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| − | '''Work By Ryne Rayburn (rrayburn)''' | + | '''''Work By Ryne Rayburn (rrayburn)''''' |
<math>x(t)=\sqrt{t}</math> | <math>x(t)=\sqrt{t}</math> | ||
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<math>E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt</math> (due to sqrt limiting to positive Real numbers) | <math>E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt</math> (due to sqrt limiting to positive Real numbers) | ||
| − | <math>E\infty=.5*t^2| | + | <math>E\infty=.5*t^2|_{-\infty}^\infty</math> |
<math>E\infty=.5(\infty^2-0^2)=\infty</math> | <math>E\infty=.5(\infty^2-0^2)=\infty</math> | ||
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<math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt</math> | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt</math> | ||
| − | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*.5t^2| | + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*.5t^2|_0^T</math> |
<math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(.5T^2)</math> | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(.5T^2)</math> | ||
<math>P\infty=lim_{T \to \infty} \ (.25T)=\infty</math> | <math>P\infty=lim_{T \to \infty} \ (.25T)=\infty</math> | ||
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<math>E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt</math> | <math>E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt</math> | ||
| − | <math>E\infty=\int_{-\infty}^\infty |1|^2\,dt=t| | + | <math>E\infty=\int_{-\infty}^\infty |1|^2\,dt=t|_{-\infty}^\infty</math> |
<math>E\infty=\infty</math> | <math>E\infty=\infty</math> | ||
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<math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt</math> | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt</math> | ||
| − | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}t| | + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*t|_{-T}^T</math> |
| − | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}(T-(-T))</math> | + | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(T-(-T))</math> |
<math>P\infty=lim_{T \to \infty} \ 1</math> | <math>P\infty=lim_{T \to \infty} \ 1</math> | ||
<math>P\infty=1</math> | <math>P\infty=1</math> | ||
| − | [ | + | [[Computing_E_infinity_and_P_infinity| Return to Computing E infinity and P infinity]] |
Latest revision as of 15:09, 25 February 2015
Work By Ryne Rayburn (rrayburn)
$ x(t)=\sqrt{t} $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ (due to sqrt limiting to positive Real numbers) $ E\infty=.5*t^2|_{-\infty}^\infty $ $ E\infty=.5(\infty^2-0^2)=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*.5t^2|_0^T $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(.5T^2) $ $ P\infty=lim_{T \to \infty} \ (.25T)=\infty $
$ x(t)=\cos(t)+\jmath\sin(t) $
$ |x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1 $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |1|^2\,dt=t|_{-\infty}^\infty $ $ E\infty=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*t|_{-T}^T $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(T-(-T)) $ $ P\infty=lim_{T \to \infty} \ 1 $ $ P\infty=1 $
