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[[Category:ECE301Spring2011Boutin]]
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= Compute the energy <math>E_\infty</math> and the power <math>P_\infty</math> of the following discrete-time signal =
[[Category:problem solving]]
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= Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the following discrete-time signal=
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  <span class="texhtml">''x''[''n''] = ''j''</span>
  <math>x[n]= j </math>
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What properties of the complex magnitude can you use to check your answer?  
 
What properties of the complex magnitude can you use to check your answer?  
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----
 
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==Share your answers below==
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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!
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== 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!  
 +
 
 
----
 
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===Answer 1===
 
<math>
 
\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}
 
</math>
 
  
So <math class="inline">E_{\infty} = \infty</math>.
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=== Answer 1 ===
  
:<span style="color:green"> Instructors comment: Good job! The answer is correct and the justification is very clear. Now can someone compute the power? --[[User:Mboutin|Mboutin]] 19:31, 13 January 2011 (UTC)  </span>
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<math>\begin{align}
--[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 [[Category:ECE301Spring2011Boutin]]
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E_{\infty}&=\lim_{N\rightarrow \infty}\sum_{n=-N}^N |j|^2 \\
===Answer 2===
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&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\
write it here.
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&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\
===Answer 3===
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&= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\
write it here.
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&=\infty. \\
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\end{align}</math>
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So <math>E_{\infty} = \infty</math>.
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<math>\begin{align}
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P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N |j|^2 \\
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&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\
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&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{-j^2})}^2  \\
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&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N 1  \\
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&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=0}^{2N} 1  \\
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&= \lim_{N\rightarrow \infty}{2N \over {2N+1}}  \\
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&= \lim_{N\rightarrow \infty}{2 \over {2}}  \\
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&= 1  \\
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\end{align}</math>
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 +
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So <math>P_{\infty} = 1</math>.
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:<span style="color: green;"> Instructors comment: Good job! The answer is correct and the justification is very clear. Now can someone compute the power? --[[User:Mboutin|Mboutin]] 19:31, 13 January 2011 (UTC)  </span>
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--[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011  
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=== Answer 2 ===
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write it here.  
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=== Answer 3 ===
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write it here.  
  
 
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]

Revision as of 04:55, 14 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_{N\rightarrow \infty}\sum_{n=-N}^N |j|^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N {(\sqrt{-j^2})}^2 \\ &= \lim_{N\rightarrow \infty}\sum_{n=-N}^N 1 \\ &=\infty. \\ \end{align} $


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

$ \begin{align} P_{\infty}&=\lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N |j|^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{jj^*})}^2 \\ &= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N {(\sqrt{-j^2})}^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 \over {2N+1}} \\ &= \lim_{N\rightarrow \infty}{2 \over {2}} \\ &= 1 \\ \end{align} $


So $ P_{\infty} = 1 $. 
Instructors comment: Good job! The answer is correct and the justification is very clear. Now can someone compute the power? --Mboutin 19:31, 13 January 2011 (UTC)

--Rgieseck 21:35, 12 January 2011

Answer 2

write it here.

Answer 3

write it here.


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