(→Average Power) |
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== Average Power == | == Average Power == | ||
| − | <math>\frac{1}{2\pi-0}\int_0^{2\pi}{| | + | <math>\frac{1}{2\pi-0}\int_0^{2\pi}{|2cos(2t)|^2dt}</math> |
| − | <math>=\frac{1}{2\pi-0}\frac{1}{2}\int_0^{2\pi}(1+cos( | + | <math>=\frac{1}{2\pi-0}4 \times \frac{1}{2}\int_0^{2\pi}(1+cos(4t))dt</math> |
| − | <math>=\frac{1}{ | + | <math>=\frac{1}{2\pi-0}2 \times (t+\frac{1}{4}sin(4t))|_{t=0}^{t=2\pi}</math> |
| − | <math>=\frac{1}{ | + | <math>=\frac{1}{2\pi-0}2 \times (2\pi+\frac{1}{4}-0-0)</math> |
| − | <math>=\frac{1}{2}</math> | + | <math>=\frac{1}{2\pi-0}4*\pi + \frac{1}{2}</math> |
Revision as of 06:34, 3 September 2008
The signal is: x(t) = 2cos(2t)
Energy
$ \int_0^{2\pi}{|2cos(2t)|^2dt} $
$ = 4 \times \frac{1}{2}\int_0^{2\pi}(1+cos(4t))dt $
$ =2 \times (t+\frac{1}{4}sin(4t))|_{t=0}^{t=2\pi} $
$ =2 \times (2\pi+\frac{1}{4}-0-0) $
$ =4*\pi + \frac{1}{2} $
Average Power
$ \frac{1}{2\pi-0}\int_0^{2\pi}{|2cos(2t)|^2dt} $
$ =\frac{1}{2\pi-0}4 \times \frac{1}{2}\int_0^{2\pi}(1+cos(4t))dt $
$ =\frac{1}{2\pi-0}2 \times (t+\frac{1}{4}sin(4t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{2\pi-0}2 \times (2\pi+\frac{1}{4}-0-0) $
$ =\frac{1}{2\pi-0}4*\pi + \frac{1}{2} $
