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Multidimensional Signal Recovery in Discrete Evolution Systems via Spatiotemporal Trade Off

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Abstract

The problem of recovering an evolving signal from a set of sam- ples taken at different time instances, has been well-studied for one-variable signals modeled by l2(Zd) and l2(Z): However, most observed time-variant signals in applications are described by at least two spatial variables. In this paper, we study the spatiotemporal sampling pattern to recover the initial signals modeled by l2(Zd1 × Zd2) and l2(Z × Z) which are evolving in a dis-crete evolution system and provide specic reconstruction results.
SAMPLING THEORY IN SIGNAL AND IMAGE PROCESSING
c
2015 SAMPLING PUBLISHING
Vol. 14, No. 2, 2015, pp. 153–169
ISSN: 1530-6429
Multidimensional Signal Recovery in Discrete Evolution Systems via
Spatiotemporal Trade Off
Roza Aceska
Department of Mathematical Sciences, Ball State University
Muncie, IN, USA
e-mail address:raceska@bsu.edu
Armenak Petrosyan
Department of Mathematics, Vanderbilt University
Nashville, TN, 37240, USA
e-mail address: armenak.petrosyan@vanderbilt.edu
Sui Tang
Department of Mathematics, Vanderbilt University
Nashville, TN, 37240, USA
e-mail address: sui.tang@vanderbilt.edu
Abstract. The problem of recovering an evolving signal from a set of sam-
ples taken at different time instances, has been well-studied for one-variable
signals modeled by 2(Zd) and 2(Z).However, most observed time-variant
signals in applications are described by at least two spatial variables. In this
paper, we study the spatiotemporal sampling pattern to recover the initial
signals modeled by 2(Zd1×Zd2) and 2(Z×Z) which are evolving in a dis-
crete evolution system and provide specific reconstruction results.
Key words and phrases : Distributed sampling, reconstruction, frames.
2010 AMS Mathematics Subject Classification94A20,94A12,42C15,15A29.
1. Introduction
The ongoing development [1, 2, 4, 8, 9, 19] in sampling theory suggests to
combine coarse spatial samples of a signal’s initial state with its later-time sam-
ples. In these cases, time-dependency among samples permits a reduction of the
number of used expensive sensors via the increase of their usage frequency. The
reconstruction of the initial distribution of a signal is achieved by the use of the
evolutionary nature of that signal under some certain constraints, which is not
fully considered in classical sampling problems [3, 5, 6, 10, 12, 14, 16, 17, 18].
The so-called dynamical sampling problem (motivated by [13, 15]) has been
well-studied in the one-variable setting [4, 8, 9, 1, 2] but there have been no
results in the multivariable setting. In industrial applications (sampling of air
pollution, wireless networks) the observed time-variant signals are described by
154 R. ACESKA, A. PETROSYAN AND S. TANG
at least two variables. In this paper, we formulate the problem of spatiotemporal
sampling for two-variable data and provide specific reconstruction results.
1.1. Stating the Dynamical Sampling Problem. In real-life situations,
physical systems evolve over time, under the influence of a family of opera-
tors {At}t0. Let f0be the initial state defined on a domain D. Dynamical
sampling problem asks when we can recover the initial state f0by the spa-
tiotemporal sampling data {SXtiAtif0:i= 1, . . . , N 1}, where SXtiis a sub-
sampling operator defined by a coarse sampling set XiDat time instances
ti,i= 0, ..., N 1. In other words, we would like to compensate for the lack
of sufficient samples of the initial state by adding coarse samples of the evolved
states {Atif0=fti, i = 0, . . . , N 1}. In this way, we can use fewer sampling
devices to save budget but lose no information.
The dynamical sampling problem is solved when conditions on the sampling
sets and time instances tiare found, such that recovery of the signal is possible,
preferably in a stable way. That is, if one (or both) of the following properties
is satisfied:
ISP Invertibility sampling property. The operators At0, ..., AtN1, the sam-
pling sets Xt0, ..., XtN1and the number of repeated sampling proce-
dures satisfy this condition within a class of signals, if any signal hin
that class is uniquely determined by its samples data set.
SSP Stability sampling property. The operators, the sampling sets and the
number of repeated samplings satisfy this condition in a fixed class of
signals, if for any two signals h,h1in that class, the two norms
hh12
2and
N
i=0
SXtiAti(hh1)2
2are equivalent.
SSP is clearly a stronger property and implies ISP. In [1, 2, 8, 9] the authors
have studied the dynamical sampling problem for the discrete spatially invariant
evolution system, in which the initial state fis defined on the domain D=
Zdor Zunder certain constraints. At time instance t=nN, the initial state f
is altered by convolution with a filter a n times to be An(f) = aa...af=anf.
At each time instance t=n, the altered state An(f) is under-sampled at a
uniform subsampling rate m. The invertibility and stability questions have been
fully answered under the specific constraints. Namely, for a uniform discrete
sampling grid X=mD D, specific conditions on aand Nare stated so that
a function fcan be recovered from the samples
{f(X), a f(X),· · · ,(aN1f)(X)},for XD. (1.1)
The multidimensional dynamical sampling problem we consider in this paper
has similarities with problems considered by some other authors. For example,
in [11], the authors work in a multivariable shift-invariant space(MSIS) setting,
and study linear systems {Lj:j= 1,· · · , s}such that one can recover any
fin MSIS by uniformly downsampling the functions {(Ljf) : j= 1,· · · , s},
i.e. taking the generalized samples {(Ljf)(Mα)}αZd,j=1,··· ,s . In dynamical
MULTIDIMENSIONAL SIGNAL RECOVERY 155
sampling, there is only one convolution operator A, and it is applied iteratively
to the function f. This iterative structure is important for our analysis of the
kernel of the arising matrix, and using that special structure we are able to add
extra samples outside of the initial uniform sampling greed and get full recovery
of the signal.
In addition, certain singularity problems, which can occur due to the specific
properties of awhen sampling on a uniform grid X, have been successfully
overcome in the cited papers by adding additional samples. Since most real-life
phenomena are described by functions of multiple variables, we find it important
to expand the dynamical sampling concept on two variable setting, i.e. D=
Zd1×Zd2and Z×Z.As we will see later, the two variable problem is more
complicated in structure and we find it more subtle to overcome the singularity
problems. Studying the stated problem in 3 (and higher) variable setting would
require similar coping techniques to the ones we use in this paper to expand the
domain from one to two dimensions.
2. Dynamical sampling on Zd1×Zd2
For a positive integer d,Zddenotes the finite group of integers modulo d. In
the finite discrete setting, we work on the domain D=Zd1×Zd2,d1, d2N+.
Let the operator Aact on the signal of interest f2(D) as a convolution with
some a1(D) given by
Af(k, l) = af(k, l) =
(s,p)D
a(s, p)f(ks, l p),for all (k, l)D. (2.1)
Note that Ais a bounded linear operator that maps 2(D) to itself. The initial
signal fis evolving in time under the repeated effect of Asuch that at time
instance t=n, the evolved signal is fn=Anf=aa · · · af(and
f=f0=A0f).
We assume that d1and d2are odd numbers, such that di=Jimifor integers
mi1, Ji1, i= 1,2. We set the sampling sensors on a uniform coarse grid
X=m1Zd1×m2Zd2to sample the initial state fand its temporally evolved
states Af,A2f, . . . , AN1f. Note that, given such a coarse sampling grid, each
individual measurement is insufficient for recovery of the sampled state.
Let SX=Sm1,m2denote the assigned subsampling operator related to the
sampling grid. Specifically,
(SXf)(k, l) = f(k, l) if (k, l)X
0 otherwise (2.2)
For some N2, our objective is to reconstruct ffrom the combined coarse
samples set
{yj=SX(Ajf)}, j = 0,1, ..., N 1.(2.3)
We denote by Fthe 2dimensional discrete Fourier transform (2d DFT) and use
the notation ˆx=F(x). After applying Fto (2.3), due to the two-dimensional
156 R. ACESKA, A. PETROSYAN AND S. TANG
Poisson’s summation formula, we obtain
ˆyn(i, j) = 1
m1m2
m11
k=0
m21
l=0
ˆan(i+kJ1, j +lJ2)ˆ
f(i+kJ1, j +lJ2) (2.4)
for (i, j)∈ I ={0,· · · , J11}×{0,· · · , J21}and n= 0,1, . . . , N 1.
Let ¯
y(i, j) = ( ˆy0(i, j) ˆy1(i, j)... ˆyN1(i, j))T, (i, j)∈ I and
¯
f(i, j) =
ˆ
f(i, j)
ˆ
f(i+J1, j)
.
.
.
ˆ
f(i+ (m11)J1, j)
ˆ
f(i, j +J2)
.
.
.
ˆ
f(i+ (m11)J1, j +J2)
.
.
.
.
.
.
ˆ
f(i, j + (m21)J2)
.
.
.
ˆ
f(i+ (m11)J1, j + (m21)J2)
.
We use the block-matrices
Al,m1m2(i, j) =
1 1 ... 1
ˆa(i,j+lJ2) ˆa(i+J1,j +lJ2)... ˆa(i+(m11)J1,j+lJ2)
.
.
...
.
..
.
..
.
.
ˆaN1(i,j+lJ2) ˆaN1(i+J1,j +lJ2)... ˆaN1(i+(m11)J1,j+lJ2)
,
where l= 0,1, ..., m21, to define the N×m1m2matrix
Am1,m2(i, j) = [A0,m1m2(i, j )A1,m1m2(i, j)...Am21,m1m2(i, j)] (2.5)
for all (i, j)∈ I . Equations (2.4) have the form of vector inner products, so we
restate them in matrix product form
¯
y(i, j) = 1
m1m2
Am1,m2(i, j)¯
f(i, j).(2.6)
By equation (2.6), we need Nm1m2to be able to recover the signal f.
