Latest revision as of 14:57, 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] $