Modern Control Engineering

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252 Chapter 5 / Transient and Steady-State Response Analyses

For stability, we require that

From the first and second conditions,Kmust be greater than 1. For K>1, notice that the term

1-CK^2 /(K-1)Dis always negative, since

Thus, the three conditions cannot be fulfilled simultaneously. Therefore, there is no value of Kthat

allows stability of the system.

A–5–18. Consider the characteristic equation given by

(5–67)

The Hurwitz stability criterion, given next, gives conditions for all the roots to have negative real

parts in terms of the coefficients of the polynomial. As stated in the discussions of Routh’s stability

criterion in Section 5–6, for all the roots to have negative real parts, all the coefficients a’s must

be positive. This is a necessary condition but not a sufficient condition. If this condition is not sat-

isfied, it indicates that some of the roots have positive real parts or are imaginary or zero. A suf-

ficient condition for all the roots to have negative real parts is given in the following Hurwitz

stability criterion: If all the coefficients of the polynomial are positive, arrange these coefficients

in the following determinant:

where we substituted zero for asifs>n. For all the roots to have negative real parts, it is neces-

sary and sufficient that successive principal minors of be positive. The successive principal

minors are the following determinants:

whereas=0ifs>n. (It is noted that some of the conditions for the lower-order determinants

are included in the conditions for the higher-order determinants.) If all these determinants are

positive, and a 0 >0as already assumed, the equilibrium state of the system whose characteristic

¢i= 5

a 1

a 0

0



0

a 3

a 2

a 1



0

p

p

p

p

a2i- 1

a2i- 2

a2i- 3



ai

5 (i=1, 2,p,n-1)

¢n

¢n= 7

a 1

a 0

0

0





0

a 3

a 2

a 1

a 0





0

a 5

a 4

a 3

a 2





0

p

p

p

p

p

0



an

an- 1

an- 2

an- 3

an- 4

0



0

0

an

an- 1

an- 2

0  0 0 0 0

an

7

a 0 sn+a 1 sn-^1 +a 2 sn-^2 +p+an- 1 s+an= 0

K- 1 - K^2

K- 1

=

- 1 +K(1-K)

K- 1

60

K 70

K- 1

K

70

1 -

K^2

K- 1

70

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