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<math>x(t) = \cos(4t)</math> | <math>x(t) = \cos(4t)</math> | ||
| − | <math>\int_{t1}^{t2} |x(t)|\, dt</math> | + | <math>E=\int_{t1}^{t2} |x(t)|^2\, dt</math> |
| + | |||
| + | <math>E=\int_{0}^{5\pi} |\cos(4t)|^2\, dt</math> | ||
| + | |||
| + | <math>E=\frac{1}{2}\int_{0}^{5\pi} |1+\cos(8t)|^2\, dt</math> | ||
| + | |||
| + | <math>E=\frac{1}{2}\times|t+\frac{1}{8}\sin(8t)|\ t = 0 to 5\pi</math> | ||
| + | |||
| + | <math>=\frac{1}{2} \times (5\pi + 0 - 0-0)</math> | ||
| + | |||
| + | <math>=\frac{5}{2}\times\pi</math> | ||
| + | |||
| + | |||
| + | == Power == | ||
| + | |||
| + | <math>P=\frac{1}{t2-t1}\times\int_{t1}^{t2} |x(t)|^2\, dt </math> | ||
| + | |||
| + | <math>P=\frac {1}{5\pi}\times\int_{0}^{5\pi} |\cos(4t)|^2\, dt </math> | ||
| + | |||
| + | <math>P=\frac{1}{10\pi}\times(t + \frac{1}{8}\sin(8t) </math> <math>t= 0 to 5\pi</math> | ||
| + | |||
| + | <math>P=\frac{1}{10\pi}\times(5\pi+0-0-0)</math> | ||
| + | |||
| + | <math>P=\frac{1}{2}</math> | ||
Latest revision as of 16:11, 4 September 2008
I assumed $ x(t)=\cos(4t) $ and time interval 0 to $ 5\pi $
Energy
$ x(t) = \cos(4t) $
$ E=\int_{t1}^{t2} |x(t)|^2\, dt $
$ E=\int_{0}^{5\pi} |\cos(4t)|^2\, dt $
$ E=\frac{1}{2}\int_{0}^{5\pi} |1+\cos(8t)|^2\, dt $
$ E=\frac{1}{2}\times|t+\frac{1}{8}\sin(8t)|\ t = 0 to 5\pi $
$ =\frac{1}{2} \times (5\pi + 0 - 0-0) $
$ =\frac{5}{2}\times\pi $
Power
$ P=\frac{1}{t2-t1}\times\int_{t1}^{t2} |x(t)|^2\, dt $
$ P=\frac {1}{5\pi}\times\int_{0}^{5\pi} |\cos(4t)|^2\, dt $
$ P=\frac{1}{10\pi}\times(t + \frac{1}{8}\sin(8t) $ $ t= 0 to 5\pi $
$ P=\frac{1}{10\pi}\times(5\pi+0-0-0) $
$ P=\frac{1}{2} $
