both sums diverge.}} \end{align} $

The answer to your initial question ("if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|>a or when |z|<a?") is in the last step. As you can see from this step, X(z) only converges if |a|<|z|. Note that, since a could be a complex number, it is important not to say a< |z|

Anybody sees anything else? Do you have more questions? Comments? Please feel free to add below.

Back to ECE438, Fall 2010, Prof. Boutin