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<center><math>y[n] + a_{0}y[n-n_{0}] + a_{1}y[n-n_{1}] + ... + a_{n-1}y[n-n_{n-1}] = x[n]\!</math></center>
 
<center><math>y[n] + a_{0}y[n-n_{0}] + a_{1}y[n-n_{1}] + ... + a_{n-1}y[n-n_{n-1}] = x[n]\!</math></center>
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== Finding the Frequency Response from a Difference Equation ==
 
== Finding the Frequency Response from a Difference Equation ==
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<math>H(e^{j\omega}) = \frac{5}{(1 + 2e^{-j\omega} - \frac{1}{2}e^{-3j\omega} + e^{-4j\omega})} \!</math>
 
<math>H(e^{j\omega}) = \frac{5}{(1 + 2e^{-j\omega} - \frac{1}{2}e^{-3j\omega} + e^{-4j\omega})} \!</math>
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== Sources ==
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Signals & Systems, 2nd edition, Oppenheim, Willsky

Revision as of 10:38, 23 October 2008

Difference Equations

DT systems described by linear constant-coefficient difference equations are very important to the practice of signals and systems. They are of special importance when implementing filters. These equations are of the form:

$ y[n] + a_{0}y[n-n_{0}] + a_{1}y[n-n_{1}] + ... + a_{n-1}y[n-n_{n-1}] = x[n]\! $


Finding the Frequency Response from a Difference Equation

If we are given a system defined by a difference equation, it is possible to find the frequency response (actually it is quite simple to find the frequency response). An example of this is given below.

Example

Find the frequency response of the following difference equation.

$ y[n] + 2y[n-1] - \frac{1}{2}y[n-3] + y[n-4] = 5x[n]\! $


$ y[n] + 2y[n-1] - \frac{1}{2}y[n-3] + y[n-4] = 5x[n]\! $

$ F(y[n] + 2y[n-1] - \frac{1}{2}y[n-3] + y[n-4]) = F(5x[n])\! $

$ Y(\omega) + 2F(y[n-1]) - \frac{1}{2}F(y[n-3]) + e^{-4j\omega}Y(\omega) = 5X(\omega)\! $

$ Y(\omega) + 2e^{-j\omega}Y(\omega) - \frac{1}{2}e^{-3j\omega}Y(\omega) + e^{-4j\omega}Y(\omega) = 5X(\omega)\! $

$ Y(\omega) (1 + 2e^{-j\omega} - \frac{1}{2}e^{-3j\omega} + e^{-4j\omega}) = 5X(\omega)\! $

$ Y(\omega) = \frac{5}{(1 + 2e^{-j\omega} - \frac{1}{2}e^{-3j\omega} + e^{-4j\omega})} X(\omega)\! $

By the definition of frequency response, $ Y(\omega) = H(e^{j\omega})X(\omega)\! $, so the frequency response in this example is

$ H(e^{j\omega}) = \frac{5}{(1 + 2e^{-j\omega} - \frac{1}{2}e^{-3j\omega} + e^{-4j\omega})} \! $


Sources

Signals & Systems, 2nd edition, Oppenheim, Willsky

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