Contents
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} $
Partial Fraction Expansion
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}=\frac{23}{10} $
The final PFE turns out to be:
$ \, \frac{\frac{4}{23}}{\frac{1}{5}x+2}+\frac{\frac{23}{10}}{-2x+3} $
Now finally to find $ \,h[n] $, take the inverse Fourier Transform:
$ \, h[n]=\mathcal{F}^{-1} \left (\frac{\frac{4}{23}}{\frac{1}{5}e^{-j\omega}+2}+\frac{\frac{23}{10}}{-2e^{-j\omega}+3}\right) $
After all this I did not get a good geometric series, but if this were in CT it would be clear how to find the inverse fourier transform of this. if my equation were of the form:
$ \, h[n]=\mathcal{F}^{-1} \left (\frac{\frac{4}{23}}{1-\frac{1}{5}e^{-j\omega}}+\frac{\frac{23}{10}}{1-2e^{-j\omega}}\right) $
Then my h[n] is simple, and is:
$ \, h[n]=\frac{4}{23}\left (\frac{1}{5} \right) ^n\,u[n] + \frac{23}{10}(2)^n\,u[n] $
Example 2
Another example shown in class was : $ \, H(e^{j\omega})=\frac{1}{1-\frac{1}{2}e^{-j\omega}} $
Solve for the difference equation.
Solution
Simply do the inverse of the previous process, without the partial fraction expansion of course:
$ \, \mathcal{Y}(\omega)=H(\omega)\mathcal{X}(\omega)=H(e^{j\omega})\mathcal{X}(\omega)=\frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal{X}(\omega) $
$ \, (1-\frac{1}{2}e^{-j\omega})\mathcal{Y}(\omega)=\mathcal{X}(\omega) $
$ \, \mathcal{Y}(\omega)-\frac{1}{2}e^{-j\omega}\mathcal{Y}(\omega)=\mathcal{X}(\omega) $
And now finally take the inverse F.T. of these values to get:
$ \, y[n]+\frac{1}{2}y[n-1]=x[n] $
