Difference between revisions of "Z transform ECE438F10" - Rhea

Revision as of 18:14, 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 e^s = 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 e^s 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.

Properties

1. Unit step

When $ x[n]=1 (n{\ge}0) $
  x[n] = 0(n < 0)

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}1\cdot z^{-n}=\frac{1}{1-z^{-1}} $ , ROC : |z|>1

2. Power series

x[n]=aⁿ,

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}a^{n} z^{-n}=\frac{1}{1-az^{-1}} $ , ROC : |z|>a

3. Exponential funtion

x[n]=e^-an,

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}e^{-an} z^{-n}=\sum_{n=0}^{\infty}[e^{-a} z^{-1}]^{n}=\frac{1}{1-e^{-a}z^{-n}} $ , ROC : |z|>e^-a

4. Sinusoidal function

x[n]=sinwn,

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}\frac{e^{jn{\omega}} -e^{-jn{\omega}}} {2j} z^{-n} $ $ =\frac{1}{2j} (\frac{1}{1-e^{j\omega}z^{-1}}-\frac{1}{1-e^{-j\omega}z^{-1}}) $ $ =\frac{1}{2j} (\frac{-e^{-j\omega}z^{-1}+e^{j\omega}z^{-1}}{1-e^{-j\omega}z^{-1}-e^{j\omega}z^{-1}+z^{-2}}) $ $ =\frac{z^{-1}sin(\omega)}{1-2z^{-1}cos(\omega)+z^{-2}} $

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.

Back to ECE438, Fall 2010, Prof. Boutin