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.
