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<math>x[n] = \sum^{\infty}_{k = -\infty} a_k e^{jk\frac{\pi}{2} n}\,</math> | <math>x[n] = \sum^{\infty}_{k = -\infty} a_k e^{jk\frac{\pi}{2} n}\,</math> | ||
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| + | <math>y[n] = \sum^{\infty}_{k = -\infty} a_k F(z) e^{jk\frac{\pi}{2} n}\,</math> | ||
Revision as of 11:35, 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^{\infty}_{k = -\infty} a_k e^{jk\frac{\pi}{2} n}\, $
$ y[n] = \sum^{\infty}_{k = -\infty} a_k F(z) e^{jk\frac{\pi}{2} n}\, $
