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
q= temperature
Q=rate of heat input
• What is the relationship between the heat supply
and the temperature of the room?
q  f (Q, t )
• What is f?
Modelling a Simple Process
• There are two ways in which heat is “used up”
dq
C
dt
heats the room
leaks out through the walls
Lq
• We combine these to give a first order differential equation
model
dq
Q  Lq  C
dt
[1]
• C is the heat capacity of the room
• L is the loss coefficient (how leaky the room is)
First order processes
• The behaviour is characteristic of what we call a
first order lag process
q
t
• We can analyse the equation to tell us:
– The final temperature
– The transient behaviour
dq
Q  Lq  C
dt
Steady State Behaviour
• If we hold Q constant the temperature will eventually
level out
dq
Q  Lq  C
dt
dq
Q
• At that point
 0 so q  
dt
L
Q
• If q t 
then q t must rise
L
Q
• If q t 
then q t must fall
L
q
is the steady state
temperature
qt
is the temperature
at time t
Transient Behaviour
dq
• If we solve Q  Lq  C
we obtain
dt
Q 
Q
q t   q 0   e
L 
L

Lt
C
[2]
q
Q
•
tells us the final temperature
L
L
•
tells us how fast we reach it
C
t
q0
qt
is the initial
temperature
is the temperature
at time t
The time constant
•  L tells us how fast we reach the steady state
C
• We can express this another way
q
C
t
L
t
t
• Where t 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
Standard Form
• Any first order lag process reaches
– ~0.63 of its steady state value when t=t
– ~0.95 of its steady state value when t=3t
– ~0.99 of its steady state value when t=5t
• Recall
dq
Q  Lq  C
dt
q
t
t
[1]
• A simple rearrangement gives us a standard form:
Q
C dq
q 
L
L dt
in general
dq
GU  q   q  t
dt
[3]
Solution by Inspection
• Once in standard form we can solve by inspection:
dq
GU  q  t
dt
[3]
• GU is the steady state
– where U is the process input (or control action)
– G is called the steady state gain
Q
C dq
q 
L
L dt
• Here
Q
GU  q  
L
C
t
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
VM  K1s  RI
• The net voltage is related to
current by Ohm’s law
[4]
K1
is a motor constant
s
is the motor speed
R
is the resistance
of the motor
I
is the current drawn
Our DC Motor example revisited
• The torque produced is proportional to the current
drawn
ds
[5]
M
 K2 I
dt
VM  K1s  RI
[4]
• We can combine these to make a
first order model
K2
is a motor constant
M
is the mass being
driven
Questions
1. Combine the equations to form the first order model
2. What is the steady state speed?
3. What is the time constant t?
Adding Gears
• We can alter the properties of the process by adding
gears
• g is the gear ratio (g >1 means a step down gear)
• input speed = g output speed
VM  K1g s  RI [4]
• g input torque = output torque
ds
M
 g K2 I
dt
[5]
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 t?
Analysing our PE Controller
• Recall our proportional error control law
VT   (sT  s)
where
sT
e
+-
e  sT  s

VT
s
G
Properties of PE Control
• The PE controller will not reach the target speed
sT
s
s
t
• The difference e  sT  s is called the steady state
error
• How big will it be?
Analysing our PE Controller
• The open loop behaviour was described by
VM
MR ds
 s 2
g K1
g K1 K 2 dt
[6]
VT   (sT  s)
[7]
• We can remodel the new system by substituting [7]
back into [6]
• For now assume that
VM  VT
Analysing our PE Controller
• The new system is described by
 sT
MR
ds
 s 2
g K1  
g K1 K 2   K 2g dt
Steady state
Time constant
• So as   the steady state error disappears
and it reaches the steady state more quickly
VT   (sT  s)
• But big  causes instability
[7]
Proportional Integral Control (PI)
• IDEA: Add a proportion of the accumulated error
to the control signal
sT
VT  1 ( sT  s )   2  (sT  s )dt
s
t
• GOOD: there is no steady state error
• BAD:
– Hard to tune
– More lag
– More unstable
PID control
• IDEA: Add a third term which looks at how
fast the error is changing
sT
de
dt
s
t
t
d ( sT  s )
VT  1 ( sT  s )   2  (sT  s )dt   3
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
sT
de
dt
s
t
• BAD:
very hard to tune
t
Summary
• Seen generality of first order model
• Used it to analyse PE control
• Seen principles of PI and PID control