Файл: Ersoy O.K. Diffraction, Fourier optics, and imaging (Wiley, 2006)(ISBN 0471238163)(427s) PEo .pdf

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48 SCALAR DIFFRACTION THEORY

where x, y, fx, fy are the sampling intervals, and n1, m1, as well as n2, m2 are integers satisfying

M1n1; m1 M1

4

4-2

Þ

M2n2; m2 M2

ð

:

 

We will write Uð xn1; yn2; zÞ and Að fxm1; fym2; zÞ as

U½n1; n2; z&

 

and

A½m1; m2; z&, respectively. In terms of discretized space and frequency variables, Eqs. (4.3-2) and (4.3-8) are approximated by

 

X1

X2

U½n1

 

; 0&e j2pð fx xn1m1þ fy yn2m2Þ

ð4:4-3Þ

A½m1; m2; 0& ¼ x y

n

n

; n2

½ 1 2 & ¼

x y

m1

m2

½

1

2 &

ð Þ

 

 

f f

X X

 

 

jzpk2 4p2 fx2m2

fy2m2

 

U n ; n ; z

 

 

A m ; m ; 0 e

1þ

2

 

 

e j2pð fx xn1m1þ fy yn2m2Þ

 

 

ð4:4-4Þ

M1 and M2 should be chosen such that the inequality (4.3-13) is satisfied if evanescent waves are to be neglected. An approximation is to choose a rectangular region in the Fourier domain such that

jkxj ¼ 2pj fxj k

ð4:4-5Þ

jkyj ¼ 2pj fyj k

ð4:4-6Þ

Then, the following must be satisfied:

 

1

ð4:4-7Þ

j fx max j ¼ j fy max j ¼

 

l

Choosing M ¼ M1 ¼ M2, f ¼ fx ¼ fy gives

 

1

 

 

ð4:4-8Þ

fmax ¼ fM ¼

 

 

l

Hence,

 

1

 

 

 

 

 

M ¼

 

 

ð4:4-9Þ

f l

In practice, it is often true that fx max and fy max Eq. (4.4-8) becomes

M ¼ fx max

1

fx

are less than 1=l. If they are known,

ð4:4-10Þ

FFT IMPLEMENTATION OF THE ANGULAR SPECTRUM OF PLANE WAVES

49

in the x-direction, and

M2 ¼

fy max

ð4:4-11Þ

fy

in the y-direction. Assuming fx max ¼ fy max , fx ¼ fy ¼ f gives M ¼ Mx ¼ My. FFTs of length N will be assumed to be used along the two directions so that

¼ x ¼ y, and f ¼ fx ¼ fy. How N is related to M is discussed below. In order to be able to use the FFT, the following must be valid:

f ¼

1

ð4:4-12Þ

N

U½n1; n2; z& and A½m1; m2; z& for any z are also assumed to be periodic with period N. Hence, for m1, m2, n1, n20, they satisfy

U½ n1; n2; z& ¼ U½N n1; n2; z&

 

U½n1;

n2; z& ¼ U½n1; N n2; z&

 

U½ n1;

n2; z& ¼ U½N n1; N n2

; z&

A½ m1; m2; z& ¼ A½N m1; m2; z&

ð4:4-13Þ

 

A½m1;

m2; z& ¼ A½m1; N m2; z&

 

A½ m1;

m2; z& ¼ A½N m1; N m2; z&

Mapping of negative indices to positive indices is shown in Figure 4.2. Using Eq. (4.4-13), Eqs. (4.4-3) and (4.4-4) can be written as follows:

X X

 

N 1 N

1

2p

 

 

 

 

A½m1; m2; 0& ¼ 2

U½n1; n2; 0&e j

N

ðn1m1þn2m2Þ

ð4:4-14Þ

n1¼ 0 n2¼ 0

 

X X

 

N 1

N 1

 

U½n1; n2; z& ¼ ð f 0Þ2

A½m1; m2; z&e j

2p

ðm1n1þm2n2Þ

ð4:4-15Þ

N

m1¼ 0 m2¼ 0

 

 

 

 

 

 

 

 

 

y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

IV

III

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

II

 

 

 

 

I

II

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

III

 

 

 

IV

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.2. Mapping of negative index regions to positive index regions to use the FFT.

