1. Spring 2018. CS 599. Instructor: Jyo Deshmukh
Autonomous Cyber-Physical Systems: Stability, Modeling with Hybrid Systems
Spring CS 599.
Instructor: Jyo Deshmukh
Acknowledgment: Some of the material in these slides is based on the lecture slides for CIS 540: Principles of Embedded Computation taught by Rajeev Alur at the University of Pennsylvania.
This lecture also uses some other sources, full bibliography is included at the end of the slides.

2. Layout Stability Analysis Hybrid Dynamical Systems
Probabilistic Models

3. Nonlinear Dynamical System
Simple Pendulum
Arc length 𝑠=ℓ𝜃
Linear velocity: 𝑣= 𝑑𝑠 𝑑𝑡 =ℓ 𝑑𝜃 𝑑𝑡
Linear acceleration: 𝑎= 𝑑𝑣 𝑑𝑡 =ℓ 𝑑 2 𝜃 𝑑 𝑡 2
𝐹=𝑚𝑎=−𝑚𝑔 sin 𝜃 −𝑏 𝜃
ℓ 𝑑 2 𝜃 𝑑 𝑡 2 =−𝑔 sin 𝜃 − 𝑏 𝑚 𝑑𝜃 𝑑𝑡 𝜃 ℓ
𝑚𝑔 sin 𝜃
Friction
𝑚𝑔 cos 𝜃 𝑚𝑔

4. Simple Pendulum Dynamics
Let 𝑥 1 =𝜃, 𝑥 2 = 𝜃 , and let ℓ=𝑔, and 𝑏 𝑚ℓ =𝑘
Rewriting previous equation:
𝑥 1 = 𝑥 2
𝑥 2 =− sin 𝑥 1 −𝑘 𝑥 2
For small angles, the above equation is almost linear (replace sin 𝑥 1 by 𝑥 1 ), and we can use techniques for linear systems to show stability
But, for any larger angle (less than 𝜋 𝑐 ), we know the pendulum eventually stops
How do we show pendulum system is stable?
Use my method!!

5. Lyapunov’s first method
Given 𝐱 =𝑓 𝐱 = 𝑓 1 𝐱 ⋮ 𝑓 𝑛 (𝐱) , Step 1: find the Jacobian matrix for 𝑓(𝐱)
𝐽 𝑓 = 𝜕 𝑓 1 𝜕 𝑥 1 … 𝜕 𝑓 1 𝜕 𝑥 𝑛 ⋮ ⋱ ⋮ 𝜕 𝑓 𝑛 𝜕 𝑥 1 … 𝜕 𝑓 𝑛 𝜕 𝑥 𝑛
Step 2: Set 𝐱= 𝐱 ∗ in 𝐽(𝑓) to get a matrix in ℝ 𝑛×𝑛 (say 𝐴)
If linear system 𝐱 =𝐴𝐱, is stable, original nonlinear system is stable locally at the equilibrium point. (Local = exists some neighborhood of equilibrium point)

6. Lyapunov’s first method for pendulum
Dynamics:
𝑥 1 = 𝑥 2 ; 𝑥 2 =− sin 𝑥 1 −𝑘 𝑥 2
𝑓= 𝑥 2 −sin 𝑥 1 −𝑘 𝑥 2
Step 1: 𝐽 𝑓 = 0 1 −cos 𝑥 1 −𝑘
Step 2:
𝐽 𝑓 ​ 𝐱=(0,0) = 0 1 −1 −𝑘 =𝐴
Step 3:
Eigenvalues 𝜆 of 𝐴 satisfy
𝜆 2 +𝑘𝜆+1=0
⇒ both eigenvalues have negative real parts
Pendulum is locally stable

7. Lyapunov’s second method
Method to prove global stability
Relies on notion of Lyapunov Functions
Intuition:
Find Lyapunov function 𝑉(𝐱) that could be interpreted as the energy of the system
Stable systems eventually lose their energy and return to the quiescent state
Prove that the energy of the system (as encoded by the Lyapunov function) decreases over time

8. Lyapunov’s Second Method: the math
Assume w.l.o.g., that equilibrium point 𝐱* is at the origin, i.e. 0
𝑉(𝐱) is a Lyapunov function over the domain 𝐷 iff:
∀𝐱∈𝐷−{𝟎} : 𝑉 𝐱 >0 [Positivity condition]
∀𝐱∈𝐷− 𝟎 : 𝑑 𝑑𝑡 𝑉 𝐱 ≤0 [Derivative negativity condition]
if 𝐱=𝟎: 𝑉 𝐱 =0, 𝑑 𝑑𝑡 𝑉 𝐱 =0
System 𝐱 =𝑓(𝐱) is stable in the sense of Lyapunov if such a 𝑉(𝐱) exists

9. Illustration and Lie-derivative
𝑉(𝐱) decreases as system evolves, i.e. for every pair of points 𝐚,𝐛 along a system trajectory, 𝑉 𝐚 ≥𝑉 𝐛
In other words, 𝑉(𝐱 t ) is a decreasing function in time, or its derivative is negative semi-definite
𝑉 𝐱 is called Lie derivative of 𝑉
By chain-rule, 𝑉 𝐱 =
𝜕𝑉 𝜕𝑥 𝑑𝐱 𝑑𝑡 = 𝜕𝑉 𝜕𝑥 𝑓(𝐱)
𝑥 2
𝐱(0) 𝟎
𝑥 1 𝐚
𝑉(𝐚) ≥ 𝐛
𝑉(𝐛)

10. Lyapunov’s second method for pendulum
Choose Lyapunov function 𝑉 𝐱 =(1−cos 𝑥 1 ) 𝑥 2 2
Consider 𝐷= − 𝜋 2 , 𝜋 2 × −1,1 ; recall, 𝑓= 𝑥 2 −sin 𝑥 1 −𝑘 𝑥 2
By observation, 𝑉 𝐱 >0, except at 𝟎, where it is 0
Let’s look at the Lie derivative
𝜕𝑉 𝜕𝑥 𝑓 𝐱 = sin 𝑥 1 𝑥 𝑥 2 −sin 𝑥 1 −𝑘 𝑥 2
= 𝑥 2 sin 𝑥 1 − 𝑥 2 sin 𝑥 1 −𝑘 𝑥 2 2
= 𝑥 2 sin 𝑥 1 − 𝑥 2 sin 𝑥 1 −𝑘 𝑥 2 2
=−𝑘 𝑥 2 2 ≤0

11. Second method for asymptotic/exponential stability
𝑉(𝐱) is a Lyapunov function over the domain 𝐷 iff:
∀𝐱∈𝐷−{𝟎} : 𝑉 𝐱 >0 [Positivity condition]
∀𝐱∈𝐷− 𝟎 : 𝑑 𝑑𝑡 𝑉 𝐱 ≤0 [Derivative negativity condition]
if 𝐱=𝟎: 𝑉 𝐱 =0, 𝑑 𝑑𝑡 𝑉 𝐱 =0
System 𝐱 =𝑓(𝐱) is stable in the sense of Lyapunov if such a 𝑉(𝐱) exists
Asymptotic stability: Change second condition to:
∀𝐱∈𝐷− 𝟎 : 𝑑 𝑑𝑡 𝑉 𝐱 <0 [Derivative negativity condition]
Exponential stability: Change second condition to:
∀𝐱∈𝐷− 𝟎 : 𝑑 𝑑𝑡 𝑉 𝐱 <−𝛼‖𝐱‖ or ∀𝐱∈𝐷− 𝟎 : 𝑑 𝑑𝑡 𝑉 𝐱 <−𝛼𝑉(𝐱)

