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Euler’s formula:
eix  cos( x)  i sin( x)
Richard Feynman: “…one of the most remarkable, almost astounding, formulas in all of
mathematics.” (Feynman Lectures on Physics, V 1)
Proof by Taylor expansion:
Taylor expansion about x = 0:
f ( x)  f (0)  f (0) x 
Expansion of eix:
e
ix
f (0) 2
f n  (0) n
x 
x 
2!
n!
2
3
5


ix 
ix  ix  ix 
 1  ix 




2!
3!
4!
5!
4
6

 

x
x
x
x3 x5 x7
 1 


   i x 


 
2! 4! 6!
3! 5! 7!

 

 cos( x)  i sin( x)
2
Electromagnetic wave equation for empty space:
 2E
 E   0 0 2  0
t
Monochromatic, plane wave solution, traveling an arbitrary direction:
2
E (r, t )  A cosk  r  t   
where :
 is the phase at r  0, t  0
r   x, y, z  is the position v ector
k  k x , k y , k z  is the wave vector in radians/me ter
Note that
and k 
kx
, etc. are direction cosines of the propagatio n direction
k
2

Hence : E r, t   A cos( k x  x  k y  y  k z  z    t )
Switching to complex notation, using Euler’s formula we get:


E  ReA exp i k x x  k y y  k z z    t  
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where the Real Part function will generally be assumed. Note that only linear operations
and the operation A  E  E  will give valid answers with complex notation. The real
part must be extracted for non-linear operations. Also note that taking the complex
conjugate of E i   i  , produces a wave traveling in the opposite direction.
2
This equation is redundant, in that
k
2
x
 k y2  k z2   k  k , so that kz can be determined
from kx and ky. Since we are interested in describing plane waves in x-y planes, we will
leave off the kz term.
We will be adding planes waves together on x-y planes. Since it doesn’t matter when we
do this, as long as it is at the same time for all waves (since only the phase of a plane
wave changes with time), we will assume that t = 0.
We will also allow the amplitude, A, to be complex and have it include the phase angle,
.
One last modification is to define variables with more geometric meaning. Hence we
k
cos  x
u x 
2

will define
, where  x ,  y are the x, y direction cosines of the wave.
k y cos  y
v

2

These variables are spatial frequencies, with dimensions of cycles per length in the x-y
plane.
With all this, the equation for a plane wave can be written as:
E  A expi 2 ux  vy
On an x-y plane, the spatial period between maxima is 1/u in the x-direction and 1/v in
the y-direction.
Plane Wave Spectrum of a Distribution:
Consider the Fourier Transform pair:
Au , v    E  x, y  exp  i 2 ux  vy dx dy
E  x, y    Au , v  exp i 2 ux  vy du dv
Since expi2 ux  vy represents a plane wave traveling in the direction with direction
cosines u , v , we can interpret this transform as converting between a (harmonic) field
distribution on an x-y plane and a set of plane waves traveling in various directions with
various amplitudes, which add up to the distribution on the given plane – i.e., the plane
wave spectrum of the distribution.
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The integrals in the Fourier Transform go from - to , but direction cosines have a valid
u v
range of –1 to 1. What is the meaning of values of ,
 1?
 
These values represent spatial frequencies in the field distribution, E(x,y), that have
spatial periods less than , and hence cannot be formed by (and cannot create)
propagating waves. These field components are known as evanescent waves, and their
amplitude decays exponentially in the +z direction.
Since the plane wave spectrum consists solely of propagating waves, the arguments of A
1
never exceed  .

If we use the Discrete Fourier Transform on these functions, the Plane Wave Spectrum,
1
2
A, will consist of a collection of waves where u, v  0,  ,  ,  That is, the plane
N
N
wave components will have 0, 1, 2, etc cycles within the discrete array of N by N
samples.
n=1 component of
discrete PWS
D
Tx
phi
lam
Let D be equal to N x, where x is the sample spacing of the field distribution, E(x,y)
and N is the number of samples in the x, y directions. Taking the Fourier Transform of
E, we get the discrete PWS, A(u,v). The elements of the PWS have spatial frequencies of
n/D, where n = 0, 1, …, N/2. Illustrated is the n=1 component of the PWS of a field
distribution over the aperture, D.
n cos x 
n
 cos x  
 sin   , so the nth component of the
In general, u n  
D

