| Line 49: | Line 49: | ||
<math>\, | <math>\, | ||
| − | H(e^{j\omega})=\frac{4}{-\frac{2}{5} | + | H(e^{j\omega})=\frac{4}{-\frac{2}{5}x^2-\frac{17}{5}x+6} |
| + | </math> | ||
| + | |||
| + | Now for the partial fraction expansion steps: | ||
| + | 1) Write a polynomial expansion(or find the roots) of the denominator: | ||
| + | |||
| + | <math>\, | ||
| + | \left (-\frac{2}{5}x^2-\frac{17}{5}x+6 \right )= \left ( \frac{1}{5}x+2 \right )(-2x+3) | ||
| + | </math> | ||
| + | |||
| + | 2) Now setup the PFE in the form of: | ||
| + | |||
| + | <math>\, | ||
| + | \frac{4}{\left ( \frac{1}{5}x+2 \right )(-2x+3)}=\frac{A}{\left ( \frac{1}{5}x+2 \right )}+\frac{B}{-2x+3} | ||
| + | </math> | ||
| + | |||
| + | 3) To find A, set the denominator of A to 0, and solve for x(This method is called the Parāvartya Sūtra method): | ||
| + | |||
| + | <math>\, | ||
| + | \frac{1}{5}x+2=0, x=-10 | ||
| + | </math> | ||
| + | |||
| + | Now plug this value into the left equation to solve for A: | ||
| + | |||
| + | <math>\, | ||
| + | A=\frac{4}{20+3}=\frac{4}{23} | ||
| + | </math> | ||
| + | |||
| + | Do the same for B: | ||
| + | |||
| + | <math>\, | ||
| + | -2x+3=0, x=\frac{3}{2} | ||
| + | B=\frac{4}{\frac{1}{5}\frac{3}{2}+2} | ||
</math> | </math> | ||
Revision as of 09:53, 24 October 2008
Definition
A system characterized by a difference equation in DT is given as:
$ \, \sum_{k=0}^N a_k\,y[n-k]=\sum_{k=0}^N b_k\,x[n-k] $
We will likely be asked to solve for the frequency response $ \,H(e^{j\omega}) $, the unit impulse response $ \,h[n] $, or the system's response to an input $ \,x[n] $.
Example 1
Find $ \,H(e^{j\omega}) $, and $ \,h[n] $ for the following system in DT domain:
$ \, -\frac{2}{5}y[n-2]-\frac{17}{5}y[n-1]+6y[n]=4x[n] $
Solution
First find $ \,H(e^{j\omega}) $:
1) Take the fourier transform of every term:
$ \, -\frac{2}{5}\mathcal{Y}(\omega)e^{-2j\omega}-\frac{17}{5}\mathcal{Y}e^{-j\omega}+6\mathcal{Y}(\omega)=4\mathcal{X}(\omega) $
2) Factor out the y terms:
$ \, \mathcal{Y}(\omega) \left (-\frac{2}{5}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6 \right )=4\mathcal{X}(\omega) $
3) Now isolate $ \,H(\omega) $
$ \, H(\omega)=H(e^{j\omega})=\frac{\mathcal{Y}(\omega)}{\mathcal{X}(\omega)}=\frac{4}{-\frac{2}{5}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6} $
2nd, find $ \,h[n] $
$ h[n]=\mathcal{F}^{-1}(H(e^{j\omega}) $
This is the rough part, as partial fraction expansions must be used :P
for simplification purposes, let $ \,x=e^{-j\omega} $ , so the fraction becomes :
$ \, H(e^{j\omega})=\frac{4}{-\frac{2}{5}x^2-\frac{17}{5}x+6} $
Now for the partial fraction expansion steps: 1) Write a polynomial expansion(or find the roots) of the denominator:
$ \, \left (-\frac{2}{5}x^2-\frac{17}{5}x+6 \right )= \left ( \frac{1}{5}x+2 \right )(-2x+3) $
2) Now setup the PFE in the form of:
$ \, \frac{4}{\left ( \frac{1}{5}x+2 \right )(-2x+3)}=\frac{A}{\left ( \frac{1}{5}x+2 \right )}+\frac{B}{-2x+3} $
3) To find A, set the denominator of A to 0, and solve for x(This method is called the Parāvartya Sūtra method):
$ \, \frac{1}{5}x+2=0, x=-10 $
Now plug this value into the left equation to solve for A:
$ \, A=\frac{4}{20+3}=\frac{4}{23} $
Do the same for B:
$ \, -2x+3=0, x=\frac{3}{2} B=\frac{4}{\frac{1}{5}\frac{3}{2}+2} $
