Fourier transform

1. Fourier Transform
and Phase Determination
Fourier Series and Synthesis
Direct Methods
Patterson Methods

2. Fourier Series
• Euler’s formula
for any real number φ,
proved by Taylor
Series Expansions
   



sin
cos i
ei


Fig.1 Euler’s formula
Image from Wikepedia

3. Fourier Series
Definition
Any periodic function, if it is
(1) piecewise continuous;
(2) square-integrable in one period,
it can be decomposed into a sum of sinusoidal and
cosinoidal component functions---Fourier Series

4. Fourier Series
e.g, f(t) is periodic with period T,
Here,
It is the nth harmonic (angular frequency) of the function f
   


2
2
cos
2 T
T
n
n dt
t
t
f
T
a     


2
2
sin
2 T
T
n
n dt
t
t
f
T
b 
T
n
n


2

     
 






1
0
sin
cos
2 n
n
n
n
n t
b
t
a
a
t
f 


5. Fourier Series
In exponential form
where, the fourier coefficients are
Relationship between an,bn and cn
  2
n
n
n ib
a
c 

  




n
t
i
n
n
e
c
t
f 
 



2
2
1 T
T
t
iw
n dt
e
t
f
T
c n

6. Fourier Series
   
 
 

,
...
)
sin(
1
2
...
)
3
sin(
2
2
sin
2
sin
2
1
3
1
2
1














x
nx
n
x
x
x
n
  x
x
f 
No cosine terms?
Because this is an odd function
Fig.2 Fourier Series for Identity function
Animation from Wikipedia

7. Fourier Series
In exponential form
where, the fourier coefficients are
Relationship between an,bn and cn
  2
n
n
n ib
a
c 

Transform pair
 



2
2
1 T
T
t
i
n dt
e
t
f
T
c n

  




n
t
i
n
n
e
c
t
f 

8. Family of Fourier Transform
Transform Time Frequency
Continuous Fourier
transform
Continuous,
Aperiodic
Continuous,
Aperiodic
Fourier series
Continuous,
Periodic
Discrete,
Aperiodic
Discrete-time Fourier
transform
Discrete,
Aperiodic
Continuous,
Periodic
Discrete Fourier
transform
Discrete, Periodic Discrete, Periodic

9. Fourier Series in XRD
Fourier series
Continuous, Periodic  Discrete, Aperiodic
Election Density Distribution  Structure Factor
ρ(x,y,z)  F(hkl)
Real Space (R3)  Reciprocal Space (Z3)

10. The Nature of F Transform
Re-express a function in one domain into another domain
(conjugate domain)
e.g. Time domain  Frequency domain
Real space  Reciprocal space
Now, to know a function, we have two ways:
(1) From the original domain
directly measure (x,f(x)), get enough sample points and
do regression. (can we measure the electron density
distribution as we measure water density in a pond?)
(2) From the conjugate domain
get Cn, and transform to original domain

11. About Cn
The meaning of C0
ω0 = 0,
Integrated mean
think that F(000) = the amount of total electrons in a cell
 


2
2
1
0
T
T
dt
t
f
T
c
 



2
2
1 T
T
t
iw
n dt
e
t
f
T
c n

12. About Cn
  2
n
n
n ib
a
c 

Assume f is real, thus an and bn are both real
When f is odd, an = 0;
When f is even, bn = 0, so Cn is real,
which means the phase angle of Cn is kπ
Center of symmetry in real space  F(hkl) is real
a-n = an, b-n = -bn, so C-n = C*n,
Friedel’s law (symmetry in reciprocal space)
 ρ(x,y,z) is real, which is definitely right
Generally speaking, reality in one domain implies
symmetry in the conjugate domain.

