Contents
Computing the Fourier series coefficients for a Discrete Time signal x[n]
The System
$ y[n] = x[n] + x[n-1] + x[n-2] + x[n-3]\, $
Unit Impulse Response
$ x[n] = \delta[n]\, $
$ h[n] = \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3]\, $
Frequency Response
$ y[n] = \sum^{\infty}_{\infty} h[n] * x[n] dn\, $ where $ x[n] = e^{jwn} \, $
$ y[n] = \sum^{\infty}_{-\infty} (\delta[n] + \delta[n-1]+ \delta[n-2] + \delta[n-3]) e^{jwn} \, $
$ y[n] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{jw(n-m)} \, $
$ y[n] = e^{jwn} \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \, $
$ H[z] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \, $
$ H[z] = e^{-jw0} + e^{-jw1}+ e^{-jw2}+ e^{-jw3}\, $
$ H[z] = 1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}\, $
