Part A - Personal Web Pages

SE1CC11 – Feedback – Part A
SE1CC11 Cybernetics and Circuits
Feedback – Part A
Dr Richard Mitchell
SE1CC11 and SE1CA11 introduce fundamental concepts and applications
of Cybernetics – Feedback theory, Artificial Intelligence and
Robotics; and they cover Electronic Circuits and Computing.
This year there are 5 lectures in Autumn, 10 in Spring on Feedback
Assessment:
Laboratory Practicals (see separate timetable)
3 hour exam with questions on Feedback, Circuits and Op-amps
Note material used to be in SE1CA5/9 and SE1EA5
On Blackboard, Books, etc.
Lecture Notes on Blackboard and directly on link from
http://www.personal.reading.ac.uk/~shsmchlr/teach.htm
This year we are not providing printed handouts – we expect you to
download file with lecture notes and then add your own comments.
There are some web pages available to help with course at
http://www.personal.reading.ac.uk/~shsmchlr/javascript/index.htm
“Custom book” ‘Cybernetics, Circuits and Computing’, Mitchell, Harwin,
Cadenas, Gong, Potter and Warwick, Pearson, ISBN 978-1-78016-067-2
I have set up a ‘discussion board’ on Blackboard
Feel free to ask a question or answer someone else’s
Tutorial Questions to do on Blackboard – from week 2
2012/13 we reorganized the module - more Feedback less Circuits
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SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
Bring calculator to lectures
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Feedback Syllabus
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
Cybernetics – a Different Perspective
Introduction to systems with feedback - show variety
Kybernetes
(Steersman)
Block diagram analysis - show what feedback does
Positive and Negative Feedback - and consequences
Robots
Systems with Limits - applications of inevitable limits
Neural Nets
Static and Dynamic Systems - dynamics and stability
Learning Robot
Frequency Response Analysis
VR
Computer Modeling of Feedback Systems
Gaia
Note, aim to cover these topics with minimal maths – though the
Maths module you do reinforces the material
All different +
use feedback
Feedback and System Analysis is also covered in lectures on
operational amplifiers and circuit theory.
Feedback also features in Artificial Intelligence and Robotics
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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Standard and Cybernetic Approaches
Standard View: Aristotelian (Greek) - Cause & Effect
Cause
System
Effect
System
Feedback Problems
System more complicated - can lead to run away disasters
The Open
Loop View
Prices Up
+
Cause
The (practical)
Cybernetic view :
Closed Loop –
Has Feeback
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Effect
+
Wages Up
Note, signs – net sign round the loop is positive
Reaction
Arms race is similar – between countries and animals
Not good – but can be useful for quick changes
Irony: Cybernetics comes from a Greek Word!
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
However …
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Feedback Advantages - for Control
This will be illustrated by first considering the steersman
Winds/Tides
No control
– like having
eyes closed
Course
Boat
Guess
As Applied To Other Systems
Winds/Hills
Speed Control
of Car (or
other vehicle)
Slow /
Speed
+
Driver
Winds/Tides
With control
– looking
where you
are going
Obstacles
Course
Boat
Left /
Right
-
+
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Sun, PCs, People
Heat /
Cool
-
Temperature
Negative Feedback produces ‘regulation’ –
If output moves from desired
feedback moves it back
+
Net sign round loop -ve
Air Con/Boiler
Positive Feedback produces change
If output moves
Sun, Illness, etc
Of Human Body
feedback moves it further
Temperature
Body
Sweat /
Shiver
Poss run away - eg inflation, arms race
A system with +ve and -ve can be good
+
Potentially quicker move from one
Action
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state to another state
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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For Instance, Daisyworld
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Daisyworld Continued
Lovelock’s Imaginary world to demonstrate Gaia principle
Namely that Life and Earth work together to mutual advantage
Grey Planet - black/white daisy seeds in soil
Daisies grow best at 22OC
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Negative and Positive Feedback
Temperature Control
Room
+
Joint Motors
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Of Rooms
Position
Robot Arm
Turn
Clock
/ Anti -
Positioning
Robot Gripper
Steersman
Net sign -ve
Speed
Car
Growth
Sun
Once 7OC:
daisies grow,
Temperature
Planet
Black /
heating or
cooling,
until too hot
White
Daisies
No grow if < 7OC or > 37OC
Temp
7
Daisyworld’s Sun is heating up
22
- like Earth’s
What happens to planet’s temperature?
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
37
Temp
C
37
if no life
22
7
if life
Time
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Note, for long
period, temp
constant –
better if more
species!
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Learning – Another Feedback Process
Learning is a feedback process:
Inputs
Each neuron sums products of each input
and weight of connection
For more advanced, need intelligent control .. Must learn
Provide inputs, calculate outputs
‘You learn by your mistakes’
But must learn weights
Trial and Error – used by our Robots / Babies
Inputs
So have training data
Use feedback!
Well done?
Outputs
Feedback Control fine for some simple systems.
Do Task
Neural Network Learning
Like brain: have network of neurons
Outputs
Network
Change
Weights
Better
Way
-
+
Learning
Refiner
Expected Outputs
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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Human Computer Interaction
Cybernetics Is In Fact Not New
Just positioning a mouse is a feedback process
Watt Steam Engine Governor
Ultimate HCI is ‘Virtual Reality’
Computer
Move
-ment
-
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Controlling Speed of Steam Engine
Image
By affecting amount of steam from boiler to engine
In fact governor borrowed from Wind Mills
+
Steam
to engine
Human
Also, Augmented Reality - mixed real and virtual world,
Tele-operation - remote control where operator given input to
suggest he/she at remote location.
Needs force feedback and touch – ‘haptics’
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
First Man Made Feedback System
Constant Flow into vessel starting at Sun up
Height of water indicates time
But need to ensure constant flow
So have second vessel, with hole, which keep full ….
valve
from Boiler
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Some Other Examples Of Feedback
If too hot, valve pushed up, air escape, pressure down
Cooking
Taste stew, if too bland, add salt, taste, ….
Feedback Seismometer
Small pendulum in coil, feedback stops pendulum moving
Earthquake
float
regulator
vessel
clock
vessel
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Coupled to
Steam Engine
Pressure Cooker - control pressure
250 BC Water Clock:
water
supply
Throttle
Valve
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Left /
Right
Position
Pendulum
-
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Coil
+
Coil output is
measure of
Earthquake
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
2: Modeling Feedback Systems
Summary
We have introduced the cybernetic principle of feedback
Thus have demonstrated the principle of feedback
And seen it in technological, animal & environmental systems
And that it can be applied to many types of system
“control and communication in the animal and the machine”, Wiener
technological as well as animal and (briefly) economic
Winds/Tides
Next week we will look at more systems
One example, a
control system :
the steersman
Specifically control systems
systems with inherent feedback
Course
Boat
Left /
Right
and introduce the generation and analysis
-
+
of models of feedback systems
Steersman
There is, however, a more instructive/useful form of block diagram
In this lecture we introduce this and how to develop such diagrams
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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Classical Feedback Control System
Block Diagram Components
Often draw block diagram as follows
Error
Desired
By way of explanation : these are components of block diagrams
Disturbance
Controller
Device
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Actual
A
B
X
A
Block : X defines how A becomes B: B = A*X
Output (‘actual’), fedback, compared with input (‘desired’).
A
Comparison : subtract actual from desired (see + / - in ‘summer’)
If Actual ≠ Desired have Error processed by Control block
makes ‘Device’ change its ‘actual’ state.
A
C
Take off : B = C = A
A
C
B
Summing
Junctions :
C
B
C
B
C=A+B
B
C=A-B
C=A-B
i.e. See what have got, if not what want, do something!
Note A, B, C are signals (voltages, positions, speeds, etc)
Device under control affected by external disturbance
In Control System, Error = Desired – Actual; for instance
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More Complete Diagram
Developing System Models
Actuator turns control signal to suitable form so can drive device.
Sensor measures Actual, converts to form so can compare.
Desired
Disturbance
Error
Controller
Actuator
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
To develop model: divide system into manageable parts
Then bring together to form whole model. e.g. A bike.
Legs
Actual
Device
Chain
Pedals
Legs
Speed
Bike
Speed
Wheel
Speed = Legs * Bike
Wind
Legs
Sensor
Speed
Bike
Human can try to ride at constant speed
NB Feedback aim: Error = 0, so
Measured Actual = Desired
Fine if measurement correct, problematic otherwise
e.g. if speedo’ says 30mph if doing 35mph or league tables
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Legs
Desired
Speed
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Human
Bike
Wind
Actual
Speed
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Friction / Going Up Hill
Another Example – a Mobile Robot
Consider robot on different surfaces on the flat or up/down hills.
Motors are connected to robot wheels – motors turn, robot moves
If put robot on floor, friction between wheel and floor slows robot
Friction is a force, Ff … smooth floor Ff = 1; carpet Ff = 3 …
Friction with surface and robot’s weight act as disturbances.
Motor
Apply voltage v to motor  force  motor turns
l
Fw
weight
Simple Model: Robot speed O = k * v
v
O = speed
k is constant – let it equal 0.125
v
but..
