Modern Control Engineering

(Chris Devlin) #1
720 Chapter 9 / Control Systems Analysis in State Space

Problems


B–9–1.Consider the following transfer-function system:

Obtain the state-space representation of this system in (a)
controllable canonical form and (b) observable canonical
form.

B–9–2.Consider the following system:

Obtain a state-space representation of this system in a di-
agonal canonical form.

B–9–3.Consider the system defined by

where

Transform the system equations into the controllable canon-
ical form.

B–9–4.Consider the system defined by

where

Obtain the transfer function Y(s)/U(s).

B–9–5.Consider the following matrix A:

Obtain the eigenvalues l 1 ,l 2 ,l 3 , and l 4 of the matrix A.
Then obtain a transformation matrix Psuch that

P-^1 AP=diagAl 1 , l 2 , l 3 , l 4 B

A= D


0

0

0

1

1

0

0

0

0

1

0

0

0

0

1

0

T


A=C


- 1

1

0

0

- 2

0

1

0

- 3

S, B= C


0

0

1

S, C=[ 1 1 0 ]


y=Cx

x#=Ax+Bu

A=B


1

- 4

2

- 3

R, B= B


1

2

R, C=[ 1 1 ]


y=Cx

x#=Ax+Bu

y%+ 6 y$+ 11 y# + 6 y= 6 u

Y(s)
U(s)

=

s+ 6
s^2 +5s+ 6

B–9–6.Consider the following matrix A:

ComputeeAtby three methods.

B–9–7.Given the system equation

find the solution in terms of the initial conditions x 1 (0),
x 2 (0), and x 3 (0).

B–9–8.Find x 1 (t)andx 2 (t)of the system described by

where the initial conditions are

B–9–9.Consider the following state equation and output
equation:

Show that the state equation can be transformed into the
following form by use of a proper transformation matrix:

Then obtain the output yin terms of z 1 , z 2 , and z 3.

B–9–10.Obtain a state-space representation of the follow-
ing system with MATLAB:

Y(s)
U(s)

=

10.4s^2 +47s+ 160
s^3 +14s^2 +56s+ 160

C


z# 1
z# 2
z# 3

S = C


0

1

0

0

0

1

- 6

- 11

- 6

SC


z 1
z 2
z 3

S + C


1

0

0

Su


y =[1 0 0]C


x 1
x 2
x 3

S


C


x# 1
x# 2
x# 3

S = C


- 6

- 11

- 6

1

0

0

0

1

0

SC


x 1
x 2
x 3

S + C


2

6

2

Su


B


x 1 (0)
x 2 (0)

R= B


1

- 1

R


B


x# 1
x# 2

R= B


0

- 3

1

- 2

RB


x 1
x 2

R


C


x# 1
x# 2
x# 3

S =C


2

0

0

1

2

0

0

1

2

SC


x 1
x 2
x 3

S


A= B


0

- 2

1

- 3

R


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