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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 {9/5} $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010