(New page: == Signal == <math>f(t) = \sin^2(t)</math> == Energy == <math>E = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> <math>E = \int_{t_1}^{t_2}\!|\sin^2(t)|^2\ dt</math> <math>E = \int_{0}^{2\pi}\...) |
(→Energy) |
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<math>E = \int_{0}^{2\pi}\!|\sin^4(t)|\ dt</math> | <math>E = \int_{0}^{2\pi}\!|\sin^4(t)|\ dt</math> | ||
+ | |||
+ | Since <math>\sin^2(t) = \frac{1-\cos(2t)}{2}</math> | ||
+ | |||
+ | <math>E = \frac{1}{4}\int_0^{2\pi}(1-\cos(2t))^2</math> | ||
+ | |||
+ | <math>E = \frac{1}{4}\int_0^{2\pi}(1-2\cos(2t)+\cos^2(2t))</math> | ||
+ | |||
+ | <math>E = \frac{1}{4}[1]^{2\pi}_{0} - \frac{1}{4}[sin(2t)]^{2\pi}_{0} + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t))</math> | ||
+ | |||
+ | <math>E = \frac{1}{2}\pi - 0 + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t))</math> | ||
+ | |||
+ | Since <math>\cos^2(2t) = \frac{1+\cos(4t)}{2}</math> | ||
+ | |||
+ | <math>E = \frac{1}{2}\pi + \frac{1}{8}\int_0^{2\pi}(1+\cos(4t))</math> | ||
+ | |||
+ | <math>E = \frac{1}{2}\pi + \frac{1}{8}[1]^{2\pi}_{0} + \frac{1}{32}[\sin(4t)]^{2\pi}_{0}</math> | ||
+ | |||
+ | <math>E = \frac{1}{2}\pi + \frac{1}{8}(2\pi) + 0</math> | ||
+ | |||
+ | <math>E = \frac{1}{2}\pi + \frac{1}{4}\pi</math> | ||
+ | |||
+ | <math>E = \frac{3}{4}\pi </math> | ||
+ | |||
+ | == Average Power == | ||
+ | |||
More to come .... | More to come .... |
Revision as of 13:25, 2 September 2008
Signal
$ f(t) = \sin^2(t) $
Energy
$ E = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $
$ E = \int_{t_1}^{t_2}\!|\sin^2(t)|^2\ dt $
$ E = \int_{0}^{2\pi}\!|\sin^4(t)|\ dt $
Since $ \sin^2(t) = \frac{1-\cos(2t)}{2} $
$ E = \frac{1}{4}\int_0^{2\pi}(1-\cos(2t))^2 $
$ E = \frac{1}{4}\int_0^{2\pi}(1-2\cos(2t)+\cos^2(2t)) $
$ E = \frac{1}{4}[1]^{2\pi}_{0} - \frac{1}{4}[sin(2t)]^{2\pi}_{0} + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $
$ E = \frac{1}{2}\pi - 0 + \frac{1}{4}\int_0^{2\pi}(\cos^2(2t)) $
Since $ \cos^2(2t) = \frac{1+\cos(4t)}{2} $
$ E = \frac{1}{2}\pi + \frac{1}{8}\int_0^{2\pi}(1+\cos(4t)) $
$ E = \frac{1}{2}\pi + \frac{1}{8}[1]^{2\pi}_{0} + \frac{1}{32}[\sin(4t)]^{2\pi}_{0} $
$ E = \frac{1}{2}\pi + \frac{1}{8}(2\pi) + 0 $
$ E = \frac{1}{2}\pi + \frac{1}{4}\pi $
$ E = \frac{3}{4}\pi $
Average Power
More to come ....