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--[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]]
 
[[Category:problem solving]]
 
[[Category:problem solving]]
 
= Compute the energy  <math class="inline">E_\infty</math> and the power  <math class="inline">P_\infty</math> of the following continuous-time signal=
 
= Compute the energy  <math class="inline">E_\infty</math> and the power  <math class="inline">P_\infty</math> of the following continuous-time signal=
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===Answer 1===
 
===Answer 1===
write it here.
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b) <math class="inline">E_{\infty}=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T |(cos(2t) + j*sin(2t))|^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T {\sqrt{(cos(2t))^2 + (sin(2t))^2}}^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx = \lim_{T\rightarrow \infty} t \Big| ^T _{-T} </math>
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<math class="inline">E_{\infty} = \infty</math>
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<math class="inline">P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |e^{(2jt)}|^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}} (-T) = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} = 1</math>
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<math class="inline">P_{\infty} = 1</math>
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<math>P_\infty</math> is larger than 0, so <math>E_\infty</math> should be infinity, and it is.
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===Answer 2===
 
===Answer 2===
 
write it here.
 
write it here.

Revision as of 15:50, 12 January 2011

--Cmcmican 19:50, 12 January 2011 (UTC)

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

$ x(t)= e^{2jt} $

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

b) $ E_{\infty}=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T |(cos(2t) + j*sin(2t))|^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T {\sqrt{(cos(2t))^2 + (sin(2t))^2}}^2 dx = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx = \lim_{T\rightarrow \infty} t \Big| ^T _{-T} $

$ E_{\infty} = \infty $

$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |e^{(2jt)}|^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}} (-T) = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} = 1 $

$ P_{\infty} = 1 $

$ P_\infty $ is larger than 0, so $ E_\infty $ should be infinity, and it is.


Answer 2

write it here.

Answer 3

write it here.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin