Файл: Ersoy O.K. Diffraction, Fourier optics, and imaging (Wiley, 2006)(ISBN 0471238163)(427s) PEo .pdf
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HELMHOLTZ EQUATION |
43 |
where
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Uf ðr; f Þ ¼ |
ð |
Uðr; tÞe j2pftdt |
ð4:2-6Þ |
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Substituting Eq. (4.2-6) into the wave equation (3.5-7) again results in the Helmholtz equation for Uf ðr; f Þ.
EXAMPLE 4.1 Show that a spherical wave, defined by e jkr=r where |
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r ¼ |
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x |
þ y2 |
þ z |
, is a solution of the Helmholtz equation. |
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Solution: |
in spherical coordinates is given by |
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@2 |
1 @ |
1 @2 |
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r2 ¼ |
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þ |
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þ |
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þ |
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@r2 |
r |
@r |
r2 |
@ 2 |
@y2 |
e jkr=r is spherically symmetric. In this case, r2 in polar coordinates is simplified to
r2 |
1 @ |
r2 |
@ |
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¼ |
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r2 |
@r |
@r |
The Helmholtz equation with a spherically symmetric wave function U(r) becomes
1 @ |
r2 |
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UðrÞ þ k2UðrÞ ¼ 0 |
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r @r |
@r |
Let U(r) be of the form f ðrÞ=r. It is not difficult to show by partial differentiation that
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r |
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1 @2f r |
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r2 |
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ð |
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¼ |
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r2 @r |
@r |
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@r2 |
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The Helmholtz equation becomes |
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@2f ðrÞ |
þ |
K2 |
f ðrÞ |
¼ |
0 |
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@r2 |
r |
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r2 |
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The solution of this equation is given by
f ðrÞ ¼ e jkr
Thus, UðrÞ ¼ e jkr=r:
44 SCALAR DIFFRACTION THEORY
EXAMPLE 4.2 Find a similar solution in cylindrical coordinates. Solution: In cylindrical coordinates, r2 is given by
r2 |
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1 @ |
1 @2 |
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¼ |
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@r2 |
r |
@r |
r2 |
@ 2 |
@z2 |
Assuming a solution with cylindrical symmetry, Uðr; ; zÞ becomes Uðr; zÞ. Cylindrical symmetry means the wavefronts are circles for constant z. In this case, there are no simple solutions. The exact solution has a Bessel-function type of dependence on r. It can also be shown that
C jkr
Uðr; zÞ p e r
approximately satisfies the wave equation. Such a wave is called a cylindrical wave.
4.3ANGULAR SPECTRUM OF PLANE WAVES
We will consider the propagation of the wave field Uðx; y; zÞ in the z-direction. The wave field is assumed to be at a wavelength l such that k ¼ 2p=l. Let z ¼ 0 initially. The 2-D Fourier representation of U(x,y,0) is given in terms of its Fourier transform
Aðfx; fy; 0Þ by
Uðx; y; 0Þ ¼ ðð |
Að fx; fy; 0Þe j2pðfxxþfyyÞdfxdfy |
ð4:3-1Þ |
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where |
ðð |
Uðx; y; 0Þe j2pðfxxþfy yÞdxdy |
ð4:3-2Þ |
Að fx; fy; 0Þ ¼ |
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Including time variation, Að fx; fy; 0Þej2pðfxxþfy yþftÞ is a plane wave |
at z ¼ 0, |
propagating with direction cosines ai given by Eqs. (3.5-15)–(3.5-17). Að fx; fy; 0Þ is called the angular spectrum of Uðx; y; 0Þ.
Consider next the wave field Uðx; y; zÞ. Its angular spectrum Að fx; fy; zÞ is given by
Að fx; fy; zÞ ¼ |
ðð Uðx; y; zÞe j2pð fxxþfyyÞdxdy |
ð4:3-3Þ |
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and its Fourier representation in terms of its angular spectrum is given by |
ð4:3-4Þ |
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Uðx; y; zÞ ¼ |
ðð Að fx; fy; zÞe j2pð fxxþfyyÞdfxdfy |
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1
ANGULAR SPECTRUM OF PLANE WAVES |
45 |
Uðx; y; zÞ satisfies the Helmholtz equation at all points without sources, namely,
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r2Uðx; y; zÞ þ k2Uðx; y; zÞ ¼ 0 |
ð4:3-5Þ |
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Substitution of Uðx; y; zÞ from Eq. (4.3-4) into Eq. (4.3-5) yields |
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ðð |
dz2 Að fx; fy; zÞ þ ðk2 |
4p2ð fx2 þ fy2ÞÞAð fx; fy; zÞ e j2pð fxxþfy yÞdfxdfy ¼ 0 |
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This is true for all waves only if the integrand is zero:
d22 Að fx; fy; zÞ þ ðk2 4p2ð fx2 þ fy2ÞÞAð fx; fy; zÞ ¼ 0 dz
This differential equation has the solution
Að fx; fy; zÞ ¼ Að fx; fy; 0Þe jmz
where
q
m ¼ k2 4p2ð fx2 þ fy2Þ ¼ kz
ð4:3-6Þ
ð4:3-7Þ
ð4:3-8Þ
ð4:3-9Þ
If 4p2ð fx2 þ fy2Þk2, m is real, and each angular spectrum component is just modified by a phase factor e jmz. Plane wave components satisfying this condition are known as homogeneous waves.
If 4p2ð fx2 þ fy2Þk2, then m can be written as
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m ¼ j 4p2ð fx2 þ fy2Þ k2 |
ð4:3-10Þ |
and Eq. (4.3-8 ) becomes |
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Að fx; fy; zÞ ¼ Að fx; fy; 0Þe mz |
ð4:3-11Þ |
This result indicates that the amplitudes of such plane wave components are strongly attenuated by propagation in the z-direction. They are called evanescent waves.
If 4p2ð fx2 þ fy2Þ ¼ k2, Að fx; fy; zÞ is the same as Að fx; fy; 0Þ. Such components correspond to plane waves traveling perpendicular to the z-axis.
Knowing Að fx; fy; zÞ in terms of Að fx; fy; 0Þ allows us to find the wave field at ðx; y; zÞ by using Eq. (4.3-11) in Eq. (4.3-4 ):
ðð1 p
Uðx; y; zÞ ¼ |
Að fx; fy; 0Þe jz k2 4p2ð fx2þfy2Þe j2pð fxxþfyyÞdfxdfy |
ð4:3-12Þ |
1
46 |
SCALAR DIFFRACTION THEORY |
Thus, if Uðx; y; 0Þ is known, Að fx; fy; 0Þ can be computed, followed by the computation of Uðx; y; zÞ. The limits of integration in Eq. (4.3-12 ) can be limited to a circular region given by
4p2ð fx2 þ fy2Þ k2 |
ð4:3-13Þ |
provided that the distance z is at least several wavelengths long so that the evanescent waves may be neglected.
Under these conditions, Eq. (4.3-8) shows that wave propagation in a homogeneous medium is equivalent to a linear 2-D spatial filter with the transfer function given by
(p
H |
ð |
fx; fy |
Þ ¼ |
e jz k2 4p2ð fxþfy2Þ |
4p2ð fx2 þ fy2Þk2 |
ð |
4:3-14 |
Þ |
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otherwise |
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This is schematically shown in Figure 4.1. The impulse response, which is the inverse Fourier transform of this transfer function, is discussed in Example 4.6.
Propagation of a wavefield in the z-direction in a source-free space is correctly described by the propagation of the angular spectrum in the near field as well as the far field. Two other ways to characterize such propagation are in terms of the Fresnel and Fraunhofer approximations, discussed in Chapter 5. However, they are valid only under certain constraints.
Let F[ ] and F 1[ ] denote the forward and inverse Fourier transform operators, respectively. In terms of these operators, Eq. (4.3-12) can be written as
hpi
Uðx; y; zÞ ¼ F 1 F½Uðx; y; 0Þ&ejkz |
1 ax2 |
ay2 |
ð4:3-15Þ |
Aðfx; fy; 0Þ ¼ F½Uðx; y; 0Þ& for particular values of |
fx |
and fy |
is the complex |
amplitude of a plane wave traveling in the direction specified by the direction cosines ax ¼ 2pfx, ay ¼ 2pfy, and az ¼ 2pfz, where fz ¼ 21p ½1 a2x a2y &1=2. The
effect of propagation is to modify the relative phases of the various plane waves by |
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without changing their amplitudes. |
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1 a2 a2 |
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EXAMPLE 4.3 Find the angular spectrum of a wave field at plane z ¼ a (constant)
if the wave field results from a plane wave Be jkx0 xe jkz0 z passing through a circular aperture of diameter d at z ¼ 0.
Solution: The wave at z ¼ 0 can be written as
Uðx; y; 0Þ ¼ Bejkx0 xcyl r
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p |
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¼ Bej2pfx0 xcyl |
x2 þ y2 |
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FFT IMPLEMENTATION OF THE ANGULAR SPECTRUM OF PLANE WAVES |
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The FTs of the two factors above are |
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Be j2pfx0 x $ Bdð fx |
fx0 ; fyÞ |
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r |
d2 |
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cyl |
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sombðd Þ |
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By convolution theorem, the angular spectrum of Uðx; y; 0Þ is given by |
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Að fx; fy; 0Þ ¼ Bdð fx |
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fx0 ; fyÞ |
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d2p |
sombðd Þ |
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¼ Bdð fx |
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fx0 ; fyÞ |
d p |
somb d |
fx2 þ fy2 |
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d2pB |
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q |
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somb d |
ð fx |
fx0 Þ2 þ fy2 |
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It is observed that the single plane wave component before the circular aperture is changed to a spectrum with an infinite number of such components after the aperture. They are mostly propagating in the direction of the incident wave field, with a spread given by the sombrero function. This spread is inversely proportional to d, the diameter of the circular aperture.
At a distance z, the angular spectrum becomes
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Að fx; fy; zÞ ¼ |
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somb d |
ð fx fx0 Þ2 þ fy2 |
e jzpk |
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4p |
ð fx |
þfy |
Þ |
4.4 FAST FOURIER TRANSFORM (FFT) IMPLEMENTATION OF THE ANGULAR SPECTRUM OF PLANE WAVES
The angular spectrum of plane waves relating Uðx; y; zÞ to Uðx; y; 0Þ can be implemented by the fast Fourier transform (FFT) algorithm [Brigham, 1974] after discretizing and truncating the space variables and the spacial frequency variables. For a discussion of the discrete Fourier transform and the FFT, see Appendix B. The discretized and truncated variables in the space domain and the spatial frequency domain are given by
x ¼ x n1
y ¼ y n2
ð4:4-1Þ
fx ¼ fx m1 fy ¼ fy m2