Modern Control Engineering

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Example Problems and Solutions 257

From this analysis, we see that

The Hurwitz conditions for asymptotic stability

reduce to the conditions

The Routh array for the polynomial

wherea 0 >0 andn=4, is given by

From this Routh array, we see that

(The last equation is obtained using the fact that ) Hence the

Hurwitz conditions for asymptotic stability become

Thus we have demonstrated that Hurwitz conditions for asymptotic stability can be reduced to

Routh’s conditions for asymptotic stability. The same argument can be extended to Hurwitz

determinants of any order, and the equivalence of Routh’s stability criterion and Hurwitz stabil-

ity criterion can be established.

A–5–21. Consider the characteristic equation

Using the Hurwitz stability criterion, determine the range of Kfor stability.

Solution.Comparing the given characteristic equation

s^4 +2s^3 +(4+K)s^2 +9s+ 25 = 0

s^4 +2s^3 +(4+K)s^2 +9s+ 25 = 0

a 17 0, b 17 0, c 17 0, d 170

a 34 =0, aˆ 44 =a 4 , and a 4 =b 2 =d 1.

a 44 =aˆ 44 -

aˆ 43

a 33

a 34 =a 4 =d 1

a 33 =a 3 -

a 1

a 22

a 23 =

a 3 b 1 - a 1 b 2

b 1

=c 1

a 22 =a 2 -

a 0

a 1

a 3 =b 1

a 11 =a 1

a 0

a 1

b 1

c 1

d 1

a 2

a 3

b 2

a 4

a 0 s^4 +a 1 s^3 +a 2 s^2 +a 3 s+a 4 = 0

a 117 0, a 227 0, a 337 0, a 447 0, p

¢ 17 0, ¢ 27 0, ¢ 37 0, ¢ 47 0, p

¢ 1 =a 11

¢ 2 =a 11 a 22

¢ 3 =a 11 a 22 a 33

¢ 4 =a 11 a 22 a 33 a 44