13. Reexamine Structure Factor
Coordinate Form
Vector Form
Where h = {h, k, l}, rj = {xj, yj, zj}
fj is the atomic scattering factor, and it is relevant
to the modulus of reflection index
 
 




N
j
j i
f
F
1
2
exp
)
(
)
( j
r
h
h
h 
 
 





N
j
j
j
j
j lz
ky
hx
i
f
hkl
F
1
2
exp
)
( 

14. Reexamine Structure Factor
F(hkl) is considered as the resultant of adding the waves
scattered in the direction of the hkl reflection from all the
atoms in the unit cell.
Assumption: the scattering power of the electron cloud
surrounding each atom could be equated to that of a
proper number of electrons (fj) concentrated at the
atomic center.
What is the scattering factor fj ?
It is a continuous Fourier transform of electron distribution
around an atom. Since we use sphere model of atoms,
the scattering factor is also isotropic, only relevant to |h|.

15. Reexamine Structure Factor
Another view
Consider all electrons in one unit cell as a whole. In this
case, F(hkl) is the sum of the wavelets scattered from all
the infinitesimal portions of the unit cell. It is the Fourier
Transform of electron density distribution in the cell.
   
 
 

 







V
dv
lz
ky
hx
i
z
y
x
hkl
F 
 2
exp
,
,
   
 
 

V
dv
i
F r
h
r
h 
 2
exp
)
(
The integration domain V is the space of one unit cell.
Also, no sphere model assumption is made in this
formula.

16. Fourier Transform pair
In Crystallography
   
 
 

V
dv
i
F r
h
r
h 
 2
exp
)
(
 
 


h
r
h
h
r ))
(
2
exp(
1
)
( i
F
V


Note
the exponential term, one has a minus sign, the other hasn’t;
the period (V) is involved in one equation.

17. Reexamine Structure Factor
If the center of symmetry exists
   
 
 
 
   
   
   
 
 
 
 























2
1
2
1
2
1
2
2
2
cos
2
2
sin
2
cos
2
sin
2
cos
)
(
N
j
j
N
j
j
N
j
i
i
j
f
i
i
f
e
e
f
F
r
h
r
h
r
h
r
h
r
h
h r
h
r
h







So F is real when center of symmetry exists.
In the same way, we can explore the effects of other
symmetry elements and systematic absences.

18. Wilson Plot
Before knowing how phases are obtained, a
general idea of how the magnitude is got
from initial intensity data is useful.
First, a series of corrections, like absorption correction, LP
correction…after these, we get Irel;
Then, we need to put these real intensities on an
approximately absolute basis, that is what the Wilson
plot deals with.

19. Wilson Plot



N
j
j
obs f
I
1
2
Theoretical average intensity
It merely depends on what is in the cell, not on where it is.
Ideally, the ratio of the two intensities should be the scaling
factor. To calculate, it is not simple.
First, fj varies with sinθ/λ. So the averaging should be
carried out in thin concentric shells, not the whole range.

20. Wilson Plot
Second, f contains thermal motion effects, which could be
expressed as
In which, fo stands for the scattering factor of a static
atom, also B is not known. If we assume same B value for
all atoms, the exponential term is the same for all foj,
  2
2
sin 

B
oe
f
f 

 




N
j
oj
B
obs f
e
I
1
2
/
sin
2 2
2



21. Wilson Plot
Now if
C is the coefficient that brings Irel to an absolute basis, it is
a constant. Now, substitude Iobs and the equation could be
transformed to
In which, B and C are what we assume constant for all
reflections. If we plot this equation, the intercept is lnC, the
slope is -2B.
obs
rel I
C
I 
 
2
2
1
2
sin
2
ln
ln


B
C
f
I
N
j
oj
rel



















22. Wilson Plot
2
2
/
sin 











N
j
oj
rel f
I
1
2
ln
Fig.3 an illustration of Wilson plot
Image from http://www.ocms.ox.ac.uk/mirrored/xplor/manual/

23. Wilson Plot
We get C by extrapolation, then we get
Note, the structure factor in this lecture, also
the term “structure factor” we generally
use, is the Fobs, the scaled F.
C
I
F rel
obs 

