Restoration by Inverse Filtering

Taking Fourier transform of the above convolution, we get

$$G(f_x)=F(f_x)H(f_x)$$

where $G$, $F$, and $H$ are the spectra of $g$, $f$ and $h$, respectively. Specifically, we have

$$H(f_x)=\int_{-\infty}^{\infty} h(x) e^{-j2\pi x f_x} dx =\int... ...j2\pi x f_x} dx = e^{-j\pi Lf_x}\frac{sin(\pi L f_x)}{\pi f_x}$$

While the inverse filtering method could be applied to restore $f(x)$ by inverse transforming $F(f_x)$

$$F(f_x)=\frac{G(f_x)}{H(f_x)}$$

we also realize that those points of $F(f_x)$ corresponding to $H(f_x)=0$ at $f_x=k/L,\;\;(k=\pm 1, \pm 2, \cdots)$ can never be restored. Interpolation from the neighboring points would not work (why?).

Moreover, this inverse filtering method is sensitive to noise that may exist in the imaging process.