(2 intermediate revisions by the same user not shown)
Line 3: Line 3:
 
The energy can be computed using the formula:
 
The energy can be computed using the formula:
  
<math>E = \int_{b}^{a}{|x(t)|^2}dt\,</math>
+
:<math>E = \int_{b}^{a}{|x(t)|^2}dt\,</math>
  
  
Suppose we want to compute the energy of the signal <math>cos(t)</math> in the interval 0 to 2
+
Suppose we want to compute the energy of the signal <math>cos(t)</math> in the interval <math>0</math> to <math>2\pi</math>.
 +
 
 +
The formula then becomes:
 +
 
 +
 
 +
:<math>E = \int_{0}^{2\pi}{|cos(t)|^2}dt\,</math>
 +
 
 +
 
 +
Using trigonometric identity, <math>cos^2(t) = \frac{1}{2} + \frac{1}{2}cos(2t)\,</math>
 +
 
 +
This implies:
 +
 
 +
 
 +
:<math>E = \frac{1}{2}\int_{0}^{2\pi}1 + cos(2t)dt\,</math>
 +
 
 +
 
 +
Integrating yields
 +
 
 +
 
 +
:<math>E = \frac{1}{2}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\,</math>
 +
 
 +
 
 +
:<math>E = \frac{1}{2}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\,</math>
 +
 
 +
 
 +
:<math>E = \frac{1}{2}\left(2\pi\right) = \pi\,</math>
 +
 
 +
 
 +
The formula for calculating average power is similar to energy:
 +
 
 +
 
 +
:<math>P = \frac{1}{T}\int_{0}^{T}{|x(t)|^2}dt\,</math>
 +
 
 +
 
 +
Applying the signal and interval, the following formula is obtained:
 +
 
 +
 
 +
:<math>P = \frac{1}{2\pi-0}\int_{0}^{2\pi}|cos(t)|^2dt\,</math>
 +
 
 +
 
 +
:<math>P = \frac{1}{4\pi}\int_{0}^{2\pi}1 + cos(2t)dt\,</math>
 +
 
 +
 
 +
:<math>P = \frac{1}{4\pi}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\,</math>
 +
 
 +
 
 +
:<math>P = \frac{1}{4\pi}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\,</math>
 +
 
 +
 
 +
:<math>P = \frac{1}{4\pi}\left(2\pi\right) = \frac{1}{2}</math>

Latest revision as of 20:14, 4 September 2008

Suppose a signal is defined by $ cos(t) $

The energy can be computed using the formula:

$ E = \int_{b}^{a}{|x(t)|^2}dt\, $


Suppose we want to compute the energy of the signal $ cos(t) $ in the interval $ 0 $ to $ 2\pi $.

The formula then becomes:


$ E = \int_{0}^{2\pi}{|cos(t)|^2}dt\, $


Using trigonometric identity, $ cos^2(t) = \frac{1}{2} + \frac{1}{2}cos(2t)\, $

This implies:


$ E = \frac{1}{2}\int_{0}^{2\pi}1 + cos(2t)dt\, $


Integrating yields


$ E = \frac{1}{2}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\, $


$ E = \frac{1}{2}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\, $


$ E = \frac{1}{2}\left(2\pi\right) = \pi\, $


The formula for calculating average power is similar to energy:


$ P = \frac{1}{T}\int_{0}^{T}{|x(t)|^2}dt\, $


Applying the signal and interval, the following formula is obtained:


$ P = \frac{1}{2\pi-0}\int_{0}^{2\pi}|cos(t)|^2dt\, $


$ P = \frac{1}{4\pi}\int_{0}^{2\pi}1 + cos(2t)dt\, $


$ P = \frac{1}{4\pi}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\, $


$ P = \frac{1}{4\pi}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\, $


$ P = \frac{1}{4\pi}\left(2\pi\right) = \frac{1}{2} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett