(Computing the Inverse Z.T.)
 
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==Relationship between Z-Transform and F.T.==
 
==Relationship between Z-Transform and F.T.==
  
*<math>X(\omega) = X(e^{j\omega})</math>
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<math>X(\omega) = X(e^{j\omega})</math>
  
*<math>X(z)=X(re^{j\omega})</math>
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<math>X(z)=X(re^{j\omega})</math>
  Then <math>X(z) = F(x[n]r^-n)</math>
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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>
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  Then <math>X(z) = F(x[n]r^{-n})</math>
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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>
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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>
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<math> x[n] = -2^nu[-n-1]</math>

Latest revision as of 15:57, 30 November 2008

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] $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn