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}) $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett