Control systems in state space
Paula Raica
Department of Automation
71-73 Dorobantilor Str., room C21, tel: 0264 - 401267
26-28 Baritiu Str., room C14, tel: 0264 - 202368
email: [email protected]
http://rocon.utcluj.ro/st
Technical University of Cluj-Napoca
Control systems in state space
Analysis of systems in state-space
Control systems in state space
State-space models
The state equation and the output equation :
ẋ(t) = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t)
x(t) is an (n × 1) state vector, where n is the number of
states or system order
u(t) is an (m × 1) input vector, where m is the number of
input functions
y(t) is a (p × 1) output vector where p is the number of
outputs
A is an (n × n) square matrix called the system matrix or
state matrix
B is an (n × m) matrix called the input matrix
C is a (p × n) matrix called the output matrix
D is a (p × m) matrix which represents any direct connection
between the input and the output. It as called the
feedthrough matrix.
Control systems in state space
Stability in state-space
The system poles are given by the solutions of the characteristic
equation that are also the eigenvalues of matrix A:
sX(s) = AX(s) + BU(s)
Y(s) = CX(s) + DU(s)
If we separate X(s) in the state equation we obtain:
X(s) = (sI − A)−1 BU(s)
and then the output is:
Y(s) = [C(sI − A)−1 B + D]U(s)
If the system has one input and one output the transfer function is:
H(s) = C
adj(sI − A)
B+D
det(sI − A)
The characteristic equation of the system is given by:
det(sI − A) = 0
Control systems in state space
Example
Given the system represented in spate-space by:
0
1
0
ẋ(t) =
x(t) +
u(t)
−2 −3
1
y (t) =
1 0 x(t)
The eigenvalues of the matrix A are obtained from:
λ −1
(λI − A) =
det(λI − A) = (λ + 1)(λ + 2) = 0
2 λ+3
λ1 = −1,
λ2 = −2
The system transfer function is computed from
"
# s+3
1
0
(s+1)(s+2)
(s+1)(s+2)
H(s) = C(sI − A)−1 B = 1 0
−2
s
1
(s+1)(s+2)
(s+1)(s+2)
H(s) =
1
,
(s + 1)(s + 2)
⇒ The poles: − 1, −2
Control systems in state space
System poles and the eigenvalues of A. Example
In special cases, it is possible that a part of the system poles are
canceled by some of the system zeros when transforming a
state-space model into a transfer function.
Example. Consider a system with the input u(t) and the output
y (t) having the state-space model:
0 1
0
ẋ(t) = Ax(t) + Bu(t) =
x(t) +
u(t)
1 0
1
y (t) = Cx(t) = −1 1 x(t)
The characteristic equation is:
s −1
det(sI − A) = det
= s 2 − 1 = (s − 1)(s + 1)
−1 s
The eigenvalues of the system matrix, or the system poles are
λ1 = 1 and λ2 = −1.
Control systems in state space
System poles and the eigenvalues of A. Example
The transfer function is computed from:
"
s
(s−1)(s+1)
−1
H(s) = C(sI − A) B = −1 1
1
(s−1)(s+1)
or
H(s) =
1
(s−1)(s+1)
s
(s−1)(s+1)
# 0
1
1
s −1
=
(s − 1)(s + 1)
s +1
The transfer function has only one pole at −1 because the other
pole (s = 1) was canceled with the zero at s = 1.
A state-space formulation will give more information about the
system than the input-output formulation described by a transfer
function.
Control systems in state space
Stability in state-space
internal stability - given by the eigenvalues of the system
matrix
external stability - given by the transfer function poles
System poles or eigenvalues of the system matrix: λi = σi + jωi .
Stability condition
Stable
Marginally stable
Unstable
Root values
σi < 0, for all i = 1, n (All the roots are in
the left-half s-plane)
σi = 0 for any i for simple roots and no
σi > 0, for i = 1, n (At least one simple
root and no multiple roots on the jω axis
and the other roots on the left-half s-plane)
σi > 0 for any i or σi = 0 for any multipleorder root, i = 1, n (At least one simple
root in the right-half s-plane or at least one
multiple-order root on the jω axis)
Control systems in state space
Stability in state-space
When we use state-space descriptions of dynamic systems we
discuss the following types of stability:
Internal stability that refers to the state variables. Stability
conditions in this case are applied for the eigenvalues of
system matrix λi = σi + jωi (or the system poles)
External stability that refers to the output signal. Stability
conditions are analyzed for the transfer function poles
pi = σi + jωi .
Transfer functions can only be used to analyze the external
stability of systems.
State-space descriptions can be used to analyze both internal
and external stability.
Control systems in state space
Example
Consider the system in state-space form:
−2 0
1
ẋ(t) =
x(t) +
u(t)
0 3
2
y (t) = 1 0 x(t)
Eigenvalues: λ1 = −2 < 0, λ2 = 3 > 0. The system is internally
unstable (λ2 > 0).
The transfer function for this system:
H(s) = C(sI-A)−1 B =
1
s −3
=
(s − 3)(s + 2)
s +2
After the pole-zero cancelation, the transfer function has only one
stable pole at −2, therefore the system is externally stable.
Control systems in state space
Example
The output equation: y (t) = x1 (t).
Figure: Block diagram of the system
Control systems in state space
Controllability and observability
The goal for control design: controlling the system with a
signal u(t) that is a function of several measurable state
variables.
If all the state variables are measurable use a full-state
feedback control law:
If not all the states can be measured: use an observer to
estimate the states connected to the full-state feedback
control law.
Control systems in state space
Controllability and observability
A system is said to be controllable at time t0 if it is possible
to transfer the system from any initial state x(0) to any other
state in finite time, by means of an unconstrained control
vector u(t).
A system is said to be observable at time t0 if, with the
system in state x(t0 ), it is possible to determine this state
from the observation of the output over a finite time interval.
(i.e. the ability to determine the state variables from the
knowledge of the input u(t) and the output y(t)).
The solution to the control problem may not exist if the
system considered is not controllable.
Control systems in state space
Controllability test
For an n-th order system:
ẋ(t) = Ax(t) + Bu(t)
Build the controllability matrix:
PC = B AB A2 B · · ·
An−1 B
The system is controllable if PC has full rank n:
rank PC = n
If the system has one input, then PC is an n × n matrix. In
this case it is easy to test the rank of PC by making sure the
determinant of PC is non-zero.
Control systems in state space
Controllability. Example
Determine if the system ẋ(t) = Ax + Bu is controllable, where
2 3
1
A=
, B=
0 5
0
Construct the matrix PC :
PC = B AB
1
where B =
and
0
1 2
⇒ PC =
0 0
2 3
AB =
0 5
1
2
=
0
0
Since det PC = 0, the rank of PC is less than 2. Hence, the system
is not controllable.
Control systems in state space
Observability test
For an n-th order system:
ẋ(t) = Ax(t) + Bu(t)
y(t) = Cx(t)
Build the observability matrix:


C
 CA 

PO = 
 ··· 
CAn−1
The system is completely observable (all states are
observable) if PO has rank n:
rank PO = n
Control systems in state space
Observability. Example
Consider a system described by the state and output equations:


 
0
1
0
0
ẋ(t) = Ax(t) + Bu(t) =  0
0
1  x(t) + 0 u(t)
−4 −3 −2
1
y (t) = Cx(t) = 0 5 1 x(t)
The observability matrix is:


C
PO =  CA  , where C = 0 5 1 , CA = −4 −3 3
CA2


0
5
1
CA2 = −12 −13 −9 ⇒ PO =  −4 −3
3
−12 −13 −9
Since the determinant det PO = −344, PO is full rank equal
to 3. The system is thus observable.
Control systems in state space
Design of state variable feedback systems
Control systems in state space
Design of state variable feedback systems
The method: pole placement for SISO systems
Assume:
All state variables are measurable and available for feedback.
The system is completely state controllable
The poles of the closed-loop system may be placed at any
desired location by means of state feedback through an
appropriate state feedback matrix.
Assign a set of closed-loop poles that will correspond to
satisfactory dynamic response in terms of transient response
specifications.
We will consider first that the reference input is zero r (t) = 0
and then, we shall discuss the case when the reference is
non-zero.
Control systems in state space
Selection of pole locations
One method: choose the closed-loop poles for a higher order
system as a desired pair of dominant second-order poles and
select the rest as real non-dominant poles.
Remember the relation
p between the pole location
s1,2 = −ζωn ± ωn 1 − ζ 2 j and the transient response
characteristics
Settling time: ts =
Peak time: tp =
Rise time: tr =
π
ωd
π−β
ωd
4
ζωn
=
=
ωn
√π
ωn
1−ζ 2
π−β
√
1−ζ 2
√
Overshoot Mp : Mp = e −πζ/
Control systems in state space
1−ζ 2
Design of full-state feedback control law by pole placement
Consider a SISO system described in state-space by:
ẋ(t) = Ax(t) + Bu(t)
y (t) = Cx(t)
The reference input is considered zero: r (t) = 0, i.e. all state
variables must approach zero in steady-state.
We shall choose the control law (control signal) to be a linear
combination of the state variables:
u(t) = −Kx(t) = −k1 x1 − k2 x2 − · · · − kn xn
Control systems in state space
Design of full-state feedback control law by pole placement
Full state feedback control law
Full state feedback control law (figure showing all states)
Control systems in state space
Design of full-state feedback control law by pole placement
The closed-loop system (with the feedback gain):
ẋ(t) = Ax(t) − BKx(t) = (A − BK)x(t)
The characteristic equation of the closed-loop system is:
det[sI − (A − BK)] = 0
(1)
When evaluated, this gives an n-th order polynomial in s
containing the elements of the gain matrix K: k1 , k2 , ..., kn .
The desired location of the closed-loop poles are:
p1 , p2 , · · · , pn and the corresponding desired characteristic
equation is:
(s − p1 )(s − p2 ) · · · (s − pn ) = 0
The required elements of K are obtained by matching
coefficients of equations (1) and (2).
Control systems in state space
(2)
Example. Inverted pendulum
Consider the problem of balancing the inverted pendulum.
The angle subsystem of the pendulum is given for some specific
values of the parameters by:
0 1
0
ẋ(t) =
x(t) +
u(t)
9 0
−2
T
where the state vector is x(t) = θ(t) θ̇(t) .
Control systems in state space
Example. Inverted pendulum
The open-loop system poles are given by:
λ −1
det(λI − A) = det
= λ2 − 9 = (λ − 3)(λ + 3) = 0
−9 λ
The system poles are: 3 and −3. The open-loop system is
unstable.
Check the controllability matrix:
0 −2
PC = B AB =
, det PC = −4
−2 0
The rank of the controllability matrix is 2, thus the system is
controllable.
The design specifications: a state variable feedback matrix K
that will give a closed-loop response with an overshoot of
about 4.3% and a settling time of about 1 second. The angle
θ and the angular velocity θ̇ will be zero in steady state.
Control systems in state space
Example. Inverted pendulum
We compute the desired poles of the closed-loop system that
will give the desired behavior:
The damping ratio that gives
√ an overshoot of 4.3% can be
computed as equal to ζ = 2/2.
The settling time is
ts =
4
=
ζωn
4
√
2
2
· ωn
⇒
8
ωn = √
2
The complex poles result as:
p1,2 = −ζωn ± jωn
p
1 − ζ 2 = −4 ± 4j
The characteristic equation for the desired poles is:
(s − p1 )(s − p2 ) = (s + 4 − 4j)(s + 4 + 4j) = s 2 + 8s + 32 = 0
Control systems in state space
Example. Inverted pendulum
The characteristic equation
of the closed-loop system with the
feedback gain K = k1 k2 is:
s 0
0 1
0 det[sI−(A − BK)] = det
−
+
k1 k2 = 0
0 s
9 0
−2
or
s
−1
s 0
0
1
det
−
= det
=0
−9 − 2k1 s − 2k2
0 s
9 + 2k1 2k2
and:
s(s − 2k2 ) − (9 + 2k2 ) = s 2 − 2sk2 − 9 − 2k1 = 0
By setting the characteristic equation of the closed loop
system equal to the characteristic equation for the desired
poles we obtain:
−2k2 = 8, −9 − 2k1 = 32, and k1 = −20.5, k2 = −4
and the feedback gain matrix is: K = −20.5 −4
Control systems in state space
Example. Inverted pendulum
A plot of the closed-loop system response to initial conditions (and
zero input) is shown below.