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p25 RJM 17/09/14
In Class Exercise
C=50
0.125
Robot Speed – no control
Ff - Fw
l
Friction+Weight
Desired
Speed
Speed
0.125
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
p26 RJM 17/09/14
So if want robot to go at constant speed, when on different surfaces
at varying angles, either keep adjusting v – or use control
Motor
v
Suppose h is 1, l = 2, m = 0.2, F is 1 and c = 0.1
If v = 8, speed is 8 * 0.125 – 0.1 * (1 + 0.2*9.8*1/2) = 0.802
Robot – Illustration of need for control
Controller
c = 0.1
Reduces speed by c * this
Speed is k * v – c * (F + m * g * h / l)
If apply 8V, robot speed = 8*0.125 = 1 unit/s clockwise
If v = -24V, motor speed is 1.5 unit/s anticlockwise,
Ff + Fw
Overall disturbance force = Ff + Fw
Speed, O
k = 0.125
If robot goes up slope, weight slows it
Surface length l, up height h, mass m, g=9.8
Weight opposing motion Fw = m*g*h/l
h
c = 0.1
v
h
0.1
Actual
Speed
Speed
0.125
Robot downhill – weight speeds up
Speed is k * v – c * (F + m * g * h / l)
Robot Speed – with control
What is speed if v = 8, h = 1, l = 2, m = 0.2, F is 1 and c = 0.1
Speed is 8 * 0.125 – 0.1 * (1 - 0.2*9.8*1/2) = 0.998
http://www.personal.reading.ac.uk/~shsmchlr/javascript/robotSpeed.html
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Two tanks connected by pipe
This is first of two examples of systems with inherent feedback
and blocks of output = input * constant are not sufficient
I
Pipe has resistance - sets
flow through it
F
O
Tanks have capacity – sets
rate height of liquid changes
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
p28 RJM 17/09/14
Block Diagram of Tanks
I
Flow F = (I-O) / Resistance
F
O
Hence, block diagram
I
Flow F means O rises
I-O
F
Pipe
Tank
O
Liquid flows due to difference in pressure at ends of pipe
System reaches final state ‘steady state’, when signals constant
This will be when F = 0, which is when O = I
NB: summer does I – O: so signals I and O same type (units m)
So difference in Height causes Flow through pipe
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Flow F is volume moving at rate : units m3 / s
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Analogous Systems
Resistor-Capacitor System
I
Often same model can apply to two different system types
Water system
R
E
Voltage input, E, from battery.
C
water flows thru pipe as pressure difference across pipe
pipe has resistance (affected by its size): restricts flow
water flows into tank; means height of water increases
speed at which height rises affected by tank’s capacity
Output V, across capacitor
V
Voltage across Resistor, E – V, determines I
I into capacitor causes V to increase
E-V
Electronic system
E
current flows thru resistor, as voltage difference across it
I
Res
V
Cap
resistor has resistance – which resists current flow
current flows into capacitor, so voltage across it increases
speed of voltage rises affected by capacitor’s capacitance
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Final, Steady value, when V = E, then I = 0
E and V measured in volts V, I measured in amps A
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Blocks and Transfer Functions
Simple Block
A
X
X is Transfer Function
B
Some Examples
Resistor
V
V (voltage)
Defines how A becomes B
Spring
Sometimes call X the ‘gain’ : eg X = 10
B is 10 times A, B and A are signals of same type
But X can change signal types – eg Resistor
B (Current) = A (Voltage) * X (1/Resistance)
V
1/R
I
V = 2V, R = 4Ω, then I = 2 * ¼ = 0.5A
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Units Current Amps,
Voltage Volts
Resistance Ohms (Ω)
1/R in Siemens
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Force
Mass
F
Extension
x
A
Mass
For a capacitor, current flowing in causes voltage out to increase
V = V + amount due to I
For motor, net Force means motor accelerates ie velocity changes
Velocity = Velocity + amount due to acceleration
(we hid that detail in robot example)
Mathematically, ‘integration’ does this sort of operation
Output = Output + amount * Input
Hence blocks can include Integrators (output constant if input 0)
Can also have blocks which reflect change : differentiation
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
x
F
Damper
Dashpot
(for
friction)
V
F=k*x
A
1/m
F
f
R
V=I*R
F
k
v
p34 RJM 17/09/14
I
I = V * 1/R
A = F * 1/m
F=v*f
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
When Output NOT Input * Constant
For water tank, the height of water increases when water flows in
Height = Height + amount due to flow F
I
1/R
Examples
Capacitor
I
1/C
V
V = 1/C * Integral I
V (voltage)
Inductor
I
V (voltage)
Mass
F
Mass
v
F
L
1/m
V
V = L * Differential I
v
v = 1/m * Integral (F)
Mainly we will consider these next term
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Summary
Lecture 2 After Class Exercise
Thus we have seen different types of system, some designed for
feedback, some with inherent feedback.
