(→Power) |
(→Power) |
||
| Line 11: | Line 11: | ||
<math>9*\int_1^5 sin(6\pi t)^2 dt</math> | <math>9*\int_1^5 sin(6\pi t)^2 dt</math> | ||
| − | <math>9*(6\pi | + | <math>9*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi *t)}{4})</math>evaluated at 5 and 1 |
| − | <math>9*(3\pi *t-sin(12\pi *t) | + | <math>9*(3\pi *t-\dfrac{sin(12\pi *t)}{4})</math>evaluated at 5 and 1 |
<math>27\pi *t-9/4*sin(12\pi *t)</math>evaluated at 5 and 1 | <math>27\pi *t-9/4*sin(12\pi *t)</math>evaluated at 5 and 1 | ||
Revision as of 08:15, 4 September 2008
Given the Signal $ x(t)=3sin(2*pi*3t) $, Find the energy and power of the signal from 0 to 5 seconds.
Power
$ \int_1^5 |x(t)|^2 dt $
$ \int_1^5 |3sin(6\pi t)|^2 dt $
$ 9*\int_1^5 sin(6\pi t)^2 dt $
$ 9*\int_1^5 sin(6\pi t)^2 dt $
$ 9*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi *t)}{4}) $evaluated at 5 and 1
$ 9*(3\pi *t-\dfrac{sin(12\pi *t)}{4}) $evaluated at 5 and 1
$ 27\pi *t-9/4*sin(12\pi *t) $evaluated at 5 and 1
$ 27\pi *5-9/4*sin(12\pi *5)-(27\pi *1-9/4*sin(12\pi *1) $
$ \int_1^5 |3sin(6\pi t)|^2 dt=108\pi $
