(Energy)
(Power)
 
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===Power===
 
===Power===
 +
:<math> y = 2e^{t} </math> from (0,10)
 +
 +
:<math>Average Power = \frac{1}{t2 - t1}\int_{t1}^{t2}x(t)^2 </math>
 +
 +
 +
:<math>Average Power = \frac{1}{10}\int_{0}^{5}4e^{2t}dt </math>
 +
 +
:<math>Average Power = \frac{1}{10}({2}e^{20} - 2) </math>

Latest revision as of 17:57, 5 September 2008

Problem

Compute the energy and power of a CT signal $ y=2e^t $ from (0,10)

Energy

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

$ =\int_0^{10}{|2e^t|^2dt} $

$ =\int_0^{10}{(4e^{2t})dt} $

$ =(2e^{2t})|_{t=0}^{t=10} $

$ =2e^{20}-2 $

$ =9.703x10^{8} $

Power

$ y = 2e^{t} $ from (0,10)
$ Average Power = \frac{1}{t2 - t1}\int_{t1}^{t2}x(t)^2 $


$ Average Power = \frac{1}{10}\int_{0}^{5}4e^{2t}dt $
$ Average Power = \frac{1}{10}({2}e^{20} - 2) $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009