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<math>H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k}</math> | <math>H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k}</math> | ||
| − | <math>H(z)= | + | <math>H(z)=1+\frac{1}{z}</math> due to the two step functions. |
| + | |||
| + | ===Part B=== | ||
| + | First, recognize that <math>\cos(\pi n)=(-1)^n</math>. | ||
| + | |||
| + | |||
| + | The response to the input signal <math>z^n</math> is <math>H(z)z^n</math>, giving | ||
| + | |||
| + | <math>(1+\frac{1}{-1})(-1)^n=0</math>. | ||
Latest revision as of 02:59, 15 October 2008
Problem 5
An LTI system has unit impulse response $ h[n]=u[n]-u[n-2] $. Compute (a) the system's function $ H(z) $ and (b) the system's response to the input $ x[n]=\cos(\pi n) $.
Part A
First, note this is discrete time. Doing the problem in continuous time gives a very different result, as I tragically learned on the test...
$ H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k} $
$ H(z)=1+\frac{1}{z} $ due to the two step functions.
Part B
First, recognize that $ \cos(\pi n)=(-1)^n $.
The response to the input signal $ z^n $ is $ H(z)z^n $, giving
$ (1+\frac{1}{-1})(-1)^n=0 $.