24. U’s and E’s
U(hkl) unitary structure factor
Definition U(hkl) = F(hkl)point/F(000)
Fpoint is the structure factor when we replace the real atoms
with point atoms, whose scattering power is
concentrated at the nucleus, and f is no longer a function
of (sinθ)/λ, but a constant equal to the atomic number Z.
The right is a common
approximation formula
of Fpoint
 
 
real
N
j
oj
B
N
j
j
F
f
e
Z
F




2
2
sin
point



25. U’s and E’s
Note F(000) is equal to total atomic number, thus

 N
j
j
f
hkl
F
hkl
U
)
(
)
(
Here, fj includes the effects of atomic vibration.
|U|<,=1, and the phase is same as that of F(hkl)

26. U’s and E’s
E(hkl) normalized structure factor




N
j
j
f
hkl
F
U
hkl
U
hkl
E
1
2
2
2
2
2
)
(
)
(
)
(



N
j
j
f
1
2
is the theoretical
average intensity. And F(hkl)
is the scaled structure factor.
ε is usually 1, but may assume other values for special
sets of reflections where symmetry causes intensities to
be abnormally high (e.g. m┴b causes (h0l) average intensity
is twice that of general (hkl), so ε = 2 for E(h0l).)

27. U’s and E’s
E shows which reflections have above or
below average intensities. It is easy to see
that the expectation of |E|2 is one.
Also, E value can help us see if center of
symmetry exists
<|E|> <|E2-1|>
Centrosymmetric 0.798 0.968
Non-… 0.886 0.736

28. The Phase Problem
As we have seen, we need F(hkl) to do the
Fourier transform, but experiments can
only give us |F|, so we need to know the
phase by some other means.
The Direct Method and Patterson Method
are two ways to give us phase information.

29. Direct Methods
Basis
Some phase info is hided in the Magnitude of
structure factors and their relationships.
Inequality
Harker and Kasper applied Cauchy inequality to
unitary structure factor equation and got
U2(hkl)<,=1/2 +1/2 U(2h,2k,2l)
Here, centrosymmetry is assumed.

30. Direct Methods
U2(hkl)<,=1/2 +1/2U(2h,2k,2l)
If U2(hkl) = 0.6, |U(2h,2k,2l)| = 0.1,
u must be positive for the inequality to hold.
If U2(hkl) = 0.3, |U(2h,2k,2l)| = 0.4,
Could be either, but to guess u is positive could probably
be right
Weakness
rare reflections are qualified, since |F| declines with sinθ
rapidly.

31. Direct Methods
Karle-Hauptman determinant
0
0
0
0
0
2
1
2
1
2
2
1
2
1
1
2
1







U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
n
n
n
n
n
n










32. Direct Methods
Iet us examine a third order determinant
0
2
1
2
2
2




 
 h
k
k
h
h
k
k
h U
U
U
U
U
U
0
1
1
1



h
k
k
k
h
h
k
h
U
U
U
U
U
U
If it is centrosymmetric
If |U|’s at the right side are large, the left must be positive
s(F) is the symbol fuction, equal to -1 when F is negative





 


 
 1
2
2
2
2
1
h
k
k
h
h
k
k
h U
U
U
U
U
U
      1
)
(
)
( 



 h
k
k
(h) F
s
F
s
F
s

33. Direct Methods
If it is non-centrosymmetric,
If U’s are all large,
  1
cos 



 h
k
k
h 


  0
cos
2
1
2
2
2






 

 h
k
k
h
h
k
k
h
h
k
k
h U
U
U
U
U
U 


  0




 h
k
k
h 



34. Direct Methods
     
 

k
k
h
k
h F
F
gV
f
F
Sayre’s Equation
(1)
This equation can be derived from Fourier Series as follows
(2)
Rewrite the equation by setting that h=L+L’, k=L’
(3)
           
 
r
L'
L
L'
L
r
L
L
r
L L
L












 
 i
F
F
V
i
F
V


 2
exp
1
2
exp
1
'
2
2
2
       
 



h k
r
h
k
h
k
r i
F
F
V

 2
exp
1
2
2

35. Direct Methods
Since squared density is also periodic, so write it into its
Fourier series form
(4)
Here G(h) is the structure factor of the squared cell.
Compare (3) and (4), it follows that
(5)
The structure factor G(h) is
(6)
     
