452 J. Opt. Soc. Am. A/Vol. 8, No. 2/February 1991

Signal reconstruction from multiple correlations:frequency- and time-domain approaches: comment

Jitendra K. Tugnait

Department of Electrical Engineering, Auburn University, Auburn, Alabama 36849

Received May 14, 1990; accepted September 18, 1990

Giannakis [J. Opt. Soc. Am. A 6, 682 (1989)] presented several approaches to reconstruction of one-dimensionaldeterministic sampled signals from their multiple correlations. I discuss the time-domain approach of Giannakisfor reconstruction of infinite-duration signals. For model selection, rank determination of a certain matrix isneeded. I give a counterexample, showing that the rank of such a matrix is not necessarily what has been claimed byGiannakis. For parameter determination, a certain other matrix is assumed to be of full rank. I give an exampleshowing that the full-rank assumption is not always true.

Giannakis' presented several approaches to the reconstruc-tion of one-dimensional deterministic sampled signals fromtheir multiple correlations. I will provide a counterexampleto show that certain claims made by Giannakis' concerningranks of certain matrices are not well founded. The coun-terexample is directed toward the time-domain approach ofRef. 1 for reconstruction of infinite duration signals.

We will consider a signal s(k)j that satisfies the linearconstant-coefficient difference equation

s(k) =-as(k-i) + (k), k > 0, lal < 1, (1)

with the initialization s(k) = 0 for k < 0. In Eq. (1) a is aconstant and 6(k) is the Dirac delta function. We can followthe notation of Ref. 1 except for using the coefficient a forthe notation a(1) used in Eq. (2b) of Ref. 1. [We will reservea for the true parameter and a(1) for its parameterization.The objective is to see whether, following the time-domainmethod of Ref. 1, we can obtain a(1) = a, as discussed below.]

The triple correlations for the model specified by Eq. (1)are given by

s3(m, n) := s(i)s(i + m)s(i + n), (2)i=o

where we use the fact that s(i) = 0 for i < 0. From Eq. (1) itfollows that

s(i + m) =-as(i + m-1) (3)for all m > 0 and for all i > 0. Substitute Eq. (3) into Eq. (2)in order to obtain

s3(m, n) = E s(i)[-as(i + m - 1)Js(i + n)i=O

= -a s(i)s(i + m - 1)s(i + n)i=O

=-as3(m-1, n) (4)

for all m > 0 and for any n. We will use Eq. (4) together with

the symmetry conditions given in Eq. (4) of Ref. 1 in whatfollows.

PARAMETER ESTIMATION

It has been claimed in Subsection 3.B of Ref. 1 that, givenknowledge of p and q, the coefficients a(i)J}P=1 and b(i))'%,can be uniquely determined from the exact triple-correla-tion samples 3(m, n) by using the method given in Subsec-tion 3.B. For the signal specified by Eq. (1), we have p = 1, q= 0, and a(1) = a. Thus only one parameter a(1) needs to beestimated, given p = 1 and q = 0. We can now proceed inshowing that, for the signal specified by Eq. (1), the matrixS, defined in Eq. (17a) of Ref. 1 by use of Eq. (16a) of Ref. 1does not have rank p(p + 3)/2; rather it has rank one. Thisimplies that Eq. (17a) of Ref. 1 involving two equations withtwo unknowns does not have a unique solution, contrary tothe claim of Ref. 1.

For the model specified by Eq. (1), Eq. (16a) of Ref. 1reduces to (set p = 1 and q = 0)

a(l)[s 3( - 1, n - 1) + s3(m + 1, n) + s3(m, n + 1)]

+ 3(0, )s3(m, n) = -[s 3( - 1, n) + s3(m, n - 1)

+ s 3(m + 1, n + 1)]. (5)

For the signal specified by Eq. (1), Ref. 1 recommends solu-tion of Eq. (5) for m = 2 with n = 1, 2. The correspondingmatrix S turns out to be

S E S3(1, 0) + s3(3, 1) + S3(2, 2) s3(2, 1)83(1, ) + s3 (3, 2) + s 3(2, 3) s3(2, 2) (6)

The rank of S must be two to permit the coefficients a(1)and A3(0, 0) to be recovered uniquely. We can now showthat the rank of is one.

Use Eq. (4) above and Eq. (4) of Ref. 1 repeatedly in orderto deduce

83(1, 1) = -as3 (1, ),

S3(3, 1) =-a 3 s3(1, ),

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S3(2, 2) = a2S3 (1, 1) = -a3 s 3 (1, 0),

S3(3, 2) = s3(2, 3) = -a3 s 3 (1, 1) = a 4s 3(1, 0),

s 3 (2, 1) = a2 s 3(1, 0).

