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.

comments:

this is not clear --mireille.boutin.1, Fri, 12 Oct 2007 16:23:04

why does Equation 4.12 follow???? Can somebody explain?