(New page: == Problem 5 == An LTI system has unit impulse response h[n]=u[n]-u[n-2]. a)Compute the system's function H(z).) |
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a)Compute the system's function H(z). | a)Compute the system's function H(z). | ||
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| + | <math>H(z) = \sum_{k=-\infty}^{\infty}h[k]z^{-k}\,</math> | ||
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| + | <math> = \sum_{k=-\infty}^{\infty}(u[k]-u[k-2])z^{-k}\,</math> | ||
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| + | <math> = \sum_{k=0}^{1}z^{-k}\,</math> | ||
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| + | <math> = 1 + \frac{1}{z}</math> | ||
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| + | b) Use your answer in a) to compute the system's response to the input x[n] = cos(<math>\pi</math>n). | ||
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| + | <math>x[n] = \sum_{k=<N>}^{}a_ke^{jk(2\pi/N)n}\,</math> | ||
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| + | Then the response is | ||
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| + | <math>y[n] = \sum_{k=<N>}^{}a_kH(e^{j2\pi k/N})e^{jk(2\pi/N)n}\,</math> | ||
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| + | <math>x[n] = cos(\pi n)\,</math> | ||
| + | <math>=\frac{e^{j\pi n}+e^{-j\pi n}}{2}\,</math> | ||
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| + | <math>output=\frac{1}{2}H(e^{j\pi})e^{j\pi n}+\frac{1}{2}H(e^{-j\pi})e^{-j\pi n}\,</math> | ||
| + | |||
| + | since <math>H(e^{j\pi})=H(e^{-j\pi})=0\,</math> | ||
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| + | <math>output=0\,</math> | ||
Latest revision as of 17:03, 15 October 2008
Problem 5
An LTI system has unit impulse response h[n]=u[n]-u[n-2].
a)Compute the system's function H(z).
$ H(z) = \sum_{k=-\infty}^{\infty}h[k]z^{-k}\, $
$ = \sum_{k=-\infty}^{\infty}(u[k]-u[k-2])z^{-k}\, $
$ = \sum_{k=0}^{1}z^{-k}\, $
$ = 1 + \frac{1}{z} $
b) Use your answer in a) to compute the system's response to the input x[n] = cos($ \pi $n).
$ x[n] = \sum_{k=<N>}^{}a_ke^{jk(2\pi/N)n}\, $
Then the response is
$ y[n] = \sum_{k=<N>}^{}a_kH(e^{j2\pi k/N})e^{jk(2\pi/N)n}\, $
$ x[n] = cos(\pi n)\, $ $ =\frac{e^{j\pi n}+e^{-j\pi n}}{2}\, $
$ output=\frac{1}{2}H(e^{j\pi})e^{j\pi n}+\frac{1}{2}H(e^{-j\pi})e^{-j\pi n}\, $
since $ H(e^{j\pi})=H(e^{-j\pi})=0\, $
$ output=0\, $
