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}\, $

Response of y[n] to the signal I defined in Question 2 using H[z] and the Fourier series coefficients

From Question 2: $ x[n] = 7sin(7\pi n + \frac{\pi}{8})\, $ where $ a_k = 7 , k = 1,3,5,7,9,11....\, $

$ x[n] = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi n}\, $

$ y[t] = \sum^{\infty}_{k = -\infty} a_k H[z] e^{jk\pi n}\, $

$ y[t] = \sum^{\infty}_{k = -\infty} a_k (1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}) e^{jk\pi n}\, $

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch