8
REGULAR LINEAR TRANSFORM OF VARIABLES
8
1
Theorem about the Fourier transform of functions related by
regular linear transform of variables
It is well–know from the experiments with the Fraunhofer diffraction that if we deform the diffraction
screen in a direction the diffraction pattern is deformed in the inverse proportional way (see Figure 1).
Theorem which relates the Fourier transforms of functions which can be made identical by a regular
linear transform of variables provides a more quantitative formulation of that phenomenon.
~ as a row
Similarly as in chapter 4 we consider the vector ~x as a column matrix and the vector X
matrix. Let the square matrix M = kmrs k specify a regular (i.e. det M 6= 0) linear transform of coordinates
~x0 = M(~x − ~x0 ).
(1)
Then the inverse transform is specified by the inverse matrix
M (−1) sr −1
M = mrs = det M ,
where Msr is the cofactor of the elements msr in the determinant det M and the inverse transform has
the form
~x = M−1 ~x0 + ~x0 .
(2)
Figure 1: The Fraunhofer diffraction from circular and elliptic apertures. The elliptic aperture resulted
from the horizontal extension of the circular aperture. As a consequence of that the diffraction pattern
is shortened in the horizontal direction by the same ratio.
8.1
The theorem
Let the functions f (~x) and f0 (~x) be related by
f (~x) = f0 M(~x − ~x0 ) .
(1)
Then their Fourier transforms are related by
~ = det M−1 exp −ik X
~ · ~x0 F0 XM
~ −1 .
F (X)
(2)
2
8
REGULAR LINEAR TRANSFORM OF VARIABLES
The proof is based on the mere substitution in the Fourier integral:
~
F (X)
N
∞Z
Z
···
= A
= AN
= AN
Z
−∞
∞Z
···
Z
−∞
∞Z
···
~ · ~x dN ~x =
f (~x) exp −ik X
~ · ~x dN ~x =
f0 M(~x − ~x0 ) exp −ik X
h
i dN ~x0
~ · M−1 ~x0 + ~x0
f0 (~x0 ) exp −ik X
=
|det M|
−∞
Z
∞Z
= |det M
−1
~ · ~x0 AN
| exp −ik X
= |det M
−1
~ · ~x F0 XM
~ −1 .
| exp −ik X
···
~ −1 · ~x0 dN ~x0 =
f0 (~x0 ) exp −ik XM
−∞
0
~ −1 ~x0 ) = (XM
~ −1 )~x0 .
In the proof we have used the relations det M = det 1M−1 and X(M
The regular linear transform 8(1) involves as special cases the translation (when M = I = M−1 and
0
~x 6= ~0), rotation and reflection (mirroring), respectively (when M is an orthogonal matrix, i.e. M−1 = MT ,
and ~x0 = ~0), and the linear deformation (when M is a general regular matrix). The translation will be
dealt with in the next chapter, the linear deformation is illustrated by examples in the rest of this
chapter. In particular we point out to Section 8.3 where the reciprocal lattice is revisited: theorem (2) is
there used to another proof that the algebraically defined reciprocal lattice is the Fourier transform of
the original lattice.
We now will pay attention just to the rotation and to the reflection, respectively, and use the proved
theorem to the formulation of the property of the Fourier transform which usually is taken as truism. In
the case of rotation and reflection, respectively, the matrix M is orthogonal, i.e. M−1 = MT , i.e. det M =
±1 (cf. e.g. [1], Sections. 96, 97). Then it follows from theorem (2) and from the assumption f (~x) = f0 (M~x)
~ = F0 (XM
~ T ) = F0 (MX
~ T )T . Thus, the rotation and the reflection, respectively, of a function
that F (X)
f and of its Fourier transform F is characterized by the same matrix M. If the object is turned by an
angle its Fourier transform is turned by the same angle. Moreover, if the object has a symmetry related
~ = F (MX
~ T )T .
with rotation, i.e. if f (~x) = f (M~x), its Fourier transform has the same symmetry F (X)
The same may be said about the reflection symmetry. If f (~x) has a reflection symmetry with respect
~ has also this reflection
to an object (a straight–line in E2 , a plain in E3 ), the Fourier transform F (X)
symmetry. Expressed by a formula:
If
M−1 = MT ,
then
f (~x) = f (M~x)
⇐⇒
~ = F (MX
~ T )T .
F (X)
(3)
Still in other words: A function f (~x) is invariant to an orthogonal transform if and only if its Fourier
~ is invariant to that transform. As the central symmetry f (~x) = f (−~x) may be considered
transform F (X)
as a special case of symmetry with respect to the orthogonal transform, the theorem 6.1(1) may be
considered as a special case of (3).
If the function f (~x) is real and possesses the reflection symmetry the theorem 6.2(1) enables us to
~ is real at points of the straight–line (in E2 ) and plane (in E3 ),
say that its Fourier transform F (X)
respectively, which is perpendicular to the straight–line or plane of reflection and passes through the
origin. It is worth–while to have it in mind while checking calculations of the Fourier transform of real
functions which do not posses the central symmetry but have a reflection symmetry (e.g. characteristic
functions of regular polygons with an odd–fold symmetry or of the regular tetrahedron.)
8.2
Example: The Fourier transform of characteristic function of a general
parallelogram
Let us calculate the Fourier transform of the characteristic function of the parallelogram in Figure 2. We
use for that the Fourier transform of characteristic function of the square with unit side which we have
already calculated in 1.3(8,9)):
8
REGULAR LINEAR TRANSFORM OF VARIABLES
3
x2
x2
W2 (- _12 , _12 )
w2
V2 (v21 , v22 =v12 )
W1 ( _12 , _12 )
v1
v2
w1
x1
0
W3
V1 (v11 ,v12)
x1
0
W4
V3
V4
f0 ( x ) = rect x1 rect x2
f( x )=f0 ( M x )
Figure 2: An example of functions which can be made identical by regular linear transform of variables:
the unit square and general parallelogram.
f0 (~x) = rect (x1 ) rect (x2 ),
~ = A2 sin(kX1 /2) sin(kX2 /2) .
F0 (X)
kX1 /2
kX2 /2
(1)
As both the parallelepiped and the square have their centers at the origin it is ~x0 = ~0 and
f (~x) = f0 (M~x).
(2)
We take advantage of the relations between the position vectors ~v1 and w
~ 1 , and ~v2 and w
~ 2 of the vortices
of the two figures and calculate the four elements of the matrix M and M−1 , respectively (cf. Figure 2).
Taking the vectors ~v1 , ~v2 , w
~ 1, w
~ 2 as column matrices we have
w
~1
w
~2
= M~v1 ,
= M~v2 ,
or
= M−1 w
~ 1,
−1
= M w
~ 2.
~v1
~v2
(3)
(−1)
We will calculate the elements mik of the inversion matrix M−1 , i.e. we use the second possibility in
(3), because their evaluation is easier than that of mik Rewriting the second possibility in coordinates
(−1)
we have two systems of two equation for mik :
2v11
2v21
(−1)
(−1)
= m11 + m12 ,
(−1)
(−1)
= −m11 + m12 ,
2v22
2v12
= 2v12
(−1)
(−1)
= m21 + m22 ,
(−1)
(−1)
= −m21 + m22 .
Hence,
(−1)
= v11 − v21 ,
= v11 + v21 ,
m11
(−1)
m12
(−1)
m21
(−1)
m22
= 0,
= 2v12 ,
so that
M−1
v11 − v21
=
0
v11 + v21 ,
2v12
det M−1 =
M=
1
v11 −v21
0
+v21
− 2v12v11
(v11 −v21 ) ,
1
2v12
1
= 2v12 (v11 − v21 ).
det M
(Notice that det M−1 is equal to the area of our parallelogram.)
The characteristic function of the parallelogram has the form
x1
v11 + v21
x2
f (~x) = f0 (M~x) = rect
− x2
rect
v11 − v21
2v12 (v11 − v21 )
2v12
(4)
(5)
(6)
4
8
REGULAR LINEAR TRANSFORM OF VARIABLES
and its Fourier transform is
~
F (X)
~ −1 ) =
= |det M−1 |F0 (XM
sin k2 X1 (v11 − v21 ) sin k2 [X1 (v11 + v21 ) + X2 2v12 ]
2
= A 2v12 (v11 − v21 ) k
.
k
2 X1 (v11 − v21 )
2 [X1 (v11 + v21 ) + X2 2v12 ]
(7)
Figure 3: The Fraunhofer diffraction patterns by squared and rhomboidal apertures. The intensity dis~ 2 and |F (X)|
~ 2 , where F0 (X)
~ is expressed by (1) and
tributions are characterized by functions |F0 (X)|
~
F (X) by (7). If we consider the parallelogram as a deformation of the square, it is symptomatic for
its diffraction pattern that the arms of the diffraction pattern remain perpendicular to the sides of the
parallelogram (the so–called Abbe theorem discussed in Section 11.3).
