Cybernetics and Circuits Feedback

p1 RJM 19/09/12 SE1CC11 Cybernetics and Circuits - Feedback A© Dr Richard Mitchell 2012

SE1CC11 Cybernetics and Circuits Feedback – Part ADr Richard Mitchell

Senior Lecturer in CyberneticsSE1CC11 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 SE1EA5This year we have reorganised the module

- more on Feedback less on circuits


On Blackboard, Books, etc.Lecture Notes on Blackboard and directly on link from

http://www.personal.rdg.ac.uk/~shsmchlr/teach.htm

BUT note it is still important to attend the lecturesas some of the handouts have deliberate gaps to fill

AND I say things that which you should write downSpare copies of notes in School Information Office G47“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’sTutorial Questions to do on Blackboard – from week 2


Feedback SyllabusIntroduction to systems with feedback - show varietyBlock diagram analysis - show what feedback doesPositive and Negative Feedback - and consequencesSystems with Limits - applications of inevitable limitsStatic and Dynamic Systems - dynamics and stabilityFrequency Response AnalysisComputer Modeling of Feedback SystemsNote, aim to cover these topics with minimal mathsFeedback and System Analysis is also covered in lectures on operational amplifiers and circuit theory.Feedback also features in Artificial Intelligence and Robotics


Kybernetes (Steersman)

Robots

Neural Nets

Learning Robot

VR

Gaia

All different + use feedback

Cybernetics – a Different Perspective


Standard and Cybernetic ApproachesStandard View: Aristotelian (Greek) - Cause & Effect

The (practical) Cybernetic view : Closed Loop –Has Feeback

Irony: Cybernetics comes from a Greek Word!

The Open Loop View

Cause EffectSystem

Cause

Reaction

EffectSystem


Feedback ProblemsSystem more complicated - can lead to run away disasters

Note, signs – net sign round the loop is positive

Arms race is similar – between countries and animals

Not good – but can be useful for quick changes

However …

Also the stock market

Wages Up

Prices Up

+ +


Feedback Advantages - for Control

No control– like havingeyes closed

With control– lookingwhere youare going

This will be illustrated by first considering the steersman

Winds/TidesCourse

BoatGuess

Winds/Tides

Steersman

CourseBoat

Left /Right +-

Net sign -ve


As Applied To Other Systems

Speed Control of Car (or other vehicle)

Positioning Robot Gripper

Obstacles

Joint Motors

PositionRobot ArmTurn

/ AntiClock

+-

Winds/Hills

Driver

SpeedCar

Slow /Speed +-


Temperature ControlOf Rooms

Of Human Body Sun, Illness, etc

Action

TemperatureBody

Sweat /Shiver +-

Sun, PCs, People

Air Con/Boiler

TemperatureRoom

Heat /Cool +-


Negative and Positive FeedbackNegative Feedback produces ‘regulation’ –

If output moves from desired

feedback moves it back

Net sign round loop -ve

Positive Feedback produces change

If output moves

feedback moves it further

Poss run away - eg inflation, arms race

A system with +ve and -ve can be good

Potentially quicker move from one

state to another state


For Instance, DaisyworldLovelock’s Imaginary world to demonstrate Gaia principle

that Life and Earth work together to mutual advantageGrey Planet - black/white daisy seeds in soil

Daisies grow best at 22OCNo grow if < 7OC or > 37OC

Daisyworld’s Sun is heating up

- like Earth’s

What happens to planet’s temperature?

7 22 37

Growth

Temp

Daisy Daisy Give me your answer do!


Daisyworld Continued

Note, for long period, temp constant –

better if more species!

Once 7OC: daisies grow,

heating or cooling,

until too hot

Sun

Daisies

TemperaturePlanet

Black /

White

if no life

if life

Time

C

7

22

37Temp


Learning – Another Feedback ProcessFeedback Control fine for some simple systems.

For more advanced, need intelligent control .. Must learn

Learning is a feedback process:

‘You learn by your mistakes’

Trial and Error – used by our Robots / Babies

Refiner

Do Task

WayBetter

Well done?


Neural Network LearningLike brain: have network of neurons

Each neuron sums products of each input and weight of connection

Provide inputs, calculate outputs

But must learn weights

So have training data

Use feedback!

Out

puts

Inpu

tsExpected Outputs

Outputs

LearningWeightsChange

Inputs

Network

+-


Human Computer InteractionJust positioning a mouse is a feedback process

Ultimate HCI is ‘Virtual Reality’

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’

Human

Computer

-ment

Image

Move +-


Cybernetics Is In Fact Not NewWatt Steam Engine Governor

Controlling Speed of Steam Engine

By affecting amount of steam from boiler to engine

In fact governor borrowed from Wind Mills

you’re in control when you know what to do with the governor’s

ballsValveThrottle

Steamto engine

from BoilerSteam EngineCoupled to


First Man Made Feedback System250 BC Water Clock:

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 ….

regulator

vesselclock

watersupply

vessel

valvefloat

Cybernetics began in the

toilet!


