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Given the following LTI DT system
 
Given the following LTI DT system
  
<math>\,s[t]=x[t]+x[t-1]\,</math>
+
<math>\,s[n]=x[n]+x[n-1]\,</math>
  
  
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<math>\,H(z)=\sum_{m=-\infty}^{\infty}h[m]z^{-m}\,</math>
 
<math>\,H(z)=\sum_{m=-\infty}^{\infty}h[m]z^{-m}\,</math>
 +
 +
<math>\,H(z)=\sum_{m=-\infty}^{\infty}(\delta[m]+\delta[m-1])z^{-m}\,</math>
 +
 +
using the sifting property
 +
 +
<math>\,H(z)=z^{0}+z^{-1}\,</math>
 +
 +
<math>\,H(z)=1+z^{-1}\,</math>
  
  
 
== Part B ==
 
== Part B ==

Revision as of 17:20, 25 September 2008

Given the following LTI DT system

$ \,s[n]=x[n]+x[n-1]\, $


Part A

Find the system's unit impulse response $ \,h[n]\, $ and system function $ \,H(z)\, $.


The unit impulse response is simply (plug a $ \,\delta[n]\, $ into the system)

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


The system function can be found using the following formula (for LTI systems)

$ \,H(z)=\sum_{m=-\infty}^{\infty}h[m]z^{-m}\, $

$ \,H(z)=\sum_{m=-\infty}^{\infty}(\delta[m]+\delta[m-1])z^{-m}\, $

using the sifting property

$ \,H(z)=z^{0}+z^{-1}\, $

$ \,H(z)=1+z^{-1}\, $


Part B

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood