
Chapter 2 Additive Error Reduction
© National Instruments Corporation 2-3 Xmath Model Reduction Module
A very attractive feature of the truncation procedure is the availability 
ofan 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, σ1 ≥ σ2 ≥ ···. Then 
the key result is,
with G, Gr the transfer function matrices of Equation2-1 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 Σ1 strictly exceeds the largest diagonal entry 
of Σ2, the reduced order system is guaranteed to be stable. 
Several other points concerning the error can be made:
• The error G(jω)–Gr(jω) 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 gr(·) denoting the impulse responses for impulse 
responses of G and Gr and with Gr of degree k, the following L1 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 Gr(s) 
has degree n–1; it can have no more than (n–2) zeros, since it is strictly 
PQΣΣ10
0Σ2
===
Gjω()Grjω()–∞2trΣ2
≤
gg
r
–142k1+()trΣ2
≤