pole balancing robot and some control strategies

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Pole balancing robot and some control strategies

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Pole balancing robot and some control strategies. |  x. f . θ. Introduction. A sketch of PBR (Pole Balancing Robot). Figure1: Pole balancing Robot. Robotic vehicle would operate on the top of the table provided (refer to Fig.2a). Fig.2.a. Pole Balancing Robot Table. - PowerPoint PPT Presentation

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Page 1: Pole balancing robot and some control strategies

Pole balancing robot and some control

strategies

Page 2: Pole balancing robot and some control strategies

SRG Symposium 2004 - Pole Balancing Robot

2

Introduction

A sketch of PBR (Pole Balancing Robot)

f

| x

θ

Figure1: Pole balancing Robot

Page 3: Pole balancing robot and some control strategies

SRG Symposium 2004 - Pole Balancing Robot

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Robotic vehicle would operate on the top of the table provided (refer to Fig.2a).

0.1 cm0.2 cm

1m 1m 1mRetro-reflectivetapes (1cm)

Region A(0.7 m)

(0. 9 m)

Region B( 1.6 m)

Region C(0.7 m)

0.1 m 0.1 m

X

X

Y

Y

0.5m

RoundedEdges

Fig.2.a. Pole Balancing Robot Table

Tabletop will have a slight gradient at the start (region A) and the end (region C) <shown in Fig.2a>

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A metallic wedge of cross section shown in Fig.2b.(not to scale) will be used as an obstacle.

The length of the wedge will match the width of the table.

The wedge will be painted to match the table surface.

A retro-reflective tape will be stuck to it at the middle to match the one on the table.

The judge will place the wedge in region B any where between the inner edges of the two innermost tapes so that the wedge is perpendicular to the path.

It will not be moved thereafter.

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SRG Symposium 2004 - Pole Balancing Robot

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Vehicle will be placed within the region A (See Fig. 2b).

L

H

5L

H= 5 mmL=10 mm

Fig.2.b Wedge Section (Enlarged View)

The operator may move the inverted pendulum to an upright position and release it upon receiving the signal from the judges.

The vehicle must balance the pole in the upright position for a minimum of 20 seconds without the vertical pole crossing the line X-X'.

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Upon completion of the task above, vehicle should move across the line X-X' once move through the region B, until the pole clears the

line Y-Y', without losing balance during transit

not hitting any part of the table or its own chassis

Upon completion of the task above, • vehicle must retrace the path, cross the line X-X'

again and get back to region A. This will complete one cycle.

• This time, during the retrace, the vehicle need not stay any length of time at region B or A, before the start of the second cycle.

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When an electronic sensing system is used for detecting the pole crossing Y-Y’ and X-X’ lines, the pole sensors at both sides will be placed such that the line of sight of the sensors will be 20 cm above the lines marked on the platform.

This may warrant that the robot moves further for the pole to intercept the line of sight of the sensors.

This is important since many robots have their poles inclined inwards towards the centre of platform at these points of turning back.

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Furthermore, no part of the robot other than the pole should be above 15 cm so that no other part of the robot (except the pole) would trigger the sensor.

The vehicle should repeat these cycles. To count these cycles as successful cycles they must be

followed by at least 20 seconds of static balancing at region A.

The robot may continue on (untouched) for more cycles, and complete them with 20 seconds of static balancing at the end, which if successful will be counted cumulatively.

If a robot is touched by the handler during the trial, it must be restarted for the next attempt.

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Pole Balancing Robot Dynamics The line diagram of a pole balancing robot is shown in

Fig.1.

The following equations can be written to describe the dynamics of the robot movement and the pole angle,θ.

[(M+m)s2 + Bs] X(s) + [(ml)s2 + (b/l)s] (s) = F(s) 1

ms2 X(s) + [mls2 + (b/l)s - mg] (s) = 0 2

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SRG Symposium 2004 - Pole Balancing Robot

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One Possible Approach:

mgsl

bslm

sm

sX

s

2

2

)(

)(

where,

M = mass of the vehicle

B = linear equivalent friction of the vehicle

m = mass of the pole

b = rotational friction of the pole

g = 9.81 m/s2

l = half length of the pole

x = distance

= angle in radians

3

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But, V(s) = s.X(s) 4

We can move a S from RHS to denominator. Hence,

mgsl

bslm

sm

sV

s

sXs

s

2)(

)(

)(

)( 5

The above Eqn.5 can be represented by Fig.3.

