Explicit dependence on t
D  n
t0
Analyze the non-autonomous systems by adapting (extending) the
tools we developed for autonomous systems.
1
Response not affected by t0
Autonomous vs. Nonautonomous
x(t) versus time (t)
x  x
t0=0
t0=3
t0=6
t0=9
x(t)
x(t0)=3
Response depends on t0
x(t) versus time (t)
x  x  t
time (t)
x(t0)=3
x(t)
Autonomous
t0=0
t0=3
t0=6
t0=9
Nonautonomous
2
time (t)
Motivation: The Tracking Problem
Tracking problem:
Given the system: x  bx  u
3
want x to follow xd  a sin(t ), a=1
Up until now we have only talked about regulation (stabilizing
the system to an equilibrium point). That problem yielded an
autonomous system even with state feedback.
Tracking problem: Define e  xd  x then design u drive e to zero so that x follows xd
Tracking problem:
e  xd  x  cos(t )  bx 3  u
Based on what we already know
a solution may look like this
V  12 e 2
V  ee

 
 e xd  (bx 3  u )  e cos(t )  bx 3  u

let u  cos(t )  bx 3  Ke with K  0
V   Ke 2
e(t )  e(0)e   t for K  0  x  xd  we have acheived tracking.
Are all signals bounded?
1. Know xd is bounded, then the closeness
Given the closed-loop system: x  cos(t )  K sin(t )  Kx
want x to follow xd  a sin(t )
of x to xd tells us that x is bounded since e goes to zero.
2. Back to the original system
x  bx 3  K ( a sin(t )  x )  x is bounded
bounded
xd  x goes to zero
 all sisgnals in the system are bounded.
Looking at the closed-loop system we see that
it is non-autonomous.
Do the tools for autonomous systems support
3
this result???
Example:
x1   x1  x2  t
x 2  2 x 2
Is there an equilibrium at f (0, t ) ?
x1  t
x2  0
 f (0, t )  0  t  t0  0  not an eq
Is there an equilibrium when x1  t0 , i.e. f ((t0 ,0), t0 ) ?
x1  0
x2  0
would seem to suggest an eq point, but
as soon as time advances
x1   x1  t0  t
x2  0
 f (( x1 ,0), t )  0 t  t0  not an eq
Difference from autonomous:
added this constraint on the
initial condition t0
Example:
x1  10 x1  x2 e3t
x2  2 x2
Find equilibrium points
0  10 x1  x2 e3t
0  2 x2
 x2  0  x1  0
 f (0, t )  0  t  t0 (for any t0 )
4
x  0 is an eq
Difference from autonomous:
added this constraint on the
initial condition t0
Evolution of the state from same initial
state but different starting times
x(0, t02 )
x(0, t01 )
t01
(Modified from Chapter 3)
5
Difference from autonomous:
added this constraint on the
initial condition t0
6
Bound on initial condition is
independent of the initial time.
Independent of t0
Independent of t0
x(0)
Independent
of t0
7
W ( x, t )
x0
semi-definite
8
x0
The function of
time and state
is dominated by the
role of the state.
Second form not useful
in Lyapunov anaysis
9
class  :  (0)=0 and
strictly increasing
Radially
unbounded  Global
Recall: x     ( x )10

From Chapter 3

11
1
If x were some small constant, this limit
is dominated by t 2 and so it is not possible
to bound W2 (x,t) as a function of x
12
(cont)
1
13
W3 ( x, t )  ( x12  x22 )(t 2  1)  V3 ( x)(t 2  1)
V3 ( x)  ( x12  x22 )  ( x12  x22 ) (t 2  1)
1
If x were some small constant, this limit
is dominated by t 2 and so it is not possible
to bound W2 (x,t) as a function of x
14
( x12  x22 )
V4 ( x) 
( x12  1)
W ( x, t ) not dependent on t
W ( x, t )  1 as x1  
15
Class-K function
1
 k1  1
2
16