Substitute Eq. (7) into Eq. (6) in order to obtain

L(1 - 2a 3 )s 3 (1, 0) a2 s 3(1, 0)1

S = (1 - 2a3)s3(1, 1) a2s3(1, 1)]

By the identities [Eq. (4) of Ref. 1], we have

s3(m, n) = s3(-m, n - m) = s3(fm, n + mn)

(7)

(8)

It is clear from Eq. (8) that the second column of S is

linearly dependent on its first column since the second col-

umn equals the first column multiplied by a constant a 2/(l -

2a3). Thus the rank of Si is one.

This shows that the signal parameters cannot be uniquelyrecovered, in general, by use of the method of Giannakis'

even if the orders p and q are known.We also note that, for the example considered here, since

S, is not of full rank, neither will be the matrix S2 as defined

in Eq. (18a) of Ref. 1. The matrix S2 is obtained from S, by

using an additional set of equations corresponding to m = 0,1, . .. , q and n = 0, 1, . .. , m, which for this example reduces

to a single equation corresponding to m = n = 0. The

claimed rank of S2 in Ref. 1 is three, implying that, after

deletion of this additional equation referred to above, therank of the so-reduced S2 must be at least two. However, we

have demonstrated above that the rank of Si is one; hence

the rank of the reduced S2 is also one. Thus the rank of S2

cannot exceed two.Finally, the above counterexample also applies when one

uses an overdetermined set of equations as specified by Eq.

(20a) of Ref. 1: continue to use Eq. (4) and Eq. (4) of Ref. 1

repeatedly.

MODEL SELECTION

It has been claimed in Section 4 of Ref. 1 that a certain

matrix So defined by the use of Eq. (16a) of Ref. 1 will be of

full rank p(p + 3)/2, provided that one takes m = -p - 1,... ,-2p - 1 and n = q - p, ... , q + p in Eq. (16a).' For

the signal specified by Eq. (1), Eq. (16a) of Ref. 1 reduces to

Eq. (5), and the lag values become m = -2, -3 and n = -1, 0,1 since nowp = 1 and q = 0. Therefore, by Ref. 1, the matrix

So for our example must have rank two. We can now pro-

ceed to show that it has rank one.

(9)

for any m and n where Fm = -m. The matrix So for the signal

specified by Eq. (1) turns out to be

s(-3, -2) + s3 (-1, -1) + S3(-2, 0) s3(-2, -1)

S3(-3, -1) + s3(-1, 0) + s3 (-2, 1) s3(-2, 0)

S = s3 (-3, 0) + s3 (-1, 1) + s3 (-2, 2) s3 (-2, 1)-Q _ - A _Q _L .e A 9 1 4- . -6 ( . () -O - 3 . 1 )

S3(-4, -1) + s3(-2, 0) + s3 (-3, 1)

s3(-4, 0) + S3(-2, 1) + S3(-3, 2)

-;J, , -,I

s3(-3, 0)

s 3 (-3, 1)

(10)

By use of Eq. (9), Eq. (10) becomes

so =

S3(3, 1) + S3(1, 0)

S3(3, 2) + S3(1, 1)

s3 (3, 3) + S3(1, 2)

s 3 (4, 2) + s 3(2, 1)

s 3 (4, 3) + s 3(2, 2)

S3(4, 4) + s 3(2, 3)

+ S3(2, 2)

+ 3(2, 3)

+ S3(2, 4)

+ 3 (3, 3)

+ 3 (3, 4)

+ s3 (3, 5)

s 3 (2, 1)

S3(2, 2)

S3(2, 3)

S3(3, 2)

S3(3, 3)

s3 (3, 4)

(11)

Finally, use Eq. (4) above and Eq. (4) of Ref. 1 repeatedly ina manner quite similar to that used for deriving Eq. (7) inorder to rewrite Eq. (11) as

So =

(1 - 2a3)s 3 (1, 0)

(1 - 2a3 )s3 (1 1)

(1 - 2a3)s3 (1, 2)

(1 - 2a3)s 3 (2, 1)

(1 - 2a3 )s 3 (2, 2)

a2s3 (1, 0)

a s 3(1, 1)

a 23(1, 2)

a 2 s3 (2, 1)

a 2S3(2, 2)

(12)

L1 - 2a3 )s3(2, 3) a3s3(2, 3)j

It is clear from Eq. (12) that S0 has rank one rather than twoas claimed in Ref. 1.

REFERENCE

1. G. B. Giannakis, "Signal reconstruction from multiple correla-tions: frequency- and time-domain approaches," J. Opt. Soc.Am. A 6, 682-697 (1989).