Abstract

The phase-retrieval problem consists of the reconstruction of an object from the modulus of its Fourier transform or, equivalently, from its autocorrelation. This paper describes a number of results relating to the reconstruction of the support of an object from the support of its autocorrelation. Methods for reconstructing the object’s support are given for objects whose support is convex and for certain objects consisting of collections of distinct points. The uniqueness of solutions is discussed. In addition, for the objects consisting of collections of points, a simple method is shown for completely reconstructing the object functions.

© 1982 Optical Society of America

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References

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  1. See, for example, H. P. Baltes, ed., Inverse Source Problems in Optics (Springer-Verlag, New York, 1978).
    [CrossRef]
  2. J. R. Fienup, “Reconstruction of an object from the modulus of its Fourier transform,” Opt. Lett. 3, 27–29 (1978); “Space object imaging through the turbulent atmosphere,” Opt. Eng. 18, 529–534 (1979).
    [CrossRef] [PubMed]
  3. R. N. Bracewell, The Fourier Transform and Its Applications, 2nd ed. (McGraw-Hill, New York, 1978).
  4. H. W. E. Jung, “Uber die kleinste Kugel, die eine raumliche Figur einschliesst,” Z. Reine Agnew. Math. 123, 241 (1901).
  5. J. Christou, “Imaging of star clusters from speckle interferometry,” Opt. Commun. 37, 331–334 (1981).
    [CrossRef]
  6. J. R. Fienup and T. R. Crimmins, “Determining the support of an object from the support of its autocorrelation,” J. Opt. Soc. Am. 70, 1581 (A) (1980).

1981 (1)

J. Christou, “Imaging of star clusters from speckle interferometry,” Opt. Commun. 37, 331–334 (1981).
[CrossRef]

1980 (1)

J. R. Fienup and T. R. Crimmins, “Determining the support of an object from the support of its autocorrelation,” J. Opt. Soc. Am. 70, 1581 (A) (1980).

1978 (1)

1901 (1)

H. W. E. Jung, “Uber die kleinste Kugel, die eine raumliche Figur einschliesst,” Z. Reine Agnew. Math. 123, 241 (1901).

Bracewell, R. N.

R. N. Bracewell, The Fourier Transform and Its Applications, 2nd ed. (McGraw-Hill, New York, 1978).

Christou, J.

J. Christou, “Imaging of star clusters from speckle interferometry,” Opt. Commun. 37, 331–334 (1981).
[CrossRef]

Crimmins, T. R.

J. R. Fienup and T. R. Crimmins, “Determining the support of an object from the support of its autocorrelation,” J. Opt. Soc. Am. 70, 1581 (A) (1980).

Fienup, J. R.

J. R. Fienup and T. R. Crimmins, “Determining the support of an object from the support of its autocorrelation,” J. Opt. Soc. Am. 70, 1581 (A) (1980).

J. R. Fienup, “Reconstruction of an object from the modulus of its Fourier transform,” Opt. Lett. 3, 27–29 (1978); “Space object imaging through the turbulent atmosphere,” Opt. Eng. 18, 529–534 (1979).
[CrossRef] [PubMed]

Jung, H. W. E.

H. W. E. Jung, “Uber die kleinste Kugel, die eine raumliche Figur einschliesst,” Z. Reine Agnew. Math. 123, 241 (1901).

J. Opt. Soc. Am. (1)

J. R. Fienup and T. R. Crimmins, “Determining the support of an object from the support of its autocorrelation,” J. Opt. Soc. Am. 70, 1581 (A) (1980).

Opt. Commun. (1)

J. Christou, “Imaging of star clusters from speckle interferometry,” Opt. Commun. 37, 331–334 (1981).
[CrossRef]

Opt. Lett. (1)

Z. Reine Agnew. Math. (1)

H. W. E. Jung, “Uber die kleinste Kugel, die eine raumliche Figur einschliesst,” Z. Reine Agnew. Math. 123, 241 (1901).

Other (2)

See, for example, H. P. Baltes, ed., Inverse Source Problems in Optics (Springer-Verlag, New York, 1978).
[CrossRef]

R. N. Bracewell, The Fourier Transform and Its Applications, 2nd ed. (McGraw-Hill, New York, 1978).

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Figures (19)

Fig. 1
Fig. 1

(a) Set S, (b) three of the translates of S that make up A, (c) autocorrelation support A = SS.

Fig. 2
Fig. 2

A symmetric set that is not an autocorrelation support.

Fig. 3
Fig. 3

Locator sets. (a) Set S, (b) A = SS, (c) locator set L = ½P, (d) formation of L = A ∩ (w + A), (e) and (f) two other members of the family of locator sets.

Fig. 4
Fig. 4

Locator sets (shaded areas) for the circle. (a) AH, (b) the square ½P, (c) A ∩ (w + A); (d) circle of radius 1 / 3.

Fig. 5
Fig. 5

Autocorrelation tri-intersection solution for convex sets. (a) Set S, (b) A = SS, (c) formation of locator set L = A ∩ (w + A), (d) formation of solution B = A ∩ (w1 + A) ∩ (w2 + A), (e) and (f) two other solutions of the form B.

Fig. 6
Fig. 6

Sphere/circle example. (a) Set S, (b) A = SS, (c) B = A ∩ (w1 + A) ∩ (w2 + A), (d) another solution for the circle combining the solutions (a) and (c).

Fig. 7
Fig. 7

Autocorrelation tri-intersection for sets consisting of a collection of distinct points. (a) Set S, (b) A = SS, (c) and (d) locators of the form L = A ∩ (w + A). Intersecting (c) or (d) with (b) yields the unique solution (a).

Fig. 8
Fig. 8

Autocorrelation intersection for redundant case. (a) Set S, (b) A = SS, (c)–(e) locators of the form L = A ∩ (w + A), (f) intersection of (c) with (d), (g) another intersection of three translates of A.

Fig. 9
Fig. 9

Redundancy types of relationships within S that would violate Condition 1. (a) and (b) three vector separations add to zero, (c) one vector separation is twice another, (d) two vector separations are equal.

Fig. 10
Fig. 10

A1 = c. hull {w1, −w1, w2, −w2, w1, −w2, w2w1}, and A1A. This and Figs. 1119 illustrate steps in the proof that A = BB.

Fig. 11
Fig. 11

The set A2.

Fig. 12
Fig. 12

C is one of the extended “notches” in A2.

Fig. 13
Fig. 13

Illustration of w2 int(D), where D = c. hull {p, w1, w2w1} is used to prove that AC = 0 and therefore AA2.

Fig. 14
Fig. 14

The lines l1, −l1, l2, −l2, l3, and −l3 are tangent to the set A at points w1, −w1, w2, −w2w1, −w2, and w2w1, respectively.

Fig. 15
Fig. 15

The set B lies between lines l1 and w1l1, between l2 and w2l2and between w2 + l3 and w1l3.

Fig. 16
Fig. 16

The entire plane is the union of the cones C0, −C0, C1, −C1, C2, and −C2.

Fig. 17
Fig. 17

The boundary of B is the union of the arcs s0, s1, and s2.

Fig. 18
Fig. 18

Illustration of the proof that p1p2∈ A when the line l intersects the arc so and p1p2 is in the cone C2 (illustrated in Fig. 16).

Fig. 19
Fig. 19

Illustration of the proof that p1p2∈ A when the line l intersects the arc s1 and p1p2 is in the cone C2.

Equations (106)

