| Line 11: | Line 11: | ||
\left\{ | \left\{ | ||
\begin{array}{ll} | \begin{array}{ll} | ||
| − | \left(\frac{5}{6}\right)^n & \text{ if } n\ | + | \left(\frac{5}{6}\right)^n & \text{ if } n\ge q 0,\\ |
0 & \text{else}. | 0 & \text{else}. | ||
\end{array} | \end{array} | ||
| Line 33: | Line 33: | ||
<math>E_{\infty} = \frac{36}{11} </math>. | <math>E_{\infty} = \frac{36}{11} </math>. | ||
| + | |||
| + | <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}} \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}} \\ | ||
| Line 39: | Line 41: | ||
&= \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 \\ | ||
| + | \end{align}</math> | ||
Revision as of 17:54, 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}} \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 \\ \end{align} $
$ P_{\infty} = 0 $
Conclusion: $ E_{\infty} = \frac{36}{11} $, $ P_{\infty} = 0 $ When $ E_{\infty} = finite number $, </math>, $ P_{\infty} = 0 $
