(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). <math>H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k}</math> <math>H(z)=\s...) |
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<math>H(z)=\sum_{k=-\infty}^{\infty}(u[k]-u[k-2])z^{-k}</math> | <math>H(z)=\sum_{k=-\infty}^{\infty}(u[k]-u[k-2])z^{-k}</math> | ||
| − | + | u[k]= 1, k > 0 and 0, else | |
| − | + | ||
| − | + | ||
| + | u[k-2]= 1, k > 0 and 0, else | ||
| + | |||
| + | h[n] is 1 between 0 and 1; and 0 elsewhere | ||
<math>H(z)=\sum_{k=0}^{1}z^{-k}</math> | <math>H(z)=\sum_{k=0}^{1}z^{-k}</math> | ||
| − | + | <math>H(z)= 1*z^{-0} + 1*z^{-1}</math> | |
b) the system's response to the input <math>x[n]=\cos(\pi n)</math>. | b) the system's response to the input <math>x[n]=\cos(\pi n)</math>. | ||
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| + | |||
| + | <math>\cos(\pi n)=(-1)^n</math> | ||
| + | |||
| + | |||
| + | <math> Y(n)= H(-1)*(-1)^n = 0</math> | ||
Latest revision as of 15:14, 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} $
$ H(z)=\sum_{k=-\infty}^{\infty}(u[k]-u[k-2])z^{-k} $
u[k]= 1, k > 0 and 0, else
u[k-2]= 1, k > 0 and 0, else
h[n] is 1 between 0 and 1; and 0 elsewhere
$ H(z)=\sum_{k=0}^{1}z^{-k} $
$ H(z)= 1*z^{-0} + 1*z^{-1} $
b) the system's response to the input $ x[n]=\cos(\pi n) $.
$ \cos(\pi n)=(-1)^n $
$ Y(n)= H(-1)*(-1)^n = 0 $
