# Fourier Transform - web.cecs.pdx.edu web.cecs.pdx.edu by dfhdhdhdhjr

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```									                                     Reciprocal Space
Fourier Transforms

Outline
 Introduction to reciprocal space
 Fourier transformation
 Some simple functions
• Area and zero frequency components
• 2- dimensions
 Separable
 Central slice theorem
 Spatial frequencies
 Filtering
 Modulation Transfer Function

22.56 - lecture 3, Fourier imaging
Reciprocal Space

real space                     reciprocal space

22.56 - lecture 3, Fourier imaging
Reciprocal Space
In[1]:=   face   , 0 , 0 , 0 , .2 , .4 , .7 , .9 , 1.2 , 2. , 2.5 , 3.1 , 3.25 , 3.2 , 3.1 , 2.9 , 2.9 , 3 , 3.1 , 2.8 ,
0
2.9 , 2.8 , 2.7 , 2.6 , 2.8 , 2.9 , 3 , 2.9 , 2.7 , 2.3 , 2.1 , 1.7 , 1.4 , 1.2 , 1 , .8 , .8 , .7 ,
.7 , .65 , .6 , .5 , .4 , .3 , .2 , .1 , 0 , 0 , 0 , 0 , 0 ;

In[33]:=   ListPlot     face , PlotJoine d      True ,   Axes   False   , PlotStyle        Thicknes s   0.01 

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5

50         100      150      200       250
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-10

In[3]:=   periodic        Join                                     0 ;
face , face , face , face , face ,  

In[5]:=   f  Fourier      periodic   ;

In[6]:=   ListPlot       RotateLeft        , 128 , PlotJoined
Re f                           True ,   PlotRange       All ,
PlotStyle          Thick nes s  0.01 

22.56 - lecture 3, Fourier imaging
Reciprocal Space

20

15

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50   100   150    200   250
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real

7.5

ListPlot   RotateLe ft    Im , 128 , PlotJoine d
f                         True ,     5
PlotRange      All ,                                              2.5
PlotStyle      Thicknes s  0.01 
50   100   150    200   250
-2.5

-5

-7.5              imaginary
22.56 - lecture 3, Fourier imaging
Reconstruction
filter     :  Table    n, 1 , If   256  n, 1 , 0 ,  , 1 , 256 
n_            If i            i                      i           ;
expand   :  ListPlot
n_                      Re Fourier  filter
Take               f         , 54 ,
n
  PlotJoined    True , PlotRange      All , Axes   False , PlotStyle     Thicknes s   0.01 ,
Dis playFunction       Identity  

8 Fourier components

16

32

64

128
22.56 - lecture 3, Fourier imaging
Fourier Transforms

For a complete story see:
Brigham “Fast Fourier Transform”
Here we want to cover the practical aspects of Fourier Transforms.

Define the Fourier Transform as:

G(k)  g(x)      g(x)eikx dx


There are slight variations on this definition (factors of π and the

Also recall that
eikx  cos(kx)  isin(kx)

22.56 - lecture 3, Fourier imaging
Reciprocal variables

k is a wave-number and has units that are reciprocal to x:
x -> cm
k -> 2π/cm
So while x describes a position in space, k describes a spatial
modulation.
Reciprocal variables are also called conjugate variables.
1
0.75
0.5
0.25

-4                        -2       -0.25              2                 4
-0.5
-0.75
-1
1
0.75
0.5
0.25

-4                        -2       -0.25              2                 4
-0.5
-0.75
-1

2
, k 

Another pair of conjugate variables are time and angular frequency.
22.56 - lecture 3, Fourier imaging
Conditions for the Fourier Transform to Exist

The sufficient condition for the Fourier transform to exist is that the
function g(x) is square integrable,


2
   g(x) dx  


g(x) may be singular or discontinuous and still have a well defined
Fourier transform.

22.56 - lecture 3, Fourier imaging
The Fourier transform is complex

The Fourier transform G(k) and the original function g(x) are both in
general complex.
g(x)  Gr (k) iGi (k)
The Fourier transform can be written as,

g(x)  G(k)  A(k)ei(k )
2
A  G  Gr  Gi2
A  amplitude spectrum, or magnitude spectrum
  phase spectrum
2
A2  G 2  Gr  Gi2  power spectrum

22.56 - lecture 3, Fourier imaging
The Fourier transform when g(x) is real

The Fourier transform G(k) has a particularly simple form when g(x)
is purely real               
Gr (k)   g(x)cos(kx)dx


Gi (k)     g(x)sin(kx)dx

So the real part of the Fourier transform reports on the even part of
g(x) and the imaginary part on the odd part of g(x).

22.56 - lecture 3, Fourier imaging
The Fourier transform of a delta function

The Fourier transform of a delta function should help to convince
you that the Fourier transform is quite general (since we can build
functions from delta functions).

 x      x eikx dx

The delta function picks out the zero frequency value,

 x  eik 0  1
x  1

x                          k
22.56 - lecture 3, Fourier imaging
The Fourier transform of a delta function

So it take all spatial frequencies to create a delta function.
In[1]:=   delta    , x_   Sum
n_            Cos  x ,   n , n, 1 
k      k,           ;

In[7]:=   Plot  delta  , x ,  ,  3 , 3  PlotRange
1         x         ,              All ,
PlotStyle      Thicknes s 0.01 

60

40
QuickTime™ an d a
Anima tion d ecompressor
are need ed to see this picture.                      20

-10   -5        5   10

22.56 - lecture 3, Fourier imaging
The Fourier transform

The fact that the Fourier transform of a delta function exists shows
that the FT is complete.

The basis set of functions (sin and cos) are also orthogonal.


 cos(k1x)cos(k2 x)dx   (k1  k2 )

So think of the Fourier transform as picking out the unique spectrum
of coefficients (weights) of the sines and cosines.

22.56 - lecture 3, Fourier imaging
The Fourier transform of the TopHat Function

Define the TopHat function as,
 x  1
1;
g(x)  
 x  1
0;


The Fourier transform is,              G(k)     g(x)eikx dx

which reduces to,
1
sin(k)
G(k)  2  cos(kx)dx  2         2sinc(k)
0                  k

22.56 - lecture 3, Fourier imaging
The Fourier transform of the TopHat Function

For the TopHat function                         x  1
1;
g(x)  
 x  1
0;

1
The Fourier transform is, G(k)  2  cos(kx)dx  2 sin(k)  2sinc(k)
0                  k
2

1.5

1

0.5

-20   -10              10   20

-0.5

22.56 - lecture 3, Fourier imaging
The Fourier reconstruction of the TopHat Function

2

1.5

1

0.5

-20        -10                         10         20

-0.5

In[33]:=   square  , x_  : 
n_
2  Sum Sin  k   Cos  x ,  , 1 , n 
2     k           k      k         ;

In[36]:=   Table  Plot square  , x ,  ,  2 , 2 
n          x         ,
PlotRange      All , PlotStyle     Thick nes s   0.01 ,
 , 0 , 128 
n           

22.56 - lecture 3, Fourier imaging
The Fourier transform of a cosine Function

Define the cosine function as,

g(x)  cos(k0 x)

where k0 is the wave-number of the original function.
The Fourier transform is,         
G(k)   cos(k0 x)eikx dx
which reduces to,                

G(k)         cos(k0 x)cos(kx)dx   (k  k0 )  (k  k0 )

cosine is real and even, and so the Fourier transform is also real and
even. Two delta functions since we can not tell the sign of the
spatial frequency.

22.56 - lecture 3, Fourier imaging
The Fourier transform of a sine Function

Define the sine function as,

g(x)  sin(k0 x)

where k0 is the wave-number of the original function.
The Fourier transform is,         
G(k)   sin(k0 x)eikx dx
which reduces to,                

G(k)  i  sin(k0 x)sin(kx)dx  i (k  k0 )  (k  k0 )

sine is real and odd, and so the Fourier transform is imaginary and
odd. Two delta functions since we can not tell the sign of the spatial
frequency.

22.56 - lecture 3, Fourier imaging
Telling the sense of rotation

Looking at a cosine or sine alone one can not tell the sense of
rotation (only the frequency) but if you have both then the sign
is measurable.

22.56 - lecture 3, Fourier imaging
Symmetry

Even/odd
if g(x) = g(-x), then G(k) = G(-k)
if g(x) = -g(-x), then G(k) = -G(-k)
Conjugate symmetry
if g(x) is purely real and even, then G(k) is purely real.
if g(x) is purely real and odd, then G(k) is purely imaginary.
if g(x) is purely imaginary and even, then G(k) is purely imaginary.
if g(x) is purely imaginary and odd, then G(k) is purely real.

22.56 - lecture 3, Fourier imaging
The Fourier transform of the sign function

The sign function is important in filtering applications, it is defined
as,                             1; x  0
sgn(x)  
1;
 x  0

The FT is calculated by expanding about the origin,

2
sgn x   i
k

22.56 - lecture 3, Fourier imaging
The Fourier transform of the Heaviside function

The Heaviside (or step) function can be explored using the result of
the sign function            1
(x)  1 sgn( x)
2
The FT is then,

1                
1       1        
x    1 sgn( x)    sgn( x)

2                
2       2        

i
x    (k) 
k

22.56 - lecture 3, Fourier imaging
The shift theorem

Consider the conjugate pair,
gx   G(k)

what is the FT of                 g x  a 


g x  a      g(x  a)eikx dx

rewrite as,                  
    g(x  a)eik (xa)eika d(x  a)

The new term is not a function of x,

gx  a  eikaG(k)
so you pick up a frequency dependent phase shift.