D
PWS (in the x-z plane) propagates at an angle of sin-1(n/D) from the z-axis.
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The only approximation made so far is to ignore vector addition of the wave’s E-fields.
This only becomes significant at fairly large angles, and then only in the plane of the Evector. Waves traveling in a plane  to the E-vector add scalarly.
This allows us to calculate the PSF from a pupil field distribution by the following
means: Suppose that the exit pupil is distance f from the image plane. Then, the pupil
wavefront map is defined to be the difference between the phase of the ray field in the
exit pupil and the phase of a sphere with radius f centered on the image plane.
Conceptually, we can replace the spherical component of the wave field with an ideal
lens of focal length f. A plane wave component (of the reduced pupil field) will be
focused by this lens to a point on the image plane where the ray in the direction of the
component, passing through the center of the lens, strikes the image plane.
D
x
phi
f

 n  
Hence, xn  f tan( n )  f tan  sin 1    . This gives us the scale of the discrete PSF
 D 

calculated from the pupil function.
At this point, it is usual, for convenience, to make the paraxial assumptions sin = tan =
n
angle and get xn 
, however if more accuracy was required (within the limits of
D
scalar diffraction), a non-uniform array could be constructed and re-sampled to a uniform
array at the image plane.
Of course, the PSF is a detected function, hence is proportional to the power in the
waves, and hence is proportional to |E|2.
This justifies the previous blatant assertion that the PSF = |F{P}|2, where P is the pupil
function, as well as giving a way to determine the sample spacings.
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Sampling Issues:
The largest (angle) component in the PWS is cos x  
N

2  
, where
N
2D
d is the sample spacing in the pupil. Hence any sampling with spacing less than /2
will result in part of the output array representing evanescent waves, which cannot
contribute to field distributions further down the z-axis. In practice, spacings anywhere
near this limit result in propagation directions at large angles to the z-axis, violating the
scalar assumption (as well as any paraxial ones!).
The spacing on the image plane is x = /D. Since both  and D are fixed, there would
seem to be no way to increase the sampling resolution of the PSF. D, however, is the size
of the sampled array, not the size of the pupil. Hence, to increase the sampling
resolution of the PSF, we embed the pupil inside a larger array whose values are zero
outside the pupil.
Issues with the FFT function:
Implementations of the FFT produce an array in which the zeroth components are at the
beginning, not the middle, of the array. In this, they differ greatly from the continuous
Fourier definitions. Care must be taken to get results that approximate the continuous
calculation.
A plot of a real, symmetric function (g = gaussian):
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Using the MATLAB command G = fft(g), we should get a real, symmetric transform,
according to Fourier theory. Instead, we get a complex function:
The FFT doesn’t consider a ‘centered’ array to be symmetrical. There are functions,
fftshift, ifftshift, which re-arrange the input and output arrays such that the normal
definitions of ‘centered’ and ‘symmetrical’ hold.
Using the statement G = fftshift( fft( ifftshift(g))), we get an array that is real and
symmetrical (except for roundoff error): (This also works for 2D arrays.)
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Some examples of sampling effects:
1) Pupil filling sampling array:
Not much PSF resolution
2) Sampling array larger than pupil:
Much better PSF resolution
This was done by reducing the number of samples across the Pupil function. If a certain
number of samples have to be used (due to the structure of the Pupil function), then the
array size can be increased with the same effect. In the second case, the array size is 10
times the pupil diameter, so the PSF sample spacing is 1/10 the first case.
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The Matlab code that produced the plots is:
%DEMO file for Pupil-PSF transforms.
%Make a circular array of ones:
N = 100; %Size of arrays
x = linspace(-1,1,N);
[X,Y] = meshgrid(x,x);
R = sqrt(X.^2 + Y.^2);
%R is now an array with values equal to the distance
% from the center of the N x N array:
%
%Create a circular, uniform pupil, filling the array:
P = zeros(N,N);
P(R<=1) = 1;
%
%Transform and square the array to get the PSF array:
psf = abs(fftshift(fft2(ifftshift(P)))).^2;
%Normalize the Pupil:
psf = psf/max(max(psf));
%Plot the results:
figure
subplot(1,2,1)
imagesc(P),axis image, colormap(gray)
title('Pupil array')
subplot(1,2,2)
mesh(psf)
title('PSF array')
%Create pupil array embedded in sampled array:
P = zeros(N,N);
P(R<0.1) = 1;
%Transform and square the array to get the PSF array:
psf = abs(fftshift(fft2(ifftshift(P)))).^2;
%Normalize the Pupil:
psf = psf/max(max(psf));
%Plot the results:
figure
subplot(1,2,1)
imagesc(P),axis image, colormap(gray)
title('Pupil array')
subplot(1,2,2)
mesh(psf)
title('PSF array')
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