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Consider <math>X(j\omega)</math> evaluated according to Equation 4.9: | Consider <math>X(j\omega)</math> evaluated according to Equation 4.9: | ||
| − | < | + | <math>X(j\omega) = \int_{-\infty}^\infty x(t)e^{-j \omega t} dt</math> |
| − | and let | + | and let x(t) denote the signal obtained by using <math>X(j\omega)</math> in the right hand side of Equation 4.8: |
| − | < | + | <math>x(t) = (1/(2\pi)) \int_{-\infty}^\infty X(j\omega)e^{j \omega t} d\omega</math> |
| − | If | + | If x(t) has finite energy, i.e., if it is square integrable so that Equation 4.11 holds: |
| − | < | + | <math>\int_{-\infty}^\infty |x(t)|^2 dt < \infty</math> |
| − | then it is guaranteed that < | + | then it is guaranteed that <math>X(j\omega)</math> is finite, i.e, Equation 4.9 converges. |
| − | Let | + | Let e(t) denote the error between <math>\hat{x}(t)</math> and x(t), i.e. <math>e(t)=\hat{x}(t) - x(t)</math>, |
| − | + | then Equation 4.12 follows: | |
| − | < | + | <math>\int_{-\infty}^\infty |e(t)|^2 dt = 0</math> |
| − | Thus if | + | Thus if x(t) has finite energy, then, although x(t) and <math>\hat{x}(t)</math> may differ significantly at individual values of t, there is no energy in their difference. |
From mireille.boutin.1 Fri Oct 12 16:23:04 -0400 2007 | From mireille.boutin.1 Fri Oct 12 16:23:04 -0400 2007 | ||
Latest revision as of 11:24, 24 March 2008
Consider $ X(j\omega) $ evaluated according to Equation 4.9:
$ X(j\omega) = \int_{-\infty}^\infty x(t)e^{-j \omega t} dt $
and let x(t) denote the signal obtained by using $ X(j\omega) $ in the right hand side of Equation 4.8:
$ x(t) = (1/(2\pi)) \int_{-\infty}^\infty X(j\omega)e^{j \omega t} d\omega $
If x(t) has finite energy, i.e., if it is square integrable so that Equation 4.11 holds:
$ \int_{-\infty}^\infty |x(t)|^2 dt < \infty $
then it is guaranteed that $ X(j\omega) $ is finite, i.e, Equation 4.9 converges.
Let e(t) denote the error between $ \hat{x}(t) $ and x(t), i.e. $ e(t)=\hat{x}(t) - x(t) $, then Equation 4.12 follows:
$ \int_{-\infty}^\infty |e(t)|^2 dt = 0 $
Thus if x(t) has finite energy, then, although x(t) and $ \hat{x}(t) $ may differ significantly at individual values of t, there is no energy in their difference.
From mireille.boutin.1 Fri Oct 12 16:23:04 -0400 2007 From: mireille.boutin.1 Date: Fri, 12 Oct 2007 16:23:04 -0400 Subject: this is not clear Message-ID: <20071012162304-0400@https://engineering.purdue.edu>
why does Equation 4.12 follow???? Can somebody explain?
