(New page: This page uses the function <math>f = \sin(t)</math> ==Power== <font size="5"> <math>P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math> <math>P = \int_0^2\pi \! |\sin(t)|^2\ dt</math> <mat...) |
|||
| Line 1: | Line 1: | ||
| − | This page uses the function <math>f = \sin(t)</math> | + | ==Initial Info== |
| + | |||
| + | This page uses the function <math>f(t) = \sin(t)</math> | ||
Latest revision as of 10:12, 2 September 2008
Initial Info
This page uses the function $ f(t) = \sin(t) $
Power
$ P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt $
$ P = \int_0^2\pi \! |\sin(t)|^2\ dt $
$ P = \int_0^2\pi \! |{(1-\cos(2t))\over 2}| dt $
$ P = {1\over 2}\int_0^2\pi \! |1-\cos(2t)| dt $
$ P = {1\over 2} ( t - {1\over 2}\sin(2t) )\mid_0^{2\pi} $
$ P = {1\over 2}t - {1\over 4}\sin(2t) )\mid_0^{2\pi} $
$ P = {1\over 2}(2\pi) - {1\over 4}\sin(2*2\pi) - [{1\over 2}(0) - {1\over 4}\sin(2*0)] $
$ P = \pi - {1\over4}\sin(4\pi) = \pi + 0 $
$ P = \pi $
Energy
$ E = {1\over(t2-t1)}\int_{t_1}^{t_2} \! |f(t)|^2 dt $
$ E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |\sin(t)|^2 dt $
$ E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |{(1-\cos(2t))\over 2}| dt $
$ E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos(2t)| dt $
$ E = {1\over{4\pi}} * [ t - {1\over2}\sin(2t) ]_0^{2\pi} $
$ E = {1\over{4\pi}} * [ 2\pi - {1\over2}\sin(4\pi) - ( 0 - {1\over2}\sin(0) ) ] $
$ E = {1\over{4\pi}} * ( 2\pi ) $
$ E = {1\over2} $