50

 

 

 

 

 

 

 

 

 

SCALAR DIFFRACTION THEORY

where

 

 

 

 

 

 

 

 

 

pð Þ ð

 

 

 

½

 

1;

2; & ¼

½ 1;

 

2;

 

&

 

ð :

 

Þ

A

m

 

m z

A m

m

 

0

 

e jz

k2 4p2 fx 2 m12þm22Þ

4

4-16

 

where m1 and m2 satisfy condition (4.4-2). A½m1; m2; z& is also assumed to be periodic with period N:

A½

m1; m2; z& ¼ A½N m1; m2; z&

ð4:4-17Þ

A½m1;

m2; z& ¼ A½m1; N m2; z&

A½

m1;

m2; z& ¼ A½N m1; N m2; z&

 

Note that the periodicity condition due to the use of the FFT causes aliasing because circular convolution rather than linear convolution is computed. In order to reduce aliasing, the input and output apertures can be zero padded to a size, say,

M0 ¼ 2M. Then, N can be chosen as follows:

 

ð4:4-18Þ

N ¼

 

2M00

þ 1

N odd

 

 

2M

 

N even

 

In addition, FFT algorithms are usually more efficient when N is a power of 2. Hence, N may actually be chosen larger than the value computed above to make it a power of 2.

Equations (4.4-15) and (4.4-16) can now be written as

¼

¼

p

X X

 

 

 

N 1

N 1

A½m1; m2; 0&e jz k2 4p2ð fxÞ2ðm12þm22Þ e j

2p

 

U½n1; n2; z& ¼ ð f Þ2 m1 0 m2 0

N

ðn1m1þn2m2Þ

ð4:4-19Þ

Equation (4.4-19) is in the form of an inverse 2-D DFT except for a normalization factor.

As ð f Þ2 ¼ N12, 2 and ð f Þ2 in Eqs. (4.4-14) and (4.4-15) can be omitted, and only one of these equations, say, Eq. (4.4-14) is multiplied by N12.

In summary, the procedure to obtain Uðx; y; zÞ from Uðx; y; 0Þ with the angular spectrum method and the FFT is as follows:

1.Generate U½n1; n2; 0& as discussed above.

2.Compute A½m1; m2; 0& by FFT according to Eq. (4.4-14).

3.Compute A½m1; m2; z& according to Eq. (4.4-16).

4.Compute U½n1; n2; z& by FFT according to Eq. (4.4-19).

5.Arrange U½n1; n2; z& according to Eq. (4.4-13) so that negative space coordinates are regenerated.

The results discussed above can be easily generalized to different number of data points along the x- and y-directions.

FFT IMPLEMENTATION OF THE ANGULAR SPECTRUM OF PLANE WAVES

51

EXAMPLE 4.4 Suppose that the following definitions are made: fftshift: operations defined by Eqs. (4.4.13) and (4.417)

fft2, ifft2: routines for 2-D FFT and inverse FFT, respectively H: transfer function given by Eq. (4.3.14)

u, U: input and output fields, respectively Using these, an ASM program can be written as u1¼fftshift ((fft2(fftshift(u))))

u2¼H u1 U¼fftshift(ifft2(fftshift(u2)))

Show that this program can be reduced to u1¼fftshift((fft2(u)))

u2¼H u1 U¼ifft2(fftshift(u2))

Solution: The difference between the two programs is reduction of fftshift in the first and third steps of the program. This is possible because of the following property of

the DFT shown in 1-D:

½

 

& ¼

½

&

 

 

 

 

If u0

½

n

& ¼

u n

2

k

e

jpk. When the fftshift in the first step of the

 

 

N , then U0

 

 

U

k

 

 

 

 

skipped, the end result is the phase shift e

jpk

. When the fftshift in the third

program is

 

 

 

 

 

 

 

 

 

step is skipped, the same phase shift causes the output U to be correctly generated.

EXAMPLE 4.5 Suppose the parameters are given as follows:

Input size: 32 32

N ¼ 512

l ¼ 0:0005

z ¼ 1000

The units above can be in mm. A Gaussian input field is given by

Uðx; y; 0Þ ¼ exp½ 0:01pðx2 þ y2Þ&

The input field intensity is shown in Figure 4.3. The corresponding output field intensity computed with the ASM is shown in Figure 4.4. If z is changed to 100, the corresponding output field is as shown in Figure 4.5. Thus, the ASM can be used for any distance z. As expected, if the input field is Gaussian, the output field also remains Gaussian.

EXAMPLE 4.6 Determine the impulse response function corresponding to the transfer function given by Eq. (4.3.14).

Solution: The impulse response is the inverse Fourier transform of the transfer function:

 

1

1

 

 

hðx; yÞ ¼

ð

ð

e jkz½1 l2fx2 l2fy2&1=2e j2pð fxxþfyyÞdfxdfy

ð4:4-20Þ

11

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