Figure.3. Angle versus velocity

-mS

mls2 + (b/l).S - mg

(s)V(s)

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Getting back to the problem at hand, for any vehicle with

Fig 4.Force acting on a vehicle

But torque is written as,

ta

bsm K

R

EVt

7

Also torque can be written as,

ww rft . 8

1

M s + B

F(s) V(s) a mass, M and friction coefficient, B, we can draw, for a given force f,

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Where r = diameter of the driving wheel

gta

bsw NK

R

EVt

9

§§ It has been assumed that there is only one motor §§

However, defining back emf as Eb,

Eb = Kb.m

Where,

m is the motor’s rotational velocity.

10.a

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v = r.w

Where, w is the driving wheel’s rotational velocity

wrgNvbK

bE

gNm

wrv

gNm

w

10.bBut,

10.c

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Connecting the above equations, a block diagram can be drawn

Kt.Ng

----------Ra r

1----------MS+B

Kb.Ng

--------------

r

VsV

Figure.5.Block diagram relating applied voltage to velocity

This can be simplified as:

2gb

2wa

wa

w

gb

NKrB)R(Ms

rB)R(Ms

r

NK

1

t

wgt

s

gt

s

K

rNK

V

V

NK

V

V

11

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With the above system as the core plant , one can produce a velocity control block diagram.

Tpwm

KtNg rMRa r

2S + BRa r2 + KtKbNg

2G

VR Vdel.Vs

Figure 6.Velocity Control of robotic vehicle

Where

Tpwm = PWM Period [half period]

del = ‘ON’ fractionG = Numerical Gain

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Block diagram can be drawn.

11 + MRa r

2S + BRar2+ KtKbNg2

(GVsKtNg r / Tpwm)

(GVsKtN r / Tpwm)

(MRa r2)S + [BRa r

2 + KtKbNg2 + (GVsKtNg r / Tpwm)]

VR(s) V(s)

V(s)VR(s)

Figure 7.Complete Block Diagrams

It all looks very complicated.

Note that it is still first order dynamics.

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Once you get the numbers, it is not as complicated as it appears.

Figure 8.Simplified Dynamics

Numerical Example:Let

G = 1000 Tpwm = 1000Kt = 0.033 Ng = 8.0 (gear ratio)r = 3 cm M = 2.5 KgRa = 6 B = 2Kb = Kt Vs = 28 Volts

Then the parameters can be easily computed.

Gm

Mm.S + Bm

VR(s) V(s)

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To get the overall picture, let us combine Fig 3 and Fig.8

Figure 9. Angle Versus Velocity Dynamics of a Pole balancing robot

Gm

Mm.S + Bm

-mS

mls2 + (b/l).S - mg

VR V(S)

(S)

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Control Options:

Common Strategy: Since there two outputs but only one manipulated

variable. In all our design we use two loop system. The position reference is in the outer loop and the

error generated is used as the angle reference to the inner loop to keep the pole vertical.

Actually many variations are possible.

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Implementation can be done using one of the following techniques:

1. Polynomial based controllers: In this controllers, one can describe the transfer function and form z-domain system and use a pole placement or LQC algorithms to derive a controllers. At times LQC controller may go unstable.

2. State space controllers: One can take a state apace model with x, v, angle, angular velocity as states. Again pole placement controllers can be implemented.

3. PD Controller: Simple proportional and derivative controllers also would work. But such system is only conditionally stable.

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Typical PBR

Fig.9. Typical PBR

In our design,

•use eZdsp™ mother board

•robot uses potentiometer, and one drop encoder.

•drivers are H-bridge drivers controlling two motors.

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CONCLUSION

During this brief talk, The basic competition event was described The model of a pole balancing robot as a

single input, multi-output system was derived Possibility of a two loop controller structure

was discussed A few controller design options were

suggested.