| Line 1: | Line 1: | ||
<math>x(t)=\sqrt{t}</math> | <math>x(t)=\sqrt{t}</math> | ||
| − | + | ---- | |
<math>E\infty=\int|x(t)|^2dt</math> (from <math>-\infty</math> to <math>\infty</math>) | <math>E\infty=\int|x(t)|^2dt</math> (from <math>-\infty</math> to <math>\infty</math>) | ||
| Line 6: | Line 6: | ||
<math>E\infty=.5*t^2|</math>(from <math>0</math> to <math>\infty</math>) | <math>E\infty=.5*t^2|</math>(from <math>0</math> to <math>\infty</math>) | ||
<math>E\infty=.5(\infty^2-0^2)=\infty</math> | <math>E\infty=.5(\infty^2-0^2)=\infty</math> | ||
| + | |||
| + | <math>P\infty=lim\bullet T\rightarrow\infty*1/(2T)\int|x(t)|^2dt</math> (from <math>-T</math> to <math>T</math>) | ||
| + | |||
| + | <math>P\infty=lim\bullet T\rightarrow\infty*1/(2T)\int|\sqrt{t}|^2dt=\int tdt</math> (from <math>0</math> to <math>T</math>) | ||
| + | <math>P\infty=lim\bullet T\rightarrow\infty*1/(2T)*.5t^2|</math>(from <math>0</math> to <math>T</math>) | ||
| + | <math>P\infty=lim\bullet T\rightarrow\infty*1/(2T)*(.5T^2)</math> | ||
Revision as of 06:51, 17 June 2009
$ x(t)=\sqrt{t} $
$ E\infty=\int|x(t)|^2dt $ (from $ -\infty $ to $ \infty $)
$ E\infty=\int|\sqrt{t}|^2dt=\int tdt $(from $ 0 $ to $ \infty $ due to sqrt limiting to positive Real numbers) $ E\infty=.5*t^2| $(from $ 0 $ to $ \infty $) $ E\infty=.5(\infty^2-0^2)=\infty $
$ P\infty=lim\bullet T\rightarrow\infty*1/(2T)\int|x(t)|^2dt $ (from $ -T $ to $ T $)
$ P\infty=lim\bullet T\rightarrow\infty*1/(2T)\int|\sqrt{t}|^2dt=\int tdt $ (from $ 0 $ to $ T $) $ P\infty=lim\bullet T\rightarrow\infty*1/(2T)*.5t^2| $(from $ 0 $ to $ T $) $ P\infty=lim\bullet T\rightarrow\infty*1/(2T)*(.5T^2) $
