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

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RESTORATION FROM PHASE

231

is that there is no point of symmetry x0 such that

uðx0 þ xÞ ¼ uðx0 xÞ

ð14:11-1Þ

The two projections of interest can be written as follows:

u x

uðxÞj a

P1u ¼ ( ð0 Þ

jotherwise

ð14:11-2Þ

8 Aðf Þ cos½fðf Þ cðf Þ&ejfðf Þ

p

jfðf Þ

cðf Þj <

P2u

2

ð

14:11-3

Þ

:

0

otherwise

$ <

where A(f) is the FT amplitude, and cðf Þ

is the FT phase. It is observed that

Aðf Þ cosðfðf Þ

cðf ÞÞ is the projection of Uðf Þ in the direction given by fðf Þ. In a

number of papers, the relative directions given by fðf Þ and cðf Þ are not considered, and P2 is expressed as

P2u $ Aðf Þejfðf Þ

ð14:11-4Þ

It is possible that fðf Þ is not known for all f. Suppose that fðf Þ is known for f in a set S. Then P2 can be expressed as

8 Aðf Þ cos½fðf Þ cðf Þ&ejfðf Þ

jfðf Þ cðf Þj

<

p

and f 2 S

P2u

2

14:11-5

>

2

<

$ >

0

otherwise; f

S

ð

Þ

>

:

Uðf Þ

f2=S

>

where Uðf Þ is the FT of u(t). It is observed that P1 is a linear operator. Then, the optimal l1k is 1. A lower bound for l2k can be derived as [Stark ]

l2k lL ¼

1

þ

jP2uk P1P2ukj2

ð

14:11-6

Þ

jP1P2uk ukj2

lL can be used as an estimate of l2k. However, lL may be greater than 2. In later iterations, it is necessary to restrict l2k to be less than 2 to guarantee convergence. For example, minðlL; 1:99Þ can be chosen.

Since l1k equals 1, one cycle of iterations can be written as

ukþ1

¼ P1½1 þ l2kðP2

1Þ&uk

ð14:11-7Þ

¼ ð1 l2kÞuk þ l2kP1P2uk

where P1 uk is replaced by uk

since uk ¼ P1T2uk

1 and therefore uk 2 D1.


232 APODIZATION, SUPERRESOLUTION, AND RECOVERY OF MISSING INFORMATION

14.12 RECONSTRUCTION FROM A DISCRETIZED PHASE FUNCTION BY USING THE DFT

The discrete signal of length N will be denoted by u½n&. The phase of the DFT of u½n& as well as the support of u½n& are assumed to be known. The support of u½n& will be defined as follows:

u½n& ¼ 0 outside 0 n < M and u½0& ¼6 0

It can be shown that N 2M for the reconstruction algorithm to be satisfactory [Hayes et al., 1980]. The DFT of u½n& is given by

U½k& ¼ A½k&ejj½k&

ð14:12-1Þ

j½k& is assumed to be known.

The procedure of the algorithm is as follows:

1. Initialize A½k& with an initial guess, A1½k&. Then, the initial U1½k& is given by

U1½k& ¼ A1½k&ejj½k&

ð14:12-2Þ

2.Compute the inverse DFT of U1½k&, yielding the first estimate u1½n&. Let i equal 1.

3.Define a new sequence vi½n& as

vi½n& ¼

u n

&

0

n < M

ð14:12-3Þ

i0½

M

n < N

4. Compute the DFT Vi½k& of vi½n&. Vi½k& can be written as

Vi½k& ¼ Ai½k&ejci½k&

ð14:12-4Þ

The rest of the procedure can be written in two ways with respect to two different algorithms as follows, respectively.

Algorithm 1 [Levi and Stark]

5. Project Vi½k& as follows:

(

Ai k

cos

j

k

ci k

ejf½k&

if j½k& ci k

&j

<

p

U

k

2

ð

14:12-5

Þ

iþ1

½ & ¼

½ &

ð

½

0&

½

otherwise ½

This operation corresponds to P2. This equation can be modified so that T2 is used by

Uiþ1½k& ¼ ½1 þ l2iðP2 1Þ&½Vi½k&&

ð14:12-6Þ

l2i is computed by using the relation (14.11-6).


RECONSTRUCTION FROM A DISCRETIZED PHASE FUNCTION BY USING THE DFT 233

6.Generate a new estimate of uiþ1½n& by computing the inverse DFT of

Uiþ1½k&.

7.Repeat steps 2–6 until convergence.

Algorithm 2 [Hayes et al.]

