glue lecture 1 slides

Smriti Chopra ECE Ph.D. Candidate Georgia Tech

Control of Mobile Robots: Glue Lectures

Instructor:

Purpose •  To connect lectures to quizzes •  To give helpful hints about the quizzes •  To clarify and repeat key concepts There will be one Glue Lecture every week

2 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1

Amy LaViers Assistant Professor University of Virginia

Last year’s instructor

Glue Lecture 1: Dynamical Models Pay attention, this lecture will help you all with Quiz 1!


An example: Position…


0 0.2 0.4 0.6 0.8 10

10

20

30

40

50

60

70

80

Time

Position

t0 = 0, x0 = 10 a = 2

x(t) = x0 ea(t�t0)

Velocity…


0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

120

140

Time

Velocity

t0 = 0, x0 = 10 a = 2

x(t) = x0 ea(t�t0)

= ax(t)

x(t) = ax0 ea(t�t0)

And further on…


0 0.2 0.4 0.6 0.8 10

50

100

150

200

250

300

Time

Acceleration

t0 = 0, x0 = 10 a = 2

x(t) = x0 ea(t�t0)

= ax(t)

= a

2x(t)

x(t) = ax0 ea(t�t0)

x(t) = a

2x0 e

a(t�t0)


x(t) = x0 e2(t�t0)In Action…

= 10e2t

x10

Using the differential equation …


x(t) = 2x

x(0) = 10

x10

9

Let’s discretize the world…

t0 0.5 1 1.5 2 2.5 3

t = k�t�t = 0.5

k = 0

Smriti Chopra - Control of Mobile Robots: Glue Lecture 1

10

t0 0.5 1 1.5 2 2.5 3

t = k�t�t = 0.5

k = 0 k = 1



11

k = 2t0 0.5 1 1.5 2 2.5 3

t = k�t�t = 0.5

k = 0 k = 1



12

k = 3k = 2t0 0.5 1 1.5 2 2.5 3

t = k�t�t = 0.5

k = 0 k = 1



13

Now… how’s the ball moving?

x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3

continuous time discrete time xk+1 = xk + �txk +

�t

2

2!xk + ..


x(t) = 2x

10

14


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3

t = k�t�t = 0.5

discrete time xk+1 = xk + �txk +�t

2

2!xk + ..


x(t) = 2x

10

15


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3

t = k�t�t = 0.5


x(t) = 2x

xk+1 = 2xk

10

16


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


xk+1 = 2xk

x1 = 2x0

10

x0 = 10

17


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


10

x1 = 20xk+1 = 2xk

20

x0 = 10

18


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


10 20

x1 = 20xk+1 = 2xkx2 = 2x1

19


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


10 20

x2 = 40

xk+1 = 2xk

x1 = 20

40

20


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


10 20 40

x2 = 40xk+1 = 2xk

x3 = 2x2

21


x

t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3


10 20 40 80

x3 = 80

x2 = 40xk+1 = 2xk

22

And …. that’s a dynamical model!


x(t) = 2x

x(0) = 10

In general, x(t) = f(x, t)

x(t⇤) = x

⇤

23

Punchline: Given the following dynamical model ….

… you know how evolves J x

… numerically


x(t) = 2x

x(0) = 10

24

Punchline: Given the following dynamical model ….

To get the mathematical expression for , we integrate! x(t)

Hint : we already know the solution


x(t) = x0 e2(t�t0) = 10e2t

x(t) = 2x

x(0) = 10

Integration

25

woohoo !! Smriti Chopra - Control of Mobile Robots: Glue Lecture 1

x(t) = 2x

x(0) = 10

26

“Candidate” solution…


x(t) = 2x

x(0) = 10

x(t) = 10e2t

perform the following two checks to confirm candidate is indeed the solution … (a)  Initial condition (b) Differential equation

27

Take home message : 1)   Dynamical models

2) Solving them (i) Numerically (ii) Analytically (integration)

3) Checking a candidate solution (a) initial condition (b) differential equation

4) Equilibrium point x(t) = 0


x(t) = 9x+ 3

x(3) = 5x = �1/3

x(t) =16

3e

9(t�3) � 1

3

Check the forums, and good luck with Quiz 1!


glue lecture 1 slides

Documents