12. Challenges with Lyapunov’s second method
How do we find a Lyapunov function?
Maybe use the physics of the system to understand what encodes “energy”
For certain nonlinear systems (those with polynomial dynamics), can some algorithmic methods
In general a hard problem
Yet, is a powerful approach to prove global stability

13. Bounded signals
‖𝑥‖
A signal 𝐱 is bounded if there is a constant 𝑐, s.t. ∀𝑡: 𝐱 t <c
Bounded signals:
Constant signal : 𝑥 𝑡 =1
Exponential signal: 𝑥 𝑡 =𝑎 𝑒 𝑏𝑡 , for 𝑏≤0
Sinusoidal signals: 𝑥 𝑡 =𝑎sin 𝜔𝑡
Not bounded:
𝑥 𝑡 =𝑎+𝑏𝑡 for any 𝑏≠0
Exponential signal: 𝑥 𝑡 =𝑎 𝑒 𝑏𝑡 , for 𝑏>0 𝑡
‖𝑥‖ 𝑡 𝑡
‖𝑥‖ 𝑡
‖𝑥‖ 𝑡
‖𝑥‖
This Photo by Unknown Author is licensed under CC BY-SA

14. BIBO stability
A system with Lipschitz-continuous dynamics is BIBO-stable if:
For every bounded input 𝐮 𝑡 , the output 𝐲(𝑡) from initial state 𝐱 0 =0 is bounded
Asymptotic stability ⇒ BIBO stability!
Simple helicopter model:
Two rotors: Main rotor gives lift, tail rotor prevents helicopter from spinning
Torque produced by tail rotor must perfectly counterbalance friction with main rotor, or the helicopter spins
Image credit: From Lee & Seshia: Introduction to Embedded Systems - A Cyber-Physical Systems Approach,

15. Helicopter Model continued
𝑢: net torque on tail of the helicopter – difference between frictional torque exerted by main rotor shaft and counteracting torque by the tail rotor
𝑦: rotational velocity of the body
Torque = Moment of inertia × Rotational acceleration
𝑢 𝑡 =𝐼 𝑦 (𝑡)
𝑦 𝑡 = 1 𝐼 0 𝑡 𝑢 𝜏 𝑑𝜏
What happens when 𝑢 𝑡 is a constant input?
𝑦(𝑡) is not bounded ⇒ helicopter model is not BIBO-stable!

16. Hybrid Dynamical Systems

17. Hybrid Process Generalization of a timed process
Instead of timed transitions, we can have arbitrary evolution of state/output variables, typically specified using differential equations on
off
𝜃≤62?
𝑑𝜃 𝑑𝑡 =− 𝑘 2
𝜃≥60
𝑑𝜃 𝑑𝑡 = 𝑘 1 70−𝜃
𝜃≤70
𝜃≥68?
60≤ 𝜃 init ≤70

18. Hybrid System: Thermostat
𝜃 =− 𝑘 2
𝜃≥60
𝜃 = 𝑘 1 70−𝜃
𝜃≤70
𝜃≤62?
𝜃≥68?
60≤ 𝜃 init ≤70
off on
State machine with two modes (on / off)
State variable 𝜃 models temperature
𝜃 can be tested and updated during discrete mode transitions
𝜃 changes continuously in a mode according to specified differential equation
Mode invariants constrain how long machine can stay in any given mode

19. Executions of Thermostat
𝜃 =− 𝑘 2
𝜃≥60
𝜃 = 𝑘 1 70−𝜃
𝜃≤70
𝜃≤62?
𝜃≥68?
60≤ 𝜃 init ≤70
off on
Initial state of the machine: (off, 𝜃 0 ), 𝜃 0 ∈[60,70]
If machine enters mode off at time 𝜏, during continuous transition in mode off, 𝜃 decreases according to:
𝜃(𝑡)=𝜃 𝜏 − 𝑘 2 (𝑡)
Mode switch enabled when 𝜃≤62, and must happen before 𝜃<60
If machine enters mode on at time 𝜏, during continuous transition in mode on, 𝜃 increases according to:
𝜃 𝑡 = 70−𝜃 𝜏 𝑒 − 𝑘 1 (𝑡−𝜏)
Mode switch to off enabled when 𝜃≥68, and must happen before 𝜃>70

