Revision as of 14:48, 5 September 2008 by Skray (Talk)

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Equations

The energy expanded from a time t1 to a time t2 in a CT signal is calculated by

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

The power over a time period t1 to t2 is calculated by

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

The equation used to calculate both energy and power will be

$ y(t) = e^{t} $

Calculating Energy

The energy of $ y(t) = e^{t} $ over the time period 0 to 2 seconds is found as follows:

$ E = \int_{0}^{2} |e^{t}|^2\ dt \! $
$  =\int_{0}^{2} e^{2t}\ dt $
$  = \frac{1}{2}[e^{2t}]_{t=0}^{t=2} \! $
$ E = \frac{1}{2}(e^4 -1)\! $


Calculating Power

The power of $ y(t) = e^{t} $ over the time period 0 to 2 seconds is found as follows:

$ P =\frac{1}{2-0} \int_{0}^{2} |e^{t}|^2\ dt \! $
$  = \frac{1}{2}\int_{0}^{2} e^{2t}\ dt \! $
$  = \frac{1}{2}*\frac{1}{2}[e^{2t}]_{t=0}^{t=2} \! $
$  = \frac{1}{4}(e^2 -1)\! $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood