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\end{align}
 
\end{align}
 
</math>
 
</math>
 +
 +
 
But <math class="inline">\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). </math>
 
But <math class="inline">\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). </math>
 +
 +
and therefore <math class="inline">\sin^2x = \frac{1-\cos(2x}{2}</math>.
  
 
So <math class="inline">E_{\infty} = \infty</math>.
 
So <math class="inline">E_{\infty} = \infty</math>.

Revision as of 09:24, 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 |\sin(2 \pi t)|^2 dt \\ &=\lim_{T\rightarrow \infty}\int_{-T}^T \sin^2(2 \pi t) dt \end{align} $


But $ \cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). $

and therefore $ \sin^2x = \frac{1-\cos(2x}{2} $.

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


Back to ECE301 Spring 2018 Prof. Boutin

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman