Chapter 2
Additive Error Reduction
©
National Instruments Corporation
2-3
A very attractive feature of the truncation procedure is the availability
of an error bound. More precisely, suppose that the controllability and
observability grammians for [Enn84] are
(2-2)
with the diagonal entries of
Σ
in decreasing order, that is,
σ
1
≥ σ
2
≥
···. Then
the key result is,
with
G
,
G
r
the transfer function matrices of Equation 2-1 and Equation 2-2,
respectively. This formula shows that small singular values can, without
great cost, be thrown away. It also is valid in discrete time, and can be
improved upon if there are repeated Hankel singular values. Provided that
the smallest diagonal entry of
Σ
1
strictly exceeds the largest diagonal entry
of
Σ
2
, the reduced order system is guaranteed to be stable.
Several other points concerning the error can be made:
•
The error
G
(
j
ω
)–
G
r
(
j
ω
) as a function of frequency is not flat; it is zero
at
ω
=
∞
, and may take its largest value at
ω
= 0, so that there is in
general no matching of DC gains of the original and reduced system.
•
The actual error
may
be considerably less than the error bound at all
frequencies, so that the error bound formula can be no more than an
advance guide. However, the bound is tight when the dimension
reduction is 1 and the reduction is of a continuous-time
transfer-function matrix.
•
With
g
(·) and
g
r
(·) denoting the impulse responses for impulse
responses of
G
and
G
r
and with
G
r
of degree
k
, the following
L
1
bound
holds [GCP88]
This bound also will apply for the L
∞
error on the step response.
It is helpful to note one situation where reduction is likely to be difficult (so
that
Σ
will contain few diagonal entries which are, relatively, very small).
Suppose
G
(
s
), strictly proper, has degree
n
and has (
n
– 1) unstable zeros.
Then as
ω
runs from zero to infinity, the phase of
G
(
s
) will change by
(2n – 1)
π
/2. Much of this change may occur in the passband. Suppose
G
r
(
s
)
has degree
n
–1; it can have no more than (
n
– 2) zeros, since it is strictly
P
Q
Σ
Σ
1
0
0
Σ
2
=
=
=
G j
ω
( )
G
r
j
ω
( )
–
∞
2
tr
Σ
2
≤
g g
r
–
1
4 2
k
1
+
(
)
tr
Σ
2
≤