20. Modeling a bouncing ball
Ball dropped from an initial height of ℎ 0 with an initial velocity of 𝑣 0
Velocity changes according to 𝑣 =−𝑔
When ball hits the ground, i.e. when ℎ 𝑡 =0, velocity changes discretely from negative (downward) to positive (upward)
I.e. 𝑣 𝑡 + ≔−𝑎𝑣(𝑡) , where 𝑡 + is just after 𝑡, and 𝑎 is a damping constant
Can model as a hybrid system!

21. Hybrid Process for Bouncing ball
𝑣 =−𝑔, ℎ =𝑣
ℎ≥0
ℎ=0→sound!𝑏𝑜𝑖𝑛𝑘 ;𝑣≔−𝑎𝑣
ℎ∈ 10,20 , 𝑣∈[0,5]
What happens as ℎ→0?

22. Zeno’s Paradox
Described by Greek philosopher Zeno in context of a race between Achilles and a tortoise
Tortoise has a head start over Achilles, but is much slower
In each discrete round, suppose Achilles is d meters behind at the beginning of the round
During the round, Achilles runs d meters, but by then, tortoise has moved a little bit further
At the beginning of the next round, Achilles is still behind, by a distance of 𝑎×𝑑 meters, where 𝑎 is a fraction 0<𝑎<1
By induction, if we repeat this for infinitely many rounds, Achilles will never catch up!
If the sum of durations between successive discrete actions converges to a constant 𝐾, then an execution with infinitely many discrete actions describes behavior only up to time 𝐾 (and does not tell us the state of the system at time 𝐾 and beyond)

23. How to deal with Zeno
An infinite execution is called Zeno if infinite sum of all the durations is bounded by a constant, and non-Zeno if the sum diverges
Any state in a hybrid process is called Zeno if:
If every execution starting in state is Zeno
Non-Zeno if there exists some non-Zeno starting in that state
Hybrid process is non-Zeno if any state that you can reach from the initial state is non-Zeno
Thermostat: non-Zeno, Bouncing ball: Zeno
Dealing with Zeno: remove Zeno-ness through better modeling

24. Non-Zeno hybrid process for bouncing ball
𝑣 =−𝑔, ℎ =𝑣
ℎ≥0
bounce
ℎ=0→sound!𝑏𝑜𝑖𝑛𝑘 ;𝑣≔−𝑎𝑣
ℎ∈ 10,20 , 𝑣∈[0,5]
ℎ=0∧𝑣<𝜖→sound!𝑠𝑝𝑙𝑎𝑡 ;𝑣≔0
𝑣 =0, ℎ =0
ℎ≥0
halt

25. Hybrid Process
Inputs, Outputs, States (both continuous and discrete), Internal actions, input and output actions exactly like the asynchronous model
Continuous action/transition:
(𝑚, 𝐱 𝜏 )
𝑚,𝐱 t+𝛿 𝛿
𝐮(𝑡)/𝐲(𝑡)
Discrete mode 𝑚 does not change
𝐱(0) = 𝐱 𝝉
𝑑𝐱 𝑡 𝑑𝑡 satisfies the given dynamical equation for mode 𝑚
Output 𝐲 satisfies the output equation for mode 𝑚: 𝐲 𝑡 = ℎ 𝑚 (𝐱 𝑡 ,𝐮 𝑡 )
At all times 𝑡∈ 0,𝛿 , the state 𝐱 𝑡 satisfies the invariant for mode 𝑚

