(New page: Compute the Energy and Power of the signal <math>x(t)=\dfrac{2t}{t^2+5}</math> between 0 and 2 seconds. ==Energy== <math>E=\int_0^{2}{\dfrac{2t}{t^2+5}dt}</math> <math>U=t^2+5</math> ...) |
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<math>E=\ln {9/5}</math> | <math>E=\ln {9/5}</math> | ||
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| + | ==Power== | ||
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| + | <math>P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt</math> | ||
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| + | |||
| + | <math>P=\dfrac{1}{2-0}\int_0^{2}{\dfrac{2t}{t^2+5}dt}</math> | ||
| + | |||
| + | <math>U=t^2+5</math> | ||
| + | |||
| + | <math>dU=2tdt</math> | ||
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| + | Limits: | ||
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| + | <math>U(0)=0^2+5=5</math> | ||
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| + | <math>U(2)=2^2+5=9</math> | ||
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| + | <math>P=\dfrac{1}{2}\int_{5}^{9}\dfrac{du}{U}</math> | ||
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| + | <math>P=\dfrac{1}{2}\ln U |_{U=5}^{U=9}</math> | ||
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| + | <math>P=\dfrac{1}{2}(\ln 9 - \ln 5)</math> | ||
| + | |||
| + | <math>P=\dfrac{1}{2}\ln{(\dfrac{9}{5})} </math> | ||
Revision as of 07:02, 5 September 2008
Compute the Energy and Power of the signal $ x(t)=\dfrac{2t}{t^2+5} $ between 0 and 2 seconds.
Energy
$ E=\int_0^{2}{\dfrac{2t}{t^2+5}dt} $
$ U=t^2+5 $
$ dU=2tdt $
Limits:
$ U(0)=0^2+5=5 $
$ U(2)=2^2+5=4+5=9 $
$ E=\int_{5}^{9}\dfrac{du}{U} $
$ E=\ln U |_{U=5}^{U=9} $
$ E=\ln 9 - \ln 5 $
$ E=\ln {9/5} $
Power
$ P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt $
$ P=\dfrac{1}{2-0}\int_0^{2}{\dfrac{2t}{t^2+5}dt} $
$ U=t^2+5 $
$ dU=2tdt $
Limits:
$ U(0)=0^2+5=5 $
$ U(2)=2^2+5=9 $
$ P=\dfrac{1}{2}\int_{5}^{9}\dfrac{du}{U} $
$ P=\dfrac{1}{2}\ln U |_{U=5}^{U=9} $
$ P=\dfrac{1}{2}(\ln 9 - \ln 5) $
$ P=\dfrac{1}{2}\ln{(\dfrac{9}{5})} $
