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<math>\begin{align} | <math>\begin{align} | ||
E_{\infty}&=\sum_{n=0}^N \left|\left(\frac{5}{6}\right)^n\right|^2 \\ | 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{5}{6}\right)^{2n} \\ |
&= \sum_{n=0}^N (\left(\frac{25}{36}\right)^{n} \\ | &= \sum_{n=0}^N (\left(\frac{25}{36}\right)^{n} \\ | ||
&= \frac{1}{1-\frac{25}{36}} \\ | &= \frac{1}{1-\frac{25}{36}} \\ | ||
Revision as of 18:03, 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|\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} &= \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}}\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 $
