| Line 11: | Line 11: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx \\ | + | 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 \\ | &= \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 {\sqrt{(cos(2t))^2 + (sin(2t))^2}}^2 dx\\ | ||
| − | & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx \\ | + | & = \lim_{T\rightarrow \infty}\int_{-T}^T 1 dx \quad {\color{OliveGreen}\surd}\\ |
| − | &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T}\\ | + | &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} \quad {\color{OliveGreen}\surd}\\ |
| − | &=\infty. | + | &=\infty. \quad {\color{OliveGreen}\surd} |
\end{align} | \end{align} | ||
</math> | </math> | ||
| Line 24: | Line 24: | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |e^{(2jt)}|^2 dx \\ | + | 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 \\ | + | &= \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} \\ | + | & = \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)\\ | + | & = \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}} \\ | + | & = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} \quad {\color{OliveGreen}\surd}\\ |
&= 1 | &= 1 | ||
\end{align} | \end{align} | ||
| Line 35: | Line 35: | ||
So <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. (<span style="color:green">instructor's comment: :) .</span>) |
--[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]] | --[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]] | ||
Revision as of 17:07, 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 \quad {\color{OliveGreen}\surd}\\ &= \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 \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: :) .) --Cmcmican 19:50, 12 January 2011 (UTC)
Answer 2
write it here.
Answer 3
write it here.