Note that for N=m1m2, matrix (2.5) is square, we denote this special square
matrix by Am1,m2(i, j) and obtain the following reconstruction result:
Proposition 1. For N=m1m2, the SSP is satisfied if and only if
det Am1,m2(i, j)̸= 0 for all (i, j)∈ I.(2.7)
In the finite dimensional case, unique reconstruction is equivalent to stable
reconstruction so SSP and ISP coincide. When (2.7) holds true, the signal is
recovered from the system of equations
¯
f(i, j) = m1m2A1
m1,m2(i, j)¯
y(i, j),(i, j )∈ I.
MULTIDIMENSIONAL SIGNAL RECOVERY 157
As expected, Proposition 1 reduces to the respective result in [8] when d=d1
and d2= 1, or d=d2and d1= 1.
2.1. Extra samples for stable spatiotemporal sampling. Proposition 1
gives a complete characterization of stable recovery from the dynamical sam-
ples (2.3). In practice, however, we may not have the ideal filter asuch that
(2.7) holds true. For instance, consider a kernel awith a so-called quadrantal
symmetry, i.e. let
ˆa(s, p) = ˆa(d1s, p) = ˆa(s, d2p) = ˆa(d1s, d2p)
for all (s, p)D. Since (2.5) is a Vandermonde matrix, it is singular if and only
if some of its columns coincide. In this case, it is easy to see that Am1,m2(0,0)
is singular, which prevents the stable reconstruction.
Motivated by the above example, we propose a way of taking extra samples
to overcome the lack of reconstruction uniqueness, whenever singularities for
matrix (2.5) occur. Let
A=
Am1,m2(0,0) 0 ... 0
0Am1,m2(1,0) ... 0
.
.
..
.
.....
.
.
0 0 ... Am1,m2(J11,J21)
and
¯
f=
¯
f(0,0)
¯
f(1,0)
.
.
.
¯
f(J11,0)
¯
f(1,1)
.
.
.
¯
f(J11,1)
.
.
.
.
.
.
¯
f(0, J21)
.
.
.
¯
f(J11, J21)
,¯
y=
¯
y(0,0)
¯
y(1,0)
.
.
.
¯
y(J11,0)
¯
y(1,1)
.
.
.
¯
y(J11,1)
.
.
.
.
.
.
¯
y(0, J21)
.
.
.
¯
y(J11, J21)
.
Then
A¯
f=¯
y(2.8)
and
ker(A) =
(i,j)∈I
ker[Am1,m2(i, j)].(2.9)
The kernels of each Am1,m2(i, j) can be viewed as generated by linearly indepen-
dent vectors ˆvj2(D) such that each ˆvjhas exactly two nonzero coordinates,
one of which is equal to 1 and the other is 1. Let’s assume that the nullity of
matrix Am1,m2(i, j) equals wi,j at each (i, j)∈ I. Then there are n=i,j wi,j
158 R. ACESKA, A. PETROSYAN AND S. TANG
of such linearly independent vectors ˆvj2(D). Let {vj:j= 1,· · · , n}be their
image under the 2Dinverse DFT. Note that {vj:j= 1,··· , n} ⊂ 2(D) is also
linearly independent.
Let Ω D\Xbe the additional sampling set, that is to say, we take extra
spatial samples of the initial state fat the locations specified by Ω. By Swe
denote the related sampling operator and Ris a || × nmatrix with rows cor-
responding to [v1(k, l),· · · , vn(k , l)]{(k,l)}. With these notations, the following
result holds true:
Theorem 2.1. The reconstruction of f2(D)from its spatiotemporal samples
{Sf, SXf , SXAf, · · · , SXAm1m21f}(2.10)
is possible in a stable manner (SSP is satisfied) if and only if rank(R) = n.
In particular, if SSP holds true, then we must have || ≥ n.
Proof. Let W=span{vj:j= 1,· · · , n}. It suffices to show that
ker(S)W={0}if and only if rank(R) = n.
Suppose wis in ker(S)W. There must exist coefficients c1, c2, .., cnso that
w=n
j=1 cjvjand Sw= 0. The last statement is equivalent to
[v1(k, l), v2(k, l),· · · , vn(k, l)] [c1c2... cn]T= 0
for each (k, l)Ω. Equivalently, we have Rc= 0. Hence, c= 0 if and only if
rank(R) = n.
Since the d1d2×nmatrix R= [v1(k, l),· · · , vn(k.l)]{(k,l)D}has column rank
n, for any kernel a, there exists a minimal choice of Ω, namely ||=nsuch that
the square matrix Ris invertible. It is hard to give a formula to specify the
extra sampling set for every kernel a2(D). On the other hand, compared
to the 1variable case [8], it is more challenging to specify the rank of R
analytically, since the entries of Rwill involve the product of sinusoids mixed
with exponentials in general.
