Modern Control Engineering

34 Chapter 2 / Mathematical Modeling of Control Systems

wherexis the state vector,uis the input, and yis the output. The Laplace transforms of

Equations (2–23) and (2–24) are given by

(2–25)

(2–26)

Since the transfer function was previously defined as the ratio of the Laplace transform

of the output to the Laplace transform of the input when the initial conditions were

zero, we set x(0)in Equation (2–25) to be zero. Then we have

or

By premultiplying to both sides of this last equation, we obtain

(2–27)

By substituting Equation (2–27) into Equation (2–26), we get

(2–28)

Upon comparing Equation (2–28) with Equation (2–22), we see that

(2–29)

This is the transfer-function expression of the system in terms of A,B,C, and D.

Note that the right-hand side of Equation (2–29) involves Hence G(s)

can be written as

whereQ(s)is a polynomial in s. Notice that is equal to the characteristic poly-

nomial of G(s). In other words, the eigenvalues of Aare identical to the poles of G(s).

EXAMPLE 2–3 Consider again the mechanical system shown in Figure 2–15. State-space equations for the system

are given by Equations (2–20) and (2–21). We shall obtain the transfer function for the system from

the state-space equations.

By substituting A,B,C, and Dinto Equation (2–29), we obtain

=[1 0]C

s

k

m

- 1

s+

b

m

S

• 1
  C

0

1

m

S

=[1 0]cB

s

0

0

s

R- C

0

-

k

m

1

-

b

m

Ss

• 1
  C

0

1

m

S + 0

G(s)=C(s I-A)-^1 B+D

∑s I-A∑

G(s)=

Q(s)

∑s I-A∑

(s I-A)-^1.

G(s)=C(s I-A)-^1 B+D

Y(s)=CC(s I-A)-^1 B+DDU(s)

X(s)=(s I-A)-^1 BU(s)

(s I-A)-^1

(s I-A)X(s)=BU(s)

s X(s)-AX(s)=BU(s)

Y(s)=CX(s)+DU(s)

sX(s)-x(0)=AX(s)+BU(s)

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