(New page: If we were asked to compute the frequency response, one thing that we need to keep in mind is that, no matter how complexed the problem might look, we have to somehow arrange it into <math...) |
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| − | If we were asked to compute the frequency response, one thing that we need to keep in mind is that, no matter how complexed the problem might look, we have to somehow arrange it into <math>Y(w) = H(w)X(w)</math> The following example would illustrate this: | + | If we were asked to compute the frequency response, one thing that we need to keep in mind is that, no matter how complexed the problem might look, we have to somehow arrange it into <math>Y(w) = H(w)X(w)</math> format. Obviously, <math>H(w)</math> is the frequency response. The following example would illustrate this: |
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| + | Find the frequency response of: <math>y[n] - \frac{3}{4}y[n-1] + \frac{1}{8}y[n-2] = 2x[n]</math> | ||
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| + | First, we do the Fourier transform on both sides, which yields: | ||
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| + | :<math>Y(w) - \frac{3}{4}e^{-jw}Y(w) + \frac{1}{8}e^{-2jw}Y(w) = 2X(w)\,</math> | ||
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| + | Now we try to collect terms in order to attempt to make the problem resemble the mentioned format: | ||
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| + | :<math>(1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})Y(w) = 2X(w)\,</math> | ||
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| + | Solve for Y(w): | ||
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| + | :<math>Y(w) = \frac{2}{(1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})}X(w)\,</math> | ||
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| + | Notice that the problem now resembles the mentioned format, we can easily see H(w), and since <math>H(e^{jw})</math> is H(w), the frequency response is: | ||
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| + | :<math>\frac{2}{(1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})}\,</math> | ||
Latest revision as of 09:50, 24 October 2008
If we were asked to compute the frequency response, one thing that we need to keep in mind is that, no matter how complexed the problem might look, we have to somehow arrange it into $ Y(w) = H(w)X(w) $ format. Obviously, $ H(w) $ is the frequency response. The following example would illustrate this:
Find the frequency response of: $ y[n] - \frac{3}{4}y[n-1] + \frac{1}{8}y[n-2] = 2x[n] $
First, we do the Fourier transform on both sides, which yields:
- $ Y(w) - \frac{3}{4}e^{-jw}Y(w) + \frac{1}{8}e^{-2jw}Y(w) = 2X(w)\, $
Now we try to collect terms in order to attempt to make the problem resemble the mentioned format:
- $ (1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})Y(w) = 2X(w)\, $
Solve for Y(w):
- $ Y(w) = \frac{2}{(1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})}X(w)\, $
Notice that the problem now resembles the mentioned format, we can easily see H(w), and since $ H(e^{jw}) $ is H(w), the frequency response is:
- $ \frac{2}{(1 - \frac{3}{4}e^{-jw} + \frac{1}{8}e^{-2jw})}\, $
