Control Theory (2)

Jeremy Wyatt

School of Computer Science

University of Birmingham

Aims

• Understand first order models

• Be able to analyse them

• See their generality

• Analyse PE control for our DC motor

• Meet PI and PID control

Modelling a Simple Process= temperatureQ=rate of heat

input

• What is the relationship between the heat supply and the temperature of the room?

• What is f?

( , )f Q t

Modelling a Simple Process• There are two ways in which heat is “used up”

heats the room leaks out through the walls

• We combine these to give a first order differential equation

model

• C is the heat capacity of the room• L is the loss coefficient (how leaky the room is)

dQ L C

dt

dCdt

L

[1]

First order processes• The behaviour is characteristic of what we call a

first order lag process

• We can analyse the equation to tell us:– The final temperature

– The transient behaviour d

Q L Cdt

t

Steady State Behaviour• If we hold Q constant the temperature will eventually

level out

• At that point so

• If then must rise

• If then must fall

dQ L C

dt

0d

dt

Q

L

t

Q

L

t

Q

L t

t

is the steady state temperature

is the temperature at time t

t

Transient Behaviour

• If we solve we obtain

• tells us the final temperature

• tells us how fast we reach it

dQ L C

dt

is the initial temperature

is the temperature at time t

0

t

0

Lt

Ct

Q Qe

L L

tQ

L

L

C

[2]

The time constant

• tells us how fast we reach the steady state

• We can express this another way

• Where is pronounced “tau” and is called the time constant of the process

• Any first order lag process reaches ~0.63 of its steady state value when t=

t

L

C

C

L

Standard Form

• Any first order lag process reaches – ~0.63 of its steady state value when t=– ~0.95 of its steady state value when t=3– ~0.99 of its steady state value when t=5

• Recall

• A simple rearrangement gives us a standard form:

t

[1]d

Q L Cdt

Q C d

L L dt

d

GUdt

[3]in general

Solution by Inspection

• Once in standard form we can solve by inspection:

• GU is the steady state– where U is the process input (or control action)

– G is called the steady state gain

• HereQ C d

L L dt

dGU

dt

[3]

QGU

L

C

L

Our DC Motor example revisited

• Our DC motor can also be modelled as a first order process

• A motor generates a back voltage proportional to its speed

• The net voltage is related to current by Ohm’s law

1MV K s RI [4]is a motor constant

is the motor speed

is the resistance of the motor

is the current drawn

1Ks

R

I

Our DC Motor example revisited

• The torque produced is proportional to the current drawn

• We can combine these to make a

first order model

2

dsM K Idt

[5]

is a motor constant

is the mass being driven

2K

M

1MV K s RI [4]

Questions

1. Combine the equations to form the first order model

2. What is the steady state speed?

3. What is the time constant ?

Adding Gears

• We can alter the properties of the process by adding gears

• is the gear ratio (>1 means a step down gear)

• input speed = output speed

• input torque = output torque

2

dsM K Idt

[5]

1MV K s RI [4]

Questions

1. Combine the equations to form the new first order model

2. What is the new steady state speed?

3. What is the new time constant ?

Analysing our PE Controller

• Recall our proportional error control law

where

( )T TV s s

-+ Ts e TV

Gs

Te s s

Properties of PE Control

• The PE controller will not reach the target speed

• The difference is called the steady state error

• How big will it be?

s

t

sT

s

Te s s

Analysing our PE Controller

• The open loop behaviour was described by

• We can remodel the new system by substituting [7] back into [6]

• For now assume that

( )T TV s s

21 1 2

MV MR dss

K K K dt [6]

[7]

M TV V

Analysing our PE Controller

• The new system is described by

• So as the steady state error disappearsand it reaches the steady state more quickly

• But big causes instability

21 1 2 2

Ts MR dss

K K K K dt

( )T TV s s [7]

Steady state Time constant

Proportional Integral Control (PI)• IDEA: Add a proportion of the accumulated error to

the control signal

• GOOD: there is no steady state error• BAD:

– Hard to tune– More lag– More unstable

1 2( ) ( )T T TV s s s s dt s

t

sT

PID control• IDEA: Add a third term which looks at how

fast the error is changing

1 2 3

( )( ) ( ) T

T T T

d s sV s s s s dt

dt

s

t

sT

t

de

dt

PID control• GOOD:

– fast initial response to increases in error (D component)– No steady state error (I component)– As error decreases

P 0 and D < 0, therefore stable

• BAD: very hard to tune

s

t

sT

t

de

dt

Summary

• Seen generality of first order model

• Used it to analyse PE control

• Seen principles of PI and PID control