(Created page with "Topic: Energy and Power Computation of a Signal </center> ---- ==Question== Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\in...")
 
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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>  
  
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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}| \\
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {{je^{3\pi jn}}\times{-je^{-3\pi jn}}} \\
 
 
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\
 
&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\
 
&=\infty. \\
 
&=\infty. \\

Revision as of 03:45, 26 November 2018

Topic: Energy and Power Computation of a Signal </center>


Question

Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following DT signal:


$ 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}| \\ &= \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 $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang