52 CHAPTER 2. MATHEMATICAL MODELS IN CONTROL
From this, we can determine the desired characteristic matrix as
Acf=
010
001
− 8 − 10 − 6
Comparing the last row of coefficients betweenAcandAcf, we obtain the feed-
back gain matrix
K=[− 6 − 13 −6]
For the examples considered here, we see a fairly straightforward approach
to implementing state-variable feedback control. However, as mentioned previ-
ously, all states of the original system must be accessible, and the system must
be completely controllable. Therefore, we need to knowapriorithat the sys-
tem is fully controllable before attemptingtoimplementstate-variablefeedback
control.
Suchaprioriknowledge can be obtained by investigating the rank of the
n◊nmcontrollability matrix
£
BABA^2 B ∑∑∑ An−^1 B
§
formed by theAandBmatrices of the original system (see page 32). Recall
that annth-order single-input system is fully controllable if and only if
Rank£
BABA^2 B ∑∑∑ An−^1 B§
=n
This says that if the rank of the controllability matrix is less thann,the
system is not fully controllable. A system being not fully controllable implies
that one or more states of the system are not directly controllable. This is due
to pole-zero cancellations in the systemtransfer function as the next example
indicates.
Example 2.7Consider the state and output equations of a second-order sys-
tem given by
∑
x ̇ 1
x ̇ 2
∏
=
∑
− 20
− 1 − 1
∏∑
x 1
x 2∏
+
∑
1
1
∏
uy =£
01
§
∑
x 1
x 2∏
Computing the transfer functionG(s)=C[sI−A]−^1 B+Dyields
G(s)=
(s+1)
(s+1)(s+2)in which the pole and zero located ats=− 1 cancel, making the system uncon-
trollable at the eigenvalueλ=− 1 .Whether or not the system is fully control-
lable can also be verified by computing the rank of the controllability matrix as
follows:
Rank∑
1
1
∑
− 20
− 1 − 1
∏∑
1
1
∏∏
=Rank