(New page: ==Week11 Quiz Question 4 Solution== ----- <math> \begin{align} X_1(z)=\sum_{n=-\infty}^{\infty}x_1[n]z^{-n}&=\sum_{n=-\infty}^{\infty}(\frac{1}{2})^nu[n]z^{-n}+\sum_{n=-\infty}^{\infty}2^n...) |
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | X_2(z)=\sum_{n=-\infty}^{\infty}x_2[n]z^{-n}&=\sum_{n=-\infty}^{\infty}6(\frac{1}{2})^nu[n]z^{-n}-\sum_{n=-\infty}^{\infty}6(\frac{3}{ | + | X_2(z)=\sum_{n=-\infty}^{\infty}x_2[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}{ | + | &=\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}{ | + | &=\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}{ | + | &=\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{-6z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{ | + | &=\frac{-6z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}\text{ ,ROC: }|z|>\frac{3}{4} |
\end{align} | \end{align} | ||
</math> | </math> | ||
| − | This system is | + | This system is stable because the ROC contains unit cycle. |
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Latest revision as of 06:49, 22 November 2011
Week11 Quiz Question 4 Solution
$ \begin{align} X_1(z)=\sum_{n=-\infty}^{\infty}x_1[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} $
This system is stable because the ROC contains unit cycle.
$ \begin{align} X_2(z)=\sum_{n=-\infty}^{\infty}x_2[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{-6z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}\text{ ,ROC: }|z|>\frac{3}{4} \end{align} $
This system is stable because the ROC contains unit cycle.
