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[[Category:problem solving]] | [[Category:problem solving]] | ||
| − | = Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the following continuous-time signal | + | [[Category:power]] |
| + | [[Category:energy]] | ||
| + | [[Category:signal]] | ||
| + | [[Category:continuous-time signal]] | ||
| + | [[Category:complex numbers]] | ||
| + | [[Category:Complex Number Magnitude]] | ||
| + | [[Category:ECE301]] | ||
| + | |||
| + | <center><font size= 4> | ||
| + | '''[[Signals_and_systems_practice_problems_list|Practice Question on "Signals and Systems"]]''' | ||
| + | </font size> | ||
| + | |||
| + | |||
| + | [[Signals_and_systems_practice_problems_list|More Practice Problems]] | ||
| + | |||
| + | |||
| + | Topic: Signal Energy and Power | ||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
| + | |||
| + | |||
| + | Compute the energy <math class="inline">E_\infty</math> and the power <math class="inline">P_\infty</math> of the following continuous-time signal | ||
<math>x(t)= e^{2jt}</math> | <math>x(t)= e^{2jt}</math> | ||
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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> | |
| + | \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} | ||
| + | </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 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} | ||
| + | </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. (<span style="color:green">instructor's comment: good observation!</span>) |
| + | --[[User:Cmcmican|Cmcmican]] 19:50, 12 January 2011 (UTC)[[Category:ECE301Spring2011Boutin]] | ||
| + | *<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> | ||
| + | *<span style="color:purple">Actually, you should be integrating over t, not x. You would lose points for that.</span> | ||
| + | ---- | ||
| + | ==Answer 3== | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| − | = | + | <math> |
| − | + | \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} | ||
| + | </math> | ||
| + | *<span style="color:blue">I don't see any mistake here. </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]] | ||
Latest revision as of 11:01, 21 April 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 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.
