(New page: ==ECE438 Lab Week10 Quiz Question 4 Solution== a. By computing X(z) and Y(z), we can obtain H(z)=Y(z)/X(z) <math> \begin{align} X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}&=\sum_{n=-\infty}...)
 
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\begin{align}
 
\begin{align}
 
X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}&=\sum_{n=-\infty}^{\infty}(\frac{1}{2})^nu[n]z^{-n}+\sum_{n=-\infty}^{\infty}2^nu[-n-1]z^{-n}  
 
X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}&=\sum_{n=-\infty}^{\infty}(\frac{1}{2})^nu[n]z^{-n}+\sum_{n=-\infty}^{\infty}2^nu[-n-1]z^{-n}  
 +
\\
 +
&=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=-\infty}^{-1}2^nz^{-n} \\
 +
&=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=1}^{\infty}2^{-n}z^{n} \\
 +
&=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=0}^{\infty}2^{-n}z^{n}-1 \\
 +
&=\frac{1}{1-\frac{1}{2}z^{-1}}+\frac{1}{1-\frac{z}{2}}-1\text{ ,if }|z|>\frac{1}{2}\text{ and }|z|<2
 
\end{align}
 
\end{align}
 
</math>
 
</math>

Revision as of 17:55, 27 October 2010

ECE438 Lab Week10 Quiz Question 4 Solution

a. By computing X(z) and Y(z), we can obtain H(z)=Y(z)/X(z)

$ \begin{align} X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}&=\sum_{n=-\infty}^{\infty}(\frac{1}{2})^nu[n]z^{-n}+\sum_{n=-\infty}^{\infty}2^nu[-n-1]z^{-n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=-\infty}^{-1}2^nz^{-n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=1}^{\infty}2^{-n}z^{n} \\ &=\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}+\sum_{n=0}^{\infty}2^{-n}z^{n}-1 \\ &=\frac{1}{1-\frac{1}{2}z^{-1}}+\frac{1}{1-\frac{z}{2}}-1\text{ ,if }|z|>\frac{1}{2}\text{ and }|z|<2 \end{align} $


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