Some Other Examples Of FeedbackPressure Cooker - control pressure

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

Coil output is measure of EarthquakeCoil

PositionPendulumLeft /

Right

Earthquake

+-


SummaryThus have demonstrated the principle of feedback

And that it can be applied to many types of system

technological as well as animal and (briefly) economic

Next week we will look at more systems

including a world of daisies … environmental

mention positive and negative feedback

consider systems with inherent feedback

and introduce the generation and analysis

of models of feedback systems


2: Modeling Feedback SystemsWe have introduced the cybernetic principle of feedbackAnd seen it in technological, animal & environmental systemsReflecting Wiener’s definition of Cybernetics: “control and

communication in the animal and the machine”Winds/Tides

Steersman

CourseBoat

Left /Right +-

One example, a control system : the steersman

There is, however, a better form of block diagram

In this lecture we introduce this and how to develop such diagrams


Classical Feedback Control System

Output (‘actual’), fedback, compared with input (‘desired’).

Comparison : subtract actual from desired (see + / - in ‘summer’)

If Actual ≠ Desired have Error processed by Control block

makes ‘Device’ change its ‘actual’ state.

i.e. See what have got, if not what want, do something!

Device under control affected by external disturbance

Often draw block diagram as follows

Desired Actual

Disturbance

Controller

Error

Device


Block Diagram ComponentsBy way of explanation : these are components of block diagrams

Note A, B, C are signals (voltages, positions, speeds, etc)

In Control System, Error = Desired – Actual; for instance

Block : X defines component changing A to B Take off : B = C = A

Summing Junctions : C = A - B

A BX A CB

A

B

C A

B

CA

B

C

C = A - BC = A + B


Fuller DiagramReally ought to include Sensor and Actuator:

Actuator turns control signal to form so can drive device.

Sensor measures Actual, converts to form so can compare.

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

Desired ActualDisturbance

Controller

Error

DeviceActuator

Sensor


Developing System ModelsTo develop model: divide system into manageable parts

Then bring together to form whole model. e.g. A bike.

Legs MovementPedals WheelChain

Legs MovementBike

Legs MovementBike

Wind

Desired Speed

Actual Speed

WindHuman

Legs

Bike

A system with inherent feedback, there are others


Example – Two tanks connected by pipe

IF

O

Pipe has resistance – affects amount of flow through it

Tanks have capacity –affects rate at which height of liquid in tank changes

Weight Density*g*VolumePressure = = Area Area

Density * g*Height * Area = = constant * HeightArea

Liquid flows due to difference in pressure at ends of pipe

So difference in Height causes Flow through pipe

Block Diagram of Tanks


IF

O

Height difference I-O → Flow F

Flow F → O to rise

I OPipe

I - O

Tank

FHence, block diagram

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)

Flow F is volume moving at rate : units m3 / s


Analogous SystemsOften same model can apply to two different system typesWater system

water flows thru pipe as pressure difference across pipepipe has resistance (affected by its size): restricts flowwater flows into tank; means height of water increasesspeed at which height incr affected by tank’s capacity

Electronic systemcurrent flows thru resistor, as voltage difference across itresistor has resistance – which resists current flowcurrent flows into capacitor, so voltage across it increasesspeed of voltage incr affected by capacitor’s capacitance


Resistor-Capacitor System

Voltage across Resistor is E – V : that over R sets I

I into capacitor causes V to increase : so block diagram

E VRes

E - V

Cap

I

VCE

I

RVoltage input, E, from battery.

Output V, across capacitor

Final, Steady value, when V = E, then I = 0

E and V measured in volts V, I measured in amps A


Motor

Vin applied to armature circuit → current → force F to move

Frictional force (f * velocity, v) opposes motion

Net force is F – f*v = motor mass * acceleration (Newton’s 2nd law)

Acceleration means velocity increases

Steady state, when v such that friction force = F as set by Vin

Signal Units: Vin in Volts, forces in Newtons (N), velocity (m/s)

v = velocity

Vin

vArmature

Net Force

Mass

friction

Vin F

fv


How Signals Change – eg Consider Tanks

Suppose, initially O = 0.

Pressure difference big; large flow F

Thus O increases a lot.

Now pressure diff (PD) less, so F less

So O increase, by less

Now PD smaller, F even less, O up by less

(If first tank big enuf, I ~ constant)

FI

O

F

tO

t

I OPipe

E = I - O

Tank

F

Same shape graphs apply for RC circuit and motor.