Here robot shown on a less steep hill.
Speed = k * v – c * (F + m * g * h / l)
l
h
We have seen three different but analogous systems.
We have looked at transfer functions of blocks
a) Suppose h is 0.5, l = 2, m = 0.1, F is 1, c = 0.1 and v = 8
What is robot speed ?
We have then said we can develop a model by sorting out the
components and then combining. But how to combine?
Next week we will see how that can be done, for a control system.
We shall see that for a control system output may not equal input,
but can be close …
Before next week try following Exercise on Blackboard
If going downhill, robot’s weight will
speed it up :
l
Speed = k * v – c * (F - m * g * h / l)
h
b) For same values as above, now what is speed downhill?
Go to Blackboard, module SE1CC11, find associated quiz and answer
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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3 : Analyzing Feedback Systems
Reminder of Component Types
We have introduced the cybernetic principle of feedback and
modeled the Classical Feedback Control System
A
Error
X is transfer function
How signal A transferred to B
B
X
aim - actual (output) equals desired (input)
Desired
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Disturbance
Controller
Actual
Device
A
A
C
B
We can have models (transfer functions) of the components, but
what is the overall model (or transfer function)? Let’s find out.
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
X
But systems have combinations of such components.
Can we combine for overall model or transfer function
I
Y
If B = 0, C = A
Then TF = 1
If A = 0, C = -B
Then TF = -1
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
What of Feedback Control System ?
First consider: two blocks in series
B
C=A-B
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Analysis of Two in Series
A
B
C=A+B
To verify this (and to find out it’s not true!) we need to be able
to analyse such feedback systems.
C
E
C
P
O
C
E=I–O
From what we have already said
O=E*C*P
B =A*X
and
So
O = (I - O) * C * P
C =B*Y
=I*C*P–O*C*P
C =A*X*Y
O+O*C*P =I*C*P
ie X and Y combined
single block model
A
X*Y
C
O * (1 + C * P) = I * C * P
But needed 7 lines
of algebra ...
Transfer Function X * Y
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 – Feedback – Part A
Better : Forward over 1 Minus Loop Rule
I
E
C
Forward transfer function, I to O,
D
O
P
I
E
ignore (0) feedback signals: C * P
I
C
P
O
I
E
(here I = 0, so O to E is -1)
C
P
C
O
P
Two + in second summing junction, so O = output of P + D
I
Forward = 1
O
‘closed loop
transfer function’
p43 RJM 17/09/14
E
If assume I = 0,
Loop transfer function, ignore (set to 0) signals entering loop
E to E (or O to O) LTF = - C * P
What about Disturbances?
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
E
I
Loop same, so -CP
D
C
E
P
D
C
P
O
O
Thus
p44 RJM 17/09/14
Complete Response
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
How good is system ?
D
I
If assume D = 0,
E
C
P
If assume I = 0,
O
NO, as 1+C*P ≠ C*P
In general, I and D wont be 0, so we combine both
This is the principle of superposition (also used in circuits), so
BUT, 1/1+CP ≠ 0
When systems operate, their values change (eg resistor)
BUT, it will
However, we can get close to what we want …
p45 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p46 RJM 17/09/14
Let’s Put Some Numbers In
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
Try with C = 50 and 500 ?
Suppose C = 5 and P = 2 … let’s investigate
If P changes by 10% to 2.2
O quite close to 1; One unit of D results in O changing by < 0.1;
Conclusion – much better when C * P is big
10% change in P results in smaller 0.84% change … ok, can do better
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SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
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SE1CC11 – Feedback – Part A
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SE1CC11 – Feedback – Part A
Numbers for complete model
Lecture 3 In Class Exercise
D
E
I
O
P
C
Suppose
C = 27
P = 37
If C = 11, P = 9, I = 2 and D = 5;
O = 99/100 * 2 + 1/100 * 5 = 2.03
a) Find 1 minus Loop
If C = 500, P = 2, I = 1 and D = 4
This image cannot currently be display ed.
This image cannot currently be display ed.
b) Find O/I assuming D = 0
O = 1000/1001 * 1 + 4 / 1001 ~ 1
If (minus) loop gain (CP) high, response close to desired as
c) Find O/D assuming I = 0
This image cannot currently be display ed.
d) Evaluate O if I = 10 and D = -5
This image cannot currently be display ed.
e) Find O/I if P changed to 40
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
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SE1CC11 Feedback - Part A
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Feedback Not Just for Control
Control Engineers want O = I, Audio Engineers O = I * G
D
I
X
F
A
O

Analysis
Suppose Aβ >> 1 {much greater than 1} ‘negligibly large’, 1 – Aβ ~ Aβ
More General
Feedback System.
Control system if
A = C * P,  = -1!
(O is independent of A)
e.g. if A = -5000, β = -0.2, Aβ = 1000, 1-Aβ = -999 ~ -Aβ
If Aβ << -1, (large and negative)
By Superposition:
p51 RJM 17/09/14
This image cannot currently be display ed.
A = -5000, β = 0.2, Aβ = -1000, 1 - Aβ = 1001, ~ - Aβ
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
p52 RJM 17/09/14
Also works ..
Example – A = 100,  = 0.21
Also works if Aβ << -1, (large and negative)
e.g. if A = -5000, β = 0.2, Aβ = -1000, 1 - Aβ = 1001, ~ - Aβ
O = -4.995 * I ~ -1/β * I, independent of A and O ~ 0*D
Feedback good if modulus of Loop Gain, |Aβ |, large
{ modulus means size irrespective of sign: | 5 | = 5 |-5| = 5 }
D
I
X
F
A
O
1 - A = 1-21 = -20
β
If I = 2, and D = 0, O = -5*2 = -10
Check: F = -10*0.21 = -2.1, so X = 2+-2.1 = -0.1; O = 100*-0.1+0 = -10
If I = 0 and D = 1, O = -0.05 * 1 = -0.05.
Then O = I times -1/feedback value, is independent of A
(and hence of changes in A) and unaffected by D.
If loop gain smaller, result not as good …
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Check: F = -0.0105 = X, so O = -0.0105*100+1 = -0.05
If I = 2 and D = 3; O = -5*2 + 3*-0.05 = -10.15
Check: F = -2.1315; X = -0.1315; O = -13.15+3 = -10.15
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SE1CC11 – Feedback – Part A
Real System – in Electronics
Operational Amplifier
V i+
A
Combine into a Complete System
VO
V i-
V o = A * ( V i + - Vi - )
Potential Divider
A very big, ~105,
In use, put feedback round op-amp :
Vi- found by potential divider = VI
Suppose R1 = 9kΩ and R2 = 1kΩ
Assume A very big, 105
Block Diagram for complete analysis
VI
VO
Vm
VO
A
V i-
If R1 = 9kΩ and R2 = 1kΩ,
Vm = Vo / 10
p55 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p56 RJM 17/09/14
Summary
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Lecture 3 After Class Exercise
We have analysed simple feedback systems :
D
used forward over one minus loop for closed loop TF
X
I
We have seen benefit of high loop gain
F
We have seen a practical (op-amp) circuit.
O
A

Suppose A = 200, β = -0.2
We will extend the analysis next week
a) Find O/I if D = 0
And consider positive and negative feedback
b) Find O/D if I = 0
Before next week do following Exercise on Blackboard
p57 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
c) Find %change in O/I if A changes by 10% to 220
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4 : More Feedback Systems Analysis
We extend our analysis; show how computers can be used;
and tackle the question of positive and negative feedback.
D
I
E
C
P
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
‘Fuller’ Control System
Here we incorporate the actuator and sensor
D
I
O
E
C
A
P
O
S
For this feedback control system: use forward/1-loop
Now Forward = C * A * P and Loop is - C * A * P * S
If C*P big, O ~ 1*I + 0*D - what control engineers want
If C*A*P*S big, O ~ I/S + 0*D – want S = 1 : true measurement
p59 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
p60 RJM 17/09/14
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
10
SE1CC11 – Feedback – Part A
Potential Divider is Feedback System!
If more than one ‘loop’ ...
