Fourier Transforms

1

• Fourier Transforms
• A mathematical tool derived from spectral
  analysis.
• Any function f(t) can be written as a sum of its
  frequency components n.
• f(t) ?-? to ? F(n) exp(2pi n t) dn
• where f(t) and F(n) may be complex.
• Note There is nothing special about this
  expression its simply another way of writing
  f(t).
• What is special about the Fourier Transform is
  that its inverse
• F(n) ?-? to ? f(t) exp(-2pi n t) dt
• has a form which is almost the same as the
  forward transformation!

2

• Proof of inverse relationship.
• f(t) ?n F(n) exp(2pi n t) dn
• ?n ?t f(t) exp(- 2pi n t) exp(2pi n
  t) dt dn
• ?t f(t) ?n exp(- 2pi n (t t)) dn dt
• ?t f(t) d(t t) dt
• f(t)
• The key result being
• ?-? to ? exp(- 2pi n (t t)) dn d(t t)
• Where d(t t) is the Dirac delta function.
• d(t t) 0 if t ? t
• d(t t) ? if t t
• Such that ?-? to ? d(t t) 1

3

• Pictorial representation
• ? -? to ? exp(- 2pi n (t t)) dn d(t t)
• Can rewrite this as a summation.
• lim dn ? 0
• Sn -? to ? exp(- 2pi dnn (t t)) dn

q
Goes to ?

• When t t, t t 0 q is always 0o
• When t ? t then 2p(t t) ? 0 and q ? 0o.
• always go around in circles.
• If you are going around in circles there is
  always another vector which is the opposite
  cancels!
• When t t the red arrows go off to ? hence
  the total integration becomes infinite.

4

• Pictorial representation
• Problem is to demonstrate that
• ? -? to ? exp(- 2pi n (t t)) dn d(t t)
• Can rewrite this as a summation.
• lim dn ? 0 Sn -? to ? exp(- 2pi dnn (t
  t)) dn
• The summation can be re-grouped to get each
  non-zero vector cancelling pair-wise when t' ? t

Goes to ?
5

• Other properties of Fourier Transforms.
• f(at) ? 1/a F(n/a) called time-scaling.
• 1/b f(t/b) ? F(bn) called frequency
  scaling.
• f(t t0) ? F(n) exp 2pi nt0 called
  time shifting.
• f(t) exp -2pi n0t ? F(n n0) called
  frequency shifting.
• A more important result is the convolution
  theorem
• If
• p(t) f(t) g(t)
• then
• P(n) F(n) G(n)
• Where the convolution is defined by
• F(n) G(n) ? F(g) G(n-g) dg

6
Proof of origin shift theorem. If F(n) then what
is F(n) where the time-origin is changed. F(n)
? f(t - t0) exp(2pi nt) dt Put v (t - t0)
dv dt F(n) ? f(v) exp(2pi n (v t0))
dv exp 2pint0 ? f(v) exp(2pi n v)) dv exp
2pint0 F(n)
f(t - t0)
t
t
t0
7

• Example 1 Fourier transform of a slit.
• f(x) 1/a H(x a/2) H(a/2 x)
• Where H(x) is the heavyside function
• H(x) 0 if x lt 0 H(x) 1 if x gt 0
• F(s) ? f(x) exp(2pi s x) dx
• ?x-a/2 to a/2 1/a exp(2pi s x) dx
• 1/(a2pin) exp 2p s xx-a/2 a/2
• 1/(2ipas) exp pi sa - exp -pi sa
• sin(pas)/pas

FT ?
x
8

• Example 2 Fourier Transform of a Grid
• For exampe N pulses evenly spaced.
• - eg. a diffraction grating.
• Use the convolution theorem.

x
b
ie. N slits in a grating can be regarded as the
sum of one slit convoluted with N evenly spaced
delta functions.
9

• If p(x) f(x) g(x) then
• G(s) F(s) G(s)
• Where the convolution is defined by
• f(x) g(x) ? f(g) g(x - g) dg
• The result is that we sum over all the
  contributions of the slit functions multiplied by
  a phase factor since
• FT d(x) ? d(x) exp(2pi nx) dx exp(2pi n
  0) 1
• FT d(x - b) exp(-2pi sb)
• where we used the origin shift result from a few
  pages back.
• - hence have N phase factors for N slits.

Sn 1 to N exp -2pinsb
10

• To evaluate the summation
• Sn 1 to N exp(-2pin sb) Sn 1 to N
  exp(-2pi sb)n
• (1 - exp(-2pi N sb)) / (1 - exp(-2pi sb))
• Where we use the geometric progression partial
  sum formula in final step.
• If this was diffraction from N-grating slit then
  the intensity is given by the amplitude squared
• I (1 - exp(-2pi N sb))/(1 - exp(-2pi sb))2
• (2-2cos(2p N sb))/ (2-2cos(2p sb))
• (sin (p N sb)/ sin (p sb))2

Pattern modulated by sin(pas)/pas The single
slit pattern.
1/a
-2/b
0
1/b
2/b
-1/b
11

• The formula
• I (sin (p N kb)/ sin (p kb))2
• Has the property that the peak intensity
• I ? N2
• The width of the line
• I ? 1/N
• Hence the integrated number of counts varies
  with N!
• - This also holds in principle for diffraction
  from a crystal.
• - In practice a crystal is never perfect, you
  have small blocks of small crystls (mosaic
  blocks) hence more like the diffraction peak
  intensities varies with the number of unit cells.

Mosaic blocks
12
Example 3 Fourier transform of a cosine
function FT (cos 2pn0t) 1/2 FT (exp 2pin0t
exp -2pin0t) 1/2 ? (exp 2pin0t exp -2pin0t)
exp(2pi nt) dt 1/2 ? exp 2pi(n0n)t (exp
2pi(-n0n)t dt 1/2 d(n n0) d(n - n0)
FT ?
p/n0
-p/n0
n
- n0 n0
13

• Three dimensional representation of a Fourier
  Transform
• To define a Fourier Transform in three
  dmensions.
• F(S) ?all space f (r) exp (2pi r S) dr
• - Where r ranges over all space.
• - f(r) is a function defined on all space.
• - S is a 3 dimensional vector which equates to a
  wavevector.
• As with a 1D Fourier transform, its inverse
• f(r) ?all wavefectors F (S) exp (-2pi r S)
  dS
• This forms the basis of X-ray crystallography
  theory since we simply replace f(r) with r(r),
  and we note the S is the change in wavevector for
  the X-rays scattering of a crystal.
• - Note in X-ray crystallography you sample a
  discrete lattice, rather than a continuous
  function
• - Use a discrete Fourier Transform.

14

• Scattering from an atom
• Remember that the expression
• fatom(S) ?r r (r) exp (2pi r S) dr
• comes from summing the X-ray scattering terms
  from the electron density of the atom with an
  appropriate phase factor.
• ie. Adding up all the scattering contributions
  of a function of electron density as a complex
  exponential leads naturally to a Fourier
  Transform.

Sum probability function with a position change
in X-ray scattering dependent phase factor exp
(2pi r S) dr
15

• Scattering from a unit cell
• In principle the scattering from the unit cell
  gives a Structure Factor
• Funit cell(S) ?r r (r) exp (2pi r S) dr
• where you sum over all the electron density
  within the cell.
• However since we know that the contribution from
  each atom
• fatom(S) ?r r (r) exp (2pi r S) dr
• where you integrate over the electrons within the
  atom.
• The Structure Factor for the unit cell can be
  written (rj position of the jth atom in the unit
  cell fj(S) the atomic scattering factor for the
  jth atom)
• Funit cell(S) Sj fj(S) exp (2pi rj S)

16

• Example Benzene
• Total scattering for a unit cell is
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• First choose the unit cell
• - Should choose right hand cell since it reflects
  higher symmetry.
• - Hexagonal rather than Orthorhombic.

17

• Scattering from Benzene
• Total scatteromg for ta unit cell is
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• Must evaluate a b
• a x b x cos 2p/3 y sin 2p/3
  - 0.5 x v(3/4) y
• Where underlined bold means a unit vector
  (stupid ppt).

b
a
18

• Atomic coordinates for Benzene
• Consider 2q X
• In this special case S ? b.
• Evaluate the structure factor
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• for every given scattering vector S
• a S - 0.5 S
• b S S

s length 1/l
S
s0 length 1/l
2q 60o
19

• Atomic coordinates for Benzene
• 6 atoms in the unit cell at fractional
  coordinates
• (1/4,1/4) (1/4,1/2) (1/2,3/4) (3/4,3/4)
  (3/4,1/2) (1/2,1/4)
• Next sum over
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• for every given scattering vector S

20

• Structure Factor for Benzene
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• Sj1 to 6 fcarbon(S) exp (2pi rj S)
• Adding over the vectors (1/4,1/4) (1/4,1/2)
  (1/2,3/4) (3/4,3/4) (3/4,1/2) (1/2,1/4)
  remembering
• a S -0.5 S,
• b S S,
• S 2 sinq/l
• fcarbon(S) exp (2pi 1/8 2 sinq/l) exp
  (2pi 3/8 2 sinq/l)
• exp (2pi 1/2 2 sinq/l) exp (2pi 3/8 2
  sinq/l)
• exp (2pi 1/8 2 sinq/l) exp (2pi 0 2
  sinq/l)
• fcarbon(S) 2 exp (pi/2 sinq/l) 2 exp
  (3pi/2 sinq/l) exp (2pi sinq/l) 1
• I left out the hydrogen atoms to make things
  simpler.

21
Structure Factor for Benzene Funit cell(S)
fcarbon(S) 1 2 exp (ip/2 sinq/l) 2 exp
(i3p/2 sinq/l) exp (2pi sinq/l)
S
Funit cell (S)
picture assumes sin q/l 1/3
22

• Example - Scattering from two atoms!
• Consider two atoms separated by distance d.
• Deflected such that deflection angle is 2q.
• Path difference is
• DP d sin 2q
• If DP nl, then two waves add in phase see a
  bright spot.
• If DP is nl 1/2 add out of phase have dark
  region

s0
S
s

• Note here d is the spacing between atoms.
• Later d is used to represent the spacing between
  lattice planes!
• They are not exactly the same (lattice planes at
  angle q relative to S) this explains why the
  formula for DP is not identical to that seen in
  later examples.

23

• Could address the same problem using previous
  formula.
• Ftwo atoms(S) Sj fj(S) exp (2pi rj S)
• f1(S) exp (2pi r1 S) f2(S) exp (2pi r2
  S)
• f(S) exp (2pi r1 S) (1 exp (2pi d S))
• Evaluate Funitcell(S)2 to predict the measured
  intensity.
• - r1 is the position to one of the atoms r2
  r1 d.

24

• Scattering from a crystal lattice
• Add the scattering from all unit cells relative
  to a single origin.
• For each unit cell have scattering
• Funit cell(S) ?r r (r) exp (2pi r S) dr
• Since its a crystal we have the condition for
  translational symmetry
• r(r) r(r) u a v b w c
• hence
• Fuvw(S) Funit cell(S) exp 2pi (ua vb
  wc)S

25

• Total scattering from the crystal sum over all
  unit cells
• K(S) Suvw Fuvw(S)
• Suvw Funit cell(S) exp 2pi (ua vb
  wc) S
• Funit cell(S) Suvw exp 2pi (ua vb
  wc) S
• Funit cell(S) Suvw exp 2piu(aS) exp
  2piv(bS) exp 2piw(cS)
• Funit cell(S) Suexp 2piu(aS) Svexp
  2piv(bS) Swexp 2piw(cS)
• Note we have now three terms in the product of
  the form
• Suexp 2piu(aS)
• If any of these are zero then K(S) is zero.
• - ie there is no scattering in the direction of
  (S).

26

• To evaluate
• K(S) Funit cell(S) Suexp 2piu(aS) Svexp
  2piv(bS) Swexp 2piw(cS)
• Use again the complex plane addition of vectors.
• - If 2p aS 2p h (ie. an integer multiple
  of 2p) have all vectors parallel add in phase.
• - If 2p aS ? 2p h then vector goes round in
  circles can group the summation so that
  everything cancels.
• Argument produces the condition that diffraction
  occurs if only if
• aS h bS k cS l

27

• Phase contribution from a lattice point
• Found diffraction if
• a S h b S k c S l
• Each lattice point is described by
• r u a v b w c
• Hence
• r S (u a v b w c) S uh vk wl
  integer
• K(S) Suvw Funit cell(S) exp 2pi (ua
  vb wc) S
• Suvw Funit cell(S) exp 2pi integer
• ie. All lattice points scatter in phase (ie. f
  n 2p) !

r
r
r
r
Lattice plane indices (2, 1).
Lattice plane given indices (1, 3)
28

• Reciprocal Lattice
• Define a lattice which is perpendicular to the
  crystal lattice which has lattice vectors
• a a 1 a b 0 a c 0
• b a 0 b b 1 b c 0
• c a 0 c b 0 c c 1

