Introduction
This page calculates the Energy and Power of the signal $ 2\sin(t)\cos(t) $
Power
$ P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt $
$ P = \int_0^{2\pi} \! |2\sin(t)\cos(t)|^2\ dt $
$ P = \int_0^{2\pi} \! |\sin(2t)|^2\ dt $
$ P = \int_0^{2\pi} \! |{(1-\cos(4t))\over 2}| dt $
$ P = {1\over 2}\int_0^{2\pi} \! |1-\cos(4t)| dt $
$ P = {1\over 2} ( t - {1\over 4}\sin(4t) )\mid_0^{2\pi} $
$ P = {1\over 2}t - {1\over 8}\sin(4t) )\mid_0^{2\pi} $
$ P = {1\over 2}(2\pi) - {1\over 8}\sin(4*2\pi) - [{1\over 2}(0) - {1\over 8}\sin(4*0)] $
$ P = \pi - {1\over8}\sin(8\pi) $
$ 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} \! |2\sin(t)cos(t)|^2 dt $
$ E = {1\over(2\pi)}\int_{0}^{2\pi} \! |\sin(2t)|^2 dt $
$ E = {1\over(2\pi)}\int_{0}^{2\pi} \! |{(1-\cos(4t))\over 2}| dt $
$ E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos(4t)| dt $
$ E = {1\over{4\pi}} * [ t - {1\over4}\sin(4t) ]_0^{2\pi} $
$ E = {1\over{4\pi}} * [ 2\pi - {1\over4}\sin(8\pi) - ( 0 - {1\over4}\sin(4\pi*0) ) ] $
$ E = {1\over{4\pi}} * [ 2\pi - {1\over4}\sin(8\pi) ] $
$ E = {1\over{4\pi}} * 2\pi $
$ E = {1\over2} $
