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ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
1
In this lecture we will compare two linearizing controller for a single-link robot:• Linearization via Taylor Series Expansion• Feedback Linearization
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
2
Linear control theory has been predominantly concerned with Linear Time Invariant (LTI) systems of the form
with x being a vector of states and A being the system matrix. LTI systems have quaite simple properties such as
• A linear system has a unique equilibrium point if A is nonsingular;• The equilibrium point is stable if all eigenvalues of A have negative real parts,
regardless of initial conditions;• In the presence of an external input u(t), i.e., with
the system has a number of interesting properties. For example a sinusoidal input leads to a sinusoidal output of the same frequency.
x Ax
x Ax Bu
Slotine, Li, 1993.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
3
Never forget
THERE IS NO LINEAR SYSTEMS
IN NATURE
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
4
One of the characteristic properties of nonlinear systems is “Multiple Equilibrium Points”
Nonlinear systems frequently have more than one equilibrium point (an equilibrium point is a point where the system can stay forever without moving). This can be seen by the following simple examples.
Consider the first order linear system x x
Solution of this differential equation is 0( ) tx t x e
Following figure shows the time variation of this solution for various initial conditions. The system clearly has a unique equilibrium point at x=0.
Slotine, Li, 1993.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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Now consider the following nonlinear systems:2x x x
Solution of this differential equation is
0
0 0
( )1
t
t
x ex t
x x e
Following figure shows the time variation of this solution. The system has two equilibrium points, x=0 and x=1, and its qualitative behavior strongly depends on its initial condition.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
6
1. Linearization of Nonlinear Systems via Taylor Series Expansion
General form of an n-dimensional nonlinear system is
( ) ( )x f x g x u
and of an n-dimensional linear time-invariant system is
x Ax Bu
The linearized form of a nonlinear system can be found as
ex x
fA
x
eu u
gB
u
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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Example: Linearize System
22 1 2
2 1 1 1 2
2 2 1 2
1 2 1 2
[0,0]
cos( )
1 sin( )
cos( ) 2 sin( )
2 1 sin( ) 1 cos( )
1 0
1 1Tx
x x xf
x x x x x
x x x xf
x x x xx
f
x
Eigenvalues = 1,1
Origin is unstable
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
Let’s linearize a single-link robotic manipulator model now.
Dynamic model is as follows:
sin( )Jq Bq N q
: motor inertia
: motor position
: viscous friction coeffient
: load constant
: applied torque.
J
q
B
N
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
By selecting the state variables as
1
2
x q
x q
we get the state-space representation as follows:
1 2
2 2 1
1sin
x x
B Nx x x
J J J
sin( )Jq Bq N q
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
By setting the control input signal, u, to zero, let’s find the equilibrium points
1 2
2 2 1s
0
0in
x x
B Nx x x
J J
From the first equation, we get
2 0ex
Finally, from the second equation, we get
1 1sin 0 0 , 0,1,2,.......e ex x k k
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
Then the equilibrium points are
1
2
0
0e
e
x k
x
Let’s linearize the system around the origin
1
2
0
0e
e
x
x
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
Remember that the system dynamics is
1 2
2 2 1
1sin
x x
B Nx x x
J J J
Then
2
1
2 2 1
( )
( ) ( ) sin
xf x
B Nf x f x x x
J J
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
The Jacobian is
1 1
1 21
2 2
1 2
0 1
cos( )
f f
x x N Bx
f f J J
x x
1 2
2 2 1
1sin
x x
B Nx x x
J J J
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
Since we want to linearize the system around the equilibrium point
1
2
0
0e
e
x
x
then
1
2
1 0
0
0 1 0 1
cos( )e
e
x
x
N B N Bx
J J J J
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
Then the linearized form of the system is
1 1
2 2
0 1 0
1x x
uN Bx x
J J J
Note that this dynamical model is a general LTI system of the form
x Ax Bu
By using MATLAB Symbolic Toolbox, we find the eigenvalues of A matrix as
2
1,2
4
2
B JNB
J
not viscous friction !
viscous friction !
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
1 1
2 2
0 1 0
1x x
uN Bx x
J J J
Let’s design a simple linear state feedback controller in the form of
u Kx
x Ax Bu
so that we get
x A BK x
In this way, by properly selecting the entries of K vector, we will be able to locate the eigenvalues of newly-created system matrix, (A-BK), to the left-half plane to get stability.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
But this stability result will be valid only around the small neighborhood of the linearization point,
1
2
0
0e
e
x
x
and we will not have a global stability result.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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There are many algorithms in the literature proposed to find the entries of K vector.
The most conventional algorithm can be implemented in MATLAB as follows .
% Desired closed-loop polesp1 = -1; p2 = -2;
% Entries of KK = place(A,B,[p1 p2]);
The control input signal u=-Kx drives the trajectories to the equilibrium point x1=x2=0. If the desired trajectory for position is xd, then the control law is modified as
11 2
2
dx xu k k
x
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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2. Linearization of Nonlinear Systems via Feedback Linearization
Feedback linearization is a nonlinear control method and the control input signal to be designed will contain a nonlinear term.
Again consider the general form of a nonlinear system
( ) ( )x f x g x u
If we can rewrite the system in the simplest form, i.e., the form that we will not need a coordinate transformation to transform the system into a linear form as
( ) ( )x Ax B x u x then
1( ) ( )u x x renders the linear time-invariant and controllable system
0. x Ax B
if (A,B) controllable and
Let’s see if we can write single link robot dynamics in this form.
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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1 2
2 2 1
1sin
x x
B Nx x x
J J J
( ) ( )x Ax B x u x
1 11
2 2
0 1 0sin1
0
x xN xB
x xJ J
1
2rank B AB
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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1 11
2 2
0 1 0sin1
0
x xN xB
x xJ J
Then
1sin( )N x
renders
1 1
2 2
0 1 0
10
x xB
x xJ J
ECE 893 Industrial Applications of Nonlinear Control Dr. Ugur HasirciClemson University, Electrical and Computer Engineering Department Spring 2013
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Compare the performances of these two controllers via simulation.