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\end{align} | \end{align} | ||
</math> | </math> | ||
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| + | Thus | ||
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| + | <math>H(z)=\frac{Y(z)}{X(z)}=\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}\text{ ,ROC: }\frac{3}{4}<|Z|<2</math> | ||
| + | |||
| + | b. | ||
---- | ---- | ||
[[ECE438_Week10_Quiz|Back to Quiz Pool]] | [[ECE438_Week10_Quiz|Back to Quiz Pool]] | ||
[[ECE438_Lab_Fall_2010|Back to Lab wiki]] | [[ECE438_Lab_Fall_2010|Back to Lab wiki]] | ||
Revision as of 18:11, 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 \\ &=\frac{\frac{3}{4}z^{-1}}{(1-\frac{1}{2}z^{-1})(z^{-1}-\frac{1}{2})}\text{ ,ROC: }\frac{1}{2}<|z|<2 \end{align} $
$ \begin{align} Y(z)=\sum_{n=-\infty}^{\infty}y[n]z^{-n}&=\sum_{n=-\infty}^{\infty}6(\frac{1}{2})^nu[n]z^{-n}-\sum_{n=-\infty}^{\infty}6(\frac{3}{4})^nu[n]z^{-n} \\ &=\sum_{n=0}^{\infty}6(\frac{1}{2})^nz^{-n}-\sum_{n=0}^{\infty}6(\frac{3}{4})^nz^{-n} \\ &=\sum_{n=0}^{\infty}6(\frac{1}{2})^nz^{-n}-\sum_{n=0}^{\infty}6(\frac{3}{4})^nz^{-n} \\ &=\frac{6}{1-\frac{1}{2}z^{-1}}-\frac{6}{1-\frac{3}{4}z^{-1}}\text{ ,if }|z|>\frac{1}{2}\text{ and }|z|>\frac{3}{4} \\ &=\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}\text{ ,ROC: }|z|>\frac{3}{4} \end{align} $
Thus
$ H(z)=\frac{Y(z)}{X(z)}=\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}\text{ ,ROC: }\frac{3}{4}<|Z|<2 $
b.
