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 |
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uðx0 þ xÞ ¼ uðx0 xÞ |
ð14:11-1Þ |
The two projections of interest can be written as follows:
u x |
uðxÞj a |
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P1u ¼ ( ð0 Þ |
jotherwise |
ð14:11-2Þ |
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8 Aðf Þ cos½fðf Þ cðf Þ&ejfðf Þ |
p |
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jfðf Þ |
cðf Þj < |
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P2u |
2 |
ð |
14:11-3 |
Þ |
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: |
0 |
otherwise |
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$ < |
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where A(f) is the FT amplitude, and cðf Þ |
is the FT phase. It is observed that |
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Aðf Þ cosðfðf Þ |
cðf ÞÞ is the projection of Uðf Þ in the direction given by fðf Þ. In a |
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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 |
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P2u |
2 |
14:11-5 |
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> |
2 |
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< |
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$ > |
0 |
otherwise; f |
S |
ð |
Þ |
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> |
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: |
Uðf Þ |
f2=S |
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> |
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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 |
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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 |
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where P1 uk is replaced by uk |
since uk ¼ P1T2uk |
1 and therefore uk 2 D1. |
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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 |
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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 |
&Þ |
ejf½k& |
if j½k& ci k |
&j |
< |
p |
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U |
k |
2 |
ð |
14:12-5 |
Þ |
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iþ1 |
½ & ¼ |
½ & |
ð |
½ |
0& |
½ |
otherwise ½ |
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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 |
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N 1 |
u½n& |
uiþ1 |
½n& 2 |
Eiþ1 ¼ n¼0 |
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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 |
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Ei ¼ |
1 N 1 |
U½k& |
Ui½k& |
2 |
|
N k¼0 |
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Since Z(k) and Zi(k) have the same phase, the last equation can be written as
X |
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Eiþ1 ¼ |
1 |
N 1 |
2 |
||
N k¼0 U½k& |
Uiþ1½k& |
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We also define
X |
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Ei0 |
¼ |
1 N 1 |
U½k& |
Vi½k& |
2 |
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N k¼0 |
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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 |
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has the property |
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Eðukþ1Þ EðukÞ |
ð14:13-3Þ |
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for every l1 and l2 which satisfy |
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0 |
li |
Ai2 þ Ai |
ð |
14:13-4 |
Þ |
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2 |
1 |
ðAi |
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Ai þ Ai |
þ BiÞ |
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2 |
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A |
jP1T2uk |
T2ukj |
ð |
14:13-5 |
Þ |
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1 |
¼ jP2T2uk |
T2ukj |
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A |
jP2uk |
ukj |
ð |
14:13-6 |
Þ |
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2 |
¼ jP1uk |
ukj |
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