Difference between revisions of "Short Cut for Z-Transform" - Rhea

Latest revision as of 07:35, 23 September 2009

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Hmm... This looks a lot like my course notes... Perhaps you want to write this somewhere, otherwise one might think that you are pretending that you wrote this yourself. --Mboutin 12:26, 23 September 2009 (UTC)

Short Cut: Completely equivalent to complex integration formula

1.) Write X(z) as a power series

$ X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n} $

2.) Observe that

$ X(z) = \sum_{n=-\infty}^{\infty}x[n]z^{-n} $

i.e.

$ X(z) = \sum_{n=-\infty}^{\infty}x[-n]z^{n} $

3.) By comparison,

$ x[-n] = c_{n} $

or

$ x[n] = c_{-n} $

Example:

$ X(z) = \frac{1}{1-z} $ Observe pole at z = 1

There are 2 possible ROC's: |z| < 1 or |z| > 1

Case 1: |z| < 1 (inside circle -- left sided function)

$ X(z) = \sum_{n=0}^{\infty}z^{n} = \sum_{k=-\infty}^{0}z^{-k} = \sum_{k=-\infty}^{\infty}u(-k)z^{-k} $

So, x[n] = u[-n] --left-sided, consistent with having inside of a circle ROC

Case 2: |z| > 1

$ X(z) = \frac{1}{1-z} = \frac{1}{z((\frac{1}{z})-z)} = \frac{-1}{z}*\frac{1}{1-(\frac{1}{z})} $

Observe, |1/z| < 1, thus we can use geometric series Continuing from above,

$ =\frac{-1}{z}\sum_{n=0}^{\infty}(\frac{1}{z})^{n}=-\sum_{n=0}^{\infty}z^{-n-1} $

Let k = n + 1

$ =-\sum_{k=1}^{\infty}z^{-k}=\sum_{k=-\infty}^{\infty}-u(k-1)z^{-k} $

By comparison with Z-Transform formula, x[n] = -u[n-1] --right-sided function, consistent with having outside of circle ROC