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| − | + | ==Problem== | |
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Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of this DT signal: | Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of this DT signal: | ||
<math>x[n] = \left(\frac{5}{6}\right)^n u[n] </math> | <math>x[n] = \left(\frac{5}{6}\right)^n u[n] </math> | ||
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| + | ==Solution== | ||
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\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 | + | \left|\left(\frac{5}{6}\right)^n u[n]\right| = \left(\frac{5}{6}\right)^n // |
\end{align}</math> | \end{align}</math> | ||
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&= \frac{36}{11} \\ | &= \frac{36}{11} \\ | ||
\end{align}</math> | \end{align}</math> | ||
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<math>E_{\infty} = \frac{36}{11} </math>. | <math>E_{\infty} = \frac{36}{11} </math>. | ||
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<math>\begin{align} | <math>\begin{align} | ||
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<math>P_{\infty} = 0 </math> | <math>P_{\infty} = 0 </math> | ||
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Conclusion: | Conclusion: | ||
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 $
