0
$\begingroup$

This is not a homework, and I have no idea whether it could be one. It is only a request for help, as I do not have any experience using Fourier transform. The origin of the problem is from physics. If this kind of more applicative question is inappropriate, please let me know.

I have the following convolution product: $g=-h*f$ where $h(x)=Ax\ /\ (x^2+B)^{3/2}$ with $A, B, l\gt 0$ and $x\notin[0,l] \Rightarrow f(x)=0$

I wish to inverse the definition, i.e., to get the function $f$ explicited, or at least to know whether there are specific constraints for my question to have meaning.

I guess one way to do it is to use the inverse $1/\hat{h}$ of the Fourier transform $\hat{h}$ of $h$. But I do not trust I can do that without error.

In practice, I suppose this can be done numerically with FFT, but I would think one first has to check that it makes sense formally.

I am also wondering how well $f$ can be approximated if the function $g$ is known only on $[0,l]$.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The deconvolution problem you want to solve is typically ill-posed. In principle you are correct that you only have to divide by the Fourier transform of the kernel, but since this will go to zero for high frequencies, inversion will amplify noise there.

However, this may not be a real problem when you have some a-priori knowledge of your function. e.g. when you know that the function is smooth (then it has no frequency components at high frequencies). There are so called regularization schemes which handle these.

One possibility of implementing such regularization or a-priori information, is if you assume the original function $f$ is representable by a e.g. spline, and then solve for the spline coefficients so that the convolution of the spline best matches your result $g$. Then you also only get a linear equation system, which is well solvable with a computer.

There is also Tikhonov regularization.

The FFT approach has another difficulty, since the FFT implements the discrete Fourier transform which is not the continuous Fourier transform, which you need. The discrete Fourier transform has this periodicity assumption in both original domain and frequency domain, which is probably not corresponding to your physical situation. If it does, the FFT is right choice, though.

$\endgroup$
4
  • $\begingroup$ Thanks. My 1st question is whether my abstract physical setup makes mathematical sense, even if abstractedly. I take it your answer means yes (right ?). The 2nd point is that fast variations of $f$ will be hard to get in practice because the kernel will amplify small errors. My initial naive idea was actually to discretize $f$ over the interval $[0,l]$ partitioned into $n$ intervals $[a_{i-1},a_i]$ where it has the value $x_i$, and make $n$ corresponding measures of $g$ to get a system of $n$ linear equations in variables $x_i$. Would that work, even if not as good as your spline suggestion. $\endgroup$
    – babou
    Commented Jul 31, 2013 at 21:55
  • $\begingroup$ Yes it makes sense. But formally you probably have to work with the Laplace transform, because the Fourier transform of the inverse will not be in the $L^2$ space. From the more applied viewpoint this makes no difference. Also note your method is just the spline attempt of zeroth order, so yes that should work. Of course linear interpolation would be more precise but then you need the convolution of $h$ with the hat function instead of the rect function which may be more complicated. $\endgroup$
    – Andreas H.
    Commented Jul 31, 2013 at 22:29
  • $\begingroup$ I just realized that I do not know the Fourier transform $\hat h$ of the kernel $h$, and you did not give it. Then how do you know it goes to zero for high frequencies? I do not have the slightest idea what $\hat h$ looks like. All I know is that the function $h$ itself goes to zero for high values. (I modified the function slightly, replacing $l^2$ by $B$, but it should not matter). $\endgroup$
    – babou
    Commented Aug 2, 2013 at 20:11
  • $\begingroup$ That is more an educated guess. The function $h$ is smooth and has a pulse shape form. It hence does not contain frequency components of arbitrary high frequency (since the pulse os not arbitrary short) and so the Fourier transform should go to zero for $f$ to infinity. Consider for example the Fourier transform of a Gaussian function, which is a Gaussian itself. Or the Fourier transform of a Lorentzian which is $exp(-abs(f))$. So two pulse functions, both exponential behavior and polznomial, have vanishing Fourier transoforms at infinity. $\endgroup$
    – Andreas H.
    Commented Aug 3, 2013 at 2:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .