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8/14/2019 Lecture9b Discontinuous Systems
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Discontinuous Systems
Harry G. KwatnyDepartment of Mechanical Engineering & Mechanics
Drexel University
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Outline Simple Examples
Bouncing ball
Heating system
Gearbox/cruise control
Simulating Hybrid Systems Simulink with Stateflow Stateflow/Simulink Bouncing ball
Power conditioning system
Inverted pendulum
Brockets Necessary Condition Some systems cannot be stabilized by smooth state feedback Extensions to BNC
Solutions to Discontinuous Differential Equations Various notions of solution may be appropriate
8/14/2019 Lecture9b Discontinuous Systems
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SIMPLE EXAMPLES
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Bouncing Ball
( ) ( )
( ) ( ) [ ]
Free fall:
0Collision:
, 0,1
y g
y t y t
y t cy t c
+
+
=
= =
=
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Bouncing Ball
0
2 00
10 0 0
1 2
1 10 0 0
1 1 1
0
1 20
2
2 2 2
, , ,
2 2 2 1
1
2 1For 0 1, lim
1
Zeno phenomenon transitions in finite time
i
i
NN N Ni i
N ii i i
NN
v v gt
vx v t gt t
g
v v v
t t c t cg g g
v v v cT t c c
g g g c
vc T
g c
= = =
=
= = =
= = =
= = = =
< =
Time for N
transitions
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Heater
( ) ( )
{ }
( ) ( )( ) ( )
continuous state: ,
50, , ,
100
discrete state: , ,
, 73
, 731 , , ,
, 77
, 77
x R
x q offx f x q f x q
x q on
q on off
on q off x
off q off xq k x q k x q
off q on x
on q on x
+ == =
+ =
= = >
+ = ==
=
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Automobile Gearbox Control
Problem 1: Automated gearbox - coordinateautomatic gearshift with throttle command
Problem 2: Cruise control automate throttle andgearbox to maintain speed
Background R. W. Brockett, "Hybrid Models for Motion Control," in Perspectives in
Control, H. L. Trentelman and J. C. Willems, Ed. Boston: Birkhauser, 1993,pp. 29-54.
S. Hedlund and A. Rantzer, "Optimal Control of Hybrid Systems," presentedat Conference on Decision and Control, Phoenix, AZ, pp. 3972-3977, 1999.
F. D. Torrisi and A. Bemporad, "HYSDEL-A Tool for GeneratingComputational Hybrid Models for Analysis and Synthesis Problems," IEEETransactions on Control Systems Technology, vol. 12, pp. 235-249, 2004.
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Transmission
( ) ( )
[ ]
( )
{ }
2
1 2 3 4 5
,
sin
0,1 throttle position
engine speedengine torque-speed characteristic
, , , , transmission state
corresponding gear ratios1, ,5
rear gear
i
i
i
q fin
q eng b veh
wheel
eng
q
fin
R RM v f u F cv M g
r
u
f u
q q q q q q
R i
R
=
=
ratio
wheel radius
brake force
wheel
b
r
F
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Cruise Control
2
2 , gear2
P I I
q fin
wheel
q
v k v k v k v
R Rv
r
+ + =
=
3
3, gear3
P I I
q fin
wheel
q
v k v k v k v
R Rv
r
+ + =
=
5
5 , gear5
P I I
q fin
wheel
q
v k v k v k v
R Rv
r
+ + =
=
4
4 , gear4
P I I
q fin
wheel
q
v k v k v k v
R Rv
r
+ + =
=
1
1, gear1
P I I
q fin
wheel
q
v k v k v k v
R Rv
r
+ + =
=
0 , neutralq
u u u u
l >
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Example 1
1
2 1
x u
x x u
=
=
1.0 0.5 0.5 1.0x1
1.0
0.5
0.5
1.0
x2
1.0 0.5 0.5 1.0F1
1.2
1.0
0.8
0.6
0.4
0.2
0.2
F2
Points on F2 axisclose to F1 axis are
outside image.
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Example 2
1 3 1
2 3 1
3 2
2 1
cos cos
sin sin
Notice that points on the axis close to axis
are not in the image of the mapping.
x
x
x v x x ud
y v x x udt
x u
F F
= = =
=
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Notions of Solution forDiscontinuous Dynamics
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Solutions of ODEs
Classical Solutions
Caratheodory Solutions
Satisfies the ode almost everywhere on [0,t]
Filippov Solutions (differential inclusion a set)
( ) ( )( ) ( ) 0, , 0x t f x t t x x= =
( ) ( )( )00
,t
x t x f x s s ds= +
( ) ( )( ) ( ) 0, , 0x t x t t x x =F
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Classical Solutions
( ) ( )( )( )
,
: is continuously differenticlassical solution able.
x t f x t t
x t
=
1 2 3 4t
0.2
0.4
0.6
0.8
1.0
v t
Example: brick on rampwith stiction.
sgn sinmv v cv mg = +Not a classical solution
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Caratheodory Solutions
( ) ( )( )
[ ]
is satisfied at almost all points on every
interval , ,
Stopping solutions for the brick on ramp problems are not
Caratheodory solutions. For these solutions the brick is
stopped on a finite
x t f x t
t a b a b
=
>
( ) [ ]( ) [ ]
interval, i.e, 0 on ,
0 on ,
sgn 0 sin 0
v t t a b
v t t a b
mg
=
=
+ =
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Brick Example try something
else
sgn sin
sgn sin
sin 0
sin , sin 0
sin 0
mv v cv mg
cv v v g
m m
cv v g v
m m
v g g vm m
cv v g v
m m
= +
= +
= + >
+ + =
= +
= F
}:),(
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Example: nearest neighbor
3 agents moving in square
Q
Rule: move diametrically
away from nearestneighbor
{ } { }{ }1 2 3
Nearest neighbor to
arg min , , \
Action
i
i i i
i i
i
i i
p
p q q Q p p p p
pp
p
=
=
N
N
N
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Example: nearest neighbor,
contd
Consider 1 agent - in which case
the only obstacles are the walls.
The nearest neighbor is easily
identified on the nearest wall.
The vector field is well defined
everywhere except on the
diagonals where it is not defined
because there are multiple
nearest neighbors.
1
1
1
p qp
p q
=
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Example: nearest neighbor,
contd
( )
( )
( )
( )
( )
2 1 2
1 2 1
1 2
2 1 2
1 2 1
0, 1
1,0,
0,1
1,0
x x x
x x xx f x x
x x x
x x x