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&= \sum_{n=-\infty}^\infty a^n(u[n-2]+u[n])z^{-n}, {\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} \\ | &= \sum_{n=2}^\infty a^n(z^{-n}) + \sum_{n=0}^\infty a^n(z^{-n}). {\color{OliveGreen}\surd} \\ | ||
| − | \text{Now let }k=-n, \\ | + | \text{Now let }k=-n,& {\color{red}\text{This change of variable is not useful, unfortunately.}} \\ |
| − | \Rightarrow X(z) &= \sum_{k=-2}^\infty (a/z)^{\color{red}n} + \sum_{k=0}^\infty (a/z)^{\color{red}n} ,{\color{red}\text{ | + | \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}^\infty | + | &=\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}+ | + | & = \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.}} |
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\end{align} | \end{align} | ||
</math> | </math> | ||
| − | + | <span style="color:green"> 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 |</span> | |
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| − | Anybody sees anything else? Do you have more questions? | + | Anybody sees anything else? Do you have more questions? Comments? |
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| + | [[2010_Fall_ECE_438_Boutin|Back to ECE438, Fall 2010, Prof. Boutin]] | ||
Revision as of 13:01, 12 September 2010
Question about computing the inverse z-transform
I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it. 0 I tried the LaTeX but it failed miserably. . Maybe it would be easier if you write it out while you are reading my LaTeX fail trying to help me, otherwise I can come in and see you on Monday.
- I fixed it. Now it works. For future references, here is a good page to bookmark: it explains how to use latex in mediawiki, and gives many examples that you can cut and paste. -pm
= Computation of inverse z-transform by a student (with question about how to obtain ROC)
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)},\\ &= \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?
