Revision as of 20:40, 23 September 2008 by Jpfister (Talk)

DT LTI System

$ y[n] = x[n] + 2x[n-1] $

h[n]

$ h[n] = \delta [n] + 2 \delta [n-1] $

H(z)

$ H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} (\delta [n] + 2 \delta [n-1]) e^{-j \omega m} = \sum_{m=0}^{\infty} \frac{1}{2}e^{-j \omega m} = \sum_{m=0}^{\infty} (\frac{1}{2 e^{j \omega}})^m = \frac{1}{1-\frac{1}{2 e^{j \omega}}} $     (geometric series $ r^n $ where $ |r| < 1 $)

Response to x[n]

Input $ x[n] $ is the following signal:

SawDTJP ECE301Fall2008mboutin.jpg

The Fourier series coefficients for $ x[n] $ are:

$ a_{0} = 1 $

$ a_{1} = -\frac{1}{2} $

$ a_{2} = 0 $

$ a_{3} = -\frac{1}{2} $

$ y[n] = \sum a_{k} F(e^{jk \omega_{o}}) e^{jk \omega_{o}n} = \sum_{k=0}^{3} a_{k} F(e^{jk \frac{\pi}{2}}) e^{jk \frac{\pi}{2}n} = \sum_{k=0}^{3} a_{k} \frac{1}{1-\frac{1}{2 e^{jk \frac{\pi}{2}}}} e^{jk \frac{\pi}{2}n} $

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch