Restoration by Inverse Filtering

Taking Fourier transform of the convolution $g(x)=f(x)*h(x)$, we get

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

where $G(f_x)={\cal F}[g(x)]$, $F(f_x)={\cal F}[f(x)]$, and $H(f_x)={\cal F}[h(x)]$ are respectively the Fourier spectra of $g$, $f$ and $h$. 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 $F(f_x)$

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

However, we realize that at the points of $F(f_x)$ corresponding to $H(f_x)=0$, i.e., $\sin(\pi L f_x)=0$ at $f_x=k/L,\;\;(k=\pm 1, \pm 2, \cdots)$, the image can never be restored, as $F(f_x)$ is multiplied by $H(f_x)=0$, i.e., the information is lost. Interpolation based on the neighboring points would not work (why?).

Moreover, this inverse filtering method is sensitive to noise that may exist in the imaging process, which may be amplified when $H(f_x)$ is very close to zero.