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<math> = 1 * z^{-0} + 1 * z^{-1}\,</math> | <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: | ||
| + | |||
| + | 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\, $
