Difference between revisions of "Signal power energy exercise CT ECE301S11" - Rhea

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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?

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Answer 1=

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

• Be careful when using the start symbol for multiplication in this context. It usually denotes convolution in electrical engineering.

Answer 2

$ \begin{align} E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx, \\ & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx , \\ &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T}, \\ &= \lim_{T\rightarrow \infty} 2T , \\ &=\infty. \end{align} $

$ \begin{align} 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}} 2T \\ & = \lim_{T\rightarrow \infty} 1 \\ &= 1 \end{align} $

• Looks pretty good!
• Actually, you should be integrating over t, not x. You would lose points for that.

Answer 3

$ \begin{align} E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dt, \\ & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dt , \\ &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T}, \\ &= \lim_{T\rightarrow \infty} 2T , \\ &=\infty. \end{align} $

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

• I don't see any mistake here.

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