(New page: == DT LTI System == <math>\ y[n] = 2x[n + 3] + x[n - 8]</math>) |
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<math>\ y[n] = 2x[n + 3] + x[n - 8]</math> | <math>\ y[n] = 2x[n + 3] + x[n - 8]</math> | ||
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| + | Unit Impulse Response: | ||
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| + | <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 == | ||
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| + | Signal from question 2: | ||
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| + | <math> x [n] = (2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi)</math> | ||
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| + | Response is defined as: | ||
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| + | <math>y[n] = \sum^{\infty}_{k = - \infty} a_{k} (H(s)) (e^{jk\pi n})</math> | ||
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| + | <math>y[n] = \sum^{\infty}_{k = - \infty} a_{k} (2e^{j \omega 3} + e^{-j \omega 8}) (e^{jk\pi n})</math> | ||
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| + | <math>y[n] = (\frac{1}{4j})(-2e^{j \frac{4 \pi}{3}n} - 2e^{- \frac{2 \pi}{3}n} - 2e^{j \frac{2 \pi}{3}n} - 2e^{-j \frac{4 \pi}{3}n} - e^{j\frac{11 \pi}{6}n} - e^{-j \frac{\pi}{6}n} - e^{j\frac{7 \pi}{6}n} - e^{-j \frac{5 \pi}{6}n}) (2e^{j \omega 3} + e^{-j \omega 8}) (e^{jk\pi n})</math> | ||
Latest revision as of 15:50, 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} $
Part B
Signal from question 2:
$ x [n] = (2 + j)cos(\frac{\pi}{3}n) + sin(\pi n - \pi) $
Response is defined as:
$ y[n] = \sum^{\infty}_{k = - \infty} a_{k} (H(s)) (e^{jk\pi n}) $
$ y[n] = \sum^{\infty}_{k = - \infty} a_{k} (2e^{j \omega 3} + e^{-j \omega 8}) (e^{jk\pi n}) $
$ y[n] = (\frac{1}{4j})(-2e^{j \frac{4 \pi}{3}n} - 2e^{- \frac{2 \pi}{3}n} - 2e^{j \frac{2 \pi}{3}n} - 2e^{-j \frac{4 \pi}{3}n} - e^{j\frac{11 \pi}{6}n} - e^{-j \frac{\pi}{6}n} - e^{j\frac{7 \pi}{6}n} - e^{-j \frac{5 \pi}{6}n}) (2e^{j \omega 3} + e^{-j \omega 8}) (e^{jk\pi n}) $
