=Paper=
{{Paper
|id=Vol-1904/paper3
|storemode=property
|title=A discrete phase problem in reconstruction of signals in space-rocket hardware
|pdfUrl=https://ceur-ws.org/Vol-1904/paper3.pdf
|volume=Vol-1904
|authors=Antonina А. Kuleshova,Evgeny A. Shchelokov
}}
==A discrete phase problem in reconstruction of signals in space-rocket hardware ==
https://ceur-ws.org/Vol-1904/paper3.pdf
A discrete phase problem in reconstruction of signals in space-rocket
hardware
А.А. Kuleshova1, E.A. Shchelokov1
1
Samara National Research University, 34 Moskovskoe Shosse, 443086, Samara, Russia
Abstract
Reconstruction of information hidden in vector signal phases does not lose its relevance. Sets of vectors i in1 , called frames, in a space
C m ( R m ) can be used for theoretical research of phase retrieval. The article shows that phase retrieval is equivalent to phaseless
and C m , for which sets of vectors i in1 that simultaneously carry out phase retrieval
m
reconstruction. Examples are considered in R
and phaseless reconstruction are constructed.
Keywords: phase retrieval; phaseless reconstruction; frame; complement property; weak phase retrieval; generic frame
1. Introduction
A search of the fast algorithms for phaseless signal reconstruction is topical now. The main property of frames, which makes
them so useful in applied tasks, is their redundancy. A well-chosen frame can provide numerical stability for signal recovery and
obtaining important characteristics of the signal [1]. A family of frames recovers the signal by absolute values of frame
coefficients in polynomial time.
It is shown that in the real case under certain conditions a generic frame consisting of (2m-1)-vectors can recover the signal
without phases. The similar result was obtained in the complex space for (4m-2)-vectors.
Along with the "phaseless reconstruction", another version of the discrete phase problem statement – "phase retrieval" – is
considered. The issue of their equivalence is raised and partially resolved.
The present work continues this line of research and gives examples of signal recovery in small-dimension spaces.
2. A Discrete Phase Problem in Reconstruction of Signals in Space-Rocket Hardware
Now the problem of reduction of the mass of cable systems in spacecraft is widely known. In this connection we offer for
consideration an option of replacement of the cable system by a radio channel [2].
Widespread introduction of wireless devices has become possible as a result of improvement and reduction in cost of
electronic components. Modern chips, which are used in construction of wireless networks, only require connection of several
passive components and program setup.
In connection with the above-mentioned it is reasonable to consider the introduction of wireless technologies into space-
rocket hardware as one of ways to reduce the mass and complexity of the cable system.
Let's consider, as an example, a signal transmitted by a radio channel with a modulation of an OFDM type. The main
advantage of the chosen method is that the signal propagation delay is much less, than time of transfer of a symbol in auxiliary
carriers as compared to other types of modulation. That allows implementing more stable transfer of information under
conditions of symbols overlapping in the course of rereflections of the signal.
Figure 1 shows a distribution model of levels of signals from a different number of access points for a case when a set of
blocks is arranged inside a spacecraft. The model is presented in a two-dimensional form, however, as is obvious from
distribution of levels of signals, 4 access points is enough to provide communication of all blocks among themselves, including
by relaying.
As is evident from figure 1, the signal levels at the border of the external and internal parts of the compartment do not exceed
minus 30 dBm (blue color:-30 to -40 dBm, green color:-40 to - 50 dBm) relative to 0 dBm at the antenna exit. If we take into
account that aluminum has a shielding factor of 70 dBm (for the thickness of 5 mm) we obtain the coefficient of signal
attenuation outside the working zone equal to 100 dBm. If an additional protection is necessary, it is enough not to allow
blocking direct visibility of the transmitting part and the compartment border, which will increase attenuation by 30-40 dB [2].
In OFDM modulation, data are distributed among a great number of auxiliary carriers, that’s why it is necessary to recover
the information lost in several subchannels for further data handling. A Search of the signal recovery algorithms is topical
now. Sampling and quantization of the analog signal lead to consideration of the signal as an element of a finite-dimensional
space V . By the orthonormalized basis (ONB) {u i }im1 , the "signal" v V can be uniquely represented in the form of the sum:
m
v v, u i u i . Actual measurements prove real, and the gap between v, ui and amplitudes of measurements v, u i proves
i 1
insuperable in signal reconstruction [1, P. 280], [4, P.281].
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Fig. 1. Distribution of the signal from one access point (on the left) and from four access points (on the right).
In recent years, significant amount of works has been devoted to solution of the following task: to construct such systems of
"measuring" vectors Ф i in1 that allow recovering a signal v V by a set of real numbers v, i .
Such task has no decision in the ONB class.
The main problem set in [3] is still far from final solution. It is to find necessary and sufficient conditions for a system of
representation vectors Ф i in1 (so-called "measuring vectors"), which provide injectivity and stability of mapping of
"amplitude measurement" of the signal x
2
( A( x))( i ) : x, i
We have proved that exact recovery of the signal (to the unimodular multiplier) is theoretically possible if
complete redundant systems are used as a representation system [2, P. 354]. Frames are such redundant systems.
In 2006, Balan/Casazza/Edidin [4,5] defined one of versions of the discrete phase problem, which they called "phaseless
reconstruction". It was shown that in the real case a generic frame consisting of (2m-1)-vectors can do phaseless reconstruction
under certain conditions. The similar result was obtained for (4m-2)-vectors in the complex space.
3. Frames
Let m be a space R m or C m .
Definition 1. A family of vectors i in1 is called a frame of a Hilbert space m if there are such constants 0 < A ≤ B <
∞ that for all x H m the following inequalities are achieved:
n 2
A || x || 2 x, f i B || x || 2 .
i 1
A and B are called frame bounds. The greatest of the lower bounds is called the optimum lower bound, and the smallest of the
upper bounds is the optimum upper bound. If A=B, than the frame is called A-tight and if A=B=1, it is called a Parseval-Steklov
frame.
n
The numbers x, i i 1 are called frame coefficients.
If all frame elements have the same norm than such frames are called uniform ones.
In the finite-dimensional space the notion of a frame is equivalent to the notion of completeness of a system, that
is to the equality span{ i }in1 H m [5].
Definition 2. Let i in1 be a frame. The linear mapping:
T : H m H n , T ( x ) x, i i 1
n
is called an analysis operator.
Definition 3. The linear mapping:
n
T * : H n H m , T * (ci in1 ) ci i
i 1
is called a synthesis operator.
The composition of T and T * defines a frame operator, which is a positive, self-conjugate reversible operator:
n
S T *T : H m H m : Sx T *Tx x, i i .
i 1
It provides the exact formula for reconstruction:
n
x x, i S 1 i .
i 1
Definition 4. A family of vectors i in1 is a uniform equiangular tight frame if
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1) 0 : || i || i 1, n ;
2) c 0 : for any pair of frame vectors j and k , j≠k, we have:
j , k c.
It is known that there is an upper bound for a number of vectors in the uniform equiangular tight frame i in1 on the m-
m ( m 1)
dimensional Hilbert space H. In the real case it is n , in the complex case it is n m 2 ([6], [7]). Creation of the
2
maximum number of vectors for the uniform equiangular tight frame is a very complex and unresolved problem in the theory of
frames.
Let us consider a non-linear mapping , which transfers the vector into a set of the absolute values of frame coefficients:
: H l 2 ( I ), ( x ) {| x, i |} in1
n
Definition 5. The frame i i 1 is called generic if i i 1 L U , where U is the Zariski open set and U Gr ( m , n ) .
n
Theorem 1 [8,9]. Let Ф i i 1 C m and the mapping A : C rm C m / T 1 R m be defined by ( A( x ))( i ) : x, i
n 2
,
i 1,.., n .
Let us consider { i i* u}in1 as vectors of the space R 2 m . Let S (u ) : spanR { i i*u}in1 . The following statements are
equivalent:
(a) A is injective.
(b) dim S (u ) 2 n 1 for every u C m \ {0} .
(c) S (u ) spanR {iu} for every u C m \ {0} .
Definition 6. The family of vectors i in1 in m has the complement property if for any I {1, . . . , n} , either
{ i }iI , or { i }iI C is complete in m [10].
Definition 7. The family of vectors i in1 R m is called a set with a full spark, if every its subset of m vectors is
complete in R m [10].
Lemma 1. Every set with the full spark i in1 in R m with n 2m 1 satisfies the complement property.
Proof. Let's assume the contrary: there is such S {1,.., n} that neither { i }iS nor { i }iS C are not complete in R m .
By definition of the full spark, from this it follows that S m 1 and S C m 1 , that is n 2m 2 , which contradicts the
condition.
Theorem 2. In the real case if i in1 in R m and n 2m 2 , then mapping A is not injective.
If n 2m 1 , then mapping A is injective if and only if when i in1 is a full spark.
C
Proof. If n 2m 2 , then the set {1,.., n} can be divided into sets S and S such that the cardinality of each would not
exceed m 1 . None of the sets { i }iS , { i }iS C can be complete.
If n 2m 1 and i in1 is a full spark, then the injectivity of A follows from lemma 1 and theorem 1.
And vice-versa, if A is injective, then i in1 is an alternatively full family. Let's take an arbitrary subset S {1,.., n}
with S m . Then S C m 1 and { i }iS C can’t be full. Therefore, { i } iS is full, and i in1 is a full spark.
The exact minimum bound is unknown for the complex case. Besides, in the real case there is a simple direct
method for checking infectivity of the mapping A for the corresponding frame [7].
Theorem 3 [4, 11].
m ( m 1)
(a) If H R m , n and i in1 is a generic frame, the nonlinear map P is injective. Then the vector x H
2
can be reconstructed (up to a sign) from the set {| x, i |} in1 of absolute values of the frame coefficients in a polynomial
number (O(m 6 )) of steps.
(b) If H C m , n m 2 and i in1 is a generic frame, the nonlinear map P is injective. Then the vector x H can be
reconstructed (up to multiplication by a root of unity) from the set {| x, i |} in1 of absolute values of the frame coefficients in a
polynomial number (O(m 6 )) of steps.
4. About Equivalence of Phase Retrieval and Phaseless Reconstruction
Let x ( a1 , a 2 ,..., a m ) and y (b1 , b 2 ,..., bm ) be vectors in a space m .
Definition 1: For the phase of number z C m , we take the value of the angle ph z i 2k , k Z , defining the
deviation of the radius vector of the point on the plane, corresponding to z C m , from the real axis in C m . In the real case, the
phase in R m is equal to 0 or .
We shall say that x, y have the same phases if:
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ph a i ph bi , i 1,2,..., m .
Definition 2. Let be a family of vectors in m (respectively, {Pi }in1 be a family of projections on m )
i in1
satisfying the following property: for every x, y the following condition is satisfied:
x, i y, i , for all i 1, 2,..., n .
(Respectively,
|| Pi x || || Pi y || , for all i 1,2,..., n). .
Then
1) If there is a 1 such that x and y have the same phases, one can say i in1 does phase retrieval (respectively,
{Pi }in1 does phase retrieval).
2) If there is a 1 such that x y , one can say i in1 does phaseless reconstruction. (Respectively, {Pi }in1 does
phaseless reconstruction.)
Definition 3. Let us call a family of vectors i in1 in m an alternatively full one, if for any I {1, . . . , n} ,
either { i }iI , or { i }iI C is complete in m .
If { i }in1 retrieves phases in m then span{ i }in1 m . This means that { i }in1 is a frame in the space m .
Otherwise, there exists 0 x m such that x , i y , i 0 , i 1,2,..., n, while phases of vectors x and 0 are
not the same.
If x ( a1 , a 2 ,..., a m ) and y (b1 , b2 ,..., bm ) have the same phases then ai 0 if and only if bi 0 , for the
phase of 0 is not determined.
Theorem 1. Let i in1 be a set of vectors in R m . The mapping A : R m /{1} R n ( n m) is defined by
2
( A( x ))( i ) : x , i , i 1,.., n . If i in1 does phaseless reconstruction, then it has a complement property. In the real case
these concepts are equivalent.
Proof. ( ) Assume that fails the complement property. Then there exists I {1,.., n} such that neither { i } iI , nor
{ i }iI C is complete in R m .
We choose nonzero vectors u, v R m such that u , i 0 for all i I and v, i 0 for all i I C . For every i we then
have:
________
2 2 2 2 2
u v, i u, i 2 u, i v, i v, i u, i v, i .
