(New page: <math>x(t)=2t^2</math> ---- <math>E_{\infty}</math> <math>E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt</math> <math>E_{\infty}=\int_{-\infty}^\infty |2t^2|^2\,dt</math> <math>E_{\inft...) |
(No difference)
|
Revision as of 02:05, 22 June 2009
$ x(t)=2t^2 $
$ E_{\infty} $
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E_{\infty}=\int_{-\infty}^\infty |2t^2|^2\,dt $
$ E_{\infty}=2\int_{-\infty}^\infty t^4\,dt $
$ E_{\infty}=\frac{2}{5}t^5|_{-\infty}^\infty $
$ E_{\infty}=\infty $
$ P_{\infty} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{T}\int_{-T}^{T}t^4\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}t^5|_{-T}^{T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}[T^5 - (-T^5)] $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^5}{5T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^4}{5} $
$ P_{\infty}=\infty $