(New page: == Definition == A system characterized by a difference is given as: <math>\,\ \sum_{k=1}^N k^2 </math>)
 
 
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== Definition ==
 
== Definition ==
  
−
A system characterized by a difference is given as:
+
A system characterized by a difference equation in DT is given as:
−
<math>\,\
+
 
 +
<math>\,
 +
\sum_{k=0}^N a_k\,y[n-k]=\sum_{k=0}^N b_k\,x[n-k]
 +
</math>
 +
 
 +
We will likely be asked to solve for the frequency response <math>\,H(e^{j\omega})</math>, the unit impulse response <math>\,h[n]</math>, or the system's response to an input <math>\,x[n]</math>.
 +
 
 +
 
 +
== Example 1 ==
 +
Find <math>\,H(e^{j\omega})</math>, and <math>\,h[n]</math> for the following system in DT domain:
 +
 
 +
<math>\,
 +
-\frac{2}{5}y[n-2]-\frac{17}{5}y[n-1]+6y[n]=4x[n]
 +
</math>
 +
 
 +
=== Solution ===
 +
First find <math>\,H(e^{j\omega})</math>:
 +
 
 +
1) Take the fourier transform of every term:
 +
 
 +
<math>\,
 +
-\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>
 +
 
 +
2) Factor out the y terms:
 +
 
 +
<math>\,
 +
\mathcal{Y}(\omega) \left (-\frac{2}{5}e^{-2j\omega}-\frac{17}{5}e^{-j\omega}+6 \right )=4\mathcal{X}(\omega)
 +
</math>
 +
 
 +
3) Now isolate <math>\,H(\omega)</math>
 +
 
 +
<math>\,
 +
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>
 +
 
 +
2nd, find <math>\,h[n]</math>
 +
 
 +
<math>
 +
h[n]=\mathcal{F}^{-1}(H(e^{j\omega})
 +
</math>
 +
 
 +
This is the rough part, as partial fraction expansions must be used :P
 +
 
 +
for simplification purposes, let <math>\,x=e^{-j\omega}</math> , so the fraction becomes :
 +
 
 +
<math>\,
 +
H(e^{j\omega})=\frac{4}{-\frac{2}{5}x^2-\frac{17}{5}x+6}
 +
</math>
 +
 
 +
 
 +
==== Partial Fraction Expansion ====
 +
 
 +
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}
 +
</math>
 +
 
 +
<math>\,
 +
B=\frac{4}{\frac{1}{5}\frac{3}{2}+2}=\frac{23}{10}
 +
</math>
 +
 
 +
The final PFE turns out to be:
 +
 
 +
<math>\,
 +
\frac{\frac{4}{23}}{\frac{1}{5}x+2}+\frac{\frac{23}{10}}{-2x+3}
 +
</math>
 +
 
 +
Now finally to find <math>\,h[n]</math>, take the inverse Fourier Transform:
 +
 
 +
<math>\,
 +
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)
 +
</math>
 +
 
 +
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:
 +
 
 +
<math>\,
 +
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)
 +
</math>
 +
 
 +
Then my h[n] is simple, and is:
 +
 
 +
<math>\,
 +
h[n]=\frac{4}{23}\left (\frac{1}{5} \right) ^n\,u[n] + \frac{23}{10}(2)^n\,u[n]
 +
</math>
 +
 
 +
 
 +
== Example 2 ==
 +
Another example shown in class was :
 +
<math>\,
 +
H(e^{j\omega})=\frac{1}{1-\frac{1}{2}e^{-j\omega}}
 +
</math>
 +
 
 +
Solve for the difference equation.
 +
 
 +
=== Solution ===
 +
Simply do the inverse of the previous process, without the partial fraction expansion of course:
 +
 
 +
<math>\,
 +
\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)
 +
</math>
 +
 
 +
<math>\,
 +
(1-\frac{1}{2}e^{-j\omega})\mathcal{Y}(\omega)=\mathcal{X}(\omega)
 +
</math>
 +
 
 +
<math>\,
 +
\mathcal{Y}(\omega)-\frac{1}{2}e^{-j\omega}\mathcal{Y}(\omega)=\mathcal{X}(\omega)
 +
</math>
  
−
\sum_{k=1}^N k^2
+
And now finally take the inverse F.T. of these values to get:
  
 +
<math>\,
 +
y[n]+\frac{1}{2}y[n-1]=x[n]
 
</math>
 
</math>

Latest revision as of 11:13, 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} $


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] $

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