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-1]-\frac{3}{5}y[n-2]+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^{-j\omega}-\frac{3}{5}\mathcal{Y}e^{-2j\omega}+6\mathcal{Y}(\omega)=4\mathcal{X}(\omega) $
2) Factor out the y terms:
$ \, \mathcal{Y}(\omega) \left (\frac{2}{5}e^{-j\omega}-\frac{3}{5}e^{-2j\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^{-j\omega}-\frac{3}{5}e^{-2j\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 :
$ \, \frac{4}{\frac{2}{5}x-\frac{3}{5}x^2+6} $
