| Line 19: | Line 19: | ||
Norm of a signal: | Norm of a signal: | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | |\left(\frac{5}{6}\right)^n u[n]| = \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> | ||
| Line 25: | Line 25: | ||
<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}\right) \\ |
| − | &= \sum_{n=0}^N (\left(\frac{25}{36}\right)^{n} \\ | + | &= \sum_{n=0}^N (\left(\frac{25}{36}\right)^{n}\right) \\ |
&= \frac{1}{1-\frac{25}{36}} \\ | &= \frac{1}{1-\frac{25}{36}} \\ | ||
&= \frac{36}{11} \\ | &= \frac{36}{11} \\ | ||
| Line 46: | Line 46: | ||
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 18:04, 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|\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}\right) \\ &= \sum_{n=0}^N (\left(\frac{25}{36}\right)^{n}\right) \\ &= \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 $
