Modern Control Engineering

Section 10–2 / Pole Placement 733

First, we need to check the controllability matrix of the system. Since the controllability matrix
Mis given by

we find that |M|=–1,and therefore, rank M=3.Thus, the system is completely state control-
lable and arbitrary pole placement is possible.
Next, we shall solve this problem. We shall demonstrate each of the three methods presented
in this chapter.

Method 1: The first method is to use Equation (10–13). The characteristic equation for the system is

Hence,

The desired characteristic equation is

Hence,

Referring to Equation (10–13), we have

whereT=Ifor this problem because the given state equation is in the controllable canonical form.
Then we have

Method 2: By defining the desired state feedback gain matrix Kas

and equating with the desired characteristic equation, we obtain

=s^3 + 14 s^2 + 60 s+ 200

=s^3 +A 6 +k 3 Bs^2 +A 5 +k 2 Bs+ 1 +k 1

= 3

s

0

1 +k 1

- 1

s

5 +k 2

0

- 1

s+ 6 +k 3

3

∑s I-A+BK∑=^3 C

s

0

0

0

s

0

0

0

s

S -C

0

0

- 1

1

0

- 5

0

1

- 6

S+ C

0

0

1

SCk 1 k 2 k 3 D^3

∑s I-A+BK∑

K=Ck 1 k 2 k 3 D

=[199 55 8]

K=[200- 1  60 - 5  14 - 6]

K=Ca 3 - a 3  a 2 - a 2  a 1 - a 1 D T-^1

a 1 =14, a 2 =60, a 3 = 200

=s^3 +a 1 s^2 +a 2 s+a 3 = 0

(s+ 2 - j4)(s+ 2 +j4)(s+10)=s^3 +14s^2 +60s+ 200

a 1 =6, a 2 =5, a 3 = 1

=s^3 +a 1 s^2 +a 2 s+a 3 = 0

=s^3 + 6 s^2 + 5 s+ 1

∑s I-A∑= 3

s

0

1

- 1

s

5

0

- 1

s+ 6

3

M=CBABA^2 BD= C

0

0

1

0

1

- 6

1

- 6

31

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