IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
On the Georgiou–Lindquist Approach to Constrained
Kullback–Leibler Approximation
of Spectral Densities
Michele Pavon and Augusto Ferrante
639
II. APPROXIMATION PROBLEM
We adopt the same notation as in [15]. Let C+ ( ) denote the
positive, continuous functions on the unit circle. Consider the rational
transfer function
G(z ) = (I
Abstract—We consider the Georgiou–Lindquist constrained approximation of spectra in the Kullback–Leibler sense. We propose an alternative iterative algorithm to solve the corresponding convex optimization problem.
The Lagrange multiplier is computed as a fixed point of a nonlinear matricial map. Simulation indicates that the algorithm is extremely effective.
Index Terms—Approximation of spectral densities, convex optimization,
fixed point, Kullback–Leibler pseudodistance, spectral estimation.
0 zA)01 B;
A
2 n2n ;
B
2 n21
where A is a stability matrix, i.e., has all its eigenvalues in the open unit
disc, and (A; B ) is a reachable pair. Let 9 2 C+ ( ) represent the a
priori estimate of the spectrum of an underlying zero-mean, wide-sense
stationary stochastic process fy (n); n 2 g. Consider the situation
where new data become available in the form of an estimate 6 of the
state covariance of the system
I. INTRODUCTION
Georgiou and Lindquist [15] have recently studied the problem of approximating a spectral density in the Kullback–Leibler sense with another spectral density that is consistent with given, asymptotic state-covariance statistics. In [15], it is shown that this problem has as special
cases various important problems of control and signal processing engineering. These are briefly reviewed in the next section. We mention
here, in particular, the Byrnes–Georgiou–Lindquist spectral estimator
(THREE) [1], [2], [18]. The latter permits to obtain higher resolution
of the spectrum in certain desired frequency bands. It also appears to
work better than usual estimation techniques, such as periodogram and
AR methods, when short observation records are available.
In [15], the (convex) optimization problem is approached via duality
theory. It is shown there that the dual problem is also convex and admits a unique solution in a certain prescribed set of Hermitian matrices
L+ (0). In general, the solution cannot be obtained in closed form. It
is then necessary to solve the dual problem numerically. In [15], it is
observed that such a numerical solution can be based on Newton’s
method, after a suitable reparametrization of L+ (0), along the lines
of techniques developed in [2], [8], and [17] for similar problems. As
observed in [15], the global coordinatization of L+ (0) needs to preserve convexity. Nevertheless, the available methods may lead to loss
of global convexity or to numerical instability because of unboundness
of the gradient at the boundary.
In this note, we present an alternative algorithm to solve the optimization problem based on a nonlinear iteration scheme for the Lagrange
multiplier matrix. Differently from [15], we do not restrict our “dual
variable” to lie in the range of a certain operator 0 [see (11)]. So our
dual problem may have infinitely many solutions. Problems connected
to the reparametrization of L+ (0) are therefore altogether avoided. Extensive simulation indicates that this iteration converges extremely fast.
The outline of the note is as follows. In Sections II and III we review,
with some minor differences, the problem setting, the variational analysis and the corresponding dual problem of Georgiou and Lindquist
[15]. In Section IV, we introduce the new algorithm and analyze its
properties. Simulation results are presented in Section V. A preliminary
version of the results of this note has been presented, without detailed
proofs, in [19].
Manuscript received September 5, 2005; revised December 5, 2005. Recommended by Associate Editor E. Jonckheere. This work was supported in part
by the MIUR-PRIN Italian Grant “Identification and Control of Industrial Systems.”
M. Pavon is with the Dipartimento di Matematica Pura ed Applicata, Università di Padova, 35131 Padova, Italy (e-mail: [email protected]).
A. Ferrante is with the Dipartimento di Ingegneria dell’Informazione, Università di Padova, 35131 Padova, Italy.
Digital Object Identifier 10.1109/TAC.2006.872755
The Hermitian matrix 6 is assumed to be positive definite. It is obtained
by feeding y to the measurement device (a bank of filters) modeled by
G until it reaches steady state, and then estimating the state covariance.
If 9 is not consistent with 6, we need to find 8 in C+ ( ) that is closest
to 9 in a suitable sense among spectra consistent with 6. In [15], the
following measure of distance for spectra in C+ ( ) is considered:
i
9 log 9 =
9(ei ) log 9(ei )
8
8(e )
0
(9k8) =
d
2
:
The Kullback–Leibler pseudodistance for probability distributions is
defined similarly. It is also known as divergence, relative entropy, information distance, etc. It is also denoted by (1; 1) and H (1; 1). If
8 = 9, we have (9k8) 0. The choice of (9k8) as a
distance measure, even for spectra with different zeroth moment, is
justified in [15, Sec. III]. Notice that minimizing (9k8) rather than
(8k9) is unusual in to the statistics-probability-information theory
world. Besides leading to a more tractable problem, however, it also
includes as special case (9 1) maximization of entropy. As in [15],
we consider the following.
Problem 2.1: (Approximation problem) Let 9 2 C+ ( ), and let
6 2 n2n satisfy 6 = 63 > 0, where star denotes transposition plus
^ that solves
conjugation. Find 8
minimize
over
(9k8)
8 2 C+ ( ) j
(1)
G 8G 3 = 6
:
(2)
The constraint (2) expresses the fact that 8 is consistent with 6. It is not
necessary to require that 8 satisfies the hermitian property 8(ei )3 =
8(ei ), 8 . Indeed, whenever 9 2 C+ ( ) is a spectrum, namely it
satisfies the parahermitian property, the solution of the approximation
problem also turns out to be a spectrum. First of all, one needs to worry
about existence of 8 2 C+ ( ) satisfying constraint (2). It was shown
in [12], [13] that such family is nonempty if and only if there exists
H 2 12n such that
6 0 A6A3 = BH + H 3 B 3
or, equivalently, the following rank condition holds:
rank
0018-9286/$20.00 © 2006 IEEE
6 0 A 6A3 B
= rank
0
B3
0
B3
B
0
:
(3)
640
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
We now show, following [15], that this problem includes as special
cases a variety of important applications. In all three examples below,
the family of spectral densities consistent with the data is nonempty if
6 0 and contains infinitely many elements if 6 > 0.
Example 1: (Covariance extension problem) Let
0 1 0 ... 0
0 0 1 ... 0
..
.
A=
..
.
..
.
..
.
B=
0
0
The (rather straightforward) variational analysis follows quite
closely [15] with some minor differences that are spelled out in
[19]. It is, therefore, kept to a minimum. For 3 2 n2n satisfying
G3 3G > 0 on all of , consider the Lagrangian function
..
.
(4)
Notice that, differently from [15], we do not require the matrix
belong to the range of the operator 0 defined on C ( ) by
6=
..
.
0(8) =
. . . cn01
. . . cn02
c1
c0
..
.
..
(5)
..
.
.
cn01 cn02 . . .
G3 3G8 0 trace (36):
= (9k8) +
the Toeplitz matrix
c0
c1
G8G3 0 6
L(8; 3) = (9k8) + trace 3
0
1
n0k
so that the k th component of G(z ) is Gk (z ) = z
. Also, let 6 be
0 0 0 ... 1
0 0 0 ... 0
III. OPTIMALITY CONDITIONS AND THE DUAL PROBLEM
G8G3 :
(10)
3 to
(11)
Consider the unconstrained minimization of the strictly convex functional L(8; 3)
c0
minimizefL(8; 3)j8 2 C+ (
)g:
(12)
where
ck := E fy(n)y(n + k)g:
Thus, the information available on the process y is the finite sequence
of covariance lags c0 ; c1 ; . . . ; cn01 . If 9 1, then Problem 2.1 is just
the maximum-entropy covariance extension problem; see, e.g., [16].
When 9 is a positive polynomial of degree n, the solution of Problem
2.1 has degree 2n [9], [11], [6], [5], [3], [4]
Example 2: In this case
0 0 ... 0
0 p2 0 . . . 0
..
.
..
.
0
0
..
..
.
.
..
.
B=
0 0 ... 0
0 0 . . . pn
(6)
1
1
In this case the problem is a Nevanlinna-Pick interpolation problem
[10], [11], [1], [2] whose solution permits spectral estimation by selective harmonic amplification [1], [2], [18], [12] (similarly, for Example
3).
Example 3: In this case
A=
..
.
..
.
..
.
..
.
0 0 0 ... 1
0 0 0 ... p
so that the k th element of G(z ) is Gk (z )
The matrix 6 is of the form
B=
0
0
..
.
(8)
0
1
(14)
where W is an upper triangular toeplitz matrix and E is the reachability
gramian of the pair (A; B ).
(15)
is the unique solution of the approximation problem (1)–(2).
^ satisThus, the original problem (1)–(2) is now reduced to finding 3
fying (13)–(14). This is accomplished in [15] via duality theory. Consider, as in [15], the dual functional
3 ! inf fL(8; 3)j8 2 C+ ( )g:
For 3 satisfying (13), the dual functional takes the form
3!L
9 ; 3 = 9 log G3 3G 0 trace (36)+ 9:
G3 3G
(16)
Consider now the maximization of the dual functional (16) over the set
L+ := f3 = 33 jG3 3G > 0
8ei 2 g:
(17)
Let, as in [15]
9 log G3 3G + trace (36):
The dual problem is then equivalent to
minimize
(9)
(13)
^ given by
Then, 8
J9 (3) := 0
= z n0k =(1 0 pz )n+10k .
6 = 1 (W E + EW 3 )
2
9
G 3 ^ G3 = 6:
G 3G
8ei 2
8^ = 39^
G 3G
so that the k -th element of G(z ) is Gk (z ) = 1=(1 0 pk z ). The matrix
6 is a Pick matrix with elements
6i;j = wi + wj
(7a)
1 0 pi pj
1 e0i! + pk 8(ei! )d!;
wk =
k =1; 2; . . . ; n: (7b)
4 0e0i! 0 pk
p 1 0 ... 0
0 p 1 ... 0
^G > 0
G3 3
1
1
p1
A=
This is a convex optimization problem. We have the following optimality condition.
^ = 3^ 3 is such that
Theorem 3.1: Suppose 3
fJ9(3)j3 2 L+ g:
(18)
The dual problem is also a convex optimization problem. The functional J9 (3), however, is not strictly convex on 3 2 L+ . We have the
following result [15].
^ 2 L+ solves (18) if and only if it satisfies (14).
Proposition 3.2: 3
In [15], existence of a (unique) minimum of J9 on L+ (0) := L+ \
range (0) is established via a profound homeomorphism result for continuous maps between open, connected sets of the same dimension [7].
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
This result implies that, under assumption (3), there exists a (unique)
3^ in L+ (0) satisfying (14). Such a 3^ then provides the optimal solution of the primal problem (1)–(2) via (15). This, of course, establishes
existence also for our dual problem (18) in view of Proposition (3.2).
Notice, however, that our dual problem has in general infinitely many
solutions in L+ . It is observed in [15] that, in general, to solve their
dual problem, it is necessary to resort to an iterative algorithm, and
that the latter, after a suitable parametrization of L+ (0) can be based
on Newton’s method. In the following section, we propose an alternative strategy.
641
where
G0
9 G03 = I
03
^ 0 G0
G 3
then
G
rank
I
0 AA3
B3
B
0
= rank
0
B
B3
0
:
9
M 1=2 G
G3 M G
G3 M 1=2 :
[2(M )] =
9
=
G3 M G
G3 M G
For x 6= 0 in , let us denote by 5x
projection onto span fxg. We have
2(M ) =
=
M 1=2 G
5M
G
9
9I
(23)
9
G
9
G3 M G
G3 > m
.
GG3 :
GG3 is the solution of the Lyapunov equa-
X
= AXA3 + BB 3 :
As it is well known, when A is asymptotically stable and the pair
(A; B ) is reachable, such a solution is positive definite.
The Algorithm: Let M0 = (1=n)I . Note that M0 2 M+ . Define
the sequence fMk g1
k=0 by
(24)
(25)
:= 2(Mk ):
(26)
2M
Notice that, by Theorem 4.1, Mk
+ for all k . Moreover, since
Mk
, k , the sequence is bounded. Hence, it has at least one
.
accumulation (limit) point in the closure of
^
Theorem 4.2: Suppose that the sequence Mk k1=0 has a limit M
^ satisfies (13)–(14) and, therefore, provides the optimal
+ . Then M
solution of (1)–(2) via (15).
^ is a fixed point of the map 2.
Proof: M
M M
f g
M
^
M
For M 2 M, we now want to show that G3 (ei! )2(M )G(ei! ) > 0 on
such that G3 (ei! )2(M )G(ei! ) = 0.
all of . Suppose there exists !
Then
=0
:
We have m > 0, since 9 is positive and G3 M G has no poles on
Then
2M8
G3 M 1=2
G3 (ei! )M 1=2 Q()M 1=2 G(ei! )d
of the continuous function
G3 M G
Mk+1
9 = 1:
9 = I:
G 3 > 0:
(22)
= x(x3 x)01 x3 the orthogonal
G3 M G
9
G3 M G
Indeed, let m be the minimum on
(21)
Theorem 4.1: 2 maps M into M and M+ into M+ .
Proof: It is apparent that 2(M ) = 2(M )3 . Moreover
trace
Recalling that 9 is everywhere positive, this implies
G3 (ei! )M 1=2 G(ei! ) = 0. Hence, M 1=4 G(ei! ) = 0 and,
consequently, G3 (ei! )M G(ei! ) = 0, a contradiction. Suppose now
that M 2 M+ . We want to show that 2(M ) > 0. Since M 1=2 is
nonsingular, it suffices to show that
It remains to observe that
tion
M := fM 2 L+ j0 M I; trace [M ] = 1g
M+ := fM 2 MjM > 0g:
For M 2 M, define the map 2 by
2
9(ei! )
= 0:
G3 (ei! )M G(ei! )
(20)
9 = 1.
We assume, moreover, w.l.o.g., that
Let
2(M ) :=
G3 (ei! )M 1=2 G(ei! )
(19)
In order to have a nonempty feasible set for problem (1)–(2), we assume
8:
= ! :
In particular, this holds for G
8^ = 039^ 0 0 :
G 3G
G3 (ei ):
G3 (ei! )M 1=2 Q()M 1=2 G(ei! ) = 0
9 G3 = 6:
^ 0 G0
G03 3
^ in (15) may also be obtained directly from G0 and 3^ 0 as
Hence, 8
G3 (ei )M G(ei )
Since the integrand of (25) is a continuous and nonnegative function
of , it follows that the integrand is identically zero, i.e.,
IV. NEW ALGORITHM
To simplify the writing, we assume, without loss of generality, that
6 = I . Indeed, if 6 6= I , it suffices to replace G by G0 := 601=2 G
and (A; B ) with (A0 = 601=2 A61=2 ; B 0 = 601=2 B ). Moreover, if
3^ 0 has been found such that
9
Q() := G(ei )
= 2(M^ ):
2
(27)
^ > 0, (13) is satisfied and (27) gives (14).
Q: E: D:
Since M
Next, we analyze the algorithm. Theorem 4.2 shows that every positive definite M which is a fixed point of the map 2 yields the optimal
8^ via (15). In general, a nonnegative definite fixed point of the map
does not lead to the optimal solution. Indeed, we identify below a large
that do not, in general, yield the opfamily of fixed points of 2 in M
^.
timal solution 8
642
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
Proposition 4.3: Let x 2 n be such that x3 G(ei ) is not identi is a fixed point of
cally zero. Then the orthogonal projection 5x 2 M
the map 2.
Proof: Recall that
2(M ) = 5M
G
9:
Notice that x3 G is a rational function. If it is not identically zero, then
it has at most finitely many zeroes on . Hence, 55 G = 5x almost
everywhere on . Thus
2(5x ) = 55 G 9 = 5x 9 = 5x :
Q:E:D:
0 the
Let us introduce M0 := fM 2 M+ j2(M ) = M g, and M
closure of M0 .
0 , is
Theorem 4.4: The set M0 , and consequently its closure M
convex.
Proof: Let M1 and M2 belong to M0 . Since they are nonsingular
and fixed points of 2, it follows that
G
9
G3 M 1 G
G3 =
G
9
G3 M 2 G
9=G3 M1 G and 82
G3 = I:
9=G3 M2 G solve the
Hence, both 81 :=
:=
approximation problem (1)–(2). Since the solution is unique, we get
G3 M1 G G3 M2 G. Hence
M := (1 0 )M1 + M2 ;
9
G3 M G
G3 M1 G. We
G3 = I:
We conclude that 2(M ) = M , and M 2 M0 .
0 . Then, there exist a sequence
Now let M be a singular matrix in M
fM3 k g1
with
elements
in
M
such
that Mk ! M . It follows that
0
k=0
G Mk G ! G3 MG at each point on . Since G3 Mk G does not vary
with k , it also follows that G3 MG G3 Mk G. We conclude that
0 yields the optimal solution of the approximation
any matrix M 2 M
problem via the formula
8o := G39MG :
Let us denote by
k0 = min (Nk ) k+ = max (Nk )
the minimum and maximum eigenvalues of Nk . We then have the following result.
Proposition 4.5:
1 Nk Nk+1 1 Nk
0
k+
(31)
Proof: Since k0 I
Nk , we get from (30)
G 9 G3 = 1 N :
Nk+1 1
k
k0
G3 M k G
k0
Q :E :D :
Similarly, for the other inequality.
Corollary 4.6: For all k 1, we have
k0 1 k+
k0
k0+1 k++1 k+
k+
:
k0
(32)
(33)
Remark: Notice that Proposition 4.5 implies the following: If, for
some k , Nk I , then Nk+1 Nk . Namely, the iteration for Nk is
monotonically nonincreasing on fN jN I g. Similarly, it also follows
from Propositon 4.5 that the iteration for Nk is monotonically nondecreasing on fN jN I g. This suggests that the iteration features some
intrinsic mechanism to converge toward the identity matrix. Corollary
4.6, nevertheless, shows that after one iteration, Nk 0 I does not possess any definiteness property. We conjecture that the mechanism of
the iteration for Nk continues to work for the portion of Nk greater
than the identity and for the portion of Nk smaller than the identity.
