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Revision as of 18:19, 16 December 2010
Z transform
Z transform is a general form of DTFT. If x[n] is a discrete periodic funtion, DFT of this function is $ x[k] = \sum_{n=0}^{N-1} x[n]e^{-j\frac{{2\pi}k n}{N}} $
Let's say x[n] is discrete nonperiodic function. Nonperiodic function is also a function with period $ \infty $. Therefore DTFT of this function can be DFT with $ N=\infty $. $ \lim_{N\to\infty}\sum_{n=0}^{N-1}x[n]e^{-j\frac{{2\pi}k n}{N}}=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n} $, where $ \omega=\lim_{N\to\infty}\frac{{2\pi}k}{N} $
$ X(w)=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n} $ ---> DTFT
The difference between DFT and DTFT is that DFT has a discrete function with k and DTFT is a continuous function with ω.
Let's generalize the DTFT. By substitution of real value frequency ω into complex frequency value s = σ + iω, DTFT is now discrete function of Laplace transform.
$ X(s)=\sum_{n=0}^{\infty}x[n]e^{-sn} $ ---> Laplace Transform
When es = z, it is a Z transform.
$ X(z)=Z[x[n]]=\sum_{n=0}^{\infty}x[n]z^{-n} $ ---> Z Transform
The first reason why we are using z transform instead of laplace is becuase it is easier to calculate the geometric series by substitution of variable from s to z. Also properties of power series with differential equation is useful.
As the variable es is substituted with z, region of convergence changes from the complex plane. The region of convergence in the complex plane for Z transform is outside region of a unit circle | z | = r = 1, where r is radius of a unit circle.
Inverse Z transform can be stated as
$ x[n]=\frac{1}{j{2\pi}}\oint{X(z)z^{n-1}dz} $
These are the examples of Z transform.
Comments:
Good job so far. You should specify the ROC for each of the signals that you applied the Z-transform on. -Sbiddand
Anybody see anything else? Do you have more questions? Comments? Please feel free to add below.
