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<math>\begin{align} | <math>\begin{align} | ||
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^2n \\ | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^2n \\ | ||
− | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \ | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \left(\frac{25}{36}\right)^n \\ |
− | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 \frac{1}{1-\frac{25}{36}} \\ |
− | + | ||
&= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ | &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ | ||
&= 0 \\ | &= 0 \\ | ||
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Conclusion: | Conclusion: | ||
− | <math>E_{\infty} = \frac{36}{11} </math>, <math>P_{\infty} = 0 </math> | + | <math>E_{\infty} = \frac{36}{11} </math>, <math>P_{\infty} = 0 </math> \\ |
When <math>E_{\infty} = \text{finite number} </math>, | When <math>E_{\infty} = \text{finite number} </math>, | ||
<math>P_{\infty} = 0 </math> | <math>P_{\infty} = 0 </math> |
Revision as of 17:59, 1 December 2018
Topic: Energy and Power Computation of a DT Exponential Signal </center>
Compute the energy $ E_\infty $ and the power $ P_\infty $ of this DT signal:
$ x[n] = \left(\frac{5}{6}\right)^n u[n] $
$ x[n] = \left\{ \begin{array}{ll} \left(\frac{5}{6}\right)^n & \text{ if } n\ge q 0,\\ 0 & \text{else}. \end{array} \right. $
Norm of a signal: $ \begin{align} |\left(\frac{5}{6}\right)^n u[n]| = \left(\frac{5}{6}\right)^n \end{align} $
$ \begin{align} E_{\infty}&=\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^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} &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^2n \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \left(\frac{25}{36}\right)^n \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 \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 $