Modern Control Engineering

(Chris Devlin) #1

Section 10–5 / State Observers 759


Since the desired characteristic equation is


by comparing Equation (10–66) with this last equation, we obtain


or


Method 3: We shall use Ackermann’s formula given by Equation (10–65):


where


Thus,


and


As a matter of course, we get the same Keregardless of the method employed.
The equation for the full-order state observer is given by Equation (10–57),


or


Finally, it is noted that, similar to the case of pole placement, if the system order nis 4 or
higher, methods 1 and 3 are preferred, because all matrix computations can be carried out by a
computer, while method 2 always requires hand computation of the characteristic equation
involving unknown parameters ke1,ke2,p,ken.


Effects of the Addition of the Observer on a Closed-Loop System. In the


pole-placement design process, we assumed that the actual state x(t)was available for


feedback. In practice, however, the actual state x(t)may not be measurable, so we will


need to design an observer and use the observed state for feedback as shown in Fig-


ure 10–12. The design process, therefore, becomes a two-stage process, the first stage


being the determination of the feedback gain matrix Kto yield the desired characteristic


equation and the second stage being the determination of the observer gain matrix Ke


to yield the desired observer characteristic equation.


Let us now investigate the effects of the use of the observed state rather than


the actual state x(t),on the characteristic equation of a closed-loop control system.


x(t),


x(t)


B


x 1
x 2

R = B


0

1

- 100

- 20

RB


x 1
x 2

R+ B


0

1

Ru+ B


120.6

20

Ry


x =AA-Ke CBx +Bu+Ke y

= B


120.6

20

412

120.6

RB


0

1

1

0

RB


0

1

R = B


120.6

20

R


Ke=AA^2 + 20 A+ 100 IBB


0

1

1

0

R



  • 1
    B


0

1

R


f(A)=A^2 + 20 A+ 100 I

f(s)=As-m 1 BAs-m 2 B=s^2 +20s+ 100

Ke=f(A)B


C

CA

R



  • 1
    B


0

1

R


Ke= B


120.6

20

R


ke1=120.6, ke2= 20


s^2 +20s+ 100 = 0
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