Control 1

Keypoints:

• The control problem

• Forward models:– Geometric– Kinetic– Dynamic

• Process characteristics for a simple linear dynamic system

The control problem

How to make a physical system (such as a robot) function in a specified manner.

Particularly when:

• The function would not happen naturally

• The system is subject to arbitrary changes

e.g. get the mobile robot to a goal, get the end-effector to a position, move a camera…

“Bang-bang” control

• Simple control method is to have physical end-stop…

• Stepper motor is similar in principal:

Forward models• Given the control signals, can we predict the motion of the

robot?

Consider trajectory of robot hand in task space – X(t)X(t) depends on the joint angles in the arm A(t)which depend on the coupling forces C(t)delivered by the transmission from the motor torques T(t)produced by the input voltages V(t)

V(t) T(t) C(t) A(t) X(t)(assuming no disturbances such as obstacles)

Motor command

Robot in environment

Problem

• In general, we have good formal methods for linear systems

Reminder:Linear system:

• In general, most robot systems are non-linear

bxaxf )(

)()()( 2121 xfxfxxf

Kinematic (motion) models

• Differentiating the geometric model provides a motion model (hence sometimes these terms are used interchangeably)

• This may sometimes be a method for obtaining linearity (i.e. by looking at position change in the limit of very small changes)

Dynamic models

• Kinematic models neglect forces: motor torques, inertia, friction, gravity…

• To control a system, we need to understand the continuous process

• Start with simple linear example:

Battery voltage

VB

Vehicle speed

s? VB

IR

e

Electric motor

• Ohm’s law• Motor generates voltage:

proportional to speed• Vehicle acceleration:

where M is motor constant• Torque, proportional to current: • Putting together:

eIRVB

ske 1

M

torque

dt

ds

Iktorque 2

skdt

ds

k

MRVB 1

2

General form

• VB – Control variable – input

• s – State variable – output• A+Bd/dt – Process dynamics• Dynamics determines the process, given an initial

state.• State variable separates past and future• Continuous process models are often differential

equations

dt

ds

k

MRskVB

21

dt

dsBAsVB

Dynamical systems

• Differ from standard computational view of systems:– Continuous coupling rather than

input processing steps output– Analog vs. digital, thus set of states describe a

state-space, and behaviour is a trajectory

• Current debate whether human cognition is better described as computation or as a dynamical system (e.g. van Gelder, 1998)

Process Characteristics

Given the process, how to describe the behaviour?

Concise, complete, implicit, obscure…dt

ds

k

MRskVB

21

Characteristics:•Steady-state: what happens if we wait for the system to settle, given a fixed input?•Transient behaviour: what happens if we suddenly change the input?•Frequency response: what if we smoothly/regularly change the inputs?

Control theory

Control theory provides tools:

• Steady-state: ds/dt =0,

• Transient behaviour (e.g. change in voltage from 0 to 7V) - get exponential decay towards steady state.

• Half-life of decay:

MEMORIZE!

dt

ds

k

MRskVB

21

11 so k

VsskV BB

212

1 7.0kk

MR

Example

Suppose: M:vehicle mass R:setting

• If robot starts at rest, and apply 7 volts: • Steady state speed• Half-life:

Time taken to cover half the gap between current and steady-state speed

dt

dssVB 207

1

117

7 mskVs B

skk

MR27

20*7.07.021

21

1k

2k

MR

Motor with gears

Battery voltage

VB

sout

? Gear ratio γ where more gear-teeth near output means γ > 1

smotor

smotor= γsout : for γ > 1, output velocity is slower

torquemotor= γ-1 torqueout : for γ > 1, output torque is higher

skdt

ds

k

MRVB

12

Thus:

Same form, different steady-state, time-constant etc.

Motor with gears

• Steady-state:

• Half-life:

i.e. for γ > 1, reach lower speed in faster time, robot is more responsive, though slower.

N.B. have modified the dynamics by altering the robot morphology.

1kVs B

212

21 7.0

kk

MR

Electric Motor Over TimeSimple dynamic example – We have a process model:

Solve to get forward model:

• Derivation using Laplace transformation

Battery voltage

VB

Vehicle speed

v?

VB

IR

e

dt

ds

k

MRskVB

21

)exp(1 21

1

tMR

kk

k

Vs B

21

21

1

1

1

7.0

)2

1ln(

)1(2

1

2

1 when for solve :Halflife

so , as :state-Steady

)1(

21

21

kk

MRt

tMR

kk

e

k

Vst

k

Vst

ek

Vs

tMR

kk

B

B

tMR

kkB