I need to clarify this and rather confused by this.
Lets say:
$x = h * g$
$x$ - measured data
$h$ - raw data
$g$ - instrument response function (convolution kernel)
So lets say I have $x$ and $h$. Can I say if I were to deconvolve $x$ with $h$ I will be able to obtain $g$?
Or if I have $x$ and $g$, deconvolving $x$ with $g$, I will be able to obtain $h$. So till here, is my understanding correct?
So can I say $g$ is both convolution and deconvolution kernel? Is this correct?
Discrete Convolution (Cyclic) is described by Circulant Matrix.
Circulant Matrices are diagonalized by the DFT Matrix (Will be denoted as $ F $).
So given a convolution kernel $ g $ with its matrix form given by $ G $ one could have its diagonalization by:
$$ G = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F $$
Where $ \mathcal{F} \left( g \right) $ is the Discrete Fourier Transform (DFT) of the kernel $ g $.
In the general case the deconvolution is given by:
$$ x = G h \Rightarrow {G}^{-1} x = {G}^{-1} G h = h $$
Where $ {G}^{-1} $ is the Deconvolution Operator (Inverse of $ G $ the convolution operator).
In our case we're looking for the case $ G = {G}^{-1} \Rightarrow G G = I $.
Utilizing the diagonalization of Discrete (Cyclic) Convolution Operators by the DFT Matrix (Which is Unitary Matrix):
$$ G G = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F = {F}^{H} \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) F $$
Namely we need $ \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) \operatorname{diag} \left( \mathcal{F} \left( g \right) \right) = I $ which means $ \forall i, \, {\mathcal{F} \left( g \right)}_{i} \in \left\{ -1, 1 \right\} $, namely the Fourier Transform of $ g $ has the values of -1 or 1.
So there is a set of kernels $ \mathcal{S} $ which is defined by:
$$ \mathcal{S} = \left\{ g \mid \forall i, \, {\mathcal{F} \left( g \right)}_{i} \in \left\{ -1, 1 \right\} \right\} $$
One should notice that the identity Kernel ($ I $) is indeed part of this set of kernels.
The above kernels obey $ \delta \left[ n \right] = \left( g \circ g \right) \left[ n \right] $ where $ \circ $ is the Circular Convolution (Which is multiplication of the DFT).
How could one generate such a kernel?
Well, just generate a vector of numbers which each is composed of $ 1 $ or $ -1 $ and use its ifft().
Pay attention that if you are after real operator the vector must obey the symmetry of the DFT for Real Numbers.
You may find a code to replicate the above in my StackExchange Signal Processing Q19646 GitHub Repository (Look at the SignalProcessing\Q19646 folder).
Generally not, there are sides of a similar coin. There is a symmetry in $x=h\ast g$, and either $h$ or $g$ can be considered a convolution kernel, and hence each is "a deconvolution" kernel in some extended sense, by applying the inverse (when its exists).
Indeed, depending on the properties of $h$ or $g$ (and somehow on $x$, recovering the other is not granted. Convolution is a direct and simple problems with unique output, deconvolution is an inverse problem with from no to several potential solutions.
So, somehow you can consider $g$ (or its inverse) as both a convolution and a deconvolution kernel.