Better control if feedback position and velocity :
E
I
Vo = R2 * (I – IL), so
D
C
P
V
IL
Vs
1/R1
I
O
Vo
R2
TF = Forward over 1 minus sum of each loop (-CP and –PV)
Rule ok in this configuration – will return to multi loops later
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
p61 RJM 17/09/14
p62 RJM 17/09/14
If more than one ‘Forward’
Put in an ‘anticipatory’ amount – so called feedforward control
E
E
I
F
C
Web Pages to help you learn
http://www.personal.reading.ac.uk/~shsmchlr/javascript/index.htm
There are many pages here which you should use throughout the
module: just click on the appropriate link
D
P
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
O
For Forward/1-Loop, follow link transfunc.html
You are presented with various systems – select one listed under
‘Static Systems’ (dynamic systems are taught next term)
Now ‘Forward’ has two paths F*P and C*P, sum each Forward
eg Control, General Feedback, Control + VelFb, Op-Amp, etc
You can investigate O/I or O/D
You see Forward, press Next for Loop, press Next for Overall
You can change values of parameters for C, P, etc…
You are strongly recommended to investigate this page
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SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
p64 RJM 17/09/14
Positive/Negative Feedback
Lecture 4 - In Class Exercise
For system with feed-forward control :
SE1CC11 – Feedback – Part A
© Dr Richard Mitchell 2014
We will correct erroneous definitions / claims often made.
Wrong to say negative feedback because - sign in ‘summer’
(Changing sign of  has same effect as changing + to -)
The important point is to have both
a)
claims for what negative feedback does
b)
a consistent definition for negative feedback
To that end the correct view is that Negative Feedback
a)
reduces effects on output of disturbances
reduces effects on output of parameter changes
b)
occurs if |closed loop gain| < |open loop gain|
NB | x | or modulus of x, means size : ignore sign
p65 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
p66 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
11
SE1CC11 – Feedback – Part A
Negative Feedback (Harold Black 1930s)
Forward (Open Loop) Gain = A
Negative Feedback & Disturbances
D
I
O
A
F
Negative Feedback

| Closed Loop Gain | < | Open Loop Gain |
Negative Feedback, when | 1 – Aβ | > 1,
Reduces effect of Disturbances D on output O
Reduces effect of changes in A on output O
p67 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p68 RJM 17/09/14
… and Changes in Parameters
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Interest Only: A → A(1 + )
Let A change by a small proportion: call it : i.e. A := A(1 + )
Feedback reduces the effect of change in A if |1 - A | > 1.
Let A = 5 and β = -4 (-ve fb) and A change by 10% to 5.5 ie δ = 0.1
Rel Change: open loop = 0.1; closed loop = 0.1/21 = 0.005 (smaller)
If instead β = 0.04 (+ve fb)
Rel Change: open loop = 0.1; closed loop = 0.1/0.8 = 0.125 (bigger)
p69 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p70 RJM 17/09/14
Effect of Changing A : 50 to 55
β=
a) -1/50
β
1-Aβ
b) -1/10
c) 1/100
Graphs for b) : A = 50, β = -0.1
d) 3/100 e) 1/10
%
diff
1/2=0.5
50/2=25
55/2.1= 26
+4%
b -1/10
1/6=0.17
50/6=8.3
55/6.5= 8.5
+1.5%
55/0.45= 122
+22%
c 1/100 1-0.5=0.5
1/0.5=2
50/0.5=100
d 3/100 1-1.5=-0.5 1/-0.5=-2
50/-0.5=-100
55/-0.65= -84.6 -15%
e 1/10
5 -/-4=-12.5
55/-4.5= -12.2
1-5=-4
1/-4=-0.25
b) highest loop gain: best at rejecting D & changes in A.
No system has high loop gain.
c) and d) have +ve feedback: O/D > 1 and % diff > 10%
p71 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
It can be useful to plot graphs of O vs I and O vs D
a -1/50 1+1=2
1+5=6
Can also do by differentiating closed loop TF w.r.t A
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
-2%
As straight lines … associated system is said to be linear
If gradient of O/D < 1 : system has negative feedback
p72 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
12
SE1CC11 – Feedback – Part A
Analysis Using MATLAB
Using MATLAB on Control Systems
MATLAB is an interactive package that you will use in labs, etc
>> C = 11; P = 9; I = 2; D = 5;
It processes matrices, which have one or many numbers
>> O = I * (C * P) / (1 + C * P)
At the prompt (>>) you can type commands to
O=
assign values to variables, and/or to perform calculations
1.9800
call functions provided by MATLAB or ones you write (like C)
>> O = D / (1 + C * P)
plot graphs, etc
O=
You can define a suitable graphical interface …
NB 35 is a number
[35, 8, 1, -8]
single row vector (matrix) 4 columns
[35, 67; 89, 88; 3, -1*5]
3 row 2 column matrix (; for new row)
p73 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Or to see effect of change in P
>> C = 50; P = 2; Pa = P * 1.1;
>> OoverI = C * P / (1 + C * P)
OoverI =
0.9901
>> OoverD = 1 / (1 + C * P)
OoverD =
0.0099
>> OoverIa = C * Pa / (1 + C * Pa)
OoverIa =
0.9910
>> PCchange = 100 * (OoverIa - OoverI) / OoverI
PCchange =
0.0901
p75 RJM 17/09/14
0.0500
>> O = I * (C * P) / (1 + C * P) + D / (1 + C * P)
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Summary
O=
{ could extend easily to cope
with V in multi loop system, or
F for feed forward }
2.0300
p74 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Can Write MATLAB Function – and call it
function FBSys (C, P, pc);
% store in file fbsys.m
% FBSYS (C, P, pc)
% Analyse system with gains C and P noting effect if P change by pc%,
% Dr Richard Mitchell 12/7/05
den = (1 + C * P);
% calculate values
OoverI = C * P / den;
OoverD = 1 / den;
Pa = P + P * pc/100;
% find new P
OoverIa = C * Pa / (1 + C * Pa);
% find new O/I
[OoverI, OoverD, OaoverI, 100*(OoverIa-OoverI)/OoverI]
% put values in matrix and print
>> fbsys(50, 2, 10)
0.9901 0.0099
0.9910
p76 RJM 17/09/14
0.0901
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Lecture 4 After Class Exercise
We have analysed feedback system : A and CP
I
We have briefly noted multiple loops
We have an appropriate definition of negative feedback
X
F
We have shown that can plot graphs of O vs I and O vs D
A
O
β
These are straight lines … means systems are linear.
