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# Fourier transform.ppt

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# Fourier transform.ppt

Fourier transform

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### Fourier transform.ppt

1. 1. Fourier Transform and Phase Determination Fourier Series and Synthesis Direct Methods Patterson Methods
2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 60. Patterson Methods Fig.6 equivalent positions of P21/c from International Table
61. 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. 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. 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. 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.