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| + | [Category:energy]] | ||
| + | [[Category:signal]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:practice problem]] | ||
| − | |||
| + | ==Problem== | ||
| + | Calculate the energy <math>E_\infty</math> and the average power <math>P_\infty</math> for the CT signal | ||
| + | <math>f(t)=5 j \sin (t)</math> | ||
| + | ---- | ||
| + | ==Solution 1== | ||
<math>E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt)</math> | <math>E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt)</math> | ||
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<math>P\infty= \frac{25}{2} + 0</math> | <math>P\infty= \frac{25}{2} + 0</math> | ||
| + | ---- | ||
| + | ---- | ||
| + | [[Signal_energy_CT|Back to CT signal energy page]] | ||
Revision as of 17:24, 25 February 2015
[Category:energy]]
Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ f(t)=5 j \sin (t) $
Solution 1
$ E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt) $
$ E_\infty = \int_{-\infty}^\infty 25sin(t)^2u(t)\,dt) $
$ E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt) $
$ E_\infty =\frac{25t}{2} + \frac{25sin(t)}{4}\bigg]_{-\infty}^\infty) $
$ E_\infty =\infty-0 = \infty $
$ P_\infty calculation $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt $
$ = lim_{T \to \infty} \frac{1}{2T}*(\frac{25t}{2} + \frac{25sin(t)}{4}|_{-T}^T) $
$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{4}-\frac{25sin(-T)}{4} $
$ = lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{2}) $
$ P\infty= \frac{25}{2} + 0 $