 


h
r
h
h
r i
G
V

 2
exp
1
2
     
 

k
k
h
k
h F
F
V
G
1
   




N
j
j i
g
G
1
2
exp j
r
h
h 

36. Direct Methods
In equation (6), gj is the scattering factor of the squared
cell. If we assume equal atom model, (3) reduces to
(7)
The normal structure for equal atoms is
(8)
Thus we can obtain
(9)
   




N
j
i
g
G
1
2
exp j
r
h
h 
   




N
j
i
f
F
1
2
exp j
r
h
h 
   
h
h F
f
g
G 

37. Derect Methods
Finally, from (9) and (5), we get Sayre’s equation
(10)
1) Don’t care the fraction coefficient outside the
summation.
2) The summation contains a large number of terms;
however, in general it will be dominated by a smaller
number of large |F(k)F(h-k)|. Considering a reflection
with large |F(h)|, it can therefore be assumed that the
terms with large |F(h)F(h-k)| have their angular part
approximately equal to the angular part of |F(h)| itself
     
 

k
k
h
k
h F
F
gV
f
F

38. Direct Methods
So we get,
Or
If it is centrosymmetric, which means phase could only be
0 or π,
Same result as deduced from Karle-Hauptman determinant
Note, these relationships are independent of origin location.
)
(
)
(
)
( k
h
k
h 

 


0
)
(
)
(
)
( 



 k
h
k
h 


      1
)
(
)
( 


 k
h
k
(h) F
s
F
s
F
s

39. Direct Methods
Does the Magnitude of F(hkl) change with
the shift of origin?
No.
Does the phase of F(hkl) change with the
shift of origin?
Yes.
However, certain linear combinations of phases don’t
Change regardless of the arbitrary assignment of cell
origins. These are called structure invariants.

40. Direct Methods
Structure Invariants
Assume atom j’s position is rj. Now the origin shifts through
a vector r, the new position of this atom is rj’. The beam,
which contributes to reflection h and is scattered by this
atom j now has a phase change equal to 2π(h•r),
independent of where the atom is.
Thus, the phase change of F(h) is 2π(h•r).
If we want the sum of several reflection phases to be
constant, which means their total phase changes should
be zero, the condition is easily derived,
h1+h2+h3+…+hn = 0
This kind of combination is a structure invariant.

41. Direct Methods
The choice of origin is not arbitrary other than P1 space
group, the symmetry elements should be considered and
the “permissible origins” are much less, so there are
many more this kind linear combinations, which are
called structure seminvariants.
e.g. In space group P(-1), the origin must be located at one
of the eight centers of symmetry in the cell, so the shift
vector r can take eight values only,
r = (s1, s2, s3), s1,2,3 = 0 or ½
Think about the phase of F(222), does it change?

42. Direct Methods
Table 1 relative sign relationships for possible origins
Origin shift eee oee eoe eeo ooe oeo eoo ooo
0 + + + + + + + +
a/2 + - + + - - + -
b/2 + + - + - + - -
c/2 + + + - + - - -
(a+b)/2 + - - + + - - +
(a+c)/2 + - + - - + - +
(b+c)/2 + + - - - - + +
(a+b+c)/2 + - - - + + + -

43. Direct Methods
Now, we have Inequality, Sayre’s equation,
which use Intensity data to guess phase;
we have triple structure invariants and
seminvariants, which show what remains
unchanged when the origin is shifted.