2 2
From this it follows that u v, i u v, i for every i , and A(u v ) A(u v ) . Moreover, since u and v are
nonzero by assumption, then u v (u v ) . Thus there is no phaseless reconstruction.
() Assume that i in1 fails phaseless reconstruction. That means there exist vectors x , y R m such that x y
and A( x ) A( y ) . Take I : {i : x, i y, i } .
We have: x y , i 0 for every i I . Otherwise if i I C , we have x, i y , i and then x y , i 0 . According
to the assumption, x y , therefore x y 0 and x y 0 . Thus, neither { i } iI or { i }iI C are complete in R m .
In R m the phase of a vector can be equal to 0 or . Coordinates of the vectors have the same phases if signs of the
coordinates of the vector x are the same as those of the vector y . At the same time the phase of 0 is not defined. That is, if
x ( a1 , a 2 ,..., a m ) and y (b1 , b2 ,..., b m ) , then x, y have the same phase in the following cases:
1) If a i 0 bi , then a i bi 0 .
2) If a i 0 , then corresponding to it bi 0 (it is symmetric: if bi 0 , then corresponding to it a i 0 ).
Otherwise, the vectors have different phases.
Then, if we are given two vectors, so as to define whether their phases are equal or not it is necessary:
1) To check equality of all indices of zero coordinates of the vectors. If all indexes of zero coordinates of the first vector
correspond to the indexes of the second vector (and vice versa), then it is necessary to check 2), otherwise, the vectors have
different phases.
2) For nonzero coordinates to check fulfillment of the following condition: if a i bi 0 the vectors have the same phases,
and if a i bi 0 then the vectors have different phases.
Definition 1 in the real case will mean:
Let i in1 be a set of vectors in R m , satisfying the following property: for every x, y the following condition is
fulfilled:
x, i y, i , i 1, 2,..., n .
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Then,
1) i in1 does phases reconstruction if there exist 1 such that
а) For 1 the vectors x and y have the same phase.
b) For 1 the vectors x and y have the same phase.
2) i in1 does phaseless reconstruction if there exists 1 such that
с) For 1 x y .
d) For 1 x y .
Theorem 2. Let i in1 be a set of vectors in R m . The mapping A : R m /{1} R n ( n m) is defined by
2
equations ( A( x))( i ) : x, i , i 1,.., n . If i in1 does phase retrieval, then it has the complement property. In the real
case these concepts are equivalent.
Proof. Assume that does phase retrieval, but fails phaseless reconstruction. Assume that the set i in1 fails
complement property, that is there exists I {1,.., n} such that neither { i } iI , nor { i }iI C is complete in R m .
Let us choose nonzero vectors x (a1 , a 2 ,..., a m ), y (b1 , b2 ,..., bm ) R m such that x, i 0 for all i I and y, i 0
for all i I C . Then for some i either x, i 0 , or y, i 0 . Fix c 0 , so that for every 1 i n
x cy, i x cy, i
Then
2 2
x cy, i x cy, i .
By assumption, does phase retrieval, and it means that there exists 1 such that ( x cy ) and ( x cy ) have the
ai 0
same phases. Let's assume that there exists 1 i 0 m such that a i 0 0 bi 0 and let c . Then
bi 0
ai 0
( x cy) i 0 ai 0 cbi 0 ai 0 bi 0 0
bi 0
and
ai 0
( x cy) i 0 ai 0 cbi 0 ai 0 bi 0 2ai 0 0 .
bi 0
But it is impossible because if x and y have the same phases, then a i 0 if and only if bi 0 .
As the vectors x and y are nonzero, then the last two equalities are possible if and only if either a i 0 , or bi 0 ,
m
1 i m . Let I {1 i m : bi 0} and {ei }i 1 be an orthonormalized basis in R m . Then
x y ai ei bi ei and x y ai ei (bi )ei .
iI iI C iI iI C
Then there exists 1 such that ( x y ) and ( x y ) have the same phases and they are equal. We have arrived at a
contradiction.
Let's consider an example in R 2 . Let x ( a1 , a 2 ) ( 0,0) and y (b1 , b2 ) ( 0,0) .
For i 3i 1 , we take the Mercedes-Benz frame in R 2 , consisting of 3 unit vectors located at an angle of 120º (Fig. 2):
3 1 3 1
1 ( 0,1), 2 ( , ), 3 ( , ) .
