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Practice Problem: 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
$ \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.
