Post on 11-Dec-2015
description
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!
3 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
An example: Position…
4 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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…
5 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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…
6 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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)
7 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = x0 e2(t�t0)In Action…
= 10e2t
x10
Using the differential equation …
8 Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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
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t0 0.5 1 1.5 2 2.5 3
t = k�t�t = 0.5
k = 0 k = 1
Let’s discretize the world…
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
11
k = 2t0 0.5 1 1.5 2 2.5 3
t = k�t�t = 0.5
k = 0 k = 1
Let’s discretize the world…
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
12
k = 3k = 2t0 0.5 1 1.5 2 2.5 3
t = k�t�t = 0.5
k = 0 k = 1
Let’s discretize the world…
Smriti Chopra - Control of Mobile Robots: Glue Lecture 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 + ..
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
10
14
Now… how’s the ball moving?
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 + ..
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
10
15
Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
t = k�t�t = 0.5
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
xk+1 = 2xk
10
16
Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
xk+1 = 2xk
x1 = 2x0
10
x0 = 10
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Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
10
x1 = 20xk+1 = 2xk
20
x0 = 10
18
Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
10 20
x1 = 20xk+1 = 2xkx2 = 2x1
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Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
10 20
x2 = 40
xk+1 = 2xk
x1 = 20
40
20
Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
10 20 40
x2 = 40xk+1 = 2xk
x3 = 2x2
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Now… how’s the ball moving?
x
t0 0.5 1 1.5 2 2.5 3k = 0 k = 1 k = 2 k = 3
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
10 20 40 80
x3 = 80
x2 = 40xk+1 = 2xk
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And …. that’s a dynamical model!
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
x(0) = 10
In general, x(t) = f(x, t)
x(t⇤) = x
⇤
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Punchline: Given the following dynamical model ….
… you know how evolves J x
… numerically
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
x(0) = 10
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Punchline: Given the following dynamical model ….
To get the mathematical expression for , we integrate! x(t)
Hint : we already know the solution
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = x0 e2(t�t0) = 10e2t
x(t) = 2x
x(0) = 10
Integration
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woohoo !! Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
x(t) = 2x
x(0) = 10
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“Candidate” solution…
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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
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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
Smriti Chopra - Control of Mobile Robots: Glue Lecture 1
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!
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