(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...)
 
Line 18: Line 18:
  
 
==Energy==
 
==Energy==
We will find the average power in one cycle of the cosine waveform.
+
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>

Revision as of 13:41, 5 September 2008

Consider the signal $ 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} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn