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<math>P=\dfrac{1}{2}(\ln 30 - \ln 14)</math> | <math>P=\dfrac{1}{2}(\ln 30 - \ln 14)</math> | ||
| − | <math>P=\ln | + | <math>P=\ln {(30/14)}^{0.5}</math> |
Revision as of 13:41, 4 September 2008
Compute the Energy and Power of the signal $ x(t)=\dfrac{2t}{t^2+5} $ between 3 and 5 seconds.
Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
$ E=\int_3^{5}{\dfrac{2t}{t^2+5}dt} $
$ U=t^2+5 $
$ dU=2tdt $
Limits:
$ U(3)=3^2+5=9+5=14 $
$ U(5)=5^2+5=25+5=30 $
$ E=\int_{14}^{30}\dfrac{du}{U} $
$ E=\ln U |_{U=14}^{U=30} $
$ E=\ln 30 - \ln 14 $
$ E=\ln {30/14} $
Power
$ P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt $
$ P=\dfrac{1}{5-3}\int_3^{5}{\dfrac{2t}{t^2+5}dt} $
$ U=t^2+5 $
$ dU=2tdt $
Limits:
$ U(3)=3^2+5=9+5=14 $
$ U(5)=5^2+5=25+5=30 $
$ P=\dfrac{1}{2}\int_{14}^{30}\dfrac{du}{U} $
$ P=\dfrac{1}{2}\ln U |_{U=14}^{U=30} $
$ P=\dfrac{1}{2}(\ln 30 - \ln 14) $
$ P=\ln {(30/14)}^{0.5} $
