Modern Control Engineering

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Example Problems and Solutions 253

equation is given by Equation (5–67) is asymptotically stable. Note that exact values of determi-

nants are not needed; instead, only signs of these determinants are needed for the stability criterion.

Now consider the following characteristic equation:

Obtain the conditions for stability using the Hurwitz stability criterion.

Solution.The conditions for stability are that all the a’s be positive and that

It is clear that, if all the a’s are positive and if the condition is satisfied, the condition

is also satisfied. Therefore, for all the roots of the given characteristic equation to have neg-

ative real parts, it is necessary and sufficient that all the coefficients a’s are positive and

A–5–19. Show that the first column of the Routh array of

is given by

where

Solution.The Routh array of coefficients has the form

1

a 1

b 1

c 1







a 2

a 3

b 2

c 2







a 4

a 5

b 3









a 6

p

p

p an

ak= 0 if k 7 n

¢r= 7

a 1

a 3

a 5







a2r- 1

1

a 2

a 4









0

a 1

a 3









0

1

a 2

















0 0 0   

ar

7 , (nr1)

1, ¢ 1 ,

¢ 2

¢ 1

,

¢ 3

¢ 2

,p,

¢n

¢n- 1

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

¢ 37 0.

¢ 270

¢ 370

=a 3 Aa 1 a 2 - a 0 a 3 B-a^21 a 470

=a 1 Aa 2 a 3 - a 1 a 4 B-a 0 a^23

¢ 3 = 3

a 1

a 0

0

a 3

a 2

a 1

0

a 4

a 3

3

¢ 2 =^2

a 1

a 0

a 3

a 2

(^2) =a 1 a 2 - a 0 a 370
a 0 s^4 +a 1 s^3 +a 2 s^2 +a 3 s+a 4 = 0