| Line 12: | Line 12: | ||
\begin{align} | \begin{align} | ||
E_{\infty}&=\lim_{T\rightarrow \infty}\int_{-T}^T |e^{(2jt)}|^2 dx \quad {\color{OliveGreen}\surd}\\ | 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 \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\\ | + | &= \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}\int_{-T}^T 1 dx \quad {\color{OliveGreen}\surd}\\ | ||
&= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} \quad {\color{OliveGreen}\surd}\\ | &= \lim_{T\rightarrow \infty} t \Big| ^T _{-T} \quad {\color{OliveGreen}\surd}\\ | ||
| 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. (<span style="color:green">instructor's comment: | + | <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: good observation!</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 16:09, 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 \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.