2 2 2 2
2
Fig. 2. The Mercedes-Benz Frame in R , consisting of 3 unit vectors located at an angle of 120º.
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3
Then fulfillment of the condition x, i y, i means that
i 1
( a1 , a 2 ), (0,1) (b1 , b2 ), (0,1) a 2 b2
3 1 3 1 3 1 3 1
( a1 , a 2 ), ( , ) (b1 , b2 ), ( , ) a1 a 2 b1 b2
2 2 2 2 2 2 2 2
3 1 3 1
( a , a ), ( 3 , 1 ) (b , b ), ( 3 , 1 ) a1 a 2 b1 b2
1 2 1 2
2 2 2 2 2 2 2 2
a 2 b2
3a1 a 2 3b1 b2 .
3a1 a 2 3b1 b2
Squaring the equations of the last system we obtain that
a1 a 2 b1b2 and a12 b12 .
From this it follows that the first equality gives coincidence of the signs (to the multiplier), and the second – coincidence of
the absolute values of coordinates. Moreover, from these equalities it also follows that zero coordinates are the same, if any.
We obtain that either x y , or x y . Then there really exists 1 such that if i 3i 1 is the Mercedes-Benz
frame, it does both phases reconstruction (because the vectors have the same signs, which means that the phases are the same
too) and phaseless reconstruction (for x y ) at the same time.
If we know the absolute values of coordinates of the vectors x ( a1 , a 2 ) and y (b1 , b2 ) , then x and y can be one of 4
vectors (Fig. 3):
a2 b2
a1 b1
Fig. 3. Possible vectors for the known absolute values of coordinates of the vectors. x, y .
After scalar multiplication by the frame coordinates, the condition a1 a 2 b1b2 imposes restrictions on the signs of
coordinates (Fig. 4):
a1 a 2 b1b2 a1 a 2 b1b2
a2
a1
a1
a2
Fig. 4. Possible vectors for the known x, i , y , i .
Let's consider an example in C 2 . In the complex case:
x ( x1 , x 2 ) ( a ib, c id ) and y ( y1 , y 2 ) ( e if , g ih) .
As i 5i 1 let's take a frame of the following type:
1 1 1 1
1 (1,0), 2 ( , ), 3 ( , ), 4 ( 1 , 1 i ), 5 ( 1 , 1 i ) .
2 2 2 2 2 2 2 2
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5
Then fulfillment of the condition x, i y , i means that
i 1
( x1 , x 2 ), (1,0) ( y1 , y 2 ), (1,0)
1 1 1 1
( x1 , x 2 ), ( , ) ( y1 , y 2 ), ( , )
2 2 2 2
1 1 1 1
( x1 , x 2 ), ( , ) ( y1 , y 2 ), ( , ) .
2 2 2 2
( x , x ), ( 1 , 1 i ) ( y1 , y 2 ), (
1
,
1
i)
1 2
2 2 2 2
1 1 1 1
( x1 , x 2 ), ( , i) ( y1 , y 2 ), ( , i)
2 2 2 2
Then x, i ( x1 , x 2 ), ( i1 , i 2 ) x j ij and x, i y , i
5
mean that:
i 1
j 1, 2
i 1,5
x1 y1 x1 x1
1 x x 2 x1 x 2 1 x x 2 y1 y 2
x1 x 2 x1 x 2 x x y y
1 2 1 2 .
x1 x 2 i x1 x 2 i x1 x 2 i y1 y 2 i
x1 x 2 i x1 x 2 i y1 y 2 i y1 y 2 i
Let's rewrite the system in the following form:
a2 b2 e2 f 2
2 2 2 2
( a c ) (b d ) ( e g ) ( f h )
( a c ) 2 ( b d ) 2 ( e g ) 2 ( f h ) 2 .
2 2 2 2
( a d ) (b c ) ( e h ) ( f g )
( a d ) 2 (b c ) 2 (e h ) 2 ( f g ) 2
From the last system we obtain
c2 d 2 g 2 h2 .
And it means that the absolute values of the second complex coordinates of the vectors x and y are equal, because:
c 2 d 2 x2 = g 2 h 2 y 2 .
