Difference between revisions of "HW1.5 Bavorndej Chanyasak ECE301Fall2008mboutin" - Rhea

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} $