More explicitly, we conjecture (but cannot quite prove) the following:
0 that is not a fixed point of 2. Then,
Let M be any matrix in MnM
the corresponding sequence fNk g produced by the iteration (29)–(30)
always converges to the identity matrix, thereby providing the optimal
solution of the approximation problem.
V. SIMULATION RESULTS
Hence, it suffices that Mk defined in (26) converges to any element of
Now, let
G 9 G3 :
G3 M k G
In the examples that follow, we have iterated the algorithm (26) with
the stopping condition kM 0 2(M )k < 1005 . Consider first the following instance of Example 1. Let
0
0 :
1
The (estimated) finite covariance sequence is c0 = 3, c1 = 1, c2 = 1,
A=
In order to facilitate the analysis of the algorithm, rewrite the recursion
(26) as follows:
Mk+1 = Mk1=2 Nk Mk1=2
M0 = 1 I
n
G 9 G3
G 9 G3 :
Nk+1 =
=
n
N
0
1
=
2
1
=
2
G3 G
G3 M k Nk M k G
8 k 0:
k
(28)
M 0 to obtain the solution of the primal problem (1)–(2).
Nk =
G 9 G3 :
G3 G
Nk 01
is a positive–definite matrix also satisfying G3 M G
get
G
Notice that Mk and Nk generated by the recursion (29)–(30) are all
positive definite. Also notice that if Nk = I , then Mk+1 = Mk and
Nk+1 = Nk , namely we have a fixed point of the dynamical system.
Moreover, if Nk = I , 2 + is a scalar matrix, then Nk+1 = I .
Finally, notice that, since M I , the Nk have the uniform lower bound
0 1 0
0 0 1
0 0 0
B=
so that
(29)
(30)
3 1 1
6= 1 3 1 :
1 1 3
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
Fig. 1.
Graphics of 9(e
643
) and 8 (e ) (bold line) as functions of ! .
and
9( ) = ( + 2)(3 01 + 2)
Also let
9( ) = ( + 1 1)( 0 1 2)(0 0671 + 1 1)( 01 0 1 2)
Notice that 9 = 1. After the renormalization described at the beginning of Section V to transform 6 into the identity, we get
0 1462 1 0237 0 0237
0 = 0 1462 0 0237 1 0237
00 0475 00 1700 00 1700
00 0866
0 = 00 0866
0 6205
^0
The algorithm converged in 13 steps and the optimal matrix o = 3
:
z
z
:
z
:
:
A
z
:
:
:
:
and notice that
:
:
:
z
z
:
A
:
:
9 = 1. After the usual renormalization, we get
0
00 1234 0 1234
0 0173 0 9922 0 0075
00 0173 00 0075 00 9922
0 9847
0 = 0 1234
0 1234
:
0=
:
:
:
:
:
:
:
B
:
:
z
:
:
:
B
z
:
:
:
:
M
is given by
The algorithm converged in four steps and the optimal matrix M o
is given by
= 3^ 0
0 0004 0 0710 0 1106
0 4328 00 1469 0 4325
o = 0 0710 049 0867
0 7108
= 00 1469 0 1345 00 1469
0
1106
0
7108
048
9779
0 4325 00 1469 0 4328
The optimal solution 8o may now be computed through (19). The func- The optimal solution 8o ( i! ) is depicted (bold line) together with
tion 8o ( i! ) is depicted in bold line together with 9( i! ) (thin line) 9( i! ) (thin line) in Fig. 2.
:
o
M
:
:
:
:
:
:
:
:
:
M
:
:
:
:
:
:
:
:
:
:
e
e
e
in Fig. 1.
Consider next one instance of Example 2. Let
0 0
0
1
= 0 0 99 0
= 1
0 0 00 99
1
1
1
1
6 = 1 50 2513 0 5050
1 0 5050 50 2513
A
B
:
:
:
:
:
:
e
We have tested the algorithm, with similar results, also in many other
cases of Examples 1–3, see [19] where some other simulation results
are illustrated. The algorithm never failed to rapidly converge to a positive definite M which, by Theorem 4.2, provides an optimal solution
to the original problem. This happened even when the algorithm was
initialized with an M0 very close to one of the one-dimensional projections that are fixed points of the map 2 (Proposition 4.3). The latter,
^ via
as observed before, do not, in general, yield the optimal solution 8
(15).
644
Fig. 2.
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 51, NO. 4, APRIL 2006
Graphics of 9(e
) and 8 (e ) (bold line) as functions of ! .
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