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<math>\begin{align} | <math>\begin{align} | ||
| − | E_{\infty}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |je^{3\pi jn}| \\ | + | E_{\infty}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |je^{3\pi jn}|^2 \\ |
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ | &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ | ||
&=\infty. \\ | &=\infty. \\ | ||
Revision as of 14:00, 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 the DT exponential signal below:
$ x[n]= e^{-j3\pi n} $
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}|^2 \\ &= \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 $