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a X + b Y { a x + b y ; x X and y Y } ,
a x + b Y { a x + b y : y Y } ,
f f ( x ) = E N f ( y ) f ( y + x ) d y
= E N f ( y ) f ( y - x ) d y .
A = y S ( S - y ) = S - S = { x - y : x , y S } .
- A = A ,
0 A
S 1 ~ S 2 ,
S 2 = v + β S 1 = { v + β x ; x S 1 } ,
t x + ( 1 - t ) y X
c . hull ( X - X ) = c . hull ( X ) - c . hull ( X ) .
v + S L .
L = A H ,
L = ½ P ,
L = A ( w + A )
L = w W ( w + A )
S p S ( p + A ) p W ( p + A ) .
S = ½ A = { x / 2 : x A } .
A = ( ½ A ) - ( ½ A ) .
S = A ( w + A ) ,
w 1 ( A ) ,             w 2 ( A ) ( w 1 + A ) .
B = A ( w 1 + A ) ( w 2 + A ) .
A = B - B .
C = A ( w 1 + A ) ( w 2 + A ) ( w 3 + A )
S t = t S 1 + ( 1 - t ) S 2
S t - S t = [ t S 1 + ( 1 - t ) S 2 ] - [ t S 1 + ( 1 - t ) S 2 ] = t S 1 - t S 1 + ( 1 - t ) S 2 - ( 1 - t ) S 2 = t A + ( 1 - t ) A = A ,
S 1 / 2 = ½ S 1 - ½ S 1 = ½ A
S = i = 1 n t i S i
L = c . hull [ { ½ w 1 , - ½ w 1 , ½ w 2 , - ½ w 2 } ] S L .
f ( x ) = m = 1 M f m δ ( x - x m )
S = { x m : m = 1 , , M } .
x 1 , x 2 , y 1 , y 2 , z 1 , z 2 , S ,             x 1 x 2 ,             x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 0 ,
B = A ( w 1 + A ) ( w 2 + A ) .
S ~ B .
f f ( x ) = E N f ( y ) f ( y + x ) d y = n = 1 M m = 1 M f n f m δ ( x - x m + x n ) ,
f f ( x ) = n = 1 M f n 2 δ ( x ) + n = 1 M m n M f n f m δ ( x - x m + x n ) ,
[ a δ ( x - x 1 ) ] [ b δ ( x - x 2 ) ] = { a b δ ( x - x 1 ) x 2 = x 1 0 , x 2 x 1
A P j k ( x ) = [ f f ( x ) ] [ f f ( x - x j + x k ) ] = [ ( n f n 2 ) δ ( x ) + n m n f n f m δ ( x - x m + x n ) ] × [ ( n f n 2 ) δ ( x - x j + x k ) + n m n f n f m × δ ( x - x m + x n - x j + x k ) ]
= ( n f n 2 ) f j f k δ ( x ) + ( n f n 2 ) f j f k δ ( x - x j + x k ) + f j f k m k , j f m 2 δ ( x - x m + x k ) + f j f k n k , j f n 2 δ ( x - x j + x n ) + ( OT ) ,
A P j k ( x ) = n m n m f n f m f n f m δ ( x - x m + x n ) × δ ( x - x m + x n - x j + x k ) ,
x = x m - x n = x m - x n + x j - x k .
A P j k ( x ) = f j f k [ f 2 ( x + x k ) + f 2 ( - x + x j ) ] + ( n k , j f n 2 ) f j f k [ δ ( x ) + δ ( x - x j + x k ) ] + ( OT ) .
A P j k j k ( x ) = [ f f ( x ) ] [ f f ( x - x j + x k ) ] × [ f f ( x - x j + x k ) ] = A P j k ( x ) [ f f ( x - x j + x k ) ] = f j f k [ ( n f n 2 ) δ ( x ) + ( n f n 2 ) δ ( x - x j + x k ) + n k , j f n 2 δ ( x - x n + x k ) + n k , j f n 2 δ ( x - x j + x n ) ] × [ ( n f n 2 ) δ ( x - x j + x k ) + n m n f n f m δ ( x - x m + x n - x j + x k ) ] = f k f j f j { n k , j , j f n 3 δ ( x - x n + x k ) + ( n f n 2 ) [ f k δ ( x ) + f j δ ( x - x j + x k ) + f j δ ( x - x j + x k ) ] }
= f k f j f j [ f 3 ( x + x k ) + ( n k f n 2 ) f k δ ( x ) + ( n j f n 2 ) f j δ ( x - x j + x k ) + ( n j f n 2 ) f j δ ( x - x j + x k ) ] .
D n f n 2 = f f ( 0 )
A = D - 1 A P j k j k ( 0 ) = f k 2 f j f j ,
B = D - 1 A P j k j k ( x j - x k ) = f k f j 2 f j ,
C = D - 1 A P j k j k ( x j - x k ) = f k f j f j 2 .
f k = ( A 3 B C ) 1 / 4 ,
f j = ( B 3 A C ) 1 / 4 ,
f j = ( C 3 A B ) 1 / 4 ,
f k f j f j = ( A B C ) 1 / 4 .
f m = [ A P j k j k ( x m - x k ) f k f j f j ] 1 / 3 .
A ( x ) = δ ( x ) + n = 1 M m n δ ( x - x m + x ) ,
x m - x n = x m - x n + x j - x k
A = y S ( S - y ) = S - S .
B ( x , r ) = { y E N : x - y r }
f r ( x ) = B ( x , r ) f ( y ) d y .
( f f ) 2 r ( x - y ) = B ( x - y , 2 r ) f f ( p ) d p = B ( 0 , 2 r ) E N f ( ω ) × f ( w + p + x - y ) d w d p = E N f ( w ) f 2 r ( w + x - y ) d w = E N f ( w + y ) f 2 r ( w + x ) d w B ( 0 , r ) f ( w + y ) f 2 r ( w + x ) d w f r ( x ) B ( 0 , r ) f ( w + y ) d w = f r ( x ) f r ( y ) > 0.
0 < ( f f ) r ( x ) = ( f f ) r ( - x ) = B ( - x , r ) f f ( p ) d p = B ( 0 , r ) E N f ( y ) f ( y + p - x ) d y d p = B ( 0 , r ) E N f ( y + p ) f ( y + x ) d y d p = E N f r ( y ) f ( y + x ) d y .
0 < B ( z , r ) f r ( y ) f ( y + x ) d y < f 2 r ( z ) B ( z , r ) f ( y + x ) d y = f 2 r ( z ) f r ( z + x ) .
B 1 = c . hull { 0 , w 1 , w 2 } .
A 1 = c . hull ( w 1 , - w 1 , w 2 , - w 2 , w 1 - w 2 , w 2 - w 1 ) .
w 1 , - w 1 , w 2 , - w 2 , w 1 - w 2 , w 2 - w 1 A ,
A 2 = A 1 ( B 1 - w 1 - w 2 ) ( B 1 + w 2 - w 1 ) ( B 1 + w 1 - w 2 ) ( w 1 + w 2 - B 1 ) ( w 1 - w 2 - B 1 ) ( w 2 - w 1 - B 1 ) .
C = { w 2 + λ 1 w 1 + λ 2 ( w 2 - w 1 ) : λ 1 > 0 and λ 2 > 0 } .
A A 2 = [ B 1 ( w 1 + w 2 - B 1 ) ] [ ( - B 1 ) ( B 1 - w 1 - w 2 ) ] [ ( B 1 - w 1 ) ( w 2 - w 1 - B 1 ) ] [ ( w 1 - B 1 ) ( B 1 + w 1 - w 2 ) ] [ ( B 1 - w 2 ) ( w 1 - w 2 - B 1 ) ] [ ( w 2 - B 1 ) ( B 1 + w 2 - w 1 ) ] .
p - w 1 ( B 1 - w 1 ) ( w 2 - B 1 ) A 1 A .
C 0 = { λ 1 w 1 + λ w w 2 :             λ 1 0 and λ 2 0 } .
E 2 = C 0 ( - C 0 ) C 1 ( - C 1 ) C 2 ( - C 2 ) .
s 0 = C 0 ( B ) , s 1 = ( w 1 + C 1 ) ( B ) , s 2 = ( w 2 + C 2 ) ( B ) .
p 1 - p 2 = q 1 - q 2 = ( q 1 - w 2 ) - ( q 2 - w 2 ) = ( 1 - λ ) ( q 1 - w 2 ) A .
G - G = { 0 , w 1 , - w 1 , w 2 , - w 2 , w 1 - w 2 , w 2 - w 1 } A .
x = x - 0 S 1 - S 1 = A .
x w 1 + A .
x w 2 + A .
G i = { 0 , x , w i } ,             i = 1 , 2.
G i - G i = { 0 , x , - x , w i , - w i , x - w i , w i - x } A .
{ v , v + β w 1 , v + β w 2 } = v + β G S
{ v i , v i + β i x , v i + β i w i } = v i + β i G i S ,             i = 1 , 2.
( v + β w i ) - v = v i - ( v i + β i w i ) ,             i = 1 , 2.
x 1 = v + β w i ,             x 2 = v ,             y 1 = v i ,             y 2 = v i + β i w i ,
v i = v + β w i ,             i = 1 , 2.
v 1 - ( v 1 + β 1 x ) = v 2 - ( v 2 + β 2 x ) .
v 1 = v 2 .
( v j + β j w j ) - v j = ( v + β w j ) - v .
v + β x = v j + β j x v j + β j G j S
x β ( S - v ) = S 1 .
{ 0 , g 1 , - g 1 , g 2 , - g 2 , g 1 - g 2 , g 2 - g 1 } = G - G A .
x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 0.
x 1 = y 2             or             x 1 = z 2 ,             and             x 2 = y 1             or             x 2 = z 1 .
y 1 - x 2 = x 1 - x 2 + y 1 - y 2 = g 1 + g 2 - g 1 = g 2
x 2 + G = { x 2 , x 2 + g 1 , x 2 + g 2 } = { x 2 , x 1 , y 1 } S .
z 1 - x 2 = x 1 - x 2 + z 1 - z 2 = g 1 - g 2
x 2 + g 1 - G = { x 2 + g 1 , x 2 , x 2 + g 1 - g 2 } = { x 1 , x 2 , z 1 } S .
x 1 = v 2 , x 2 = v 1 , y 1 = v 1 , y 2 = u 2 , z 1 = u 1 , z 2 = v 1 .
x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 0.
z 2 - x 1 = v 1 - v 2 = u 1 - u 2 0.
x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 0.
x 1 - x 2 = z 2 - z 1 .
x 1 - x 2 = y 2 - y 1 .
g 1 - g 2 = x 1 - x 2 + y 1 - y 2 = z 2 - z 1 S - S = A ,
{ v , v + g 1 , v + g 2 } = v + β G S .
( x 2 + g 1 ) - ( x 2 + g 2 ) = g 1 - g 2 = z 2 - z 1 .
{ v , v - g 1 , v - g 2 } = v + β G S .
( x 1 - g 1 ) - ( x 1 - g 2 ) = g 2 - g 1 = z 1 - z 2 .