22.56 - lecture 3, Fourier imaging
The shift theorem
In[18]:=   d :  Table   Table Cos  x  n 2 Pi 16 ,  , 0 , 15 
2                    n          ,
 ,  10 , 10 , 20 63 
x                      ;

In[35]:=   f  Table RotateLe ft   Fourier       
d  n  , 32 ,  , 1 , 16 
n          ;

In[19]:=   ListPlot3D   , 
d   PlotRange           
   1 , 1  Mesh   False
,                 

2
15
0

-2
10

20
5
1                                                                                    40

0.5                                                                                              60
0                                               15
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-1                                          10

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60
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22.56 - lecture 3, Fourier imaging                                                  60
The similarity theorem

Consider the conjugate pair, gx   G(k)

what is the FT of                      g ax 

                                      k
1           i ax
ikx
g ax                g(ax)e          dx      g(ax)e a d(ax)
                         a 

1 k
g ax   G( )
a a
so the Fourier transform scales inversely with the scaling of g(x).

22.56 - lecture 3, Fourier imaging
The similarity theorem
In[3]:=   square  Table 
Table    64  n , 0 , If   64  n , 0 , 1 ,  , 1 , 128 
If x                 x                      x           ,
 , 1 , 16 
n          ;

In[13]:=   f  Table 
RotateLe ft   Fourier   RotateLe ft    square    
 n  , 64 , 64 ,
 , 1 , 16 
n          ;

1                                                          3
0.8                                                          2
0.6                                                  15                                           15
0.4                                                         1
0.2                                                             0
0                                           10                                           10

50                         5                              50              5

100                                                     100

22.56 - lecture 3, Fourier imaging
The similarity theorem
8
1
6
2a
4

2

-3      -2        -1          1        2        3

-a           a
-2

1
a

1
0.8
0.6                                           15   3
0.4
0.2                                               2
0                                    10                                                        15
1
0
50                                                                               10
5

100
50                    5

100
22.56 - lecture 3, Fourier imaging
Rayleigh’s theorem

Also called the energy theorem,

               
2               2
   g(x) dx       G(k) d(k)
              
The amount of energy (the weight) of the spectrum is not changed
by looking at it in reciprocal space.

In other words, you can make the same measurement in either real or
reciprocal space.

22.56 - lecture 3, Fourier imaging
The zero frequency point

Also weight of the zero frequency point corresponds to the total
integrated area of the function g(x)
                          
g(x) k0        g(x)eikx dx             g(x)dx
                k0       

22.56 - lecture 3, Fourier imaging
The Inverse Fourier Transform

Given a function in reciprocal space G(k) we can return to direct
space by the inverse FT,

1 
g(x)  1G(k)       G(k)eikx dk
2 

To show this, recall that                 G(k)     g(x)eikx dx


1                  1        
 G(k)eikx 'dk   dx g(x)  eik (x'x )dk
2                2      
2 ( x' x)
g( x')

22.56 - lecture 3, Fourier imaging
The Fourier transform in 2 dimensions

The Fourier transform can act in any number of dimensions,

 
ik y y ik x x
g(x, y)x,y    g(x, y)e       e        dxdy
 

It is separable
g(x, y)x,y  g(x, y)x g(x, y)y

and the order does not matter.

22.56 - lecture 3, Fourier imaging
Central Slice Theorem

The equivalence of the zero-frequency rule in 2D is the central slice
theorem.
              ik y y ik x x
g(x, y)x,y                          g(x, y)e        e        dxdy
k x 0        
or                                                                                     k x 0

 ik y       
g(x, y)x,y                     e   y
dy  g(x, y)dx
k x 0                               k x 0

So a slice of the 2-D FT that passes through the origin corresponds
to the 1 D FT of the projection in real space.

22.56 - lecture 3, Fourier imaging
Filtering

We can change the information content in the image by manipulating
the information in reciprocal space.

Weighting function in k-space.

22.56 - lecture 3, Fourier imaging
Filtering

We can also emphasis the high frequency components.

Weighting function in k-space.

22.56 - lecture 3, Fourier imaging
Modulation transfer function

i(x, y)  o(x, y)  PSF(x, y)  noise

I(k x , k y )  O(k x , k y ) MTF(kx , k y ) noise

22.56 - lecture 3, Fourier imaging

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