Difference between revisions of "ECE438 Week8 Quiz Q2sol" - Rhea

Latest revision as of 10:20, 13 October 2010

Solution to Q2 of Week 8 Quiz Pool

First, find the impulse response of $ h_1[n] $. (we assumed that $ h_1[n]=0 $ when $ n<0 $)

$ \begin{align} & h_1[n] = 0.25 h_1[n-1] + \delta[n] \\ & h_1[0]=1 \\ & h_1[1]=0.25h_1[0]=0.25 \\ & h_1[2]=0.25h_1[1]=\left(0.25\right)^2 \\ & \ldots \\ & h_1[n] = \left(0.25\right)^n u[n] \\ \end{align}\,\! $

In order to satisfy $ x[n]=h_2[n]\ast h_1[n]\ast x[n] $ for any discrete-time signal $ x[n] $,

$ h_2[n] $ must satisfy $ h_2[n]\ast h_1[n] = \delta[n] $.

Therefore, their Z-transform must satisfy $ H_1(z) H_2(z) = 1 $.

Since $ H_1(z)=\frac{1}{1-0.25z^{-1}} $, it follows that

$ H_2(z)=\frac{1}{H_1(z)}=1-0.25z^{-1} $

By its casual assumption, $ h_2[n]=\delta[n]-0.25\delta[n-1]\,\! $.

Then, the difference equation of the LTI system with the impulse reponss of $ h_2[n] $ is,

$ y[n]=x[n]-0.25x[n-1]\,\! $

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