Difference between revisions of "Computation of Energy and Power of a DT signal" - Rhea

Revision as of 16:01, 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}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |\left(\frac{5}{6}\right)^n}|^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N \left(\frac{5}{6}\right)^{2n} \\ &=\infty. \\ \end{align} $

$ \begin{align} P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N |je^{3\pi jn}|^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N 1 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^{2N} 1 \\ &= \lim_{N\rightarrow \infty}{2N+1 \over {2N+1}} \\ &= \lim_{N\rightarrow \infty}{1}\\ &= 1 \\ \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} = \infty $.

&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^N (\frac{1}{2})^n \\ &= \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{2-\frac{1}{2^N}}{2N+1} \right) \\ &= \frac{2}{\infty}\\ &= 0 \\

$ P_{\infty} = 1 $

Conclusion:

Therefore, $ E_{\infty} = \infty $, $ P_{\infty} = 1 $