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= [[:Category:Problem_solving|Practice Problem]]: 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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'''[[Signals_and_systems_practice_problems_list|Practice Problem on "Signals and Systems"]]'''
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Topic: Signal Energy and Power
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----
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==Question==
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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 continuous-time signal
 
  <math>x(t)= e^{2jt}</math>
 
  <math>x(t)= e^{2jt}</math>
  

Revision as of 15:13, 26 November 2013


Practice Problem on "Signals and Systems"

Topic: Signal Energy and Power


Question

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

$ \begin{align} E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx \quad {\color{OliveGreen}\surd}\\ &= \lim_{T\rightarrow \infty}\int_{-T}^T |(cos(2t) + j*sin(2t))|^2 dx \quad {\color{OliveGreen}\text{ (You could skip this step.)}}\\ &= \lim_{T\rightarrow \infty}\int_{-T}^T {\sqrt{(cos(2t))^2 + (sin(2t))^2}}^2 dx \quad {\color{OliveGreen}\text{ (You could skip this step.)}}\\ & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx \quad {\color{OliveGreen}\surd}\\ &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} \quad {\color{OliveGreen}\surd}\\ &=\infty. \quad {\color{OliveGreen}\surd} \end{align} $

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

$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |e^{(2jt)}|^2 dx \quad {\color{OliveGreen}\surd}\\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 1 dx \quad {\color{OliveGreen}\surd}\\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} t \Big| ^T _{-T} \quad {\color{OliveGreen}\surd}\\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} T - {1 \over {2T}} (-T) \quad {\color{OliveGreen}\surd}\\ & = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} \quad {\color{OliveGreen}\surd}\\ &= 1 \end{align} $

So $ P_{\infty} = 1 $.

$ P_\infty $ is larger than 0, so $ E_\infty $ should be infinity, and it is. (instructor's comment: good observation!) --Cmcmican 19:50, 12 January 2011 (UTC)

Answer 2

write it here.

Answer 3

write it here.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

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

Buyue Zhang