Modern Control Engineering

Section 10–5 / State Observers 751

we get

Sinceu(q)=0,we have, from Equation (10–33),

and so

Hence, for r=1,we have

It is noted that, as in any design problem, if the speed and damping are not quite satisfactory,

then we must modify the desired characteristic equation and determine a new matrix Computer

simulations must be repeated until a satisfactory result is obtained.

10–5 State Observers

In the pole-placement approach to the design of control systems, we assumed that all

state variables are available for feedback. In practice, however, not all state variables are

available for feedback. Then we need to estimate unavailable state variables.

Kˆ.

j(q)=1.1

j(q)=

1

kI

CKx(q)D=

1

kI

k 3 x 3 (q)=

- 56.0652

- 50.9684

r=1.1r

u(q)= 0 =-Kx(q)+kI j(q)

u(q)= 0

0

0.2

x1 versus t

062 4

t Sec

0

2

• 1

1

x3 versus t

062 4

0.5

1.5

x5 versus t

(^00624)
1
t Sec
t Sec
x^1
x^3
x^5
0
0.5

• 0.5

x2 versus t

062 4

t Sec

0

2

• 1

1

x4 versus t

062 4

t Sec

x^2

x^4

Figure 10–10

Curvesx 1 versust, x 2

versust, x 3

(=outputy) versus

t, x 4 versust,and

x 5 (=j) versus t.