(New page: == Number 5 == An LTI system has unit impulse response <math>h[n] = u[n] - u[n - 2]\,</math>. a)Compute the system's function H(z). b)Use the answer from a) to compute the system's r...)
 
 
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== Number 5 ==
 
== Number 5 ==
  
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b)Use the answer from a) to compute the system's response to the input <math>x[n] = cos(\pi n)\,</math>
 
b)Use the answer from a) to compute the system's response to the input <math>x[n] = cos(\pi n)\,</math>
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== Answer ==
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a)
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<math>H(z) = \sum_{k=-\infty}^{\infty}h(n)z^{-n}\,</math>
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After graphing out the two spikes:
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<math> = 1 * z^{-0} + 1 * z^{-1}\,</math>
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<math> = 1 + \frac{1}{z}\,</math>
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b)
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<math>x[n] = cos(\pi n) = (-1)^{n}\,</math>
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Given this, z = -1, so:
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System response <math> = H(z)z^{n} = (1 + \frac{1}{-1}) = 0\,</math>

Latest revision as of 14:16, 15 October 2008

Number 5

An LTI system has unit impulse response $ h[n] = u[n] - u[n - 2]\, $.

a)Compute the system's function H(z).

b)Use the answer from a) to compute the system's response to the input $ x[n] = cos(\pi n)\, $


Answer

a)

$ H(z) = \sum_{k=-\infty}^{\infty}h(n)z^{-n}\, $

After graphing out the two spikes:

$ = 1 * z^{-0} + 1 * z^{-1}\, $

$ = 1 + \frac{1}{z}\, $


b)

$ x[n] = cos(\pi n) = (-1)^{n}\, $

Given this, z = -1, so:

System response $ = H(z)z^{n} = (1 + \frac{1}{-1}) = 0\, $

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