NB Principle of Superposition only true for linear systems
Next week we build on this …
and consider limits …
which make systems non-linear
Before next week do following exercise on blackboard
p77 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Here, I = 2, A = 10.
Find O and state whether positive feedback if:
a) β = -1
b) β = +1
c) β = -0.02
d) β = +0.02
p78 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
13
SE1CC11 – Feedback – Part A
5 : Feedback Systems and Limits
Limits
Signals cannot be infinite … they are limited … we determine effect.
D
Recall last week’s system with
A = 50, β = -0.1
I
X
1 - Aβ = 1+5 = 6
F
A

O
O/I graph implies that as I increases, so O increases
Not true in practical systems
e.g. output of a component can’t exceed its power supply, etc
The output has limits; and we can incorporate them.
Below are shown limits graphically and as a block diagram
Out
+L
Limit Block
In
In
Straight line graphs –
system is linear
Out
If -L ≤ In ≤ L, Out = In;
|1 - Aβ| > 1 : -ve fb
if In < -L, Out = -L; if In > L, Out = L
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p80 RJM 17/09/14
Limits in Feedback Systems
D
I
+L
-L
X
F
A
Y
-L
+L Z
O

Let A = 50;  = -0.1; +L = +25 and -L = -25.
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
What happens when pass Limit?
A = 50
D
I
 = -0.1
X
L = 25
A
F
At limit
Y
-L
+L Z
O

I = 3, O = 25
Consider O vs I assuming D = 0.
If I increases, X and Y increase, but Z and O stay at 25
When no limiting, limit box transfer function = 1.
Use same argument for limit -25;
or say ‘by symmetry’ when I = -3, O = -25,
if I more –ve O stay -25
Hence (non linear) graph of OvI
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
General Case – and other example
p82 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
O vs D : a different response
D
I
X
F
A
Y
-L
+L Z
O

Suppose A = 100; β = -0.1 +L = 10 and –L = -10
System linear when -10 ≤ Y ≤ 10, then
Strategy, find D when just limit: Y = Z = +L: when
Do similarly for when Y = Z = -L : or use symmetry
p83 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
p84 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
14
SE1CC11 – Feedback – Part A
So, when A = 100;  = -0.1
Lecture 5 In Class Exercise
D
I
X
Y
A
F
-L
O
+L Z

A = 1000
 = -1/5
+L = 20
-L = -20
Calc O & D at L = -20 to label the graph
Then, if D ↓ 1 to -12, O = -12 + 10 = -2. ie O ↓ 1 : O/D now 1
By symmetry, Y = Z = -10 when D = 11;
If D then ↑ 1 to 12, O up by 1
If -11 ≤ D ≤ 11, effects of D on O
reduced by feedback,
otherwise no feedback, no reduction.
p85 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p86 RJM 17/09/14
Limits and Hole in the Ozone Layer
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Limits when have Positive Loop Gain
We will now see different response when loop gain > 0 …
The ozone layer is a feedback system
X
I
Must be (according to Gaia) so correct amount of u.v. getting to
Earth’s surface:
F
Too much u.v → cancer; too little → rickets
If ozone layer too thin, u.v. gets through : finds oxygen; turns
it to ozone, thickens ozone layer: feedback!
