Modern Control Engineering

Example Problems and Solutions 713

These msimultaneous equations determine the akvalues (where k=0,1,2,p,m-1). Noting

thatf(A)= 0 because it is a minimal polynomial, we have f(A)as follows:

Hence, referring to Equation (9–105), we have

(9–112)

where the akvalues are given in terms of fAl 1 B,f¿Al 1 B,f–Al 1 B,fAl 4 B,fAl 5 B,p,fAlmB. In terms of

the determinant equation,f(A)can be obtained by solving the following equation:

(9–113)

Equation (9–113) shows the desired modification in the form of the determinant. This equation

gives the form of Sylvester’s interpolation formula when the minimal polynomial of Ainvolves

three equal roots. (The necessary modification of the form of the determinant for other cases will

be apparent.)

A–9–14. Using Sylvester’s interpolation formula, compute eAt, where

Solution.Referring to Problem A–9–9, the characteristic polynomial and the minimal polynomial

are the same for this A. The minimal polynomial (characteristic polynomial) is given by

Note that l 1 =l 2 =2andl 3 =1. Referring to Equation (9–112) and noting that f(A)in this

problem is eAt, we have

wherea 0 (t),a 1 (t), and a 2 (t)are determined from the equations

a 0 (t)+a 1 (t)l 3 +a 2 (t)l^23 =el^3 t

a 0 (t)+a 1 (t)l 1 +a 2 (t)l^21 =el^1 t

a 1 (t)+ 2 a 2 (t)l 1 =tel^1 t

eAt=a 0 (t) I+a 1 (t) A+a 2 (t) A^2

f(l)=(l-2)^2 (l-1)

A= C

2

0

0

1

2

3

4

0

1

S

f–Al 1 B

2

f¿Al 1 B

fAl 1 B

fAl 4 B







fAlmB

f(A)

= 0

0 0 1 1    1 I

0

1

l 1

l 4







lm

A

1

2 l 1

l^21

l^24







l^2 m

A^2

3 l 1

3 l^21

l^31

l^34







l^3 m

A^3

p p p p p p

(m- 1 )(m- 2 )

2

lm 1 -^3

(m- 1 )lm 1 -^2

lm 1 -^1

lm 4 -^1







lmm-^1

Am-^1

f(A)=a(A)=a 0 I+a 1 A+a 2 A^2 +p+am- 1 Am-^1

f(A)=g(A)f(A)+a(A)=a(A)