# Inverse of filtering and inverse of equalization

Is comb filtering a reversible operation? If applied to a signal, can I restore the original signal from the filtered image?

What about the histogram equalization operation? Is it reversible?

Let's do a simple 1-D discrete combfilter.

The impulse response is given

$$h[n] = x[n] + g\cdot x[n-M]$$

I.e. a direct part and a reflection at time $$M$$ with a gain of $$g$$. The z-transform of that system is

$$H(z) = 1 + g\cdot z^{-M}$$

and the inverse is

$$G(z) = H^{-1}(z) = \frac{1}{1 + g\cdot z^{-M}}$$

This has exactly $$M$$ poles at

$$p_m = g^{1/M}\cdot e^{j2\pi \frac{m}{M}}, m = 0...M-1$$

For $$|g| < 1$$, the poles are all inside the unit circle and the system stable. As soon the magnitude of the reflection is one or greater the poles are on or outside the unit circle and the system is unstable.

The intuitive explanation is easy enough. As soon as the reflection can completely cancel the direct component there are frequency where $$H(\omega) = 0$$ and information at these frequencies gets completely destroyed and cannot be recovered. If you try to invert it you end up with $$\frac{1}{0}$$

A comb filter with zeros cannot be inverted. At least not in a linear fashion without using som assumption about the signal.

I don’t think that histogram equalization can generally be inverted either. Information about the true scene is thrown away in a way that depends on that very same information.

• Could you please give me more explanation about inverse comb filtering, why it is not possible, or send me a suitable reference that explains why a comb filter with zeros cannot be inverted.
– Noha
Dec 16, 2021 at 4:13
• A deep zero means that possible frequency components are totally erased, not only attenuated.You cannot regain that which is unknown. Dec 16, 2021 at 5:06