(Problem 5)
(Problem 5)
 
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<math> = 1 + \frac{1}{z}</math>
 
<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>
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<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>
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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\, $

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