26. Hybrid Process Discrete action/transition: 𝑚′,𝑟 𝐱 𝜏 (𝑚, 𝐱 𝜏 )
(𝑚, 𝐱 𝜏 )
𝑚′,𝑟 𝐱 𝜏
𝑔(𝐱)/𝐱≔𝑟 𝐱
Discrete action/transition:
Happens instantaneously
Changes discrete mode 𝑚 to 𝑚 ′
Can execute only if 𝑔( 𝐱 𝜏 ) evaluates to true
Changes state variable value from 𝐱 𝜏 to 𝑟 𝐱 𝜏
𝑟 𝐱 𝜏 should satisfy mode invariant of 𝑚′
Some definitions make 𝑔 a function of 𝐱 and 𝐮
Output will change from ℎ 𝑚 𝐱 𝜏 to ℎ 𝑚 ′ 𝑟 𝐱 𝜏

27. Executions of a hybrid process
(0,3)
𝑚 4
At each cell boundary, assume that the guard is the equation of the boundary, and the reset is 𝑥≔𝑥+𝜖, 𝑦≔𝑦+𝜖, where 𝜖=0.01
Invariant for each cell/mode is the open set enclosed by its boundaries
Suppose initial state is (0,0.9)
𝑚 3
𝑚 9
𝑥 =1
𝑦 =0.5
𝑦 =−1
𝑥 =0
𝑦 =0
𝑦 =1
𝑥 =−1
𝑥 =0.5
(0,2)
𝑚 2
𝑚 5
𝑚 8
(0,1)
𝑚 1
𝑚 6
𝑚 7
𝑦=1?
𝑥,𝑦 +=𝜖
( 𝑚 1 ,0,0.9)
𝑚 1 ,0.1,1
𝑚 𝟐 ,0.11,1.01
0.1 2
(0,0)
𝑦=2?
𝑥,𝑦 +=𝜖
(1,0)
(2,0)
(3,0)
0.99
𝑚 𝟑 ,0.12,2.01
𝑚 𝟐 ,0.11,2

28. Stability of hybrid systems
Hybrid systems can have surprising results with respect to stability
No uniform method like Lyapunov analysis for analyzing all hybrid systems
Example: Piecewise Linear (PWL) Dynamical System
Special class of hybrid system, in which each mode has linear dynamics, guards, resets are all linear/affine
Each mode in the PWL system can have stable dynamics (by doing eigen- value analysis), but resulting hybrid system may be unstable!

29. Example Dynamics in each mode are stable! 𝑥 2 =−0.2 𝑥 1 ?
𝑥 2 =−0.2 𝑥 1 ?
𝐱 = −1 − −1 𝐱
𝐱 = −1 10 −100 −1 𝐱
𝑥 2 =5 𝑥 1 ?
Dynamics in each mode are stable!

30. Simulation results
𝑥 2
𝑥 2 =5 𝑥 1
𝑥 2 =−0.2 𝑥 1
Initial State
𝑥 1

31. Stability analysis for hybrid systems
Two main approaches1,2:
Find a common Lyapunov function that works for all modes
I.e. for each mode 𝑚, dynamics are 𝐱 = 𝑓 𝑚 𝐱 , find 𝑉 𝐱 such that: ∀𝑚∀ 𝐱 ≠𝟎 : 𝜕𝑉 𝜕𝐱 𝑓 𝑚 𝐱 <0, and ∀ 𝐱 ≠𝟎 𝑉 𝐱 >0
Usually hard to find a “one-size-fits-all” Lyapunov function
Find multiple Lyapunov functions, one for each mode
In each mode its Lyapunov function value decreases, and at the switching instant the destination mode’s Lyapunov function value does not increase
𝑉 𝑚 (𝐱) value decreases every time mode 𝑚 is entered
Finding Lyapunov functions satisfying these conditions is again hard

32. Design Application: Autonomous Guided Vehicle
Objective: Steer vehicle to follow a given track
Control inputs: linear vehicle speed 𝑣 , vehicle angular velocity (𝜔), start/stop
Constraints on control inputs:
𝑣∈ 𝑣 max , 𝑣 max /2,0
𝜔∈{−𝜋,0,𝜋}
Designer choice: 𝑣= 𝑣 max only if 𝜔=0, otherwise 𝑣= 𝑣 max 2

