Introduction of Lyapunov-Based Control
1An Example of Nonlinear Systems
Linear System ˙x=Ax+Bu
y=Cx
(1) it has the superposition property.Besides,the stability of the linear system completely depends on its parameters.
Nonlinear System ˙x=f(x,u)
y=g(x)
(2)
superposition does not hold for nonlinear systems,and the stability of a nonlinear system depends on both system parameters and initial conditions.
Example:The dynamic model for a2-DOF overhead crane system(see Figure??)can be presented as follows
M(q)¨q+V m(q,˙q)˙q+G(q)=u(3)
q=[x(t)θ(t)]T(4) where x(t)∈R1denotes the gantry position,θ(t)∈R1denotes the payload angle with respect to the vertical,and M(q)∈R2×2,V m(q,˙q)∈R2×2,G(q)∈R2,and u(t)∈R2are defined as follows
M(q)= m c+m p−m p L cosθ
−m p L cosθm p L2 ,
V m(q,˙q)= 0m p L sinθ˙θ
00 ,
G(q)= 0m p gL sinθ T,u(t)= F0 T,(5)
where m c,m p∈R1represent the gantry mass and the payload mass,respectively,L∈R1represents the length of the rod to the payload,g∈R1represents the gravity coefficient,and F(t)∈R1 represents the control force input acting on the gantry(see Figure??).
2Common Nonlinear Systems Behaviors
2.1Multiple Equilibrium Points
For the system
˙x=f(x)(6)
1
Figure1:
To obtain equilibrium points,solve the following equation
f(x s)=0
if it has multiple solutions,then the system has multiple equilibrium points.
Example:
˙x=−x+x2(7)
−x+x2=0=⇒x s=1or x s=0
has too equilibrium points:x s=1and x s=0.
Question:Which one is stable?Why?
Rewrite the equation as follows
˙x=x(x−1)
then if x>1,˙x>0,x increases with time.For the case of x<1,(x−1)<0,thus
˙x<0,for1>x>0
˙x>0,for x<0
therefore,x s=1is unstable and x s=0is stable.
We can solve the system equation to obtain the response as follows
x(t)=
x0e−t
1−x0+x0e−t
=1−
1−x0
1−x0+x0e−t
From this formula,it can be seen that if x0<1,the denominator1−x0+x0e−t>0,thus
lim
t→∞
x(t)=0.
The response of x0=0.99and x0=1.0001is demonstrated in Figure??.
2
Figure2:
2.2Limit Cycle
Limit Cycle:Oscillation offixed amplitude andfixed frequency without external excitation.
Example:
¨x+(x2−1)˙x+x=0
if|x|>1,x2−1>0,the damper consumes energy.If|x|<1,x2−1<0,the damper produces energy.Therefore,the system has a tendency to make x oscillate around some region.
Response of
x0=5,˙x0=5
and
x0=−1,˙x0=−3
is shown in Figure3
2.3Chaos
The system output is extremely sensitive to initial conditions.
Example:
¨x+0.1˙x+x5=6sin(t)
Figure4shows the response of the system to two almost identical initial conditions
x0=2,˙x0=3
and
x0=2.01,˙x0=3.01.
Due to the high nonlinearity in x5,the response become apparently different after a certain time.
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