(New page: ==Introduction== This page calculates the Energy and Power of the signal <math>2\sin(t)\cos(t)<\math> ==Power== <font size="5"> <math>P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math> <math>P...) |
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==Introduction== | ==Introduction== | ||
− | This page calculates the Energy and Power of the signal <math>2\sin(t)\cos(t)< | + | This page calculates the Energy and Power of the signal <math>2\sin(t)\cos(t)</math> |
+ | |||
==Power== | ==Power== | ||
<font size="5"> | <font size="5"> | ||
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− | <math>P = \int_0^2\pi \! |\sin(t)|^2\ dt</math> | + | <math>P = \int_0^{2\pi} \! |2\sin(t)\cos(t)|^2\ dt</math> |
− | <math>P = \int_0^2\pi \! | | + | <math>P = \int_0^{2\pi} \! |\sin(2t)|^2\ dt</math> |
− | <math>P = | + | <math>P = \int_0^{2\pi} \! |{(1-\cos(4t))\over 2}| dt</math> |
− | <math>P = {1\over 2} | + | <math>P = {1\over 2}\int_0^{2\pi} \! |1-\cos(4t)| dt</math> |
− | <math>P = {1\over 2}t - {1\over 4}\sin( | + | <math>P = {1\over 2} ( t - {1\over 4}\sin(4t) )\mid_0^{2\pi}</math> |
− | <math>P = {1\over 2} | + | <math>P = {1\over 2}t - {1\over 8}\sin(4t) )\mid_0^{2\pi}</math> |
− | <math>P = \pi - {1\ | + | <math>P = {1\over 2}(2\pi) - {1\over 8}\sin(4*2\pi) - [{1\over 2}(0) - {1\over 8}\sin(4*0)]</math> |
+ | |||
+ | |||
+ | <math>P = \pi - {1\over8}\sin(8\pi) </math> | ||
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− | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |\sin(t)|^2 dt</math> | + | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |2\sin(t)cos(t)|^2 dt</math> |
+ | |||
+ | |||
+ | <math>E = {1\over(2\pi)}\int_{0}^{2\pi} \! |\sin(2t)|^2 dt</math> | ||
− | <math>E = {1\over(2\pi | + | <math>E = {1\over(2\pi)}\int_{0}^{2\pi} \! |{(1-\cos(4t))\over 2}| dt</math> |
− | <math>E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos( | + | <math>E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos(4t)| dt</math> |
− | <math>E = {1\over{4\pi}} * [ t - {1\ | + | <math>E = {1\over{4\pi}} * [ t - {1\over4}\sin(4t) ]_0^{2\pi}</math> |
− | <math>E = {1\over{4\pi}} * [ 2\pi - {1\ | + | <math>E = {1\over{4\pi}} * [ 2\pi - {1\over4}\sin(8\pi) - ( 0 - {1\over4}\sin(4\pi*0) ) ]</math> |
Revision as of 08:15, 4 September 2008
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 ) $
$ E = {1\over2} $