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<math>x(t)= e^{-2\pi jt} </math> | <math>x(t)= e^{-2\pi jt} </math> | ||
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==Answer 1=== | ==Answer 1=== | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | E_{\infty}&=\int_{-\infty}^\infty | | + | E_{\infty}&=\int_{-\infty}^\infty |e^{-2\pi jt}|^2 dt \\ |
| − | &=\int_{-\infty}^\infty | + | &=\int_{-\infty}^\infty e^{-2\pi jt} * e^{2\pi jt} dt \\ |
| − | + | &=\int_{-\infty}^\infty (1) dt \\ | |
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| − | &=\int_{-\infty}^\infty | + | |
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&=\infty | &=\infty | ||
\end{align} | \end{align} | ||
Revision as of 10:16, 22 January 2018
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^{-2\pi jt} $
Answer 1=
$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty |e^{-2\pi jt}|^2 dt \\ &=\int_{-\infty}^\infty e^{-2\pi jt} * e^{2\pi jt} dt \\ &=\int_{-\infty}^\infty (1) dt \\ &=\infty \end{align} $
So $ E_{\infty} = \infty $.
$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\pi t)|^2 dt \quad \\ \text{Similar to math above, the expression can be derived towards}\\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1}{2} dt - \int_{-T}^T \frac{1}{2} * \cos(4\pi t) dt) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2} t \Big| ^T _{-T} - \frac{1}{8\pi} \int_{-T}^T \cos(4\pi t) d(4\pi t)) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \Big| ^T _{-T}) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (T - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ &= \frac{1}{2} \quad \\ \end{align} $
So $ P_{\infty} = \frac{1}{2} $.
