(New page: The function that we are using in this example to compute the signal power and energy is: <math> f(x)=sin(x) \!</math> == Power Calculation == <math>P = \int_0^2\pi \! |\sin(x)|^2\ dx ...) |
(→Energy Derivation) |
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<math>P = \pi</math> | <math>P = \pi</math> | ||
− | == Energy | + | == Energy Calculation == |
<math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |\sin(x)|^2 dx</math><br> | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |\sin(x)|^2 dx</math><br> | ||
<math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |{(1-\cos(2x))\over 2}| dx</math><br> | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |{(1-\cos(2x))\over 2}| dx</math><br> |
Latest revision as of 18:37, 3 September 2008
The function that we are using in this example to compute the signal power and energy is: $ f(x)=sin(x) \! $
Power Calculation
$ P = \int_0^2\pi \! |\sin(x)|^2\ dx $
$ P = \int_0^2\pi \! |{(1-\cos(2x))\over 2}| dx $
$ P = {1\over 2}\int_0^2\pi \! |1-\cos(2x)| dx $
$ P = [{1\over 2} (x - {1\over 2}\sin(2x) )]\mid_0^{2\pi} $
$ P = [{1\over 2}x - {1\over 4}\sin(2x)]\mid_0^{2\pi} $
$ P = \pi - {1\over4}\sin(4\pi) $
$ P = \pi $
Energy Calculation
$ E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |\sin(x)|^2 dx $
$ E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |{(1-\cos(2x))\over 2}| dx $
$ E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos(2x)| dx $
$ E = {1\over{4\pi}} * [ x - {1\over2}\sin(2x) ]_0^{2\pi} $
$ E = {1\over{4\pi}} * [ 2\pi - {1\over2}\sin(4\pi) - ( 0 - {1\over2}\sin(0) ) ] $
$ E = {2\pi\over{4\pi}} $
$ E = {1\over2} $