33. On/Off control for Path following𝑦
Turn right
Go straight
𝑥 =( 𝑣 max /2) cos 𝜃
𝑦 = 𝑣 max 2 sin 𝜃
𝜃 =−𝜋
𝑑≥𝜖
𝑥 = 𝑣 max cos 𝜃
𝑦 = 𝑣 max sin 𝜃
𝜃 =0
−𝜖≤𝑑≤𝜖
𝑑≤𝜖? 𝜃 𝑥
𝑑≥𝜖? 𝑑
𝑠𝑠?𝑠𝑡𝑜𝑝
𝑠𝑠?𝑠𝑡𝑎𝑟𝑡∧
𝑑≥𝜖?
Track
𝑠𝑠?𝑠𝑡𝑜𝑝
𝑑≤−𝜖?
𝑑≥−𝜖?
𝑠𝑠?𝑠𝑡𝑎𝑟𝑡∧
−𝜖≤𝑑≤𝜖?
𝑥 =0
𝑦 =0
𝜃 =0
𝑥 =( 𝑣 max /2) cos 𝜃
𝑦 = 𝑣 max 2 sin 𝜃
𝜃 =𝜋
𝑑≤−𝜖
When 𝑑∈ −𝜖,+𝜖 , controller decides that vehicle goes straight, otherwise executes a turn command to bring error back in the interval
𝑠𝑠?𝑠𝑡𝑜𝑝
𝑥≔ 𝑥 0
𝑦≔ 𝑦 0
𝜃≔ 𝜃 0
𝑠𝑠?𝑠𝑡𝑎𝑟𝑡∧
𝑑≤−𝜖?
Stationary
Turn left

34. Design Application: Robot Coordination
Autonomous mobile robots in a room, goal for each robot:
Reach a target at a known location
Avoid obstacles (positions not known in advance)
Minimize distance travelled
Design Problems:
Cameras/vision systems can provide estimates of obstacle positions
When should a robot update its estimate of the obstacle position?
Robots can communicate with each other
How often and what information can they communicate?
High-level motion planning
What path in the speed/direction-space should the robots traverse?

35. Path planning with obstacle avoidance𝑦
Assumptions:
Two-dimensional world
Robots are just points
Each robot travels with a fixed speed
Dynamics for Robot 𝑅 𝑖 :
𝑥 𝑖 =𝑣 cos 𝜃 𝑖 ; 𝑦 𝑖 =𝑣 sin 𝜃 𝑖
Design objectives:
Eventually reach 𝑥 𝑓 , 𝑦 𝑓
Always avoid Obstacle1 and Obstacle 2
Minimize distance travelled
𝑅 2
𝑣, 𝜃 2
Obstacle 2
𝑝 𝑜2 = 𝑥 𝑜2 , 𝑦 𝑜2
𝑝 2 = 𝑥 2 , 𝑦 2
𝑥 𝑓 , 𝑦 𝑓
Goal
Obstacle 1
𝑝 𝑜1 = 𝑥 𝑜1 , 𝑦 𝑜1
𝑅 1
𝑣, 𝜃 1
𝑝 1 = 𝑥 1 , 𝑦 1 𝑥

36. Divide path/motion planning into two parts
Computer vision tasks
Actual path planning task
Assume computer vision algorithm identifies obstacles, and labels them with some easy-to-represent geometric shape (such as a bounding boxes)
In this example, we will assume a sonar-based sensor, so we will use circles
Assuming the vision algorithm is correct, do path planning based on the estimated shapes of obstacles
Design challenge:
Estimate of obstacle shape is not the smallest shape containing the obstacle
Shape estimate varies based on distance from obstacle

