Download State-Space Representations. Linearization

MCE441: Intr. Linear Control Systems
Lecture 16: State-Space Realizations
TF-SS Conversion
Linearization of a nonlinear system about an equilibrium
point
Dorf Ch.3
Cleveland State University
Mechanical Engineering
Hanz Richter, PhD
Instructor
MCE441 – p.1/22
State-Space Modeling
The state-space approach takes into account
intermediate variables between input and output:
state variables
The state variables may correspond to actual physical
quantities or may be just mathematical constructs
The state vector is composed of n state variables,
where n is the system order
Unlike transfer functions, state-space models can
include nonlinearities. Here we focus on linear,
time-invariant state-space systems
Digital controllers are based on state-space
representations of the control TF
It is possible to make conversions between TF and SS.
MCE441 – p.2/22
State-Space Modeling...
To go from TF to SS is to obtain a SS realization of the TF
A given TF admits an infinite number of SS realizations
A general (possibly nonlinear) nth-order state-space model with m inputs and
p outputs has the form:
ẋ1
=
f1 (x1 , x2 , ...xn , u1 , u2 , ...um )
ẋ2
=
f2 (x1 , x2 , ...xn , u1 , u2 , ...um )
ẋn
=
fn (x1 , x2 , ...xn , u1 , u2 , ...um )
y1
=
h1 (x1 , x2 , ...xn , u1 , u2 , ...um )
=
hp (x1 , x2 , ...xn , u1 , u2 , ...um )
..
.
..
.
yp
MCE441 – p.3/22
State-Space Modeling...
With SS, we replace a single n − th order ODE by n 1st
order ones.
The column vector x = [x1 , x2 , ...xn ]T is called the state
vector
The set where x belongs is the state space
For MCE441 we use Rn as the state space. Others can
be used too.
The column vector u = [u1 , u2 ...um ]T is the input vector
The column vector y = [y1 , y2 , ...yp ]T is the output vector
MCE441 – p.4/22
Linear State-Space Systems
In a linear SS system, the fi and hi are linear functions,
so that:
ẋ = Ax + Bu , y = Cx + Du
A is an n-by-n matrix, B is n-by-m, C is p-by-n and D is
p-by-m.
Conversion to TF (prove it using Laplace transform)
Y (s)
U (s)
= G(s) = C(sI − A)−1 B + D
It can be shown that the poles of G(s) are the
eigenvalues of A.
MCE441 – p.5/22
Example
b
k
m
u
Mass-spring-damper system. Choosing position and
velocity as states, we find the following representation:
# "
#"
#
# "
"
0
x1
0
1
ẋ1
+
u
=
b
1
x2
− kb − m
ẋ2
m
MCE441 – p.6/22
Example
Following our I/O approach we would have found:
mÿ + bẏ + ky = u
which gives a transfer function
G(s) =
Y (s)
1
=
U (s)
ms2 + bs + k
We can check that G(s) = C(sI − A)−1 B + D
MCE441 – p.7/22
Another Example
Find a ss description of the the following RLC circuit. Find
the TF between the input and output voltages.
L
R
Vo
i
C
Vi
MCE441 – p.8/22
Solution
MCE441 – p.9/22
Solution
MCE441 – p.10/22
Realization of Transfer Functions
If there are no zeros, obtain the I/O differential equation
and define the n states to be succesive derivatives of
the output, starting with the 0th order (the output itself)
until the derivative of order (n − 1).
Write equations of the form ẋi = xi+1 for i = 1, 2..n − 1.
Find ẋn from the I/O equation itself
Example: Realize G(s) =
2
s2 +s+1
MCE441 – p.11/22
Solution
MCE441 – p.12/22
Realization of Transfer Functions
Often, zeros are present. That is:
b0 sn + b1 sn−1 + ... + bn−1 s + bn
G(s) = n
s + a1 sn−1 + ... + an−1 s + an
Use a canonical realization (one choice among infinite)
State definition:
x1 = y − β0 u
x2 = ẏ − β0 u̇ − β1 u = ẋ1 − βu
..
.
xn = y (n−1) − β0 u(n−1) − ... − βn−1 u = ẋn−1 − βn−1 u
MCE441 – p.13/22
Realization of Transfer Functions...
Coefficient definition:
β0 = b0
β1 = b1 − a1 β0
..
.
βn = bn − a1 βn−1 − ... − an β0
System Matrices:




A=



0
0
..
.
1
0
..
.
0
1
..
.
...
...
..
.
0
0
..
.
0
0
0
... 1
−an −an−1 −an−2 ... −a1








MCE441 – p.14/22
Realization of Transfer Functions...
System Matrices...
B = [β1 β2 , ...βn−1 , βn ]T
C = [1, 0, ...0]
D = β0 = b0
System Description:
ẋ = Ax + Bu
y = Cx + Du
MCE441 – p.15/22
Example (Dorf 3.1)
Find a state-space realization of the transfer function
2s2 + 8s + 6
G(s) = 3
s + 8s2 + 16s + 6
MCE441 – p.16/22
Solution
MCE441 – p.17/22
Solution
MCE441 – p.18/22
State-Space to TF Conversions in Matlab
Old-fashioned method:
To go from tf to ss: [A,B,C,D]=tf2ss(num,den)
To go from ss to tf: [num,den]=ss2tf(A,B,C,D)
New method:
Create a tf object: >>tf_sys=tf(num,den)
Convert to ss: >>sys_ss=ss(tf_sys)
Extract system matrices: [A,B,C,D]=ssdata(sys_ss)
Go back to tf: tf_sys=tf(sys_ss)
Note: The ss realization in the old method uses the control
canonical form. The new method uses an efficient numerical procedure. The second method is preferred since it is
required for conversions to and from discrete time.
MCE441 – p.19/22
Linearization - Equilibrium Points
An equilibrium point or fixed point of a control system under
constant control u0 is defined by the property
ẋ = f (x, u0 ) = 0
In nonlinear systems, multiple equilibrium points can exist. For
example, the system
ẋ1 = x21 − u
ẋ2 = x1 x22 − x2 u
has 4 equilibrium points for u0 = 1. Find them.
MCE441 – p.20/22
Linearization - Equilibrium Points
We may linearize a nonlinear system about an equilibrium point by
using the Jacobian of f . In doing so, we obtain a linear state-space
˙ = A∆x + B∆u. As an
system for state and control deviations: ∆x
example, we linearize the above system about u0 = 1 and
x0 = [1, 0]T
∂f
∂f
∂f
1
1 1
˙ 1 =
∆u
∆x
+
∆x
+
∆x
1
2
∂x1 u0 ,x0
∂x2 u0 ,x0
∂u u0 ,x0
∂f
∂f
∂f
2
2
2
˙ 2 =
∆u
∆x
+
∆x
+
∆x
1
2
∂x1 ∂x2 ∂u u0 ,x0
u0 ,x0
u0 ,x0
MCE441 – p.21/22
Linearization - Equilibrium Points
The matrix A is obtained by taking all partial derivatives of f with
∂fi
), evaluated at (u0 , x0 ). The
respect to all variables (A = {aij } = ∂x
j
matrix B is obtained by taking partial derivatives w.r.t. all control
∂fi
.
components: B = {bi } = ∂u
i
Note that the linearized system is valid only in close proximity to
(u0 , x0 ).
MCE441 – p.22/22