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<math>X(z) = \sum^{\infty}_{n = -\infty} x[n]z^{-n}\!</math> | <math>X(z) = \sum^{\infty}_{n = -\infty} x[n]z^{-n}\!</math> | ||
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| + | <math>= \sum^{\infty}_{n = -\infty} x[n](re^{j\omega})^{-n}\!</math> | ||
Revision as of 16:32, 3 December 2008
The Z-Transform
Similar to the Laplace Transform, the Z-Transform is an extension of the Fourier Transform, in this case the DT Fourier Transform. As previously defined, the response, $ y[n]\! $, of a DT LTI system is $ y[n] = H(z)z^n\! $, where $ H(z) = \sum^{\infty}_{n = -\infty} h[n]z^{-n}\! $. When $ z = e^{j\omega}\! $ with $ \omega\! $ real, this summation equals the Fourier Transform of $ h[n]\! $. When $ z\! $ is not restricted to this value, the summation is know as the Z-Transform of $ h[n]\! $. To be exact,
where $ z\! $ is a complex variable. This is sometimes denoted as $ X(z) = Z(x[n])\! $.
Relationship between Z-Transform and Fourier Transform
The Fourier Transform at $ \omega\! $ is equal to the Z-Transform at $ e^{j\omega}\! $, as shown below.
$ X(\omega) = X(e^{j\omega})\! $
If we look at the unit circle with radius $ r\! $ and $ X(z) = X(re^{j\omega})\! $, then
$ X(z) = \!F(x[n]r^{-n})\! $ because
$ X(z) = \sum^{\infty}_{n = -\infty} x[n]z^{-n}\! $
$ = \sum^{\infty}_{n = -\infty} x[n](re^{j\omega})^{-n}\! $