The only different step is the following: 8. Project Vi½k& as follows:

Uiþ1½k& ¼ Ai½k&ejf½k&

ð14:11-7Þ

All the other steps are the same as in Algorithm 1.

EXAMPLE 14.5 Let the error Ei at the ði þ 1Þth iteration be defined as

X

N 1

u½n&

uiþ1

½n& 2

Eiþ1 ¼ n¼0

where u½n& is the desired signal to be reconstructed. Prove that Ei is nonincreasing as i increases.

Solution: By Parseval’s theorem, the following is true:

X

Ei ¼

1 N 1

U½k&

Ui½k&

2

N k¼0

Since Z(k) and Zi(k) have the same phase, the last equation can be written as

X

Eiþ1 ¼

1

N 1

2

N k¼0 U½k&

Uiþ1½k&

We also define

X

Ei0

¼

1 N 1

U½k&

Vi½k&

2

N k¼0

The triangle equation for vector differences can be used to note the following:

Eiþ1 Ei0

with equality if Uiþ1½k& ¼ Vi½k&. The inequality above is due to the fact that Uiþ1½k& has the same phase as U½k& whereas Vi½k& has some other phase.


234 APODIZATION, SUPERRESOLUTION, AND RECOVERY OF MISSING INFORMATION

Again using the Parseval’s theorem on this inequality results in

X

N

1

Ei

ju½n& vi½n&j2

n¼0

Since vi½n& is before the projection to create uiþ1½n&, the above equation shows that the projection reduces the error.

14.13GENERALIZED PROJECTIONS

There are many signal recovery problems involving nonconvex constraints. For example, digitizing a signal and constraining a signal or its Fourier transform to have a specified magnitude are nonconvex projections.

When Si is a nonconvex set, there may be more than one point satisfying the definition of a projector Pi. This discrepancy can be removed by specifying additional constraint(s). Another problem with nonconvex sets is that a proof for the existence of a projection has not been found.

In the discussion below, the number of projections M will be limited to 2. In this case, the following error measure is useful:

EðukÞ ¼ jP1uk ukj þ jP2uk ukj

ð14:13-1Þ

The following theorem determines the convergence properties of generalized projections with m ¼ 2 [Levi–Stark]:

Theorem: The recursion given by

ukþ1 ¼ T1T2uk;

u0 arbitrary

ð14:13-2Þ

has the property

Eðukþ1Þ EðukÞ

ð14:13-3Þ

for every l1 and l2 which satisfy

0

li

Ai2 þ Ai

ð

14:13-4

Þ

2

1

ðAi

Ai þ Ai

þ BiÞ

2

A

jP1T2uk

T2ukj

ð

14:13-5

Þ

1

¼ jP2T2uk

T2ukj

A

jP2uk

ukj

ð

14:13-6

Þ

2

¼ jP1uk

ukj


RESTORATION FROM MAGNITUDE

235

B1

¼

ðP2T2uk

T2uk; P1T2uk

T2ukÞ

ð14:13-7Þ

2

jP2T2uk

T2ukj

B2

¼

ðP1uk

uk; P2uk

ukÞ

ð14:13-8Þ

2

jP1uk ukj

It can be shown that the upper limit in Eq. (14.13-4) includes 1. Hence, the theorem applies with P1; P2 instead of T1; T2.

The projections P1 and P2 can involve multiple constraints. Quite often, one set of constraints is in the signal domain, and the other set is in the transform domain.

EXAMPLE 14.6 A projector P maps a function uðxÞ such that the resulting function gðxÞ satisfies

gðxÞ ¼

where sðxÞ for all x. Specify P.

Solution:

P Satisfies

sðxÞ x 2 D D0 0 x 2= D0

j

Pu

uj ¼

v jv

uj

inf

Since gðxÞ ¼ PuðxÞ, we have

jPu uj ¼ jg uj ¼

ð

jgðxÞ

uðxÞj2dx þ

ð

c

jgðxÞ

uðxÞj2dx

x2=D0

ð

x2D0\D

ð

þ

jgðxÞ uðxÞj2dx þ

jgðxÞ uðxÞj2dx

x2D;sðxÞ>0

x2D;sðxÞ<0

jg uj is minimum if gðxÞ is chosen as

¼

>

0

x=D0

x2

>

P1u

8 u

x

D0

Dc

> s

ðx Þ

x 2

D; x\t > 0

>

ð Þ

2

ð Þ

<

>

:sðxÞ x 2 D; xðtÞ 0

14.14RESTORATION FROM MAGNITUDE>

This problem is also called phase retrieval. It occurs in a number of fields such as astronomy, optics, and x-ray crystallography, where only the intensity equal to the