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I think this comes down to the grammatical interpretation of Herman's sentence, thus if one reads it as:

"... the diffraction pattern is the transform of (the unit cell convoluted with the crystal lattice), ..."

then it is correct.  However the additional comma after "cell" in the original does render the interpretation ambiguous.

On the other hand if one reads it as:

"... the diffraction pattern is (the transform of the unit cell) convoluted with (the transform of the crystal lattice), ..."

then clearly it is not correct: as you point out "convoluted with" needs to be replaced with "multiplied by".

Knowing Herman, I'm pretty sure he intended the first interpretation and it was all down to that pesky comma!

Cheers

-- Ian


On 28 July 2016 at 08:17, "Weiergräber, Oliver H." <[log in to unmask]> wrote:
Well, to be precise, the molecular transform is _multiplied_, rather than convoluted, with the _reciprocal_ lattice.
The crystal structure corresponds to the convolution of the unit cell content with the crystal lattice (represented by a set of delta functions).
Then, as per the convolution theorem, the FT of the crystal structure, i.e., the diffraction pattern, equals the product (and not the convolution) of the unit cell FT (the molecular transform) and the lattice FT (the reciprocal lattice).

Cheers,
Oliver

================================================
  PD Dr. Oliver H. Weiergräber
  Institute of Complex Systems
  ICS-6: Structural Biochemistry
  Tel.: +49 2461 61-2028
  Fax: +49 2461 61-9540
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________________________________________
From: CCP4 bulletin board [[log in to unmask]] on behalf of [log in to unmask] [[log in to unmask]]
Sent: Thursday, July 28, 2016 9:03 AM
To: [log in to unmask]
Subject: [ccp4bb] AW: Making Molecular Transforms

Dear Jacob,

I was taught that the diffraction pattern is the molecular transform of the unit cell, convoluted with the crystal lattice, e.g. we see the molecular transform of the unit cell only at the position of the reflections and not anywhere else due to the bragg law.

So, putting your single molecule in a fake P1 unit cell and calculating F’s should do the trick. Using a large unit cell will give you a fine sampling and a small unit cell will give you a course sampling.

Best,
Herman



Von: CCP4 bulletin board [mailto:[log in to unmask]] Im Auftrag von Keller, Jacob
Gesendet: Mittwoch, 27. Juli 2016 18:27
An: [log in to unmask]
Betreff: [ccp4bb] Making Molecular Transforms

Dear Crystallographers,

Does anyone know how to make calculated single-molecule molecular transforms from pdb files?

Jacob


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