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<math>h[n] = \delta[n-5] + \delta[n-3]\,</math> | <math>h[n] = \delta[n-5] + \delta[n-3]\,</math> | ||
| − | Then we find the frequency response: | + | Then we find the frequency response: |
| + | |||
<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math> | <math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math> | ||
| − | <math>F(z) = | + | find m value to make the value inside the bracket zero |
| + | |||
| + | m = -5 for the first set and 3 for the second set | ||
| + | |||
| + | <math>F(z) = e^{-5j\omega} + e^{3j\omega} \,</math> | ||
| + | |||
| + | |||
=Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal= | =Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal= | ||
Revision as of 08:45, 26 September 2008
Obtain the input impulse response h[n] and the system function H(z) of your system
Defining a DT LTI:
$ y[n] = x[n+5] + x[n-3]\, $
So, we have the unit impulse response:
$ h[n] = \delta[n-5] + \delta[n-3]\, $
Then we find the frequency response:
$ F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\, $
find m value to make the value inside the bracket zero
m = -5 for the first set and 3 for the second set
$ F(z) = e^{-5j\omega} + e^{3j\omega} \, $