44. Direct Methods
General Process
(1) Pick out all stronger reflections (like |E|>2.0) as a set,
find all triple relationships among them (indices sum to
zero), compute probabilities of each triple.
(2) Select three reflections for origin determination. These
reflections are the most often and most reliably
interconnected. The phases of these three are set
freely, but do check the table on last page to avoid
conflicts. After the initial set of phases, origin is fixed.
(3) Phase propagation (phase variables may be involved)

45. Direct Methods
General Process
(4) Get enough phases, do Fourier Transform, a
preliminary electron density model is obtained.
Usually, we can get positions of several heavier atoms.
(5) Calculate phases based on this model, and connect
the calculated phases with observed |F(hkl)|, and
transform again, the result is usually better, so more
atom positions could be located, so our cell model is
improved.
(6) Repeat (5), and also gradually lower the |E| cutoff,
include more reflections in the calculation, until all the
atoms are located.

46. Direct Methods
Let’s see a phase propagation process.
E.g. a crystal with space group P21/c
(1) Assign “+” phases to F(3 1 -17), F(3 4 11), F(5 0 14),
thus the origin is set;
(2) Three starting reflections combine to imply two new
ones,
s(6 5 -6) = s(3 1 -17) • s(3 4 11) = +
s(8 1 -3) = s(3 1 -17) • s(5 0 14) = +
This example is from George Stout, Lyle Jensen, X-ray Structure Determination---A
Practical Guide, second edition, page 271-273

47. Direct Methods
(3) Based on the symmetry, in the case of P21/c,
if k+l is even, F(h k l) = F(h -k l) = F(-h k -l)
if k+l is odd, F(h k l) = -F(h -k l) = -F(-h k -l)
so, s(-3 4 -11) = -s(3 4 11) = -
s(8 -1 -3) = s(8 1 -3) = +
combine,
s(5 3 -14) = s(-3 4 -11) • s(8 -1 -3) = -
At this point, let’s assume we have used up all relationships
we can find, thus to continue this process, we need to
assign phase variable to another useful reflection.

48. Direct Methods
(4) Assign s(6 5 -12) = a (a could be + or -)
s(2 -6 9) • s(6 5 -12) = s(8 -1 -3)
s(2 -6 9) • a = +
s(2 -6 9) = a
s(3 6 5) = s(-3 1 17) • s(6 5 -12) = a
Now we have a checking relationship, all three reflections’
phases are known,
s(2 -6 9) • s(3 6 5) = s(5 0 14)
a • a = +
which is correct and lends confidence to the work so far.

49. Direct Methods
Also, s(9 1 -1) = s(3 -4 11) • s(6 5 -12) = -a
In this relationship
s(2 -6 9) • s(7 7 -10) = s(9 1 -1)
two are known in terms of the variable
a • s(7 7 -10) = -a
But s(7 7 -10) is negative absolutely.
In this way, more and more structure factor phases are
determined or assumed, and Fourier transform could be
carried out when phase information is enough.

50. Patterson Methods
• In 1935, A. L. Patterson pointed out that if
we do Fourier Synthesis with |F|2 instead
of F as coefficients, the result shows info
on all of the interatomic vectors.
• In Patterson map, if (u, v, w) is a peak, it
indicates there are atoms exist in the cell
at (x1, y1, z1,) and (x2, y2, z2) such that
u=x1-x2, v=y1-y2, w=z1-z2.

51. Patterson Methods
Formula
The above is actually the electron density function
convoluted with its inverse. This may give a better
understanding why it shows the vectors between atoms.
     
r
r
u 

 

P
     
u
h
h
h
u 


 i
e
F
P 
2
2

52. Patterson Methods
It illustrates that you can think of a Patterson as being a sum
of images of the molecule, with each atom placed in turn on
the origin. Also note the patterson cell and the crystal unit
cell has the same size.
Unit Cell Patterson Cell
Fig.4 Compare Patterson cell with unit cell
Figure from http://www-structmed.cimr.cam.ac.uk/Course/

53. Patterson Methods
• How many peaks are there in one
Patterson cell?
If we have N atoms in one cell, there are N2 interatomic
vectors, in which N vectors are 0 (just one atom to itself),
so we have N2-N+1 peaks in a Patterson cell if there are
no more overlaps.
The huge origin peak, is usually useless and may cover
other peaks near it. It is possible to remove it by
subtracting the average value of |F|2 from each term. If
we use |E|2, it will be easier. (why easier?)