In [8], the authors studied a typical low pass filter with symmetric properties
and gave a choice of a minimal extra sampling set Ω, since symmetry reflects
the fact that there is often no preferential direction for physical kernels and
monotonicity is a reflection of energy dissipation. Similarly, we consider a kernel
awith a so-called strict quadrantal symmetry: for a fixed (k, l)D, ˆa(s, p) =
ˆa(k, l) if and only if
(s, p)∈ {(k, l),(d1k, l),(k, d2l),(d1k, d2l)}.(2.11)
Since Am(i, j) is a Vandermonde matrix, it has singularity if and only if some
of its columns coincide. We can compute the singularity of each Am(i, j), as we
make use of its special structure.
Lemma 2.2. If the filter asatisfies the symmetry assumptions (2.11), then
dim(ker(A)) = d1(m21)
2+d2(m11)
2(m11)(m21)
4.
MULTIDIMENSIONAL SIGNAL RECOVERY 159
Clearly, we need an extra sampling set Ω Dwith size dim(ker(A)). Based
on Theorem 2.1, we provide a minimal Ω:
Theorem 2.3. Assume that the kernel asatisfies the strict quadrantal symmetry
assumptions (2.11) and let
Ω = {(k, l) : k= 1 · · · m11
2, l Zd2}∪{(k, l) : kZd1, l = 1,· · · ,m21
2}.
Then, any f2(D)is recovered in a stable way from the expanded set of
samples
{Sf, SXf , SXAf, ··· , SXAm1m21f}.(2.12)
Remark 2.4.Note that in this case
||=d1(m21)
2+d2(m11)
2(m11)(m21)
4,
so by Theorem 2.1 and Lemma 2.2 we can not do better in terms of its cardinality.
Proof. Set
n=d1(m21)
2+d2(m11)
2(m11)(m21)
4.
Recall that the kernels of singular blocks Am1,m2(i, j) are generated by vectors
{ˆvk:k= 1,· · · n}, such that each ˆvkhas exactly two non-zero components, 1
and 1 (corresponding to each pair of identical columns). Then the formula of
2Dinverse DFT gives
vj(k, l) =
d11
s=0
d21
p=0
ˆvj(s, p)e2πisk
d1e
2πipl
d2,(k, l)Zd1×Zd2.(2.13)
We define a row vector F1(k) = 1, e2π ik
d1,··· , e
2πi(d11)k
d1for all kZd1. For
each l= 0,1,· · · , d21, we define a row vector ¯
F2(l) of length d2m21
2, which
is derived from vector
[1, e 2πil
d2,· · · , e
2πi(d21)l
d2]
after deleting the entries that correspond to {sJ2+ 1 : 1 sm21
2}, i.e. we
omit the entries e
2πsJ2
d2for 1 sm21
2. We reorder the vectors vjso that
[v1(k, l),· · · , vn(k, l)] equals
2isin( 2π1l
m2)F1(k),··· ,sin( 2π(m21)l
2m2)F1(k),sin( 2π1k
m1)¯
F2(l)··· ,sin( 2π(m11)k
2m1)¯
F2(l)
for every (k, l)Ω. By Theorem 2.1, the proof is complete if we show that these
n=||row vectors of size nare linearly independent.
We define a row vector R(k, l) corresponding to (k, l)Ω given by
2isin( 2π1l
m2)F1(k),··· ,sin( 2π(m21)l
2m2)F1(k),sin( 2π1k
m1)¯
F2(l)··· ,sin( 2π(m11)k
2m1)¯
F2(l).
160 R. ACESKA, A. PETROSYAN AND S. TANG
Suppose that for some coefficients {c(k, l) : (k, l)}, it holds
(k,l)
c(k, l)R(k, l) = 0.
We need to show that all c(k, l) = 0. Note that, for a fixed k, the vector R(k, l)
is compartmentalized into two components with lengths m21
2and m11
2. By
construction, {F1(k)|kZd1}are linearly independent row vectors. Then, the
coefficients related to F1(k) for the first component should be zeros. Related to
the first component of length m21
2, for every fixed kZd1such that (k, l)
for some l, the following m2equations hold true
(k,l)
c(k, l) sin 2πsl
m2= 0 for s= 0,1, ..., m21.(2.14)
Case I: if km1+1
2or k= 0, then (k, l)Ω if and only if l= 1,· · · m21
2.
We restate the system of equations (2.14) in the matrix form:
sin( 2π
m2) sin( 4π
m2). . . sin(π(m21)
m2)
sin( 4π
m2) sin( 8π
m2). . . sin(2π(m21)
m2)
.
.
..
.
.....
.
.
sin(π(m21)
m2) sin(2π(m21)
m2). . . sin(π(m21)(m21)
2m2)
c(k, 1)
c(k, 2)
.
.
.
c(k, m21
2)
=0.
The matrix on the left-hand side is invertible, since
{sin(2πx), sin(4πx), ..., sin((m21)πx)}
is a Chebyshev system on [0,1](see[7]); Hence we have c(k, l) = 0 for
l= 1,· · · m21
2.
Case II: if 1 km11
2, then (k, l)Ω if and only if l= 0,· · · , d21.
Then (2.14) is equivalent to the system of equations
d21
l=0
c(k, l) sin 2πsl
m2= 0 for s= 1,2, ..., (m21)/2.(2.15)
Related to the second component of length m11
2, and combined with
the fact that c(k, l) = 0 if kis in case I, for all s= 1,2, ..., m11
2we have
d21
l=0
m11
2
k=1
c(k, l) sin 2πsk
m1¯
F2(l)
= 0.(2.16)
Let ¯
F2= [ ¯
F2(0)T,· · · ,¯
F2(d21)T], where ¯
F2(l)Tdenotes the transpose
of each row vector ¯
F2(l); ¯
F2is a (d2m21
2)×d2matrix. Using matrix
MULTIDIMENSIONAL SIGNAL RECOVERY 161
notation, the first equation in (2.16) can be restated as a product, namely
¯
F2·
m11
2
k=1
sin 2πk
m1c(k, 0)
m11
2
k=1
sin 2πk
m1c(k, 1)
.
.
.
m11
2
k=1
sin 2πk
m1c(k, d21)
= 0
As an easy consequence of equation (2.15), for each 1 jm21
2, it
holds m11
2
k=1
sin 2πk
m1d21
l=0 sin(2πlj
m2
)c(k, l)= 0,(2.17)
which is equivalent to
m11
2
k=1
d21
l=0
sin 2πlj
m2sin 2πk
m1c(k, l) = 0,
i.e.
d21
l=0
sin 2πlj
m2
m11
2
k=1
sin 2πk
m1c(k, l) = 0.(2.18)
We define a m21
2×d2matrix Eas follows:
E=
sin(2π·0
m2) sin(2π·1
m2). . . sin(2π(d21)
m2)
sin(4π·0
m2) sin(4π∗·1
m2). . . sin(4π(d21)
m2)
.
.
..
.
.. . . .
.
.
sin(π(m21)·0
m2) sin(2π(m21)
m2). . . sin(π(m21)(d21)
m2)
.
Due to (2.18), we have
E·
m11
2
k=1
sin(2πk
m1
)c(k, 0)
m11
2
k=1
sin(2πk
m1
)c(k, 1)
.
.
.
m11
2
k=1
sin(2πk
m1
)c(k, d21)
=0.(2.19)
162 R. ACESKA, A. PETROSYAN AND S. TANG
Let F2=E
¯
F2. Then
F2·
m11
2
k=1
sin(2πk
m1
)c(k, 0)
m11
2
k=1
sin(2πk
m1
)c(k, 1)
.
.
.
m11
2
k=1
sin(2πk
m1
)c(k, d21)
=0.
Note that the d2×d2matrix F2is invertible, since it is the image of a
series of elementary matrices acting on the d2×d2DFT matrix (one row
minus another row). Hence we have
m11
2
k=1
sin(2πk
m1
)c(k, 0)
m11
2
k=1
sin(2πk
m1
)c(k, 1)
.
.
.
m11
2
k=1
sin(π(m11)k
m1
)c(k, d21)
=0.(2.20)
After analyzing the rest of the equations in (2.16), we obtain:
m11
2
k=1
sin(2πjk
m1
)c(k, s) = 0 for j= 2,· · · ,m11
2,s= 0,1, ..., d21.
In a similar manner, for each l= 0,· · · , d21 we obtain the matrix equation
sin( 2π
m1) sin( 4π
m2). . . sin(π(m11)
m1)
sin( 4π
m1) sin( 8π
m2). . . sin(2π(m11)
m1)
.
.
..
.
.....
.
.
sin(π(m11)
m1) sin(2π(m11)
m1). . . sin(π(m11)(m11)
2m1)
c(1, l)
c(2, l)
.
.
.
c(m11
2, l)
= 0.
As the matrix on the left hand side is invertible, we must have c(k, l) = 0 for
k= 1,· · · ,m11
2.
We have demonstrated that c(k, l) = 0 for all (k, l)Ω. Therefore the n
row vectors {R(k, l)}(k,l)are linearly independent i.e. stability of the signal
recovery is achieved.
MULTIDIMENSIONAL SIGNAL RECOVERY 163
3. Dynamical sampling in 2(Z×Z)
In this section, we aim to generalize our results to signals of infinite length.
Somewhat surprisingly, there is not much difference between the techniques used
in these two settings and we feel that we can gloss over a few details in the second
part without overburdening the reader.
Let D=Z×Z. We study a signal of interest f2(D) that evolves over time
under the influence of an evolution operator A. The operator Ais described by
a convolution with a1(D), namely
Af(p, q) = af(p, q) =
kZ
lZ
a(k, l)f(pk, q l) at all (p, q)D.
Clearly, Ais a bounded linear operator, mapping 2(D) to itself. Given integers
m1,m21, we assume m1and m2are odd number. We introduce a coarse
sampling grid X=m1Z×m2Z. We make use of a uniform sampling operator SX,
defined by (SXf)(k, l) = f(m1k, m2l) for (k, l)D. The goal is to reconstruct
ffrom the set of coarse samples
y0=SXf
y1=SXAf
...
yN1=SXAN1f.
(3.1)
Similar to the work done in section 2, we study this problem on the Fourier
domain. Due to Poisson’s summation formula, we have the Lemma below.
Lemma 3.1. The Fourier transform of each ylin (3.1) at (ξ, ω)T×Tis
ˆyl(ξ, ω) = 1
m1m2
m21
j=0
m11
i=0
ˆalξ+i
m1
,ω+j
m2ˆ
fξ+i
m1
,ω+j
m2.(3.2)
Expression (3.2) allows for a matrix representation of the dynamical sampling
problem in the case of uniform subsampling. For j= 0,1,· · · , m21, we define
N×m1matrices
Aj,m1,m2(ξ, ω) = ˆakξ+l
m1
,ω+j
m2k,l
,
where k= 0,1,· · · , N 1, l= 0,1,··· , m11 and denote by Am1,m2(ξ, ω) the
block matrix
[A0,m1,m2(ξ, ω)A1,m1,m2(ξ, ω)... Am21,m1,m2(ξ, ω)].(3.3)
Let ¯
y(ξ, ω) = (ˆy0(ξ, ω) ˆy1(ξ, ω)...ˆyN1(ξ, ω) )Tand
164 R. ACESKA, A. PETROSYAN AND S. TANG
¯
f(ξ, ω) =
ˆ
f(ξ
m1,ω
m2)
ˆ
f(ξ+1
m1,ω
m2)
.
.
.
ˆ
f(ξ+m11
m1,ω
m2)
ˆ
f(ξ
m1,ω+1
m2)
.
.
.
ˆ
f(ξ+m11
m1,ω+1
m2)
.
.
.
.
.
.
ˆ
f(ξ
m1,ω+m21
m2)
.
.
.
ˆ
f(ξ+m11
m1,ω+m21
m2)
.(3.4)
Due to (3.2), it holds
¯
y(ξ, ω) = 1
m1m2
Am1,m2(ξ, ω)¯
f(ξ, ω).(3.5)
Proposition 2. ISP is satisfied if and only if Am1,m2(ξ , ω)as defined in (3.3)
has full column rank m1m2at a.e. (ξ , ω)T×T, where T= [0,1) under
addition modulo 1. SSP is satisfied if and only if Am1,m2(ξ, ω)is full rank for
all (ξ, ω)T×T.
By Proposition 2, we conclude that Nm1m2. In particular, if N=m1m2,
then Am1,m2(ξ, ω) is a square matrix, we denote by Am1,m2(ξ, ω) this square
matrix.
Corollary 1. When N=m1m2, the invertibility sampling property is equivalent
to the condition:
det Am1,m2(ξ, ω)̸= 0 for a.e. (ξ, ω)T×T.
Since Am1,m2(ξ, ω)has continuous entries, the stable sampling property is equiv-
alent to
det Am1,m2(ξ, ω)̸= 0 for all (ξ, ω)T×T.
From here on we assume N=m1m2. By its structure, Am1,m2(ξ, ω) is a
Vandermonde matrix, thus it is singular at (ξ, ω)T×Tif and only if some of
its columns coincide. In case Am1,m2(ξ, ω) is singular, no matter how many times
we resample the evolved states Anf,n > N 1, on the grid Ωo=m1Z×m2Z,
the additional data is not going to add anything new in terms of recovery and
stability. In such a case we need to consider adding extra sampling locations to
overcome the singularities of Am1,m2(ξ, ω).
MULTIDIMENSIONAL SIGNAL RECOVERY 165
3.1. Additional sampling locations. If Am1,m2(ξ, ω) is singular at some (ξ, ω),
then by Corollary 1 the recovery of f2(Z2) is not stable. To remove the sin-
gularities and achieve stable recovery, some extra sampling locations need to
be added. The additional sampling locations depend on the positions of the
singularities of Am1,m2(ξ, ω) that we want remove. We propose a quasi-uniform
way of constructing the extra sampling locations and give a characterization
specifying when the singularity will be removed. Then, we use this method to
remove the singularity of a strict quadrantally symmetric convolution operator.
Let the additional sampling set be given by
Ω = {X+ (c1, c2)|(c1, c2)WZm1×Zm2}.(3.6)
Let Tc1,c2denote the translation operator on 2(Z2), so that Tc1,c2f(k, l) = f(k+
c1, l +c2) for all (k, l)Z2. We employ a shifted sampling operator SXTc1,c2to
take extra samples at the initial time instance; this means that our subsampling
grid is shifted from X=m1Z×m2Zto (c1, c2) + Xand the extra samples are
given as
hc1,c2
m1,m2=Sm1,m2Tc1,c2f, (c1, c2).(3.7)
Set
uc1,c2(s, p) = e2πi c1s
m1e2πi c2p
m2,
for (s, p)Zm1×Zm2.
