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|\left(\frac{5}{6}\right)^n u[n]| = \left(\frac{5}{6}\right)^n | |\left(\frac{5}{6}\right)^n u[n]| = \left(\frac{5}{6}\right)^n | ||
\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
E_{\infty}&=\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^2 \\ | 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{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}} \\ | ||
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\end{align}</math> | \end{align}</math> | ||
| − | <math>E_{\infty} = \ | + | <math>E_{\infty} = \frac{36}{11} </math>. |
| − | + | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N |\left(\frac{5}{6}\right)^n|^2n \\ | |
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| − | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N (\frac{ | + | |
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \frac{1(1-(\frac{1}{2})^{N+1})}{1-\frac{1}{2}} \\ | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} \frac{1(1-(\frac{1}{2})^{N+1})}{1-\frac{1}{2}} \\ | ||
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 (1-(\frac{1}{2})^{N+1}) \\ | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 (1-(\frac{1}{2})^{N+1}) \\ | ||
&= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N}) \\ | &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N}) \\ | ||
| − | &= \lim_{N\rightarrow \infty} \left(\frac{ | + | &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ |
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&= 0 \\ | &= 0 \\ | ||
| − | <math>P_{\infty} = | + | <math>P_{\infty} = 0 </math> |
Conclusion: | Conclusion: | ||
| − | + | <math>E_{\infty} = \frac{36}{11} </math>, <math>P_{\infty} = 0 </math> | |
| − | + | When <math>E_{\infty} = finite number </math>, </math>, <math>P_{\infty} = 0 </math> | |
Revision as of 17:01, 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)\right^n & n \geq 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} $.
&= \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}} \frac{1(1-(\frac{1}{2})^{N+1})}{1-\frac{1}{2}} \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}2 (1-(\frac{1}{2})^{N+1}) \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}} (2-\frac{1}{2^N}) \\ &= \lim_{N\rightarrow \infty} \left({ \frac{\frac{36}{11}}{2N+1}} \right) \\ &= 0 \\
$ P_{\infty} = 0 $
Conclusion: $ E_{\infty} = \frac{36}{11} $, $ P_{\infty} = 0 $ When $ E_{\infty} = finite number $, </math>, $ P_{\infty} = 0 $
