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?