It can be seen from Figure 3 that the arms of the Fraunhofer diffraction patterns both from the
square and from the parallelogram are perpendicular to the straight–line boundaries of the diffraction
aperture. This illustrates the so–called Abbe theorem, which will be dealt with in Chapter 11. Now we
just will analyze expressions (6) and (7). We put the arguments of the functions rect ( ) in (6) equal to
±1/2 and get the equations of the straight–lines bearing the sides of the parallelogram. The slopes of
these straight–lines are
k1 =
2v12
,
v11 − v21
k2 = 0.
Putting zero the arguments of sines in (7) we get the equations of the straight–lines corresponding to
the arms of the diffraction pattern. Their slopes are
k10 = ∞,
k20 = −
v11 + v21
.
2v12
We see that the arms of the diffraction pattern are perpendicular to the sides of the parallelogram
(k1 = −1/k20 ). (This is advantageous to have in mind when we have to find mutual orientation of the
diffraction pattern and of the aperture.)
8.3
Example: Reciprocal lattice revisited
We have shown in Section 4.3 that the Fourier transform of a lattice function is proportional to the
lattice function of the reciprocal lattice. The proof of that was relatively simple in the case of the lattice
8
REGULAR LINEAR TRANSFORM OF VARIABLES
5
function of single variable (Section 4.3.1), but rather complicated in the case of more dimensional lattices.
Theorem 8.1 offers another proof of the statement which someone may consider more simple. Therefore,
we will give it here. At that we give further example how to find the transformation matrix M.
We choose the lattice function of the cubic lattice with unit lattice parameter as the function f0 (~x):
X
δ(~x − ~n).
(1)
f0 (~x) =
~
n∈inf
Here, again, ~n ∈ inf means that all the components n1 , n2 , . . . , nN of the multiindex ~n take all integer
values.
The only fact which we take over from Section 4.3 is the statement that the lattice function of single
variable
f0 (x) =
∞
X
δ(x − n)
(2)
n=−∞
has the Fourier transform
F0 (X) =
∞
2π
1 X
δ X−
h ,
B
k
(3)
h=−∞
as it follows from 4.3(5) for a = 1.
We exploit it while calculating the Fourier transform of the lattice function (1) of N variables.
According to A.5(2), every Dirac distribution of N variables in (1) may be factorized into the product
of N Dirac distributions of single variable, so that
X
f0 (~x) =
N
X Y
δ(~x − ~n) =
δ(xr − nr ).
(4)
~
n∈inf r=1
~
n∈inf
The N –multiple series of the lattice function (1) can be rewritten into the product of N latice functions
of single variable by interchange of summation and multiplication in (4)
f0 (~x) =
X
δ(~x − ~n) =
∞
X
N
Y
δ(xr − nr ),
(5)
r=1 nr =−∞
~
n∈inf
The kernel of the Fourier transform can also be factorized,
N
Y
~ · ~x =
exp −ik X
exp −ikXr xr ,
r=1
and, therefore, the Fourier transform of the lattice function (1) has also the form of the product of the
Fourier transforms of lattice functions of single variable:
( ∞
)
N
Y
X
~ =
F0 (X)
FT
δ(xr − nr ) ,
(6)
r=1
nr =−∞
Using (3) we have
~ =
F0 (X)
N
Y
1
BN
r=1
∞
X
hr =−∞
2π
δ Xr −
hr .
k
(7)
Exchanging again the order of multiplication and summation we get the resulting shape of the Fourier
~
transform F0 (X):
~ = 1
F0 (X)
BN
∞
N
X
Y
2π
1 X
2π ~
~
δ Xr −
hr = N
δ X−
h ,
k
B
k
r=1
hr =−∞
(8)
~
h∈inf
(Of course, this is in agreement with 4.3(17) for ar = a +
r = 1, r = 1, 2, . . . , N , VU = a1 a2 · · · aN = 1.)
As we know from 4.1(1) a general (i.e. non orthogonal) N –dimensional lattice has the lattice function
6
8
f (~x) =
X
REGULAR LINEAR TRANSFORM OF VARIABLES
δ ~x − n1~a1 − n2~a2 − · · · − nN ~aN .
(9)
~
n ∈ inf
We will derive its Fourier transform with the use of the theorem 8.1 and the result (8). We must, however,
find the deformation matrix M in 8.1(1).
To gain experience with the treatment of the formulae we find first the matrix for the case of the
lattice function of the one–dimensional lattice with the parameter a
f (x) =
∞
X
δ(x − na).
(10)
n=−∞
Such a lattice may be considered as deformed lattice with unit parameter (2). In theorem 8.1 the matrix
M characterizing the deformation multiplies the variable ~x. Hence, we have to rewrite the function (10)
into the similar shape:
f (x) =
∞
X
δ(x − na) =
n=−∞
∞
x
x
1 X x
1
.
−n =
δ
−n =
f0
δ a
a
|a| n=−∞
a
|a|
a
n=−∞
∞
X
(11)
In the case of the one–dimensional lattice the matrix M is a1 = a+ . From theorem 8.1 it then follows the
Fourier transform of the lattice function (10) in the shape
∞
∞
1 X
2π
1 X
2π h
F (X) = F0 (Xa) =
δ aX −
h =
δ a X−
=
B
k
B
k a
h=−∞
h=−∞
∞
X
2π h
1
δ X−
=
B|a|
k a
(12)
h=−∞
in agreement with 4.3(5).
Similarly the general N –dimensional lattice may be regarded as deformed orthogonal lattice (1) with
unit parameter. Matrix M of the deformation in question can be found in the similar way as in the one–
dimensional case. Alike in Section 4.1 we consider the multiindex ~n(n1 , n2 , . . . , nN ) to be the column
matrix and form the square matrix A from the coordinates ars of the basis vectors ~ar in orthonormal
basis. Then the following treatment of the lattice function (9) can be made:
f (~x)
=
X
δ ~x − n1~a1 − n2~a2 − · · · − nN ~aN
=
~
n ∈ inf
=
X
δ x1 − n1 a11 − n2 a21 − · · · − nN aN 1 , x2 − n1 a12 − n2 a22 − · · · − nN aN 2 , . . .
~
n ∈ inf
. . . , xN − n1 a1N − n2 a2N − · · · − nN aN N =
X
X =
δ ~x − AT ~n =
δ AT (AT )−1 ~x − ~n
=
~
n ∈ inf
~
n ∈ inf
X
= det A−1 δ (AT )−1 ~x − ~n = det A−1 f0 (AT )−1 ~x .
(13)
~
n ∈ inf
−1
Thus, the matrix AT
= A+ the rows of which are the coordinates a +
a+
st of the basis vectors ~
s of the
algebraically defined reciprocal lattice plays the role of the deformation matrix M (cf. 4.2(1), 4.2(4)).
According to the theorem 8.1 and with the use of (8) the Fourier transform of the lattice function
(9) has the form (the multiindex ~h(h1 , h2 , . . . , hN ) is the row matrix (cf. Section 4.3) and the absolute
value of the outer product is the volume of the unit cell, i.e. |det A| = VU ):
8
REGULAR LINEAR TRANSFORM OF VARIABLES
7
x2
x2
W1 (12 (a11+ a21), 12 (a12+ a22))
V( , )
_ _
1
1
1 2 2
1
2
1
2 2
i
i
1
2 1
0
x1
a2
0
1
2
a1
x1
(b)
(a)
Figure 4: The ”unit” square (a) and a general parallelogram (b) specified by position vectors of the
centres of sides.
~
F (X)
2π ~
1 X
T
T
~
~
= F0 (XA ) = N
δ XA −
h =
B
k
~
h ∈ inf
−1 1 X
~ − 2π ~h AT
=
X
AT
=
δ
BN
k
~
h ∈ inf
X 1
2π ~ T −1
1
~
=
δ X−
h A
=
B N |det A|
k
~
h ∈ inf
X 2π ~ +
1
1
~
δ X−
hA
=
=
B N |det A|
k
~
h ∈ inf
1 1 X
~ − 2π h1~a + + h2~a + + · · · + hN ~a +
=
δ
X
,
1
2
N
B N VU
k
(14)
~
h ∈ inf
which is expression 4.3(17).
