(New page: == Signal == <math> f(t) = \cos^2(t) </math> == Energy == <math>E = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> <math>E = \int_{t_1}^{t_2}\!|\cos^2(t)|^2\ dt</math> <math>E = \int_{0}^{2\pi...) |
|||
| Line 13: | Line 13: | ||
Since | Since | ||
| − | + | <math>\sin^2(t) = \frac{1+\cos(2t)}{2}</math> | |
<math>E = \frac{1}{4}\int_0^{2\pi}(1+\cos(2t))^2</math> | <math>E = \frac{1}{4}\int_0^{2\pi}(1+\cos(2t))^2</math> | ||
| Line 42: | Line 42: | ||
<math>P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> | <math>P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> | ||
| − | <math>P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|\ | + | <math>P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|\cos^2(t)|^2\ dt</math> |
| − | <math>P = \frac{1}{2\pi - 0}\int_{0}^{2\pi}\!|\ | + | <math>P = \frac{1}{2\pi - 0}\int_{0}^{2\pi}\!|\cos^4(t)|\ dt</math> |
<math>P = \frac{1}{2\pi}E</math> | <math>P = \frac{1}{2\pi}E</math> | ||
Revision as of 21:36, 4 September 2008
Signal
$ f(t) = \cos^2(t) $
Energy
$ E = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $
$ E = \int_{t_1}^{t_2}\!|\cos^2(t)|^2\ dt $
$ E = \int_{0}^{2\pi}\!|\cos^4(t)|\ dt $
Since
$ \sin^2(t) = \frac{1+\cos(2t)}{2} $
$ E = \frac{1}{4}\int_0^{2\pi}(1+\cos(2t))^2 $
$ E = \frac{1}{4}\int_0^{2\pi}(1+2\cos(2t)+\cos^2(2t)) $
$ E = \frac{1}{4}[1]^{2\pi}_{0} + \frac{1}{4}[sin(2t)]^{2\pi}_{0} + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $
$ E = \frac{1}{2}\pi + 0 + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $
Since
$ \cos^2(2t) = \frac{1+\cos(4t)}{2} $
$ E = \frac{1}{2}\pi + \frac{1}{8}\int_0^{2\pi}(1+\cos(4t)) $
$ E = \frac{1}{2}\pi + \frac{1}{8}[1]^{2\pi}_{0} + \frac{1}{32}[\sin(4t)]^{2\pi}_{0} $
$ E = \frac{1}{2}\pi + \frac{1}{8}(2\pi) + 0 $
$ E = \frac{1}{2}\pi + \frac{1}{4}\pi $
$ E = \frac{3}{4}\pi $
Average Power
$ P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|f(t)|^2\ dt $
$ P = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\!|\cos^2(t)|^2\ dt $
$ P = \frac{1}{2\pi - 0}\int_{0}^{2\pi}\!|\cos^4(t)|\ dt $
$ P = \frac{1}{2\pi}E $
$ P = \frac{1}{2\pi}(\frac{3}{4}\pi) $
$ P = \frac{3}{8} $
