Modern Control Engineering

52 Chapter 2 / Mathematical Modeling of Control Systems

and

Solution.From the definition of state variables x 2 andx 3 ,we have

(2–61)

(2–62)

To derive the equation for we first note from Equation (2–58) that

Since

we have

Hence, we get

(2–63)

Combining Equations (2–61), (2–62), and (2–63) into a vector-matrix equation, we obtain Equa-

tion (2–59). Also, from the definition of state variable x 1 ,we get the output equation given by

Equation (2–60).

A–2–7. Obtain a state-space equation and output equation for the system defined by

Solution.From the given transfer function, the differential equation for the system is

Comparing this equation with the standard equation given by Equation (2–33), rewritten

y%+a 1 y$+a 2 y#+a 3 y=b 0 u%+b 1 u$+b 2 u# +b 3 u

y%+4y$+5y# +2y=2u%+u$+u#+2u

Y(s)

U(s)

=

2s^3 +s^2 +s+ 2

s^3 +4s^2 +5s+ 2

x

3 =-a 3 x 1 - a 2 x 2 - a 1 x 3 +b 3 u

=-a 1 x 3 - a 2 x 2 - a 3 x 1 +b 3 u

=-a 1 x 3 - a 2 x 2 - a 3 x 1 +Ab 3 - a 1 b 2 - a 2 b 1 - a 3 b 0 Bu

+Ab 2 - b 2 - a 1 b 1 - a 2 b 0 Bu

+Ab 3 - a 1 b 2 - a 2 b 1 - a 3 b 0 Bu

=-a 1 x 3 - a 2 x 2 - a 3 x 1 +Ab 0 - b 0 Bu%+Ab 1 - b 1 - a 1 b 0 Bu$

+b 0 u%+b 1 u$+b 2 u# +b 3 u-b 0 u%-b 1 u$-b 2 u#

• a 2 Ay# -b 0 u#-b 1 uB-a 2 b 0 u# -a 2 b 1 u-a 3 Ay-b 0 uB-a 3 b 0 u

=-a 1 Ay$-b 0 u$-b 1 u# -b 2 uB-a 1 b 0 u$-a 1 b 1 u#-a 1 b 2 u

=A-a 1 y

$

• a 2 y

• a 3 yB+b 0 u

%

+b 1 u

$

+b 2 u

+b 3 u-b 0 u

%

• b 1 u

$

• b 2 u

x# 3 = y%-b 0 u%-b 1 u$-b 2 u#

x 3 =y$-b 0 u$-b 1 u# -b 2 u

y%=-a 1 y$-a 2 y# -a 3 y+b 0 u%+b 1 u$+b 2 u

+b 3 u

x# 3 ,

x# 2 =x 3 +b 2 u

x# 1 =x 2 +b 1 u

b 3 =b 3 - a 1 b 2 - a 2 b 1 - a 3 b 0

b 2 =b 2 - a 1 b 1 - a 2 b 0

b 1 =b 1 - a 1 b 0

b 0 =b 0

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