Revision as of 14:50, 12 January 2011 by Cmcmican (Talk | contribs)

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?


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

Answer 2

write it here.

Answer 3

write it here.


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