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| − | + | ==Problem== | |
| − | </ | + | Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the DT exponential signal below: |
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| + | <math>x[n]= e^{-j3\pi n} </math> | ||
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| + | ==Solution== | ||
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Norm of a signal: | Norm of a signal: | ||
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<math>\begin{align} | <math>\begin{align} | ||
|je^{3\pi jn}| = {{je^{3\pi jn}}\times{-je^{-3\pi jn}}} | |je^{3\pi jn}| = {{je^{3\pi jn}}\times{-je^{-3\pi jn}}} | ||
| − | = {{-j^2}{e^{3\pi jn - 3\pi jn}}} | + | &= {{-j^2}\times{e^{3\pi jn - 3\pi jn}}} |
| + | &= 1 | ||
\end{align}</math> | \end{align}</math> | ||
<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. \\ | ||
\end{align}</math> | \end{align}</math> | ||
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<math>E_{\infty} = \infty</math>. | <math>E_{\infty} = \infty</math>. | ||
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<math>\begin{align} | <math>\begin{align} | ||
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&= 1 \\ | &= 1 \\ | ||
\end{align}</math> | \end{align}</math> | ||
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<math>P_{\infty} = 1 </math> | <math>P_{\infty} = 1 </math> | ||
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| + | Conclusion: | ||
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| + | <math>E_{\infty} = \infty</math>, <math>P_{\infty} = 1 </math> | ||
Latest revision as of 18:48, 1 December 2018
Problem
Compute the energy $ E_\infty $ and the power $ P_\infty $ of the DT exponential signal below:
$ x[n]= e^{-j3\pi n} $
Solution
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:
$ E_{\infty} = \infty $, $ P_{\infty} = 1 $
