Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 160

Example 3.46

¼ e At if Find f (A)

01

A ¼

2 3

S OLUTION
The characteristic polynomial of A is

jlI Aj¼ l

1

¼ ( l þ 1)( l þ 2) ¼ 0

2 l þ 3

Hence the eigenvalues are

l 1 ¼ 1 and l 2 ¼ 2

Since they are simple eigenvalues, we have

f ( l 1 ) ¼ R( l 1 ) ¼ r 0 þ b 1 l 1

f ( l 2 ) ¼ R( l 2 ) ¼ r 1 þ b 1 l 2

or in matrix form

r 0
r 1

¼

e t
e 2t

1 1

1 2

Solving for r 0 and r 1 , we have

¼

1

¼

e t
e 2t

¼

e t
e 2t

2e t e 2t

r 0
r 1

1 1 2 1

e t e 2t 1 2 1 1

Therefore,

r 0 0 0 r 1 r 0 r 1

e At

¼ r 0 I þ r 1 A ¼ þ ¼

0 r 0 2r 1 3r 1 2r 1 r 0 3r 1

Substituting for r 0 and r 1

r 0 r 1

2e t e 2t 2e t e 2t

e At

¼ ¼

2r 1 r 0 3r 1 2e t e 2t 2e t e 2t

3.11.4.2 Modal-Matrix Technique

Assume that matrix A has distinct eigenvalues and let M be the modal matrix of A, then

A ¼ M L M 1

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