A DT-LTI System

$ y[n] = x[n] + x[n-3]\, $

Unit Impulse Response

$ y[n] = x[n] + x[n-3]\, $

$ x[n] = \delta[n]\, $

$ h[n] = \delta[n] + \delta[n-3]\, $

Frequency Response

$ y[n] = \sum^{\infty}_{\infty} h[n] * x[n] dn\, $ where $ x[n] = e^{jwn} \, $

$ y(t) = \sum^{\infty}_{-\infty} (\delta[n] + \delta[n-3]) * e^{jwn} dn\, $

$ y(t) = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-3]) e^{jw(n-m} dm\, $

$ y(t) = e^{jwn} \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-3]) e^{-jwm} dm\, $

$ H(s) = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-3]) e^{-jwm} dm\, $

$ H(s) = e^{-jw0} + e^{-jw3}\, $

$ H(s) = 1 + e^{-jw3}\, $

Response of the DT system defined in Q1

CT Periodic Signal : $ x[n] = 5\cos(6\pi n + \pi) + 7\cos(3\pi n)\, $

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi n}\, $ where $ a_k = -5 , k = 2,4,6,8,10....\, $ and $ a_k = 7 , k = 1,3,5,7,9,11....\, $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi n}\, $

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

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett