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| − | + | ==Problem== | |
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| − | <math>x[n]= ({5} | + | Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of this DT signal: |
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| + | <math>x[n] = \left(\frac{5}{6}\right)^n u[n] </math> | ||
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| + | ==Solution== | ||
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\left\{ | \left\{ | ||
\begin{array}{ll} | \begin{array}{ll} | ||
| − | \left(\frac{5}{6}\right)^n & \text{ | + | \left(\frac{5}{6}\right)^n & \text{ if } n\geq 0,\\ |
0 & \text{else}. | 0 & \text{else}. | ||
\end{array} | \end{array} | ||
\right. | \right. | ||
</math> | </math> | ||
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Norm of a signal: | Norm of a signal: | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | | | + | \left|\left(\frac{5}{6}\right)^n u[n]\right| = \left(\frac{5}{6}\right)^n \\ |
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\end{align}</math> | \end{align}</math> | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | E_{\infty}&=\ | + | E_{\infty}&=\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ |
| − | &= \ | + | &= \sum_{n=0}^N \left(\frac{5}{6}\right)^{2n} \\ |
| − | &=\ | + | &= \sum_{n=0}^N \left(\frac{25}{36}\right)^{n} \\ |
| + | &= \frac{1}{1-\frac{25}{36}} \\ | ||
| + | &= \frac{36}{11} \\ | ||
\end{align}</math> | \end{align}</math> | ||
| − | <math>E_{\infty} = \ | + | <math>E_{\infty} = \frac{36}{11} </math>. |
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<math>\begin{align} | <math>\begin{align} | ||
| − | P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n= | + | P_{\infty} &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ |
| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \left(\frac{25}{36}\right)^n \\ |
| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \frac{1}{1-\frac{25}{36}} \\ |
| − | &= \lim_{N\rightarrow \infty}{ | + | &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ |
| − | + | &= 0 \\ | |
| − | &= | + | \end{align}</math> |
| − | \end{align}</math> | + | |
| + | <math>P_{\infty} = 0 </math> | ||
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Conclusion: | Conclusion: | ||
| + | <math>E_{\infty} = \frac{36}{11} </math>, <math>P_{\infty} = 0 \\</math> | ||
| − | + | When <math>E_{\infty} = \text{finite number} </math>, <math>P_{\infty} = 0 </math> | |
Latest revision as of 19:39, 1 December 2018
Problem
Compute the energy $ E_\infty $ and the power $ P_\infty $ of this DT signal:
$ x[n] = \left(\frac{5}{6}\right)^n u[n] $
Solution
$ x[n] = \left\{ \begin{array}{ll} \left(\frac{5}{6}\right)^n & \text{ if } n\geq 0,\\ 0 & \text{else}. \end{array} \right. $
Norm of a signal:
$ \begin{align} \left|\left(\frac{5}{6}\right)^n u[n]\right| = \left(\frac{5}{6}\right)^n \\ \end{align} $
$ \begin{align} E_{\infty}&=\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ &= \sum_{n=0}^N \left(\frac{5}{6}\right)^{2n} \\ &= \sum_{n=0}^N \left(\frac{25}{36}\right)^{n} \\ &= \frac{1}{1-\frac{25}{36}} \\ &= \frac{36}{11} \\ \end{align} $
$ E_{\infty} = \frac{36}{11} $.
$ \begin{align} P_{\infty} &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \left(\frac{25}{36}\right)^n \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \times \frac{1}{1-\frac{25}{36}} \\ &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ &= 0 \\ \end{align} $
$ P_{\infty} = 0 $
Conclusion:
$ E_{\infty} = \frac{36}{11} $, $ P_{\infty} = 0 \\ $
When $ E_{\infty} = \text{finite number} $, $ P_{\infty} = 0 $