Response to Initial Conditions
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time (sec)
Figure: Inverted pendulum response to initial conditions (closed-loop
system)
From the figure, the settling time is about 1 second. The
”overshoot” is, in this case, the difference between the minimum
value and the steady state value (i.e. 0) and, as shown in the
figure, it is about 4%.
Control systems in state space
State feedback for tracking systems
Consider the system described by the state-space model:
ẋ(t) = Ax(t) + Bu(t)
y (t) = Cx(t)
The goal is to drive the output y (t) to a given reference input
r , with zero-steady-state error.
One solution is to scale the reference input r and choose the
control law to be a linear combination of the states (with r
and N scalar values for SISO)
u(t) = Nr − Kx(t) = Nr − k1 x1 − k2 x2 − · · · − kn xn
Control systems in state space
State feedback for tracking systems
The closed-loop system is then:
ẋ(t) = Ax(t) + B(Nr − Kx(t)) = (A − BK)x(t) + BNr
y (t) = Cx(t)
The feedback matrix K is determined by pole placement.
At steady-state ẋ = 0. xss , yss = the steady-state values of
the state vector and the output
0 = (A − BK)xss + BNr ⇒ xss = −(A − BK)−1 BNr
yss
= Cxss = −C(A − BK)−1 BNr
The steady-state error is zero if yss = r , ⇒ N can be
computed from:
yss = −C(A − BK)−1 BNr = r , ⇒ N = −
1
C(A − BK)−1 B
Control systems in state space
State feedback for tracking systems. Example
Consider the system described
2
ẋ =
1
y (t) = 1
in state-space by:
3
1
x(t) +
u(t)
0
0
−1 x(t)
Design a state-feedback control system so that the closed-loop
poles are p1,2 = −1 ± j and the steady-state error for a step input
is zero.
We check controllability first. The controllability matrix, the
determinant and the rank are:
1 2
PC = B AB =
, det PC = 1, ⇒ rankPC = 2
0 1
The system is controllable.
Control systems in state space
State feedback for tracking systems. Example
The feedback gain matrix K is obtained by pole placement. If
the desired poles of the closed-loop system are p1,2 = −1 ± j,
the characteristic equation for these poles is:
(s + 1 − j)(s + 1 + j) = s 2 + 2s + 2 = 0
The characteristic equation
of the closed-loop system with the
feedback gain K = k1 k2 is:
s − 2 + k1 −3 + k2
det[sI − (A − BK)] =
=0
−1
s
or:
s(s − 2 + k1 ) − (−1)(−3 + k2 ) = s 2 + (−2 + k1 )s − 3 + k2 = 0
By setting the characteristic equation of the closed loop
system equal to the characteristic equation for the desired
poles we obtain:
−2 + k1 = 2, −3 + k2 = 2, and k1 = 4, k2 = 5
Control systems in state space
State feedback for tracking systems. Example
The feedback gain matrix is:
K= 4 5
This matrix will only ensure the desire transient characteristics
and not the zero steady-state error for a step input.
Compute N from:
1
N=−
C(A − BK)−1 B
2 3
1 −2 −2
(A − BK) =
−
4 5 =
1 0
0
1
0
0
1
(A − BK)−1 =
−1/2 −1
1
0
1
1
−1
C(A − BK) B = 1 −1
= 1/2 1
= 1/2
−1/2 −1 0
0
1
N=−
= −2
1/2
Control systems in state space
State feedback for tracking systems. Example
The step response of the closed-loop system:
Step Response
Amplitude
1
0.5
0
−0.5
0
1
2
3
4
5
6
7
8
Time (sec)
Figure: Step response of the closed-loop system
The output follows the unit step, thus yss = 1 and the steady-state
error is zero: ess = 0.
Control systems in state space