By taking the Fourier transform of the samples on the additional sampling set
Ω, we obtain
ˆ
hc1,c2
m1,m2(ξ, ω) = e2πi(c1ξ
m1+c2ω
m2)
m1m2
m11
s=0
m21
p=0
uc1,c2(s, p)ˆ
fξ+s
m1
,ω+p
m2.(3.8)
where
uc1,c2(s, p) = e2πi c1s
m1e2πi c2p
m2.
For each (c1, c2)W, we define a row vector
uc1,c2={uc1,c2(s, p)}(s,p)X
with terms arranged in the same order as the terms in vector ¯
f(ξ, ω) in (3.4).
We organize the vectors uc1,c2in a matrix ¯
U= (uc1,c2)(c1,c2)Wand extend the
data vector ¯
y(ξ, ω) in (3.5) into a big vector Y(ξ, ω) by adding
{e2πi c1ξ
m1e2πi c2ω
m2(Sm1,m2Tc1,c2f(ξ, ω)}(c1,c2)W.
Then (3.2) and (3.8) can be combined into the following matrix equation
Y(ξ, ω) = 1
m1m2¯
U
Am1,m2(ξ, ω)¯
f(ξ, ω).(3.9)
166 R. ACESKA, A. PETROSYAN AND S. TANG
Proposition 3. If a left inverse for
¯
U
Am1,m2(ξ, ω)
exists for every (ξ, ω)T2, then the vector fcan be uniquely and stably recovered
from the combined samples (3.1) and (3.6) via (3.9).
If the following property holds true:
ker( ¯
U)ker(Am1,m2(ξ, ω)) = 0 (3.10)
for every (ξ, ω) in T2,we say that Wremoves the singularities of Am(ξ, ω); In
such a case, the assumption in Proposition 3 is satisfied.
Corollary 2. If Wremoves the singularities of Am(ξ, ω)then
|W| ≥ dim(ker(Am1,m2(ξ, ω)))
for every (ξ, ω).
3.2. Strict quadrantal symmetric convolution operator. We consider a
filter a, such that ˆahas the strict quadrantal symmetry property, i.e. ˆa(ξ1, ω1) =
ˆa(ξ2, ω2) for (ξ1, ω1), (ξ2, ω2)T×T=T2if and only if one of the following
conditions is satisfied:
1. ξ1=ξ2, ω1+ω2= 1
2. ξ1+ξ2= 1, ω1=ω2
3. ξ1+ξ2= 1, ω1+ω2= 1.
The following result is a direct consequence of the symmetries assumptions listed
in conditions 1 3.
Proposition 4. If ˆa(ξ, ω)has the strict quadrantal symmetry property, then we
have det Am1,m2(ξ, ω)=0when ξ= 0 or ω= 0. Moreover, the kernel of each
Am1,m2(ξ, ω)is a subspace of the kernel of one of the following four matrices:
Am1,m2(0,0) , Am1,m21
2,0, Am1,m20,1
2, Am1,m21
2,1
2.
From Proposition 4, for a strict quadrantally symmetric kernel we need to
consider only the points (ξ, ω)(0,0) ,0,1
2,1
2,0,1
2,1
2 and construct
the set W, such that it removes the singularities of the above four matrices.
Proposition 5. If ˆahas the strict quadrantal symmetry property, then
dim(Am1,m2(ξ, ω)) = (m11)m2
2+m21
2
m1+ 1
2
for every (ξ, ω)(0,0) ,0,1
2,1
2,0,1
2,1
2.
Proof. We discuss here in depth only the case ξ=ω=1
2. The proof in the other
three cases are analogous to what we present here. Because Am1,m2(1
2,1
2) is a
MULTIDIMENSIONAL SIGNAL RECOVERY 167
Vandermonde matrix, the rank is equal to the number of its different columns.
It is easy to show that
ˆa1
2+s
m1
,
1
2+p
m2= ˆa1
2+k
m1
,
1
2+l
m2
is satisfied if and only if one of the following holds true:
(1) s=k, p +l=m21
(2) p=l, s +k=m11
(3) s+k=m11, p +l=m21
using which we can easily compute that
dim(Am1,m2(1
2,1
2)) = (m11)m2
2+m21
2
m1+ 1
2=n.
Let
W=W1W2(3.11)
where
W1={1,· · · m11
2} × {0,· · · , m21},
W2={0,· · · , m11}×{1,· · · ,m21
2}}.
Remark 3.2.When Wis defined as in (3.11), we have
|W|=(m11)m2
2+m21
2
m1+ 1
2;
By Corollary 2, Whas the minimal possible size.