37. Estimation error
Robot 𝑅 1 maintains radii 𝑒 1 and 𝑒 2 that are estimates of obstacle sizes
Every 𝜏 seconds, 𝑅 1 executes following update to get estimates of shapes of each obstacle:
𝑒 1 ≔ min 𝑒 1 , 𝑟 1 +𝑎 𝑝 1 − 𝑝 𝑜1 − 𝑟 1
𝑒 2 ≔ min 𝑒 2 , 𝑟 2 +𝑎 𝑝 1 − 𝑝 𝑜2 − 𝑟 2
Computation of 𝑅 2 is symmetric
𝑝 𝑜1 = 𝑥 𝑜1 , 𝑦 𝑜1 𝑒
𝑒 ′
𝑟 1
Estimated shape
from distance 𝑑′
from distance 𝑑
Smallest shape
bounding obstacle
Estimated radius (from distance d) 𝑒=𝑟+𝑎(𝑑−𝑟),
where 𝑎∈[0,1] is a constant

38. Path planning Choose shortest path 𝜌 3 to target (to minimize time)
If estimate of obstacle 1 intersects 𝜌 3 , calculate two paths that are tangent to obstacle 1 estimate
If estimate of obstacle 2 intersects 𝜌 3 , or obstacle 1, calculate tangent paths
Plausible paths: 𝜌 1 and 𝜌 2
Calculate shorter one as the planned path 𝑦
𝜌 1
( 𝑥 𝑓 , 𝑦 𝑓 )
𝑝 𝑜1
𝜌 4
𝜌 3
𝜌 2
𝑝 𝑜2
𝑝 1 𝑥

39. Dynamic path planning Path planning inputs: Current position of robot
Target position
Position of obstacles and estimates
Output:
Direction for motion assuming obstacle estimates are correct
May be useful to execute planning algorithm again as robot moves!
Because estimates will improve closer to the obstacles
Invoke planning algorithm every 𝜏 seconds

40. Communication improves planning
Every robot has its own estimate of the obstacle
𝑅 2 ’s estimate of obstacle might be better than 𝑅 1 ’s
Strategy: every 𝜏 seconds, send estimates to other robot, and receive estimates
For estimate 𝑒 𝑖 , use final estimate = min 𝑒 𝑖 , 𝑒 𝑖 𝑟𝑒𝑐𝑣
Re-run path planner

41. Improved path planning through communication
𝜌 1
𝜌 2
𝜌 3
𝜌 4 𝑥 𝑦
𝑝 1
𝑝 𝑜1
𝑝 𝑜2
( 𝑥 𝑓 , 𝑦 𝑓 ) 𝑦
New path available
because estimate of obstacle 1 improved after receiving estimate from 𝑅 2
𝜌 1
( 𝑥 𝑓 , 𝑦 𝑓 )
𝜌 4 ′
𝑝 𝑜1
𝑝 1 (0)
𝜌 3
𝜌 2
𝑝 𝑜2
𝑝 1 (𝜏)
Old path 𝑥

42. Hybrid State Machine for Communicating Robot

43. Advantage of using hybrid processes
Hybrid process combines computation, communication and control
Elegant machine that exemplifies the basic operation of a cyber-physical system
Allows design-space exploration through simulations and reachability analysis
We can check effect of parameter choices on the behavior of the system

44. Bibliography
Alur, Rajeev. Principles of cyber-physical systems. MIT Press, 2015.
H. Lin and P. J. Antsaklis, "Stability and Stabilizability of Switched Linear Systems: A Survey of Recent Results," in IEEE Transactions on Automatic Control, vol. 54, no. 2, pp , Feb
Branicky, Michael S. "Multiple Lyapunov functions and other analysis tools for switched and hybrid systems." IEEE Transactions on automatic control (1998):
Slotine, Jean-Jacques E., and Weiping Li. Applied nonlinear control. Vol No. 1. Englewood Cliffs, NJ: Prentice hall, 1991.