(New page: Emily Thompson: <math>x(t)=\sqrt(5t)</math> <math>E\infty</math> ---- <math>E\infty=\int_{-\infty}^\infty |x(t)|^2dt</math> <math>E\infty=\int_{-\infty}^\infty |\sqrt(5t)|^2dt</math>...) |
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| − | Emily Thompson | + | Emily Thompson erthomps@purdue.edu |
<math>x(t)=\sqrt(5t)</math> | <math>x(t)=\sqrt(5t)</math> | ||
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<math>E\infty=\int_{-\infty}^\infty |5t|dt</math> | <math>E\infty=\int_{-\infty}^\infty |5t|dt</math> | ||
| − | <math>E\infty= | + | <math>E\infty=\int_{0}^\infty 5tdt</math> |
| − | <math>E\infty= \ | + | <math>E\infty=\frac{5}{2}(t^2|_{0}^{\infty})</math> |
| − | <math>E\infty=0</math> | + | <math>E\infty= \infty-0</math> |
| − | + | <math>E\infty=\infty</math> | |
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<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|5t|dt</math> | <math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|5t|dt</math> | ||
| − | <math>P\infty=\lim_{T \to \infty}\frac{\frac{5}{2} | + | <math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt</math> |
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| + | <math>P\infty=\lim_{T \to \infty}\frac{\frac{5}{2}|t^2|_{0}^{T}}{2*T}</math> | ||
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| + | <math>P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T}</math> | ||
| − | <math>P\infty=\lim_{T \to \infty}\frac{ | + | <math>P\infty=\lim_{T \to \infty}\frac{5}{4}T</math> |
| − | <math>P\infty= | + | <math>P\infty=\infty</math> |
Latest revision as of 18:59, 21 June 2009
Emily Thompson erthomps@purdue.edu
$ x(t)=\sqrt(5t) $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt(5t)|^2dt $
$ E\infty=\int_{-\infty}^\infty |5t|dt $
$ E\infty=\int_{0}^\infty 5tdt $
$ E\infty=\frac{5}{2}(t^2|_{0}^{\infty}) $
$ E\infty= \infty-0 $
$ E\infty=\infty $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |\sqrt(5t)|^2dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|5t|dt $
$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt $
$ P\infty=\lim_{T \to \infty}\frac{\frac{5}{2}|t^2|_{0}^{T}}{2*T} $
$ P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T} $
$ P\infty=\lim_{T \to \infty}\frac{5}{4}T $
$ P\infty=\infty $
