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<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+1]e^{jm\omega}\,</math> | <math>F(z) = \sum^{\infty}_{m=-\infty} h[m+1]e^{jm\omega}\,</math> | ||
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<math>F(z) = e^{j\omega} \,</math> | <math>F(z) = e^{j\omega} \,</math> | ||
Revision as of 17:29, 26 September 2008
4. Define a DT LTI system.
a) Obtain the unit impulse response h[n] and the system function H(z) of your system.
Defining a DT LTI:
$ y[n] = x[n+1]\, $
Unit impulse response:
$ h[n] = \delta[n+1]\, $
Then we find the frequency response:
$ F(z) = \sum^{\infty}_{m=-\infty} h[m+1]e^{jm\omega}\, $
$ F(z) = e^{j\omega} \, $
b) 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.
Signal defined in Question 1:
$ X[n] = 6\cos(3 \pi n + \pi)\, $
$ x[n] = \sum^{2}_{k = -1} a_k e^{jk\frac{\pi}{2} n}\, $
$ X[0] = -6 \, $
$ X[1] = 6 \, $
$ X[2] = -6 \, $
$ X[-1] = 6 \, $
The pattern of k can be seen since it forms a wave.
$ y[n] = \sum^{2}_{k = -1} a_k F(z) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = \sum^{2}_{k = -1} a_k (e^{-5j\omega} + e^{3j\omega}) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = 6 (e^{-5j\omega} + e^{3j\omega}) e^{j(-1)\frac{\pi}{2} n} + (-6) (e^{-5j\omega} + e^{3j\omega}) e^{0}\, $
