| Line 2: | Line 2: | ||
==Energy== | ==Energy== | ||
| − | <math>E=\int_{t_1}^{t_2}x(t)dt | + | <math>E=\int_{t_1}^{t_2}x(t)dt</math> |
| − | + | ||
<math>E=\int_3^{5}{\dfrac{2t}{t^2+5}dt}</math> | <math>E=\int_3^{5}{\dfrac{2t}{t^2+5}dt}</math> | ||
Revision as of 14:36, 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 <math>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=\int_{14}^{30}\dfrac{du}{U} $
$ P=\ln U |_{U=14}^{U=30} $
$ P=\ln 30 - \ln 14 $
$ P=\ln {30/14} $
