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| − | <math> Y(n)= H(-1)*(-1)^n </math> | + | <math> Y(n)= H(-1)*(-1)^n = 0</math> |
Latest revision as of 16:14, 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 $
