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{(\dfrac{9}{5})} $

Power

$ U=\int_{t_1}^{t_2}x(t)dt $


$ U=\int_0^{2}{\dfrac{2t}{t^2+5}dt} $


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})} $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett