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| − | Signal Energy and Power | + | ==Signal Energy and Power== |
| − | + | Define a signal (either CT or DT) and compute its energy and its power. Post your answer on Rhea. Give your page a descriptive title. | |
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| + | Computed- Energy and Power of the signal <font size="4"><math>2\sin(t)\cos(t)</math> | ||
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| + | ==Energy== | ||
| + | <font size="4"> | ||
| + | <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2} \! |f(t)|^2 dt</math> | ||
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| + | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |2\sin(t)cos(t)|^2 dt</math> | ||
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| + | <math>E = {1\over(2\pi)}\int_{0}^{2\pi} \! |\sin(2t)|^2 dt</math> | ||
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| + | <math>E = {1\over(2\pi)}\int_{0}^{2\pi} \! |{(1-\cos(4t))\over 2}| dt</math> | ||
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| + | <math>E = {1\over{2\pi}}*{1\over 2}\int_0^{2\pi} \! |1-\cos(4t)| dt</math> | ||
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| + | <math>E = {1\over{4\pi}} * [ t - {1\over4}\sin(4t) ]_0^{2\pi}</math> | ||
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| + | <math>E = {1\over{4\pi}} * [ 2\pi - {1\over4}\sin(8\pi) - ( 0 - {1\over4}\sin(4\pi*0) ) ]</math> | ||
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| + | <math>E = {1\over{4\pi}} * [ 2\pi - {1\over4}\sin(8\pi) ]</math> | ||
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| + | <math>E = {1\over{4\pi}} * 2\pi</math> | ||
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| + | <math>E = {1\over2}</math> | ||
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| + | ==Power== | ||
| + | <math>P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math> | ||
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| + | <math>P = \int_0^{2\pi} \! |2\sin(t)\cos(t)|^2\ dt</math> | ||
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| + | <math>P = \int_0^{2\pi} \! |\sin(2t)|^2\ dt</math> | ||
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| + | <math>P = \int_0^{2\pi} \! |{(1-\cos(4t))\over 2}| dt</math> | ||
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| + | <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 4}\sin(4t) )\mid_0^{2\pi}</math> | ||
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| + | <math>P = {1\over 2}t - {1\over 8}\sin(4t) )\mid_0^{2\pi}</math> | ||
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| + | <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> | ||
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| + | <math>P = \pi - {1\over8}\sin(8\pi) </math> | ||
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| + | <math>P = \pi</math> | ||
| + | </font> | ||
Latest revision as of 13:04, 5 September 2008
Signal Energy and Power
Define a signal (either CT or DT) and compute its energy and its power. Post your answer on Rhea. Give your page a descriptive title.
Computed- Energy and Power of the signal $ 2\sin(t)\cos(t) $
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} $
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 $