Following theorems
apply to the origin.
17
x
t
Scalar function
Need to find V1 ( x)  W ( x, t ) where V1 is PD
Need to find W ( x, t )  V3 ( x) where V3 is PSD
Need to find W ( x, t )  V2 ( x) x in D
x=0
x doesn't "blow up", may not be true for all t0
W ( x, t ) (must now remember to include
W
)
t
18
W ( x, t ) (must now remember to include
W
)
t
19
V1 ( x)
V2 ( x)
This is rarely satisfied.
20
21
Linear subsystem
Note: Designing this function
Positive definite
1  term  2
W(x,t) is radially unbounded
is the hard part.
Radially unbounded

uniformly
(for any t0 )
22
2
Rewrite last term
Always negative so
we can remove and find
a smaller upper bound
(still negative)
23
?
How can the properties of f ( x, t ) be matched to the problem of g ( x, t )
24
Difficulty is to find this W( x, t )
W ( x, t ) ignoring g ( x, t )
Real difficulty is to pick W( x, t )
so that W( x, t ) is useful
bounded by a line
Gradient since W
is a scalar function (of a vector)
25
26
Example 1 of Perturbation Analysis
x1  2 x1  sin( x1 )
g(x, t)
x=  x1 
Notes:
1) sin( x1 )  1 x1
k5
2) Autonomous system
3) Ignoring sin( x1 ) :
W ( x, t )  ( 12 ) x12 then
1
4
k1
x  W ( x, t )  1 x
2
2
k2
W ( x, t )  x1 x1
 2 x12
 2 x
2
k3
Theorem 3  GES
(for this approximated system)
Analysis using Theorem 5
27
Example 1 of Perturbation Analysis (cont)
Ignoring the perturbation
W ( x, t )   2 x
2
k3
W ( x, t )  ( 12 ) x12
Alternate

W ( x, t )  W ( x, t )   x1 
x
Full W ( x, t ) for W ( x, t )  ( 12 ) x12 :
W ( x , t )  x  1 x
2
1
2
k4
W ( x, t )  x1 x1  x1  2 x1  sin( x1 ) 
 2 x12  x1 x1
  2 x1  1 x1 1 x1
2
Notes:
1) sin( x)  1 x
k5
k3  k 4 k5  0
2  1*1  0
 origin is ES
k3
k4
k5
  2 x1  1 x1   x1
2
k3
2
k4 k5
2
28
Example 1 of Perturbation Analysis (cont)
Perturbed system
Unperturbed system
Time
29
Example 2 of Perturbation Analysis
x1  x2  x1
x 
x=  1 
 x2 
+0
g(x,t)
x2  4 x1  2 x2   x2 where   0 is unknown
Notes:
1)  x2   x
In general: x  x12  x22  x32  ...
2) Autonomous system
Triangle Inequality: x1  x2
2
2
 x1  x2
2
2
3) Ignoring  x2 :
W ( x, t )  4( 12 ) x12  ( 12 ) x22 then
1
2
x  W ( x, t )  2 x
2
2
k2
k1
W ( x, t )  4 x1 x1  x2 x2
 4 x1 x2  4 x12  4 x1 x2  2 x22

 4 x  2 x  2 x  x
2
1
2
2
2
1
2
2
 2 x
k3
2
Theorem 3  GES
(for this approximated system)
Analysis using Theorem 5
30
Example 2 of Perturbation Analysis (cont)
Ignoring the perturbation
W ( x, t )   2 x
2
k3
W ( x, t )  4( 12 ) x12  ( 12 ) x22
W ( x , t ) 

W ( x, t )   4 x1
x
x2 
W ( x, t )  16 x12  x22  16 x12  16 x22  4 x12  x22  4 x
k4
Notes:
1)  x2   x
k5
k3  k 4 k5  0
2  4  0
 origin is ES if
2
 0.5  
4
31
Example 2 of Perturbation Analysis (cont)
 0
Our Result  origin is ES if 0.5  
  1.5
  2.5
The condition we calculated appears to be:
1) Valid
2) Conservative (i.e. large  appear to be ok)
 3
32
Z
Z 
is the set of all integers {... -1,-2,-3,0,1,2,3,...}

is the set of all nonnegative integers {0,1,2,3,...}
33
Discrete
Cont.
Cont.
Discrete

