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Revision as of 14:38, 1 December 2018
Topic: Energy and Power Computation of a DT geometric signal </center>
Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following DT signal:
$ x[n]= (\frac{5}{6})u[n] $
$ x[n] = \left\{ \begin{array}{ll} \left(\frac{5}{6}\right)^n & \text{ } n>=0,\\ 0 & \text{else}. \end{array} \right. $
Norm of a signal: $ \begin{align} |je^{3\pi jn}| = {{je^{3\pi jn}}\times{-je^{-3\pi jn}}} &= {{-j^2}\times{e^{3\pi jn - 3\pi jn}}} &= 1 \end{align} $
$ \begin{align} E_{\infty}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |je^{3\pi jn}| \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ &=\infty. \\ \end{align} $
$ E_{\infty} = \infty $.
$ \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} $
$ P_{\infty} = 1 $
Conclusion:
Therefore, $ E_{\infty} = \infty $, $ P_{\infty} = 1 $
