(Power)
(Power)
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== Power ==
 
== Power ==
  
Average signal power between <math>[t_1,t_2]\!</math> is <math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>.
+
<math>P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \!</math>.
  
 
<math>P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{t}|^2\ dt \!</math>
 
<math>P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{t}|^2\ dt \!</math>

Revision as of 14:48, 5 September 2008

Signal

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

Energy

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

$ 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)\! $

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