54. Patterson Methods
Two other properties of Patterson Peaks
(1) The peak spread is the sum of the two corresponding
Fourier density peak; and also peak amount is nearly
squared, but cell size is the same. These two lead to a
great overlap. But we can create a sharpened
Patterson map, if we use |Fpoint|2.
(2) The peak height in Patterson map is approximately the
multiple of two Fourier density peaks which generate it.
Thus the vectors connecting heavy atoms will stand
out sharply,

55. Patterson Methods
• Patterson Symmetry
The Patterson space group is derived by
original space group by replacing all
translational elements by the
correspondent nontranslational elements
(axes, mirrors) and by adding a center of
symmetry if it is not already present. Their
lattice types don’t change.

56. Patterson Methods
Though translational elements are replaced, traces are left.
Harker lines and planes
---unusually high average intensity of certain lines or planes
Think about a m vertical to b axis in real space. In
Patterson map, a dense line is along b axis.
If above is not a mirror, but a c glide, can we tell the
difference based on the patterson map?
In the same way, think about axes and screw axes.

57. Patterson Methods
Fig.5 Harker section
This image is from
http://journals.iucr.org/d/issues/2001/07/00/gr2131/gr2131fig2.html
Which symmetry element generate the above harker section?

58. Patterson Methods
On a Patterson Map, we can see strongest
peaks, which are related to heavy atoms;
we can also see Harker lines and planes,
which are related to symmetry elements.
Therefore, we can probably tell heavy
atoms’ positions relative to symmetry
elements in real space. Since the origin is
also set or restricted by the symmetry
elements in each space group, we can tell
where these heavy atoms are.

59. Patterson Methods
Now, let’s look at an example of one heavy
atom in P21/c
1) List out all equivalent positions (4 in this case)
2) Caculate vectors between these positions (16 vectors,
4 of which contribute to origin peak, other overlaps
lead to stronger peaks in special positions)
3) Compare with observed harker section, and get heavy
atom coordinate

60. Patterson Methods
Fig.6 equivalent positions of P21/c
from International Table

61. Patterson Methods
x y z -x –y –z -x ½+y ½-z x ½-y ½+z
x y z 0 0 0 -2x -2y -2z -2x ½ ½ -2z 0 ½-2y ½
-x -y -z 2x 2y 2z 0 0 0 0 ½+2y ½ 2x ½ ½+2z
-x ½+y ½-z 2x ½ -½+2z 0 -½-2y ½ 0 0 0 2x -2y 2z
x ½-y ½+z 0 -½+2y ½ -2x -½ -½-2z -2x 2y -2z 0 0 0
Table 2 vectors between equivalent positions
also the peak positions in Patterson Maps
Same colors indicate overlap positions, but four
black coordinates are just four different positions

62. Patterson Methods
Now, collect redundant peaks
peak position relative weight
0 0 0 4
0 ½+2y ½ 2 along harker line
0 ½-2y ½ 2 along harker line
2x ½ ½+2z 2 in harker plane
-2x ½ ½-2z 2 in harker plane
2x 2y 2z 1
-2x -2y -2z 1
2x -2y 2z 1
-2x 2y -2z 1

63. Patterson Methods
So, on the Harker line, which is parallel to b
axis, equal to shift b axis up ½ unit along
c, we may find two strong peaks, for
example, the coordinates are (0,0.3,1/2)
and (0,0.7,1/2), thus we can get y=0.1 by
solving a simple equation. In the same
way, we can get x and z from Harker
plane. So, this heavy atom is found.

64. Patterson Method
After this, we get a simple unit cell model
with only heavy atoms, which usually
dominate the reflection phases.
Following is just routine repetition until all
atoms are solved.