(New page: == Introduction == This page calculates the energy and power of the <math>2\sin(t)\cos(t)</math> signal. == Power == <font size="5"> <math>P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math> <ma...) |
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<math>P = {1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt</math> | <math>P = {1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt</math> | ||
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− | |||
<math>P = {1\over 2}t - {1\over 8}\sin(4t) \mid_0^{2\pi}</math> | <math>P = {1\over 2}t - {1\over 8}\sin(4t) \mid_0^{2\pi}</math> | ||
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<math>P = \pi </math> | <math>P = \pi </math> | ||
+ | </font> | ||
+ | |||
+ | |||
+ | == Energy == | ||
+ | <font size="5"> | ||
+ | <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2} \! |f(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\over2\pi}\int_0^{2\pi} \! |sin(2t)|^2\ dt</math> | ||
+ | |||
+ | <math>E = {1\over2\pi}\int_0^{2\pi} \! sin^2(2t)\ dt</math> | ||
+ | |||
+ | <math>E = {1\over2\pi}\int_0^{2\pi} \! {(1-\cos(4t))\over 2} dt</math> | ||
+ | |||
+ | <math>E = {1\over2\pi}*{1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt</math> | ||
+ | |||
+ | <math>E = {1\over 4\pi} ( t - {1\over 4}\sin(4t) )\mid_0^{2\pi}</math> | ||
+ | |||
+ | <math>E = {1\over 4\pi}t - {1\over 32\pi}\sin(4t) \mid_0^{2\pi}</math> | ||
+ | |||
+ | <math>E = {1\over 4\pi}(2\pi) - {1\over 32\pi}\sin(4*2\pi) - [{1\over 4\pi}(0) - {1\over 32\pi}\sin(4*0)]</math> | ||
+ | |||
+ | <math>E = {1\over 2} - {1\over32\pi}\sin(8\pi)</math> | ||
+ | |||
+ | <math>E = {1\over 2}</math> |
Latest revision as of 08:03, 4 September 2008
Introduction
This page calculates the energy and power of the $ 2\sin(t)\cos(t) $ signal.
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} \! sin^2(2t)\ 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 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\over2\pi}\int_0^{2\pi} \! |sin(2t)|^2\ dt $
$ E = {1\over2\pi}\int_0^{2\pi} \! sin^2(2t)\ dt $
$ E = {1\over2\pi}\int_0^{2\pi} \! {(1-\cos(4t))\over 2} dt $
$ E = {1\over2\pi}*{1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt $
$ E = {1\over 4\pi} ( t - {1\over 4}\sin(4t) )\mid_0^{2\pi} $
$ E = {1\over 4\pi}t - {1\over 32\pi}\sin(4t) \mid_0^{2\pi} $
$ E = {1\over 4\pi}(2\pi) - {1\over 32\pi}\sin(4*2\pi) - [{1\over 4\pi}(0) - {1\over 32\pi}\sin(4*0)] $
$ E = {1\over 2} - {1\over32\pi}\sin(8\pi) $
$ E = {1\over 2} $