Signal

$ x(t)=e^t $ [0,1]

Energy

$ E=\int_{t_1}^{t_2}|x(t)|^2dt $

$ E = \int_{0}^{1} |e^{t}|^2\ dt \! $



$ = \int_{0}^{2} e^{2t}\ dt \! $

$ = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \! $ $ = \frac{1}{2}(e^2 -1)\! $

Power

$ P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \! $.

$ P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{t}|^2\ dt \! $

$ = \int_{0}^{1} e^{2t}\ dt \! $

$ = \frac{1}{2}[e^{2t}]_{t=0}^{t=1} \! $

$ = \frac{1}{2}(e^2 -1)\! $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood