Modern Control Engineering

whereb 3 ,b 2 , and b 1 are determined as follows. By multiplying both sides of this last

equation by (s+1)^3 , we have

(B–2)

Then letting s=–1, Equation (B–2) gives

Also, differentiation of both sides of Equation (B–2) with respect to syields

(B–3)

If we let s=–1in Equation (B–3), then

By differentiating both sides of Equation (B–3) with respect to s, the result is

From the preceding analysis it can be seen that the values of b 3 ,b 2 , and b 1 are found

systematically as follows:

=

1

2

(2)= 1

=

1

2!

c

d^2

ds^2

As^2 +2s+ 3 Bd

s=- 1

b 1 =

1

2!

e

d^2

ds^2

c(s+1)^3

B(s)

A(s)

df

s=- 1

= 0

=(2s+2)s=- 1

= c

d

ds

As^2 +2s+ 3 Bd

s=- 1

b 2 = e

d

ds

c(s+1)^3

B(s)

A(s)

df

s=- 1

= 2

=As^2 +2s+ 3 Bs=- 1

b 3 = c(s+1)^3

B(s)

A(s)

d

s=- 1

d^2

ds^2

c(s+1)^3

B(s)

A(s)

d =2b 1

d

ds

c(s+1)^3

B(s)

A(s)

d

s=- 1

=b 2

d

ds

c(s+1)^3

B(s)

A(s)

d =b 2 +2b 1 (s+1)

c(s+1)^3

B(s)

A(s)

d

s=- 1

=b 3

(s+1)^3

B(s)

A(s)

=b 1 (s+1)^2 +b 2 (s+1)+b 3

870 Appendix B / Partial-Fraction Expansion

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