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== Difference Equations == | == 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: | 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: | ||
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== Sources == | == Sources == | ||
Signals & Systems, 2nd edition, Oppenheim, Willsky | Signals & Systems, 2nd edition, Oppenheim, Willsky | ||
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| + | [[Main_Page_ECE301Fall2008mboutin|Back to ECE301 Fall 2008 Prof. Boutin]] | ||
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Revision as of 05:14, 18 November 2011
Contents
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:
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). To find the frequency response, one need only apply the equation, $ Y(\omega) = H(e^{j\omega})X(\omega)\! $. Given a difference equation of the above form, one can apply simple mathematical and Fourier Transform properties to get an equation of the form $ Y(\omega) = H(e^{j\omega})X(\omega)\! $ where $ H(e^{j\omega})\! $ is 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
