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<math>\ h[n] = 2 \delta[n + 3] + \delta[n - 8]</math> | <math>\ h[n] = 2 \delta[n + 3] + \delta[n - 8]</math> | ||
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| + | Frequency Response: | ||
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| + | <math>\ H(s) = \sum_{- \infty}^{\infty}h[n]e^{j \omega n}</math> | ||
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| + | Plugging in the values: | ||
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| + | <math>\ H(s) = \sum_{- \infty}^{\infty}(2 \delta[n + 3] + \delta[n - 8])(e^{j \omega n})</math> | ||
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| + | <math>\ = 2e^{j \omega 3} + e^{-j \omega 8}</math> | ||
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| + | == Part B == | ||
Revision as of 14:36, 26 September 2008
DT LTI System
$ \ y[n] = 2x[n + 3] + x[n - 8] $
Unit Impulse Response:
$ \ h[n] = 2 \delta[n + 3] + \delta[n - 8] $
Frequency Response:
$ \ H(s) = \sum_{- \infty}^{\infty}h[n]e^{j \omega n} $
Plugging in the values:
$ \ H(s) = \sum_{- \infty}^{\infty}(2 \delta[n + 3] + \delta[n - 8])(e^{j \omega n}) $
$ \ = 2e^{j \omega 3} + e^{-j \omega 8} $
