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= Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the following continuous-time signal= | = Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the following continuous-time signal= | ||
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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math> | |
| + | \begin{align} | ||
| + | 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}\\ | ||
| + | &=\infty. | ||
| + | \end{align} | ||
| + | </math> | ||
| − | <math class="inline">E_{\infty} = \infty</math> | + | So <math class="inline">E_{\infty} = \infty</math>. |
| − | <math | + | <math> |
| + | \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}} 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 | ||
| + | \end{align} | ||
| + | </math> | ||
| − | <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. | <math>P_\infty</math> is larger than 0, so <math>E_\infty</math> should be infinity, and it is. | ||
Revision as of 16:02, 12 January 2011
Contents
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?
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 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}\\ &=\infty. \end{align} $
So $ E_{\infty} = \infty $.
$ \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}} 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 \end{align} $
So $ 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.
