(New page: Consider the signal <math>x(t)=cos(5t)</math>. ==Energy== First we find the energy for one complete cycle <math>E=\int_0^{2\pi}{|cos(5t)|^2dt}</math> <math>=\frac{1}{2}\int_0^{2\pi}(1+c...) |
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| − | + | Suppose the signal to be <math>x(t)=cos(5t)</math>. | |
==Energy== | ==Energy== | ||
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| − | + | now we find the energy of the wave for one complete cycle | |
<math>E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(5t)|^2dt}</math> | <math>E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(5t)|^2dt}</math> | ||
Latest revision as of 13:42, 5 September 2008
Suppose the signal to be $ x(t)=cos(5t) $.
Energy
First we find the energy for one complete cycle $ E=\int_0^{2\pi}{|cos(5t)|^2dt} $
$ =\frac{1}{2}\int_0^{2\pi}(1+cos(10t))dt $
$ =\frac{1}{2}(t+\frac{1}{10}sin(10t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{2}(2\pi+0-0-0) $
$ =\pi $
Energy
now we find the energy of the wave for one complete cycle
$ E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(5t)|^2dt} $
$ =\frac{1}{2\pi-0}\frac{1}{2}\int_0^{2\pi}(1+cos(10t))dt $
$ =\frac{1}{4\pi}(t+\frac{1}{10}sin(10t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{4\pi}(2\pi+0-0-0) $
$ =\frac{1}{2} $
