INDIAN INSTITUTE OF TECHNOLOGY ROORKEE

Introduction to Industrial Control Systems :

Reduced Order Modelling

Yogesh Vijay Hote

Professor, Electrical Engineering Dept.,
IIT, Roorkee
Routh Hurwitz criterion:

The Routh stability criteria states that a polynomial would be

stable if and only if the following conditions are satisfied.

(i) All the coefficients of the polynomial should

be present and of the same sign.

(ii) All the elements in the first column of the

stability table should be of the same
sign(necessary and sufficient condition)
Routh, E. J. (1877). A Treatise on the Stability of a Given State of Motion: Particularly
Steady Motion. Macmillan.

2
C ( s ) = 1 + G ( s ) = a0 s n + a1s n −1 + ... + an −1s + an = 0

Applying Routh criterion to C(s), we get

sn a0 a2 a4 a6
s n −1 a1 a3 a5 a7
s n −2 b1 b2 b3 b4
s n −3 c1 c2 c3 c4

s2 e1 e2
s1 f1
s0 g1

3
sn a0 a2 a4 a6
s n −1 a1 a3 a5 a7
( a1a2 − a0a3 )
s n −2 b1 -b2 b3 b4 b1 =
a1
(a1a4 − a0a5 )
s n −3 c1 c2 c3 c4 b2 = ; a1a4  a0a5
a1

Stable/Unstable
s2 e1 e2
s1 f1
s0 g1

4
Research papers on Routh criterion:

[1] J. Juan et al. “Stability Analysis of a 1 DOF Haptic Interface Using

the Routh–Hurwitz Criterion,” IEEE Trans. Control System
Technology, vol. 12, no. 4, pp. 583-588, 2004.

[2] J. Aweya, “Design and stability analysis of a rate control algorithm

using the Routh- Hurwitz criterion,” IEEE/ACM Transactions on
networking, vol. 12, no. 4, 2004.

[3] M. A. Choghadi and H. A. Talebi, “The Routh Hurwitz Criterion,

Revisited: The Case of Multiple Poles on Imaginary axis,”
IEEE Transactions on Automatic Control, vol. 58,no.7,
pp.1866-1869, 2013.

5
• A closed loop control system has the characteristic equation
given by
s 3 + 4.5s 2 + 3.5s + 1.5 = 0
• Routh criterion

s3 1 3.5
s2 4.5 1.5
(4.5  3.5-11.5)
s1 =3.16 0
4.5
(3.16 1.5 − 0)
s0 =1.5 0
3.16

6
• Model Order Reduction

7
Importance of Model Order Reduction :

[1] To have low-order models so as to simplify the

understanding of a system.

[2] To reduce computational efforts in simulation problem.

[3] To decrease computational efforts and so to make the

design of the controller numerically more efficient.

[4] To obtain simpler control laws.

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9
10
11
Model Reduction Using the Routh
Stability criterion:
[1] This method was proposed by V.
Krishanmurthy in 1978.
IEEE Transactions on Automatic Control
[2] This method is based on Routh Stability
criterion.
[3] The reduced order transfer function is
determined by directly from elements in the
Routh stability arrays of the higher-order
denominator and numerator.
Let the transfer function of the high-
order system be

b11s m + b21s m−1 + b12 s m−2 + b22 s m−3 + ...

H (s) = n −1 n−2 n −3
,m  n
a11s + a21s + a12 s + a22 s + ...
n
The Routh stability array for the numerator and
denominator polynomial are shown in table 1 and 2,
respectively.
• Numerator stability array:
_____________________________
sm b11 b12 b13 b14
s m −1 b21 b22 b23 b24
_____________________________
sm−2 b31 b32 b33 b34 . . . . .
s m −3 b41 b42 b43 b44 . . . . .
. . . . . .
. . . . . .
b m,1
bm +1,1
______________________________
• Denominator stability array:
_____________________________
sn a11 a12 a13 a14
s n −1 a21 a 22 a 23 a 24
_____________________________
s n − 2 a31 a 32 a 33 a 34 . . . . .
s n −3 a41 a 42 a 43 a 44 . . . . .
. . . . . .
. . . . . .
an ,1
an +1,1
______________________________
• The transfer function constructed from
second and third rows of each table is

b21s m + b31s m−1 + b22 s m−2 + b32 s m−3 + ...

H n−1 ( s) = n −1 n−2 n −3
,m  n
a21s + a31s + a22 s + a32 s + ...
n
 Let the transfer function of an eighth
order system be
35s 7 + 1086s 6 + 13285s 5 + 82404s 4 + 278376s 3 + 511812s 2 + 482964s + 194480
H ( s) =
s8 + 33s 7 + 437 s 6 + 3017 s 5 + 11870s 4 + 27470s 3 + 37492s 2 + 28880s + 9600

Numerator Table
s7 35 13285 278376 482964
s6 1086 82402 511812 194480
s5 10629 261881 476696.1
s4 55645.5 463107.8 194480.0
s3 173419.1 439546.9
s2 322069.0 194480.0
s 334828.5
s0 194480.0
Denominator Table
s8 1 13285 278376 482964
s7 33 82402 511812 194480
s6 345.6 261881 476696.1
s5 1963.0 463107.8 194480.0
s4 6817.2 439546.9
s3 14847.1 194480.0
s2 20123.7
s1 18116.2
s0 9600.0
• Reduced order transfer function are

55645.5s 4 + 173419.1s 3 + 463107.8s 2 + 439546.9s + 194480

H 5 ( s) =
1963s 5 + 6817.2s 4 + 23973.4s 3 + 31694s 2 + 27963s + 9600

334828.5s + 194480
H 2 (s) =
20123.7 s 2 + 18116.2 s + 9600
Step response of original model and
reduced order model
Step Response
40
Original model
30
Amplitude

20

10

0
0 0.5 1 1.5 2 2.5 3 3.5 4
Time (sec)

Step Response
40

30 Second order reduced order model

Amplitude

20

10

0
0 2 4 6 8 10 12 14 16 18
Time (sec)

Step Response
40
Fifth order reduced order model
30
Amplitude

20

10

0
0 1 2 3 4 5 6 7 8 9 10
 Step Response
25
Original model

Second order reduced order model

Fifth order reduced order model

20

15
Amplitude

10

0
0 2 4 6 8 10 12 14 16 18
Time (sec)
Reduction of unstable system :

[1] For the case of unstable systems, the method may fail
since some of the first column elements of the
Routh stability table will be negative.

[2] Reduced order model be unstable if the higher-order

system is unstable.
[3] There is need to segregate unstable poles
and retained in the reduced model.
Let H(s) be of the form:
p( s) 1
H ( s) = = H s ( s)
v( s )q ( s ) v( s )

Where, p(s), q(s) , and v(s) are polynomials in s.

V(s) be the product of all the poles with zero
or positive real part.
Let H ( s ) being the asymptotically stable
s

transfer function, So, the reduced order

transfer function is
1
H R (s) = H Rs ( s )
v( s )
Advantage:

This method is versatile, since it is

computationally simple and requires no
algorithm to reconstruct the reduced-order
transfer function.