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So <math class="inline">P_{\infty} = 1 </math>.
 
So <math class="inline">P_{\infty} = 1 </math>.
  
<math>P_\infty</math> is larger than 0, so <math>E_\infty</math> should be infinity, and it is. (<span style="color:green">instructor's comment: good observation!</span>)
 
--[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]]
 
  
*<span style="color:blue">Be careful when using the start symbol for multiplication in this context. It usually denotes convolution in electrical engineering.</span>
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==Answer 2==
 
==Answer 2==

Revision as of 10:10, 19 January 2018


Practice Question on "Signals and Systems"


More Practice Problems


Topic: Signal Energy and Power


Question

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

$ x(t)= \sin (2 \pi t) $


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 dt \quad \\ &= \lim_{T\rightarrow \infty}\int_{-T}^T |(cos(2t) + j*sin(2t))|^2 dt \quad \\ &= \lim_{T\rightarrow \infty}\int_{-T}^T {\sqrt{(cos(2t))^2 + (sin(2t))^2}}^2 dt \quad \\ & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dt \quad \\ &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} \quad \\ &=\infty. \quad \end{align} $

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

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

So $ P_{\infty} = 1 $.



Answer 2


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