Thus, we have given another and independent proof of the fact, that the Fourier transform of the
lattice function is proportional (with the coefficient B1N V1U ) to the lattice function of the reciprocal
lattice with the reciprocal constant K = 2π
k .
8.4
Example: The Fourier transform of characteristic function of a general
N –dimensional parallelepiped
We shall now calculate the Fourier transform of the characteristic function of a general N –dimensional
parallelepiped. This may be useful for the formulation of the sampling theorem in EN , if it is necessary
to sample at points which do not form an orthogonal net. We will not proceede as in Section 8.2 where
the vortices have been used for the specification of the deformation of the square to the parallelogram.
Now we use for it the centres of faces (sides in Figure 4). This enables us to use the deformation matrix
−1
M = AT
derived in Section 8.3 (cf. 8.3(13)).
A general N –dimensional parallelepiped can be obtained by linear deformation of the N –dimensional
cube of unit edge. We shall therefore choose the characteristic function of that cube as the function f0 (~x)
in theorem 8.1. If the centre of the cube is at the origin O of the Cartesian coordinate system and if the
axes xj pass through the centres of the faces (see Figure 4(a)) the characteristic function of the cube is
8
8
REGULAR LINEAR TRANSFORM OF VARIABLES
f0 (~x) =
N
Y
rect (xj ).
(1)
j=1
Its Fourier transform is the product (cf. Section 1.3.5)
~ = AN
F0 (X)
N
Y
sin
j=1
1
2
kXj
.
1
2 kXj
(2)
Let us denote by ± 21 ~aj the position vectors of the centres of faces of the general parallelepiped (Figure
4(b)) The vectors ~aj , j = 1, 2, . . . , N , form a basis in EN and it is evident that the N –dimensional cube
turns to the parallelepiped under consideration by the same linear regular transformation as the Cartesian
basis ~ıj , j = 1, . . . , N , to the basis ~aj , j = 1, 2, . . . , N . In the same way as in the previous Section 8.3 we
−1
constitute the matrix A from the coordinates ajk of the vectors ~aj , calculate the matrix AT
= A+
(which is the deformation matrix M); and make up the reciprocal vectors ~a +
j . The characteristic function
f (~x) of the general parallelepiped is — according to 8.3(13) and (1) — the product
N
Y
+
+
f (~x) = f0 A+ ~x = f0 ~a +
·
~
x
,
~
a
·
~
x
,
.
.
.
,
~
a
·
~
x
=
rect ~a +
x .
1
2
j ·~
N
(3)
j=1
(The last expression (3) follows also from the fact that ~a +
x = ±1/2 are equations of planes of the
j ·~
faces of the parallelepiped.) According to the theorem 8.1 the Fourier transform of the characteristic
function (3) is the product
N
~ = |det A| F0 (X
~ AT ) = AN VU F0 X
~ · ~a + , X
~ · ~a + , . . . , X
~ · ~a
F (X)
1
2
N = A VU
+
N sin
Y
j=1
1
2
1
2
~ · ~aj
kX
~ · ~aj
kX
. (4)
~ 2 decreases most slowly with increasing X in the
It is evident from it that the squared modulus F (X)
~ · ~aj = 0, i.e. in the directions perpendicular to the position vectors of the face centres and
directions X
hence to the faces.
Particularly (N = 2), the parallelogram in Figure 4(b) with vectors ~a1 (a11 , a12 ), ~a2 (a21 , a22 ) has the
characteristic function
a22 x1 − a21 x2
−a12 x1 + a11 x2
f (x1 , x2 ) = rect
rect
.
(5)
a11 a22 − a12 a21
a11 a22 − a12 a21
(We have used expression 4.2(5),(6) for the vectors ~a +
j in (3).) Its Fourier transform is
1
sin 2 k(X1 a11 + X2 a12 ) sin 21 k(X1 a21 + X2 a22 )
2
F (X1 , X2 ) = A |a11 a22 − a12 a21 |
.
1
1
2 k(X1 a11 + X2 a12 )
2 k(X1 a21 + X2 a22 )
(6)
References
[1] Bydžovský B.: Úvod do teorie determinantů a matic a jejich užití. Jednota československých matematiků a fyziků, Praha 1947.