Discrete fourier transform and convolution -- scaling factor?

Question

I am trying to convolve a square wave (a diffraction grating phgrating with slits size \[CapitalLambda]) with a point spread function (PSF) to account for the finite size of the pixels.

I should get a "smoother" square wave -- indeed, this is what I get. But it always ends up being rescaled depending on the paramters of phgrating and PSF in ways I do not understand.

Definitions of PSF and phgrating:

PSF[rx_, x_, \[Gamma]_] := Exp[-(x^2/(2 rx^2))^\[Gamma]]

phgrating[x_, \[CapitalLambda]_, \[Phi]_] := \[Phi]/2 + 
  Sum[(\[Phi])*2/\[Pi] 1/(2 n - 1)*
    Sin[(\[Pi] (2 n - 1) x)/\[CapitalLambda]], {n, 1, 500}]

Function that generates discrete sets from PSF and phgrating for faster computation (psf and gr respectively):

PSFdiff[\[CapitalLambda]_, \[Phi]_, Nslits_, rx_, \[Gamma]_] := 
 Module[{x},
  gr = Table[{x, 
     phgrating[
      x, \[CapitalLambda], \[Phi]]}, {x, -Nslits \[CapitalLambda], 
     Nslits \[CapitalLambda], \[CapitalLambda]/100}];
  psf = Table[{x, PSF[rx, x, \[Gamma]]}, {x, -Nslits \[CapitalLambda],
       Nslits \[CapitalLambda], \[CapitalLambda]/100}] // Quiet;
  {gr, psf}
  ]

Example 1:

{gr, psf} = PSFdiff[1, \[Pi], 3, 0.1, 1];

The grating, as expected, goes from 0 to \[Pi] with each "slit" size 1:

ListLinePlot[gr]

The point spread function also makes sense:

ListLinePlot[psf, PlotRange -> Full]

Now I try to compute the convolution via the convolution theorem, and I get this:

ll = InverseFourier[
    Fourier[Transpose[psf][[2]]]*Fourier[Transpose[gr][[2]]]] // Chop;
ListLinePlot[ll]

Apart from the fact that it is shifted with respect to the original phase grating, the scaling is fine.

Example 2:

Same parameters as before, just chaning \[CapitalLambda] from 3 to 5:

{gr, psf} = PSFdiff[1, \[Pi], 5, 0.1, 1];

and this is the result:

shifted again but now also scaled!

I tried to play around with FourierParameters but did not manage to make sense of all the outcomes.

Answer 1

You are missing two scaling factors: (I) Factor in convolution theorem; (II) Factor in Kernel.

Factor (I)

To compute the factor in a linear transform (Fourier, convolution, etc.), it is helpful to first try the delta function. In the discrete case here, it is Kronecker delta.

The default Fourier transform (FT) in Mathematica has a $1/\sqrt{n}$ factor beside the summation. After product of two FT and an inverse FT, we left factor of $1/\sqrt{n}$.

n = 10;
kd = Table[Boole[i==1],{i,n}];
fkd = InverseFourier[Fourier[kd]*Fourier[kd]];
fkd[[1]] - 1/Sqrt[n]  (* = 0 *)

Conceptually it is (FT and invFT are non-scaled Fourier transform)

$$ \frac{1}{\sqrt{n}} \textrm{invFT}\left[\frac{1}{\sqrt{n}} \textrm{FT}[a] \cdot \frac{1}{\sqrt{n}}\textrm{FT}[b] \right] = \frac{1}{\sqrt{n}} \left( \frac{1}{n} \textrm{invFT}\left[ \textrm{FT}[a] \cdot \textrm{FT}[b] \right] \right). $$

The RHS shows the factor $\frac{1}{\sqrt{n}}$ and the usual convolution theorem.

Factor (II)

Usually, one would wish the PSF do not change the signal amplitude, particularly the DC signal. So simply normalize the kernel to summation one would do it.

Result

Put these two factors together we have the correctly scaled convolution theorem:

ll = Sqrt[Length[gr]]/
    Total[Transpose[psf][[2]]] InverseFourier[
     Fourier[Transpose[psf][[2]]]*Fourier[Transpose[gr][[2]]]] // Chop;
ListLinePlot[ll]

This correspond to your example 2.