Revision as of 14:30, 4 September 2008 by Drmorris (Talk)

Compute the Energy and Power of the signal $ x(t)=\dfrac{2t}{t^2+5} $ between 3 and 5 seconds.

Energy

$ 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} $

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