(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q2 of Week 8 Quiz Pool == ---- First, find the impulse response of <math>h_1[n]</math>. (we assumed that <math>h_1[n]=0</math>...) |
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& h_1[2]=0.25h_1[1]=\left(0.25\right)^2 \\ | & h_1[2]=0.25h_1[1]=\left(0.25\right)^2 \\ | ||
& \ldots \\ | & \ldots \\ | ||
| − | & | + | & h_1[n] = \left(0.25\right)^n u[n] \\ |
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Revision as of 11:49, 8 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]\,\! $
Credit: Prof. Charles Bouman
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