Abstract

Binary computer-generated holograms are similar to interferograms with fringe patterns hardclipped by a photographic process. Therefore the fringe locations in the binary hologram can be determined by solving a grating equation. However, there are two difficulties in using this approach to make binary Fourier transform holograms. First the discrete Fourier transform provides only data at discrete sampling locations. Second, the phase angles thus calculated are given in terms of the residues of the original phase angles after multiples of 2π rad are removed. In this paper an accurate numerical method which circumvents these two difficulties is described. Also discussed are three different techniques for storing amplitude information in the binary computer-generated holograms. The different solution methods discussed in this paper are further illustrated by a number of computer-generated holograms and their reconstructed images.

© 1979 Optical Society of America

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References

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1979

1978

1976

1975

H. Becker, W. J. Dallas, Opt. Commun. 15, 50 (1975).
[CrossRef]

1974

1972

K. G. Birch, F. J. Green, J. Phys. D 5, 1982 (1972).
[CrossRef]

1969

B. R. Brown, A. W. Lohmann, IBM J. Res. Dev. 13, 160 (1969).
[CrossRef]

1967

1966

1962

Becker, H.

H. Becker, W. J. Dallas, Opt. Commun. 15, 50 (1975).
[CrossRef]

Birch, K. G.

K. G. Birch, F. J. Green, J. Phys. D 5, 1982 (1972).
[CrossRef]

Brown, B. R.

B. R. Brown, A. W. Lohmann, IBM J. Res. Dev. 13, 160 (1969).
[CrossRef]

B. R. Brown, A. W. Lohmann, Appl. Opt. 5, 967 (1966).
[CrossRef] [PubMed]

Chavel, P.

Chu, D. C.

D. C. Chu, J. R. Fienup, Opt. Eng. 13, 189 (1974).
[CrossRef]

Dallas, W. J.

H. Becker, W. J. Dallas, Opt. Commun. 15, 50 (1975).
[CrossRef]

W. J. Dallas, Appl. Opt. 13, 2274 (1974).
[CrossRef] [PubMed]

Fienup, J. R.

D. C. Chu, J. R. Fienup, Opt. Eng. 13, 189 (1974).
[CrossRef]

Green, F. J.

K. G. Birch, F. J. Green, J. Phys. D 5, 1982 (1972).
[CrossRef]

Hsueh, C. K.

Hugonin, J. P.

Lanczos, C.

C. Lanczos, Applied Analysis (Prentice-Hall, Englewood Cliffs, N.J., 1956), Chap. 7.

Lee, W.-H.

W.-H. Lee, in Progress in Optics, E. Wolf, Ed. (North-Holland, Amsterdam, 1978), Vol. 16.
[CrossRef]

Lee, W-H.

Leith, E. N.

Lohmann, A. W.

O’Neill, P. K.

Paris, D.

Sawchuk, A. A.

Upatnieks, J.

Wyant, J. C.

Appl. Opt.

IBM J. Res. Dev.

B. R. Brown, A. W. Lohmann, IBM J. Res. Dev. 13, 160 (1969).
[CrossRef]

J. Opt. Soc. Am.

J. Phys. D

K. G. Birch, F. J. Green, J. Phys. D 5, 1982 (1972).
[CrossRef]

Opt. Commun.

H. Becker, W. J. Dallas, Opt. Commun. 15, 50 (1975).
[CrossRef]

Opt. Eng.

D. C. Chu, J. R. Fienup, Opt. Eng. 13, 189 (1974).
[CrossRef]

Other

W.-H. Lee, in Progress in Optics, E. Wolf, Ed. (North-Holland, Amsterdam, 1978), Vol. 16.
[CrossRef]

C. Lanczos, Applied Analysis (Prentice-Hall, Englewood Cliffs, N.J., 1956), Chap. 7.

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Figures (11)

Fig. 1
Fig. 1

(a) Detour phase hologram of a plane wave. The overlapping apertures due to sampling are indicated by arrows. (b) The regular interferogram of the same plane wave as in (a).

Fig. 2
Fig. 2

Binary pulses in (b) are generated by hardclipping a sinusoidal function offset by a bias function as shown in (a).

Fig. 3
Fig. 3

A phase function is shown with its residue function. Except for a constant value, the phase function can be recovered from its residue function.

Fig. 4
Fig. 4

The hologram in (a) is made by using method 1 for recording the amplitude information. The reconstructed image is shown in (b).

Fig. 5
Fig. 5

The hologram in (a) is similar to Fig. 4(a) except for the doubling in the carrier frequency. The reconstructed image in (b) does not have the noisy background as in Fig. 4(b).

Fig. 6
Fig. 6

The hologram in (a) contains four periods of the discrete Fourier transform of the object. The reconstructed image in (b) displays the data points forming the object.

Fig. 7
Fig. 7

This hologram is similar to Fig. 6(a) except for the increase in spatial frequency.

Fig. 8
Fig. 8

The hologram in (a) is made by using method 2.

Fig. 9
Fig. 9

The hologram is made using method 3. The interlacing of the two phase holograms is along the horizontal direction.

Fig. 10
Fig. 10

The hologram is made using method 3, but the two phase holograms are interlaced in the y direction.

Fig. 11
Fig. 11

(a) A phase hologram of a 1-D signal. (b) An image reconstructed from the hologram.

Equations (28)

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2 π x / T + ϕ ( x , y ) = 2 π n .
t ( x , y ) R exp ( - j 2 π x / T ) + A ( x , y ) exp [ j ϕ ( x , y ) ] 2 = R 2 + A 2 ( x , y ) + 2 R A ( x , y ) cos [ 2 π x / T + ϕ ( x , y ) ] ,
f ( x , y ) = { 1 for cos [ 2 π x / T + ϕ ( x , y ) ] 0 , 0 otherwise ,
cos [ 2 π x / T + ϕ ( x , y ) ] = 1.
cos [ 2 π x / T + ϕ ( x , y ) ] = cos π q ( x , y ) ,
2 π x / T + ϕ ( x , y ) = 2 π n π q ( x , y ) ,
f ( x , y ) = m = - [ sin π m q ( x , y ) π m ] × exp { j m [ 2 π x / T + ϕ ( x , y ) ] } .
cos 2 π Y / Δ = cos π A ( x , Y )             or             Y = n Δ ± Δ A ( x , Y ) / 2.
Y n ( x ) = n Δ ± A ( x , n Δ ) / 2.
T ( x , y ) = f 1 ( x , y ) f 2 ( x , y ) ,
f 1 ( x , y ) = p = - sin π p A ( x , y ) π p exp ( j 2 π p y / Δ ) ,
f 2 ( x , y ) = m = - sin m π / 2 m π exp { j m [ 2 π x / T + ϕ ( x , y ) ] } .
sin π q ( x , y ) = A ( x , y ) ,
h ( x , y ) = exp [ j Φ 1 ( x , y ) ] + exp [ j Φ 2 ( x , y ) ] .
Φ 1 ( x , y ) = ϕ ( x , y ) + θ ( x , y ) , Φ 2 ( x , y ) = ϕ ( x , y ) - θ ( x , y ) + π ,
h ( x , y ) = 2 j sin θ ( x , y ) exp [ j ϕ ( x , y ) ] .
x / T + ϕ ( x ) / 2 π ± q ( x ) / 2 + ψ / 2 π = n .
k / M + ϕ ( k T / M ) / 2 π ± q ( k T / M ) 2 + ψ / 2 π = n ,
Mod M [ k + M ϕ ( k T / M ) / 2 π ± M q ( k T / M ) / 2 + ψ M / 2 π ] = 0.
A m = n = 0 N - 1 a n exp ( j 2 π n m / N ) ,
Φ k = ϕ k / 2 π ± q k / 2 + ψ / 2 π .
Δ ϕ = δ 0 + δ 1 + δ 2 2 ,
Δ ϕ = n .
x n = ( p + n ) Δ x ,
ϕ k = 2 π N 2 k / 128 + ϕ k             k = 0 , 1 , 2 , , 127.
2 π x / T ± ϕ ( y ) = 2 π n             or             x = n T T / 2 π ϕ ( y ) ,
g ( y ) = n = 0 cos ( 2 π n α y + ψ n ) ,
ϕ ( y ) = sin - 1 g ( y ) .

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