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<math>X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math> | <math>X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math> | ||
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Where <math>\sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math> is the F.T! | Where <math>\sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math> is the F.T! | ||
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| + | ==Properties of the ROC== | ||
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| + | Refer to [[Xujun Huang: Properties of ROC_ECE301Fall2008mboutin]] | ||
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| + | ==Computing the Inverse Z.T.== | ||
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| + | <math>X(z) = \frac{1}{1-2z^{-1}} , |z| < 2</math> | ||
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| + | <b>Warning <math>|2z^{-1}| = \frac{2}{z} > 1</math>!!</b> | ||
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| + | <math> = \frac{1}{1-\frac{2}{z}} = \frac{z}{z-2} = \frac{z}{-2(1-\frac{z}{2})}</math> | ||
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| + | <math> = -\frac{z}{2}\frac{1}{1-\frac{z}{2}} = \frac{-z}{2} \sum_{n = 0}^\infty (\frac{z}{2})^n</math> | ||
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| + | <math> \sum_{n = 0}^\infty \frac{-z}{2} \frac{1}{2^n} z^n = \sum_{n = 0}^\infty \frac{-z^{n+1}}{2^{n+1}}</math> | ||
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| + | let <math>-k = n + 1</math> | ||
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| + | <math> \sum_{n = \infty}^{-1} -2^kz^{-k}</math> | ||
| + | |||
| + | <math> x[n] = -2^nu[-n-1]</math> | ||
Latest revision as of 14:57, 30 November 2008
Contents
Z Transform
Discrete analog of Laplace Transform
$ X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} $
Where z is a complex variable.
Relationship between Z-Transform and F.T.
$ X(\omega) = X(e^{j\omega}) $
$ X(z)=X(re^{j\omega}) $ Then $ X(z) = F(x[n]r^{-n}) $ $ X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n} $ Where $ \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n} $ is the F.T!
Properties of the ROC
Refer to Xujun Huang: Properties of ROC_ECE301Fall2008mboutin
Computing the Inverse Z.T.
$ X(z) = \frac{1}{1-2z^{-1}} , |z| < 2 $
Warning $ |2z^{-1}| = \frac{2}{z} > 1 $!!
$ = \frac{1}{1-\frac{2}{z}} = \frac{z}{z-2} = \frac{z}{-2(1-\frac{z}{2})} $
$ = -\frac{z}{2}\frac{1}{1-\frac{z}{2}} = \frac{-z}{2} \sum_{n = 0}^\infty (\frac{z}{2})^n $
$ \sum_{n = 0}^\infty \frac{-z}{2} \frac{1}{2^n} z^n = \sum_{n = 0}^\infty \frac{-z^{n+1}}{2^{n+1}} $
let $ -k = n + 1 $
$ \sum_{n = \infty}^{-1} -2^kz^{-k} $
$ x[n] = -2^nu[-n-1] $