Worked til too much CFC – which destroys ozone
CFCs are disturbances, and normally feedback reduces effects
of CFCs
But when too much, system not cope as then has no feedback to
reduce any extra disturbance
100
Y
+10
O
-10
0.1
Suppose O = 10 and I = 1; F = 1, X = 2, Y = 200, so O = 10
If I reduced to -0.9: X = 0.1, Y = 10, O = 10 No change.
If I now -0.91:
X = -0.91+1 = 0.09, Y = 9, O = 9
Then
X = -0.01, Y = -1, O = -1
And then
X = -1.01, Y = -101, O = -10
Very rapidly, O flipped +10 to -10 when I passed -0.9.
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SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p88 RJM 17/09/14
How to Flip Back
Limits and Positive Loop Gain
I
X
F
100
Y
+10
O
-10
0.1
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
If, however, I upped to 0.91,
X = I + O/10 = -0.09
O = 100 * X = -9
Then X = 0.01, O = 1; Then X = 1.01, Y = 101, O = 10
O
Thus, when I exceeds 0.9,
O flips back to + 10
This change from +10 to -10 is represented graphically as
O
0.9
If I reduced now no change.
I
O
But if I increased to 0
-0.9
I
X = -1, O = -10 still
Even if I upped to +0.9:
X = -0.1, O = -10, still, but..
p89 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
O stays at 10 until I again < -0.9.
O depends on I and on O!
-0.9
0.9
I
Hysteresis: figure to right
p90 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
15
SE1CC11 – Feedback – Part A
Application: Square Wave Generator
How To Find I Where O Flips
I
X
F
100
Y
+10
O
-10
0.1
Again find I
when system
just limits,
To introduce dynamic systems – consider simple use of integrator
NB if input to integrator is constant, output changes at constant rate
-C ∫
ie Y = O = 10
I
X
Y
100
+10
-10
‘hysteresis’ loop input,
O
I = -C * ∫ O
0.1
I changes at rate -CO
t = 0, I = +Flip, O = +Lim
As O > 0, I is integral of –ve constant, so I decreases
At time t, I = Flip - C*Lim*t
If I ≥ -0.9, O stays at 10, but if I < -0.9, O will flip to -10
Input to ∫ C*Lim (>0), I increases constantly
Could be found by same analysis.
At t = 4*Flip / C*Lim, I = +Flip, O := +Lim
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
p92 RJM 17/09/14
MATLAB Code to Plot Limits
[-lim, -lim, lim, lim, lim, -lim, -lim]);
elseif whatplot == 0
% -ve loop gain – plot O/I
plot ([-Ilim-5, -Ilim, Ilim, Ilim+5], [-lim, -lim, lim, lim]);
else
% -ve loop, plot O/D
Olim = lim / (A*b); % O when just limits
Dlim = Olim * OmAb;
% D when just limits
Example Output + General Summary
Graphs A = 100, β = ± 0.1, lim = 10
10
5
0
-5
-10
10
5
0
-5
-10
5
O
function fblims (A, b, lim, whatplot); % plot limit system
OmAb = 1 - A*b;
% 1 minus A * b
Ilim = lim * OmAb / A;
% value of I when just limiting
if A*b > 0
% +ve loop gain – plot hysteresis
plot([-Ilim-5, Ilim, Ilim, Ilim+5, -Ilim, -Ilim, -Ilim-5], ...
4*Flip / C*Lim
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
O
p91 RJM 17/09/14
I
At t = 2*Flip / C*Lim, I = -Flip, O := -Lim
O
By symmetry, I must exceed +0.9 for O to flip to +10
O
0
-5
0
Aβ < 0
I
-5
5
-10
0 D 10
-5
0
I
5
Aβ > 0
plot([-Dlim-5, -Dlim, Dlim, Dlim+5], [-Olim-5, -Olim, Olim, Olim + 5]);
end
p93 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Summary
We have seen effect of limits on feedback systems
p94 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
Lecture 5 After Class Exercise
D
I
If negative loop gain, get different responses O/I O/D
X
F
For positive loop gain, get hysteresis
A
Y
-L
+L Z
O

Saw how with integrator, could make square waves
Next term
We start to consider dynamic analysis
Where we will use integrators more
and (building on what you will learn in circuits), √-1
Before next week, try following exercise on blackboard
p95 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
© Dr Richard Mitchell, 2014
Suppose A = 100,  = 0.2, +L = 50, -L = 50.
Initially, I = 0 and O = -50.
a) What must happen to I to make O change to +50
b) What must then occur to I for O to return to -50?
p96 RJM 17/09/14
SE1CC11 Feedback - Part A
© Dr Richard Mitchell 2014
16

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