29

• Determining the reciprocal lattice vectors
• a b c /(a b c) b c /(volume)
• b c a /(a b c) c a /(volume)
• c a b /(a b c) a b /(volume)
• Since the diffraction condition is that
• a S h b S k c S l
• we can write
• S h a k b l c

S 2 a 1 b ie. Lattice vector (2,1)
a
a
b
b
30

• Ewald sphere construction.
• The identification
• S h a k b l c
• tells us that the diffraction condition will be
  satisified if S is a reciprocal lattice point.
• We know also that (from the definition of S)
• S 2 sin q/l.
• First draw a sphere with radius 1/l which
  touches the origin a reciprocal lattice point.

Here S intersect reciprocal lattice points
hence diffraction.
S
2q
O
31

• Predicting X-ray diffraction spots
• Use the Ewald sphere to predict where spots will
  be recovered.

Direction of scattered X-ray
S
S 2 sin q/l
s 1/l
q
Origin
q
s0 1/l
32

• Data collection
• Rotate the crystal in real space.
• - This rotates the real lattice a, b c.
• - Simultaneously rotates the reciprocal lattice
  a, b c.
• Can draw it as rotating the Ewald sphere on a
  fixed reciprocal lattice.
• As you rotate the Ewald sphere it will intersect
  some lattice points leave behind others.
• - Crystal rotating samples diffraction space.

S
1/l
2q
1/l
Origin
33

• Diffraction from a protein crystal
• A large number of reciprocal lattice points are
  sampled simultaneously.

34

• Vectors vector normals
• If you happen to meet an expression like
• r n constant
• then this should immediately remind you of
  vector-algebra where you learnt that this is an
  equation for a plane (in Cartesian coordinates)
• a x b y c z constant
• and that the vector (a,b,c) is the normal to the
  plane.
• - ie. for any constant there are a set of points
  in space which simultaneously satisfy this
  equation, defining a plane.
• - for a different constant you have a different
  plane.
• Hence n is a vector normal to some appropriately
  defined plane.

n is normal to the plane.
r
plane
35

• Relationship between S and Miller indices
• Allow the vector
• S ha kb lc
• to define the normal to a plane hence
• r S constant is the equation for a plane.
• Writing r x a y b z c (ie within the first
  unit cell of a lattice) we can ask what is the
  value of x where r crosses the a axis?
• r S x a S x a (ha kb lc) x h
  constant.
• - The plane defined by r S 1 crosses the a
  axis at x 1/h
• Likewise it crosses the b axis at x 1/k
  crosses the c axis at z 1/l.
• The equation r S 1 describes the Miller
  plane with index (h,k,l).

36

• Bragg's Law
• Choose the system of coordinates such that the
  phase angle for reflection from a plane through
  the origin is 0.
• The phase angle for the next lattice plane is 2p
  etc.
• Since S is perpendicular to the lattice plane
  with indices (h,k,l) then the distance from one
  plane to the next will be
• d r S/S
• If we go eg to r 1/h a (first point where
  (h,k,l) intersect a axis)
• d 1/h a S / S 1/h h /S 1/S
• Since S 2 sin q/l (defined earlier) 1/S
  d
• 2 sin q/l S 1/d or
• 2d sin q l which is Bragg's Law.

S
r
S
r
(2,1)
(1,3)
37

• Direction Amplitude
• Braggs Law, 2d sin q l, or equivalently
• S ha kb lc
• Tells you where to look to find the diffraction
  spots, but it doesnt tell you what the intensity
  of the spot is.
• The structure factor
• Funit cell(S) Sj fj(S) exp (2pi rj S)
• Tells you what the intensity is for the
  scattering vector S.

38

• Electron Density
• From before the structure factor amplitude was
  defined as
• Funit cell(S) ?r r (r) exp (2pi r S) dr
• Using the inverse Fourier Transform
• r (r) ?r F(S) exp (-2pi r S) dS
• In practice you make a discrete inverse Fourier
  Transform
• r (r) Shkl Funit cell(Shkl) exp (-2pi r
  Shkl)
• Shkl Fhkl exp (-2pi Fhkl) exp (-2pi r
  Shkl)
• Note that you measure the X-ray diffraction
  intensities
• Ihkl ? Fhlk(Shkl)2
• The fact that you cannot measure Fhkl directly
  is called
• THE PHASE PROBLEM