Modern Control Engineering

We see that all the expanded terms drop out with the exception of ak. Thus the residue

akis found from

Note that, since f(t)is a real function of time, if p 1 andp 2 are complex conjugates, then

the residues a 1 anda 2 are also complex conjugates. Only one of the conjugates,a 1 ora 2 ,

needs to be evaluated, because the other is known automatically.

Since

f(t)is obtained as

fort 0

EXAMPLE B–1 Find the inverse Laplace transform of

The partial-fraction expansion of F(s)is

wherea 1 anda 2 are found as

Thus

fort 0

EXAMPLE B–2 Obtain the inverse Laplace transform of

Here, since the degree of the numerator polynomial is higher than that of the denominator poly-

nomial, we must divide the numerator by the denominator.

G(s)=s+ 2 +

s+ 3

(s+ 1 )(s+ 2 )

G(s)=

s^3 + 5 s^2 + 9 s+ 7

(s+ 1 )(s+ 2 )

= 2 e-t-e-^2 t,

=l-^1 c

2

s+ 1

d+l-^1 c

- 1

s+ 2

d

f(t)=l-^1 CF(s)D

a 2 =c(s+2)

s+ 3

(s+1)(s+2)

d

s=- 2

= c

s+ 3

s+ 1

d

s=- 2

=- 1

a 1 =c(s+1)

s+ 3

(s+1)(s+2)

d

s=- 1

=c

s+ 3

s+ 2

d

s=- 1

= 2

F(s)=

s+ 3

(s+1)(s+2)

=

a 1

s+ 1

+

a 2

s+ 2

F(s)=

s+ 3

(s+1)(s+2)

f(t)=l-^1 CF(s)D=a 1 e-p^1 t+a 2 e-p^2 t+p+an e-pn^ t,

l-^1 c

ak

s+pk

d =ak e-pk^ t

ak= cAs+pkB

B(s)

A(s)

d

s=-pk

868 Appendix B / Partial-Fraction Expansion

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