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| − | = Compute the energy <math>E_\infty</math> and the power <math>P_\infty</math> of the following discrete-time signal | + | <center><font size= 4> |
| + | '''[[Signals_and_systems_practice_problems_list|Practice Question on "Signals and Systems"]]''' | ||
| + | </font size> | ||
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| + | [[Signals_and_systems_practice_problems_list|More Practice Problems]] | ||
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| + | Topic: Signal Energy and Power | ||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
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| + | Compute the energy <math>E_\infty</math> and the power <math>P_\infty</math> of the following discrete-time signal | ||
<span class="texhtml">''x''[''n''] = ''j''</span> | <span class="texhtml">''x''[''n''] = ''j''</span> | ||
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So <math>E_{\infty} = \infty</math>. | So <math>E_{\infty} = \infty</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> | ||
<math>\begin{align} | <math>\begin{align} | ||
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&= \lim_{N\rightarrow \infty}{1 \over {2N+1}}\sum_{n=-N}^N 1 \\ | &= \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}{1 \over {2N+1}}\sum_{n=0}^{2N} 1 \\ | ||
| − | &= \lim_{N\rightarrow \infty}{2N \over {2N+1}} \\ | + | &= \lim_{N\rightarrow \infty}{2N+1 \over {2N+1}} \\ |
| − | &= \lim_{N\rightarrow \infty}{ | + | &= \lim_{N\rightarrow \infty}{1}\\ |
&= 1 \\ | &= 1 \\ | ||
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So <math>P_{\infty} = 1</math>. | So <math>P_{\infty} = 1</math>. | ||
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--[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 | --[[User:Rgieseck|Rgieseck]] 21:35, 12 January 2011 | ||
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[[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
| − | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | + | [[Category:ECE301Spring2011Boutin]] |
| + | [[Category:Problem_solving]] | ||
| + | [[Category:signal power]] | ||
| + | [[Category:signal energy]] | ||
| + | [[Category:discrete-time signal]] | ||
| + | [[Category:complex numbers]] | ||
| + | [[Category:Complex Number Magnitude]] | ||
| + | [[Category:ECE438]] | ||
Latest revision as of 16:18, 26 November 2013
Practice Question on "Signals and Systems"
Topic: Signal Energy and Power
Question
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?
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 $.
- 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)
$ \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+1 \over {2N+1}} \\ &= \lim_{N\rightarrow \infty}{1}\\ &= 1 \\ \end{align} $
So $ P_{\infty} = 1 $.
--Rgieseck 21:35, 12 January 2011
Answer 2
write it here.
Answer 3
write it here.
