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<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>
+
<math>\rightbar</math>

Revision as of 19:52, 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 $

$ \rightbar $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman