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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! | 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=== | |
<math> | <math> | ||
\begin{align} | \begin{align} | ||
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--[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]] | --[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]] | ||
| − | ===Answer 2=== | + | *<span style="color:blue">Be careful when using the start symbol for multiplication in this context. It usually denotes convolution in electrical engineering.</span> |
| − | + | ---- | |
| − | === | + | ==Answer 2== |
| − | + | <math> | |
| + | \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} | ||
| + | </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}} 2T \\ | ||
| + | & = \lim_{T\rightarrow \infty} 1 \\ | ||
| + | &= 1 | ||
| + | \end{align} | ||
| + | </math> | ||
| + | *<span style="color:blue">Looks pretty good!</span> | ||
---- | ---- | ||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Revision as of 14:50, 25 February 2015
Practice Question on "Signals and Systems"
Topic: Signal Energy and Power
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?
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)
- 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!