1 1 1 1
So, if we take a frame of the type 1 (1,0), 2 ( , ), 3 ( , ), 4 ( 1 , 1 i ), 5 ( 1 , 1 i ) ,
2 2 2 2 2 2 2 2
then the absolute values of the corresponding complex coordinates of the vectors x and y are equal, i.e.:
x1 y1 r1 and x 2 y 2 r2 .
Now, let us write down the coordinates of the vectors in the polar form, taking into account the equality of the absolute
values of the coordinates:
x ( x1 , x 2 ) ( r1e i1 , r2 e i2 ) and y ( y1 , y 2 ) ( r1e i 1 , r2 e i 2 )) .
5
Then x, i y , i will look as follows:
i 1
( r1 e i1 , r2 e i 2 ), (1,0) ( r1 e i 1 , r2 e i 2 ), (1,0)
1 1 1 1
( r1 e i1 , r2 e i 2 ), ( , ) ( r1 e i 1 , r2 e i 2 ), ( , )
2 2 2 2
1 1 1 1
( r1 e i1 , r2 e i 2 ), ( , ) ( r1 e i 1 , r2 e i 2 ), ( , )
2 2 2 2 .
i 1 i 2 1 1 1 1
( r1 e , r2 e ), ( 2 , 2 i ) ( r1 e i 1 , r2 e i 2 ), ( , i)
2 2
i 1 1 1 1
( r1 e 1 , r2 e
i 2
), ( , i) ( r1 e i 1 , r2 e i 2 ), ( , i)
2 2 2 2
From this it follows that 1 1 and 2 2 . We obtain that phases of the vectors x and y are equal to 2k , k .
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5. Weak phase retrieval
Definition 1. Two vectors x ( a1 , a 2 ,..., a m ) and y (b1 , b 2 ,..., bm ) do weak phase retrieval if there is a 1 such that
phase a i phase bi , for all i 1, 2,..., m , such that a i 0 bi .
In the real case, if 1 , we say that x, y have weakly like signs and if 1 they have weakly opposite signs.
Definition 2. A family of vectors i in1 in m does weak phase retrieval if from equalities
x, i y, i , i 1,2,..., n .
It follows that there exists a 1 , such that
phase xi phase y i , for all i 1,2,..., m , so that a i 0 bi .
The weak phase retrieval differs from the phase retrieval described in Definition 2 of Section 4 in that there can be both
a i 0 and bi 0 .
Let us consider an example where the weak phase retrieval is done but the phase retrieval by Definition 2 of Section 4 is
failed. Let us consider in R m a set of vectors i im11 , which coordinates form the following matrix columns:
1 1 1 1 1
1 1 1 1 1
A .
1 1 1 1 1 m( m 1)
2 2
Then for every x ( a1 , a 2 ,..., a m ) and y (b1 , b 2 ,..., bm ) , if x, i y, i , it follows that ai a j bi b j for all i j .
This set of ( m 1) -vectors in R m will do weak phase retrieval.
Theorem 1: Let x ( a1 , a 2 ,..., a m ) and y (b1 , b 2 ,..., bm ) be two vectors in R m . Then the following statements are
equivalent:
1) sgn(ai a j ) sgn(bi b j ) , for all 1 i j m .
2) x, y have either weakly like signs or weakly opposite signs.
6. Conclusion
In case the phase information is not available, the signal recovery is theoretically possible if redundant systems called frames
are used as the system of representation. A well-chosen frame can provide numerical stability for recovery of the signal and
obtaining important characteristics of the signal.
In the real case under certain conditions a generic frame consisting of (2m-1)-vectors can do phaseless reconstruction. In the
complex space a generic frame consisting of (4m-2)-vectors can do the same under certain conditions. A family of frames
recovers the signal by the absolute value of frame coefficients in polynomial time. The issue of the equivalence of phases
retrieval and phaseless reconstruction is raised and partially resolved. Examples of signal recovery in small-dimension
spaces are given.
A search and theoretical justification of new methods of recovery of information hidden in phases of transmitted signals and
unavailable for measurements by publicly available physical instruments is conducted. The technique is based on the latest
achievements in the research of complete linearly dependent systems called space frames.
Acknowledgements
The authors thank S.Ya. Novikov for his fruitful discussions.
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