34
x(t)
t
Example: Digital Motor Controller
voltage
position Encoder
D/A +
Motor
(A/D)
Amplifier
Micro Controller
(control algorithm)
Normally we make the
assumption that if the
sample rate is high
then it acts like a
continuous system.
35
Approximation of slope at t=kT
System at kT
x(kT  T )  x(kT )
 f ( x(kT ), u (kT ))
T
x(kT  T )  x(kT )  T f ( x(kT ), u (kT ))
run
slope
rise
x(kT  T )
x  f ( x, t )
f ( x(kT ), u (kT ))
x(kT )
T
36
Assume T  1
Where x(k  1)  0
37
38
39
40
41
Rewrite V :
1
2
x12 (k )  2 x1 (k ) x2 (k )  4 x22 (k )  14 x12 (k ) 
 14 x12 (k ) 

1
2
x1 (k )  2 x2 (k )


1
4
x12 (k )  2 x1 (k ) x2 (k )  4 x22 ( k )

2
The V is not an explicit function of k , rewrite to see this:
V x 
1
4
2
1

1
2
x1  2 x2

2
thus we can use V1  14 x12 

1
2
x1  2 x2

2
 V where V1 is PD in x
42
Example 7 (cont)
Original System
1
43
Example 7 (cont)
V   32 x22  2ax14  6ax13 x2  4a 2 x16

  3x

 2ax 
  32 x22  2ax14  9 x22  3 x2  2ax13
   32  9  x22  2ax14
2
3
1
2
2
44
Extra – Barbalat’s Lemma (Slotine page 124-128)
If a differentiable function f (t ) has a finite limit as t  ,
and if f (t ) is uniformly continuous,
then f (t )  0 as t  .
Sufficient condition is that
the derivative of f (t ), i.e. f (t ), is bounded.
Leads to the Lyapunov-like Theorem:
If W ( x, t ) satifies
i) W ( x, t ) is lower bounded
ii) W ( x, t ) is negative semi-definite
iii) W ( x, t ) is uniformly continuous in time
Then W ( x, t )  0 as t  
45
Example - Using Barbalat's Lemma
System:
x1   x1  x2 (t )
x2   x1 (t )
Analysis:
W= 12 x12  12 x22
where  (t ) is a bounded, continuous function.
 lower bounded by zero 
W=x1   x1  x2 (t )   x2   x1 (t ) 
  x12  0
Negative Semidefinite 
W  2 x1 x1  2 x1   x1  x2 (t ) 
 2 x12  2 x1 x2 (t )
Can conclude that x1 and x2 are stable.
Can tell about the convergence to zero.
Assumed to be bounded at start.
Know are bounded from above analysis.
 W is bounded
 W is uniformly continuous
 W ( x, t )  0 as t  
Note that since W ( x, t ) is function of x1 ,  x1  0
Can only still say that x2 is bounded.
46
Conclusion
• Have extended idea of Lyapunov function
analysis to nonautonomous systems.
• Perturbation analysis examined the effect of
disturbances (of a specific form).
• Have extended idea of Lyapunov function
analysis to discrete time systems.
• Extended the Lyapunov analysis using
Barbalat’s Lemma to make a statements about
the derivative of the Lyapunov function.
Homework 4.1
Marquez Problems 4.4, 4.5
Homework 4.1 (sol)
Doesn’t
exist
Homework 4.1 (sol)
Homework 4.1 (sol)
Homework 4.1 (sol)
Homework 4.2
Marquez Problems 4.6, 4.9
53
Homework 4.2 (sol)
54
Homework 4.2 (sol)
55