Modern Control Engineering

716 Chapter 9 / Control Systems Analysis in State Space

Pis equal to that of Q.] If the rank of Pism, then CeATx(0)spans the m-dimensional output

space. This means that if the rank of Pism, then Cx(0)also spans the m-dimensional output space

and the system is completely output controllable.

Conversely, suppose that the system is completely output controllable, but the rank of Pisk,

wherek<m. Then the set of all initial outputs that can be transferred to the origin is of

k-dimensional space. Hence, the dimension of this set is less than m. This contradicts the as-

sumption that the system is completely output controllable. This completes the proof.

Note that it can be immediately proved that, in the system of Equations (9–114) and (9–115),

complete state controllability on 0tTimplies complete output controllability on 0tT

if and only if mrows of Care linearly independent.

A–9–16. Discuss the state controllability of the following system:

(9–119)

Solution.For this system,

Since

we see that vectors BandABare not linearly independent and the rank of the matrix [BAB]

is 1. Therefore, the system is not completely state controllable. In fact, elimination of x 2 from

Equation (9–119), or the following two simultaneous equations,

yields

or, in the form of a transfer function,

Notice that cancellation of the factor (s+2.5)occurs in the numerator and denominator of the

transfer function. Because of this cancellation, this system is not completely state controllable.

This is an unstable system. Remember that stability and controllability are quite different things.

There are many systems that are unstable, but are completely state controllable.

A–9–17. A state-space representation of a system in the controllable canonical form is given by

(9–120)

y=[0.8 1 ]B (9–121)

x 1

x 2

R

B

x# 1

x# 2

R =B

0

- 0.4

1

- 1.3

RB

x 1

x 2

R +B

0

1

Ru

X 1 (s)

U(s)

=

s+2.5

(s+2.5)(s-1)

x$ 1 +1.5x# 1 - 2.5x 1 =u#+2.5u

x# 2 =-2x 1 +1.5x 2 +4u

x# 1 =-3x 1 +x 2 +u

AB=B

- 3

- 2

1

1.5

RB

1

4

R= B

1

4

R

A=B

- 3

- 2

1

1.5

R, B= B

1

4

R

B

x

1

x# 2

R = B

- 3

- 2

1

1.5

RB

x 1

x 2

R + B

1

4

Ru

Openmirrors.com