A First Course in FUZZY and NEURAL CONTROL

2.6. STATE-VARIABLE FEEDBACK CONTROL 49

The characteristic polynomial therefore is(λ+1)(λ−2) = 0. Clearly this
system is unstable because(λ−2) = 0yields a pole in the right-halfs-plane.
The objective now is to obtain a new system such that the closed-loop system
has poles that are only in the left-halfs-plane.

Let us now consider the case in which all system statesxof the system are
fed back through a feedback matrixK,whereris the input as illustrated in
Figure 2.17.

Figure 2.17. State variable feedback control system

For a single-input system withnstates, the matrixKis a row vector of
dimension ( 1 ◊n), and a control law can be formulated whereu=r+Kx.
Consequently, the state equations can now be written asx ̇=Ax+B(r+Kx),
which yieldsx ̇=(A+BK)x+Br. Of course the output remains unchanged.
The characteristic polynomial now isdet (λI−(A+BK)) = 0. Letting the
vectorK=

£

k 1 k 2

§

for our example and substituting the matrix elements
forAandB,weobtain

det

μ

λ

∑

10

01

∏

−

μ∑

− 13

02

∏

+

∑

1

1

∏

£

k 1 k 2

§

∂∂

=0

which can be simplified to

det

μ∑

λ+1−k 1 − 3 −k 2

−k 1 λ− 2 −k 2

∏∂

=0

The characteristic polynomial is therefore

λ^2 +(− 1 −k 1 −k 2 )λ+(− 2. 0 −k 1 −k 2 )=0

Suppose now that the desired characteristic equation is

(λ+1)(λ+2)=0