06b) plane wave spectrum_1_28

7/27/2019 06b) Plane Wave Spectrum_1_28

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ECEN 4616/5616 1/28/2013

pg. 1

Euler’s formula:

)sin()cos( xi xeix

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:

nn

xn

f x

f x f f x f

!

)0(

!2

)0()0()0()( 2

Expansion of eix:

)sin()cos(

!7!5!3!6!4!21

!5!4!3!21

753642

532

xi x

x x x xi x x x

ixixixixixeix

Electromagnetic wave equation for empty space:

02

00

2

2t

EE

Monochromatic, plane wave solution, traveling an arbitrary direction:

)cos(, :Hence

2 and

directionn propagatiotheof cosinesdirectionareetc.,that Note

ter radians/meinvector wavetheis,,

ector position vtheis ,,

0,0at phasetheis

:where

cos),(

t z k yk xk At E

k

k k k

z y x

t

t At E

z y x

x

z y x

r

k

k

k

r

r

rk r

Switching to complex notation, using Euler‟s formula we get:

t z k yk xk i A E z y x expRe



ECEN 4616/5616 1/28/2013

pg. 2

where the Real Part function will generally be assumed. Note that only linear operations

and the operation E E A

2will 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 ii , produces a wave traveling in the opposite direction.

This equation is redundant, in that k k k k z y x k 222

, so that k z can be determined

from k x and k y. Since we are interested in describing plane waves in x-y planes, we will

leave off the k z 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

will define

y y

x x

k v

k u

cos

2

cos

2

, where y x , are the x, y direction cosines of the wave.

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:

vyuxi A E 2exp

On an x-y plane, the spatial period between maxima is 1/u in the x-direction and 1/v inthe y-direction.

Plane Wave Spectrum of a Distribution:

Consider the Fourier Transform pair:

dvduvyuxivu A y x E

dydxvyuxi y x E vu A

2exp,,

2exp,,

Since vyuxi 2exp represents a plane wave traveling in the direction with direction

cosines vu , , 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 withvarious amplitudes, which add up to the distribution on the given plane – i.e., the plane

wave spectrum of the distribution.



ECEN 4616/5616 1/28/2013

pg. 3

The integrals in the Fourier Transform go from - to , but direction cosines have a valid

range of – 1 to 1. What is the meaning of values of 1,

vu?

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

never exceed

1 .

If we use the Discrete Fourier Transform on these functions, the Plane Wave Spectrum,

A, will consist of a collection of waves where ,2

,1

,0, N N

vu That is, the plane

wave components will have 0, 1, 2, etc cycles within the discrete array of N by Nsamples.

D

lam

Tx

phi

n=1 component of discrete PWS

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.

In general,

sincos

cos

D

n

D

nu x

xn , so the n

thcomponent of the

PWS (in the x-z plane) propagates at an angle of sin-1

(n/D) from the z-axis.



ECEN 4616/5616 1/28/2013

pg. 4

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 E-

vector. 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 followingmeans: 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 theexit 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

phi

f

x

Hence,

D

n f f x nn

1

sintan)tan( . This gives us the scale of the discrete PSF

calculated from the pupil function.

At this point, it is usual, for convenience, to make the paraxial assumptions sin = tan =

angle and get D

n xn

, however if more accuracy was required (within the limits of

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.



ECEN 4616/5616 1/28/2013

pg. 5

Sampl ing Issu es:

The largest (angle) component in the PWS is D N

N

x

2

2cos

, where

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 cannotcontribute to field distributions further down the z-axis. In practice, spacings anywherenear 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 wouldseem 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.

Issu es with the FFT funct ion :

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 continuouscalculation.

A plot of a real, symmetric function (g = gaussian):



ECEN 4616/5616 1/28/2013

pg. 6

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.)



ECEN 4616/5616 1/28/2013

pg. 7

Some examples of sampling effects:

1) Pupil filling sampling array:

Not much PSF resolution

2) Sampling array larger than pupil:

Much better PSF resolutionThis 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 thearray 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.



ECEN 4616/5616 1/28/2013

pg. 8

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')

06b) plane wave spectrum_1_28

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