Line 10: Line 10:
  
 
<math>H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \frac{1}{2}u[m] 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</math>
 
<math>H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \frac{1}{2}u[m] 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</math>
 +
 +
<math>\bar</math>

Revision as of 18:51, 23 September 2008

DT LTI System

$ y[n] = \sum_{n=-\infty}^{\infty}\frac{1}{2}x[n] \; \; $     (scaled DT integral)

h[n]

$ h[n] = \sum_{n=-\infty}^{\infty}\frac{1}{2}\delta [n] = \frac{1}{2}u[n] $

H(z)

$ H(z) = \sum_{m=-\infty}^{\infty}h[m] e^{-j \omega m} = \sum_{m=-\infty}^{\infty} \frac{1}{2}u[m] 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 $

$ \bar $

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch