(New page: == Problem 3 == An LTI system has unit impulse response <math> h[n] = u[-n] </math>. Compute the system's response to the input <math> x[n] = 2^nu[-n] </math>. (simplify your answer until...) |
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An LTI system has unit impulse response <math> h[n] = u[-n] </math>. Compute the system's response to the input <math> x[n] = 2^nu[-n] </math>. (simplify your answer until all <math> \sum </math> signs disappear). | An LTI system has unit impulse response <math> h[n] = u[-n] </math>. Compute the system's response to the input <math> x[n] = 2^nu[-n] </math>. (simplify your answer until all <math> \sum </math> signs disappear). | ||
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| + | == Solution == | ||
| + | |||
| + | <math> y[n] = x[n]*h[n] \!</math> | ||
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| + | <math> y[n] = \sum^{\infty}_{k = -\infty} x[k] h[n-k] \!</math> | ||
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| + | <math> y[n] = \sum^{\infty}_{k = -\infty} 2^ku[-k] u[-(n-k)] \!</math> | ||
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| + | We know <math> u[-k] = 1 \!</math> when <math> K \leq 0 </math><font color=red>, else <math>u[-k]=0</math>, therefore </font> | ||
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| + | <math> y[n] = \sum^{0}_{k = -\infty} 2^k u[-n+k] \!</math> | ||
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| + | <font color=red> Let </font> <math> r = -k \!</math><font color=red>, then </font> | ||
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| + | <math> y[n] = \sum^{\infty}_{r = 0} (\frac{1}{2})^k u[-n-r] \!</math> | ||
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| + | We know <math> -n-r \geq 0 </math> when <math> -n \geq r </math> | ||
| + | |||
| + | <math> y[n] = \left\{ \begin{array}{ll} \sum^{-n}_{r = 0} (\frac{1}{2})^k \! & \text{ when } -n \geq 0, \\ | ||
| + | 0 & \text{ else. } | ||
| + | \end{array}\right. </math> | ||
| + | |||
| + | So | ||
| + | |||
| + | <math> y[n] = \left\{ \begin{array}{ll} \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \! & \text{ when } -n \geq 0, \\ | ||
| + | 0 & \text{ else. } | ||
| + | \end{array}\right. | ||
| + | </math> | ||
Latest revision as of 06:52, 22 October 2008
Problem 3
An LTI system has unit impulse response $ h[n] = u[-n] $. Compute the system's response to the input $ x[n] = 2^nu[-n] $. (simplify your answer until all $ \sum $ signs disappear).
Solution
$ y[n] = x[n]*h[n] \! $
$ y[n] = \sum^{\infty}_{k = -\infty} x[k] h[n-k] \! $
$ y[n] = \sum^{\infty}_{k = -\infty} 2^ku[-k] u[-(n-k)] \! $
We know $ u[-k] = 1 \! $ when $ K \leq 0 $, else $ u[-k]=0 $, therefore
$ y[n] = \sum^{0}_{k = -\infty} 2^k u[-n+k] \! $
Let $ r = -k \! $, then
$ y[n] = \sum^{\infty}_{r = 0} (\frac{1}{2})^k u[-n-r] \! $
We know $ -n-r \geq 0 $ when $ -n \geq r $
$ y[n] = \left\{ \begin{array}{ll} \sum^{-n}_{r = 0} (\frac{1}{2})^k \! & \text{ when } -n \geq 0, \\ 0 & \text{ else. } \end{array}\right. $
So
$ y[n] = \left\{ \begin{array}{ll} \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \! & \text{ when } -n \geq 0, \\ 0 & \text{ else. } \end{array}\right. $