Theorem 3.3. Let a1(D)be the filter such that the evolution operator is
given by Ax =ax. Suppose ˆasatisfies the strict quadrantal symmetric property
defined at the beginning of subsection 3.2. Let be as in (3.6) with Wspecified in
(3.11). Then, any f2(D)can be recovered in a stable way from the expanded
set of samples
{Sf, SXf , · · · , SXAm1m21f}.(3.12)
Proof. It suffices to show that for every (ξ, ω)T×T, it holds
ker(¯
U)ker(Am1,m2(ξ , ω)) = 0.(3.13)
By Proposition 4, we only need to study the kernels of these four matrices
Am1,m2(0,0), Am1,m2(1
2,0), Am1,m2(0,1
2), Am1,m2(1
2,1
2).(3.14)
We discuss here in depth for the case ξ=ω=1
2. Z := ker(Am1,m2(1
2,1
2)) is a
subspace in Cm1m2. By Proposition 5, the dimension of Zis n. Taking advantage
of the fact that Am1,m2(1
2,1
2) is a Vandermonde matrix, we can choose a basis
{vj:j= 1,· · · , n}for Z, such that each vjhas only two nonzero entries 1 and
168 R. ACESKA, A. PETROSYAN AND S. TANG
1. Let vker( ¯
U)Z, there exists c= (c(i))i=1,··· ,n such that v=
n
i=1
c(i)vi.
Define a n×nmatrix Rwith the row corresponds to a fixed (c1, c2)Wis
[(e
2πi(m11)c1
m1e
2πi0c1
m1)F2(c2),· · · ,(e
2πi(m1+1)c1
2m1e
2πi(m13)c1
2m1)F2(c2),
(e
2πi(m21)c2
m2e
2πi0c2
m2)¯
F1(c1),· · · ,(e
2πi(m2+1)c2
2m2e
2πi(m23)c2
2m2)¯
F1(c1)].
Then
¯
Uv= 0,which is equivalent to Rc= 0.
By the use the same strategy as in the proof of Theorem 2.3, it can be demon-
strated that these nrow vectors of Rare linearly independent. With slight
adaptations of the strategy used so far,we can come to the same conclusion for
the other three matrices in (3.14). As a consequence of Proposition 3, stability
is achieved.
4. Conclusion
In this paper, we seek the spatiotemporal trade off in the two variable discrete
spatially invariant evolution system driven by a single convolution filter in both
finite and infinite case. We characterize the spectral properties of the filters to
specify when we can recover the initial state from the uniform undersampled
future states and a way to add extra spatial sampling locations to stably recover
the signal when the filters fail the certain constraints. Compared to one variable
case, the singularity problems caused by the structure of filters are more com-
plicated and tough to solve. We give the explicit constructions of extra spatial
sampling locations to resolve the singularity issue caused by the strict quadran-
tal symmetric filters. Our results can be adapted to the general multivariable
case. Different kinds of symmetry assumptions can be imposed on the filters.
The problem of finding the right additional spatiotemporal sampling locations
for other types of filters remains open and requires a further study.
ACKNOWLEDGEMENT
We would like to thank Akram Aldroubi for his helpful discussions and com-
ments. The research of Armenak Petrosyan and Sui Tang are partially supported
by NSF Grant DMS-1322099.
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We study the spatial-temporal sampling of a linear diffusion field, and show that it is possible to compensate for insufficient spatial sampling densities by oversampling in time. Our work is motivated by the following issue often encountered in sensor network sampling, namely increasing the temporal sampling density is often easier and less expensive than increasing the spatial sampling density of the network. For the case of sampling a diffusion field, we show that, to achieve trade-off between spatial and temporal sampling, the spatial arrangement of the sensors must satisfy certain conditions. We provide in this paper the precise relationships between the achievable reduction of spatial sampling density, the required temporal oversampling rate, the spatial arrangement of the sensors, and the bound for the condition numbers of the resulting sampling and reconstruction procedures.
Conference Paper
The problem of recovering an evolving signal from a set of samples taken at different time instances, has been well-studied for one-variable signals on both discrete and continuous domains [4]-[8]. Most sampling problems in applications involve spatial coordinates, that is, the observed time-variant signals are described by at least two spatial variables. We state the problem of spatio-temporal sampling for two-variable data and provide specific reconstruction results in the l 2 (Z × Z) case.
Article
Regular and irregular sampling theorems for multivariate shift invariant subspaces are studied. We give a characterization of regular points and an irregular sampling theorem, which covers many known results, e. g. Kadec’s 1/4-theorem. We show that some subspaces may not have a regular point. We also present a reconstruction algorithm which is slightly different from the known one but is more efficient. We study the aliasing error and prove that every smooth square integrable function can be approximated by its sampling series.
Article
We consider the problem of spatiotemporal sampling in which an initial state of an evolution process is to be recovered from a set of samples at different time levels. We are particularly interested in lossless trade-off between spatial and temporal samples. We show that for a special class of signals it is possible to recover the initial state using a reduced number of measuring devices activated more frequently. We present several algorithms for this kind of recovery and describe their robustness to noise.