| Line 14: | Line 14: | ||
<math>\, | <math>\, | ||
| − | \frac{2}{5}y[n- | + | -\frac{2}{5}y[n-2]-\frac{17}{5}y[n-1]+6y[n]=4x[n] |
</math> | </math> | ||
| Line 23: | Line 23: | ||
<math>\, | <math>\, | ||
| − | \frac{2}{5}\mathcal{Y}(\omega)e^{- | + | -\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) |
</math> | </math> | ||
| Line 29: | Line 29: | ||
<math>\, | <math>\, | ||
| − | \mathcal{Y}(\omega) \left (\frac{2}{5}e^{- | + | \mathcal{Y}(\omega) \left (-\frac{2}{5}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6 \right )=4\mathcal{X}(\omega) |
</math> | </math> | ||
| Line 35: | Line 35: | ||
<math>\, | <math>\, | ||
| − | H(\omega)=H(e^{j\omega})=\frac{\mathcal{Y}(\omega)}{\mathcal{X}(\omega)}=\frac{4}{\frac{2}{5}e^{- | + | 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} |
</math> | </math> | ||
| Line 49: | Line 49: | ||
<math>\, | <math>\, | ||
| − | \frac{4}{\frac{2}{5} | + | H(e^{j\omega})=\frac{4}{-\frac{2}{5}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6} |
</math> | </math> | ||
Revision as of 09:31, 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}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6} $
