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<math>
 
<math>
 
\begin{align}
 
\begin{align}
E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |j|^2 dx
+
E_{\infty}&=\lim_{T\rightarrow \infty}\sum_{n=-T}^T |j|^2  
&= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{jj*})}^2 dx
+
&= \lim_{T\rightarrow \infty}\sum_{n=-T}^T {(\sqrt{jj*})}^2
&= \lim_{T\rightarrow \infty}\int_{-T}^T {(\sqrt{-j^2})}^2 dx
+
&= \lim_{T\rightarrow \infty}\sum_{n=-T}^T {(\sqrt{-j^2})}^2  
& = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx
+
& = \lim_{T\rightarrow \infty}\sum_{n=-T}^T 1  
&= \lim_{T\rightarrow \infty} t \Big| ^T _{-T}
+
 
&=\infty.  
 
&=\infty.  
 
\end{align}
 
\end{align}
Line 23: Line 22:
 
So <math class="inline">E_{\infty} = \infty</math>.
 
So <math class="inline">E_{\infty} = \infty</math>.
  
<math>
 
\begin{align}
 
P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |j|^2 dx
 
&= \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 1 dx
 
& = \lim_{T\rightarrow \infty} {1 \over {2T}} t \Big| ^T _{-T}
 
& = \lim_{T\rightarrow \infty} {1 \over {2T}} T - {1 \over {2T}}
 
& = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}}
 
&= 1
 
\end{align}
 
</math>
 
 
So <math class="inline">P_{\infty} = 1 </math>.
 
  
 
--[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 [[Category:ECE301Spring2011Boutin]]
 
--[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 [[Category:ECE301Spring2011Boutin]]

Revision as of 03:23, 13 January 2011

Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following discrete-time signal

$ x[n]= j  $

What properties of the complex magnitude can you use to check your answer?


Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


Answer 1

$ \begin{align} E_{\infty}&=\lim_{T\rightarrow \infty}\sum_{n=-T}^T |j|^2 &= \lim_{T\rightarrow \infty}\sum_{n=-T}^T {(\sqrt{jj*})}^2 &= \lim_{T\rightarrow \infty}\sum_{n=-T}^T {(\sqrt{-j^2})}^2 & = \lim_{T\rightarrow \infty}\sum_{n=-T}^T 1 &=\infty. \end{align} $

So $ E_{\infty} = \infty $.


--Rgieseck 21:35, 12 January 2011

Answer 2

write it here.

Answer 3

write it here.


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