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=\dfrac{1}{2}\ln{(\dfrac{15}{7})} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood