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Given the Signal <math>x(t)=3sin(2*pi*3t)</math>, Find the energy and power of the signal from 0 to 5 seconds. | Given the Signal <math>x(t)=3sin(2*pi*3t)</math>, Find the energy and power of the signal from 0 to 5 seconds. | ||
| + | == Energy == | ||
| + | <math>E=\int_1^5 |x(t)|^2 dt</math> | ||
| + | |||
| + | <math>E=\int_1^5 |3sin(6\pi t)|^2 dt</math> | ||
| + | |||
| + | |||
| + | <math>E=9*\int_1^5 sin(6\pi t)^2 dt</math> | ||
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| + | |||
| + | <math>E=9*\int_1^5 sin(6\pi t)^2 dt</math> | ||
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| + | <math>E=9*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4})\mid_1^5</math> | ||
| + | |||
| + | <math>E=9*(3\pi t-\dfrac{sin(12\pi t)}{4})\mid_1^5</math> | ||
| + | |||
| + | <math>E=27\pi *t-\dfrac{9sin(12\pi *t)}{4}\mid_1^5</math> | ||
| + | |||
| + | <math>E=27\pi *5-\dfrac{9sin(12\pi *5)}{4}-(27\pi *1-\dfrac{9sin(12\pi *1)}{4}</math> | ||
| + | |||
| + | <math>E=\int_1^5 |3sin(6\pi t)|^2 dt=108\pi</math> | ||
| + | |||
== Power == | == Power == | ||
| − | <math>\int_1^5 |x(t)|^2 dt</math> | + | <math>P=\dfrac{1}{t2-t1}\int_1^5 |x(t)|^2 dt</math> |
| − | <math>\int_1^5 |3sin(6\pi t)|^2 dt</math> | + | <math>P=\dfrac{1}{5-1}\int_1^5 |3sin(6\pi t)|^2 dt</math> |
| − | <math>9*\int_1^5 sin(6\pi t)^2 dt</math> | + | <math>P=\dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt</math> |
| − | <math>9*\int_1^5 sin(6\pi t)^2 dt</math> | + | <math>P=\dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt</math> |
| − | <math>9*(6\pi | + | <math>P=\dfrac{9}{4}*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4})\mid_1^5</math> |
| − | <math>9*(3\pi | + | <math>P=\dfrac{9}{4}*(3\pi t-\dfrac{sin(12\pi t)}{4})\mid_1^5</math> |
| − | <math>27\pi *t- | + | <math>P=\dfrac{27}{4}\pi *t-\dfrac{9sin(12\pi *t)}{16}\mid_1^5</math> |
| − | <math>27\pi *5- | + | <math>P=\dfrac{27}{4}\pi *5-\dfrac{9sin(12\pi *5)}{16}-(\dfrac{27}{4}\pi *1-\dfrac{9sin(12\pi *1)}{16}</math> |
| − | <math>\int_1^5 |3sin(6\pi t)|^2 dt= | + | <math>P=\int_1^5 |3sin(6\pi t)|^2 dt=27\pi |
Latest revision as of 08:36, 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.
Energy
$ E=\int_1^5 |x(t)|^2 dt $
$ E=\int_1^5 |3sin(6\pi t)|^2 dt $
$ E=9*\int_1^5 sin(6\pi t)^2 dt $
$ E=9*\int_1^5 sin(6\pi t)^2 dt $
$ E=9*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4})\mid_1^5 $
$ E=9*(3\pi t-\dfrac{sin(12\pi t)}{4})\mid_1^5 $
$ E=27\pi *t-\dfrac{9sin(12\pi *t)}{4}\mid_1^5 $
$ E=27\pi *5-\dfrac{9sin(12\pi *5)}{4}-(27\pi *1-\dfrac{9sin(12\pi *1)}{4} $
$ E=\int_1^5 |3sin(6\pi t)|^2 dt=108\pi $
Power
$ P=\dfrac{1}{t2-t1}\int_1^5 |x(t)|^2 dt $
$ P=\dfrac{1}{5-1}\int_1^5 |3sin(6\pi t)|^2 dt $
$ P=\dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt $
$ P=\dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt $
$ P=\dfrac{9}{4}*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4})\mid_1^5 $
$ P=\dfrac{9}{4}*(3\pi t-\dfrac{sin(12\pi t)}{4})\mid_1^5 $
$ P=\dfrac{27}{4}\pi *t-\dfrac{9sin(12\pi *t)}{16}\mid_1^5 $
$ P=\dfrac{27}{4}\pi *5-\dfrac{9sin(12\pi *5)}{16}-(\dfrac{27}{4}\pi *1-\dfrac{9sin(12\pi *1)}{16} $
$ P=\int_1^5 |3sin(6\pi t)|^2 dt=27\pi $
