Hi Yeti,
thanks for the quick response and the further hints! Just to make sure that
I am not making a mistake: if my sample has an isotropic surface, Kx and Ky
corresponding to the maximum of the PSDF (horizontal and vertical,
respectivelly I guess) are necessarily broadly the same, and in that sense
shouldn't the respective K on the radial PSDF be the same as Kx and Ky?
Regards,
Maria
On Tue, Jan 31, 2017 at 11:52 AM, David Nečas (Yeti) <[email protected]>
wrote:
> On Tue, Jan 31, 2017 at 10:42:16AM +0100, Maria Joao Almeida wrote:
> > I am trying to get a characteristic length of a film surface by analysing
> > the PSDF. By comparing the values obtained from the PSDF and the radial
> > PSDF there seems to be a difference by a factor of 2, which I find a bit
> > confusing as I would expect them to be kind of consistent (in the sense
> > that K=Kx*cos(phi)).
> > If you could give a bit more detail on the exact discrete calculations of
> > the rPSDF would be great, at least to understand if using the rPSDF
> really
> > is appropriate i my case, provided that the change in coordinates is
> > somehow consistent.
>
> The coordinates are consistent, in both cases it is spatial frequency K
> (angular, i.e. including the 2π – like ω=2πf for temporal frequencies).
>
> But forget about discretisation. Gwyddion calculates spectral densities
> that specifically do not depend on sampling. If you have the same
> real-space surface and sample it differently, by changing the scanned
> area, pixel size, ..., Gwyddion will calculate exactly the same PSDF
> from the different images (or, more precisely, estimates of the same
> PSDF, because obviously the data change so there will be some variation).
>
> All PSDFs are actual densities; if you integrate them you always get σ².
> So for one-dimensional PSDF:
>
> ∫ W₁(Kx) dKx = σ²
>
> radial PSDF:
>
> ∫ Wr(K) dK = σ²
>
> two-dimensional PSDF[*]:
>
> ∫ W(Kx,Ky) dKx dKy = σ²
>
> Hence W₁(Kx) and Wr(K) are just different projections of W(Kx,Ky):
>
> W₁(Kx) = ∫ W(Kx,Ky) dKy
> Wr(K) = K ∫ W(Kx,Ky) dφ
>
> where φ is the polar angle φ = atan₂(Ky,Kx) and K = √(Kx²+Ky²).
>
> Note that the factor K originating from the Jacobian is already included
> in Wr(K). You may want to use Wr(K)/K instead of Wr(K) if it makes more
> sense in your case, i.e. if you want angularly averaged W(Kx,Ky), not a
> density that integrates to σ². You need to insert a ‘weight’ K to the
> integral to get σ² then.
>
> Also note that Gwyddion plots only a half of W₁(Kx) because it is
> symmetric – but the integral above is for the complete function. If you
> integrate just the curve you see plotted you will get σ²/2.
>
> Hope it helps,
>
> Yeti
>
>
> [*] The output of the 2D FFT module does not follow this convention,
> mostly for historical reasons.
>
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