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Signal: <math>x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\,</math> | Signal: <math>x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\,</math> | ||
| − | <math>x[n] = \sum^{ | + | <math>x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\,</math>, where |
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| − | <math>y[n] = \sum^{ | + | <math>y[n] = \sum^{3}_{k = 0} a_k F(z) e^{jk\frac{\pi}{2} n}\,</math> |
| − | <math>y[n] = \sum^{ | + | <math>y[n] = \sum^{3}_{k = 0} a_k (1 + e^{-j\frac{\pi}{2}}) e^{jk\frac{\pi}{2} n}\,</math> |
Revision as of 13:07, 25 September 2008
Contents
Define a DT LTI system
$ y[n] = x[n+1] + x[n]\, $
Obtain the Unit Impulse Response h[n]
By definition, to obtain the unit impulse response from a system defined by $ y[n] = x[n]\, $, simply replace the $ x[n]\, $ by $ \delta[n]\, $.
$ h[n] = \delta[n+1] + \delta[n]\, $
Obtain the System Function $ F(z)\, $ of the System
$ F(z) = \sum^{\infty}_{m=-\infty} h[m]e^{jm\omega} \, $
$ F(z) = \sum^{\infty}_{m=-\infty} (\delta[m+1] + \delta[m])e^{jm\omega} \, $
$ F(z) = \sum^{\infty}_{m=-\infty} \delta[m+1]e^{jm\omega} + \delta[m]e^{jm\omega} \, $
Since the delta function is only valid when its input is zero,
$ F(z) = e^{-j\omega} + e^{0j\omega} \, $
$ F(z) = 1 + e^{-j\omega} \, $
Compute the Response of My Signal from Question 2
Signal: $ x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\, $
$ x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\, $, where
$ a_0 = 0 , k = ..., -8, -4, 0, 4, 8, ... \, $
$ a_1 = 2j , k = ..., -7, -3, 1, 5, 9, ... \, $
$ a_2 = 2 , k = ..., -6, -2, 2, 6, 10, ... \, $
$ a_3 = -2j , k = ..., -5, -1, 3, 7, 11, ... \, $
$ y[n] = \sum^{3}_{k = 0} a_k F(z) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = \sum^{3}_{k = 0} a_k (1 + e^{-j\frac{\pi}{2}}) e^{jk\frac{\pi}{2} n}\, $