Blocks and Transfer FunctionsOur block diagrams include blocks, such as:

X is Transfer Function,

Defines how signal A is transferred to signal B : X = B/A

= *X (or X * ) or = XBB A AA

A BX

We now start to define how A becomes B

Gains and Transfer Functions


Sometimes call Transfer Function X the ‘gain’ of the block,

X could be a number say 10;

B is 10 times A, B and A are signals of same type

But X can change signal types –

eg for Resistor: A = Voltage; B is Current ; X is 1/Resistance

Current (in Amps) = Voltage (in Volts) * (1/Resistance)

V I1/R

Still like a ‘gain’, as I is number * V

eg suppose V = 2V, R = 4Ω, then I = 2 * ¼ = 0.5A

NB resistance measured in Ω1/R in seimens

Some Examples


V (voltage)

V I1/R

I VR

Extensionx

Forcex F

k

V = I * RI = V * 1/R

F = k * x

Resistor

Spring

Mass F1/m

AA = F * 1/m

F = v * f

Damper Dashpot(for friction)

v Ff

FMass

A

On Integrators


For a capacitor, current flowing in causes voltage out to increase

V = V + amount due to I

Similarly, for water tank

Height = Height + amount due to flow F

Also, for motor, net Force means motor accelerates

Velocity = Velocity + amount due to acceleration

Mathematically, ‘integration’ does

Output = Output + amount * Input

Hence blocks can include Integrators (output constant if input 0)

Can also have blocks which reflect change : differentiation

Examples


V (voltage)

V = 1/C * Integral II V1/C

Capacitor

V (voltage)V = L * Differential I

InductorI V

L

Mass F1/m

vv = 1/m * Integral (F)

FMass

v

NB need also to include in block integration or differentiation – do later

Analogous Components and Signals


Used to model mechanical system with electrical one

Force equiv to Voltage Velocity equiv to Current

NB Position is Integral of velocity and acceleration differential of v

Damper = Resistor Spring = Capacitor Mass = Inductor

V (voltage)V = I * R

F = v * f

V (voltage)V = Integral I * 1/C

F = Differential v * 1/m

V (voltage)

V = Differential I * Lposition

Force

F = Integral v * k


SummaryThus we have seen different types of system, some designed for

feedback, some with inherent feedback.

We have seen systems with negative and positive feedback.

We have seen three different but analogous systems.

We have looked at transfer functions of blocks

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


Lecture 2 After Class Exercise

For this motor, you are to find steady state velocity v, as follows:

a)

b)

c)

F = 2 * 5 = 10N

F - X = 0, so X = 10

0.2 * v = 10, so v = 50 m/s

Go to Blackboard , module SE1CC11, find associated quiz and answer

vArmature

TF = 5

Net Force = F - X

Mass

FrictionTF = 0.2

Vin = 2V F

Frictional force = X

Given Vin and TF of armature, what is F?

What is X, when v at steady state?

Hence what is v then?


3 : Analyzing Feedback Systems

We have introduced the cybernetic principle of feedback and modeled the Classical Feedback Control System

aim - actual (output) equals desired (input)

To verify this (and to find out it’s not true!) we need to be able to analyse such feedback systems.

We can have models (transfer functions) of the components, but what is the overall model (or transfer function)? Let’s find out.

Desired Actual

Disturbance

Controller

Error

Device


Reminder of Component Types

bMeans b = a*X (or X *a) or = Xa

BA X

A

B

CA

B

C

C = A - BC = A + B

X is transfer functionHow signal A transferred to B

But systems have combinations of such components.

Can we combine for overall model or transfer function


Analysis of Two in SeriesFirst consider: two blocks in series

From what we have already said

B = A * X

and C = B * Y

ie X and Y combined

single block model

Transfer Function X * Y

So C = A * X * Y

A BX CY

A CX*Y


What of Feedback Control System ?

E = I – O

O = E * C * P

O = (I - O) * C * P

= I * C * P – O * C * P

O + O * C * P = I * C * P

O * (1 + C * P) = I * C * P

But needed 7 lines of algebra ...

C * PO = I * 1 C*P

C * PO or = I 1 C*P

From summing junction

Now substitute for E

Gathering ‘like’ terms

Now divide by 1 + C * P,

I OC

EP


Better : Forward over 1 Minus Loop Rule

Forward transfer function, I to O,

ignore (0) feedback signals: C * P

Loop transfer function, ignore (set to 0) signals entering loop

E to E (or O to O) LTF = - C * P

(as I = 0, so O to E is -1)

O Forward C*P C*PThus, = = = I 1-Loop 1-- C*P 1 C*P

I OC

EP

I OCE P

I OCE P

‘closed loop transfer function’


What about Disturbances?

Two + in second summing junction, so O = output of P + D

O Forward = = D 1-Loop

Thus

1 1 = 1--C*P 1 C*P

OD

IC

EP

OD

CE

P-1If assume I = 0,

Redraw as

Forward = 1; Loop same, so -CP

Complete Response


This is the principle of superposition (used in circuits also), so

C*P 1O = * I * D

1 C*P 1 C*P

In effect, we assumed D = 0 and found

C*P C*PO = which is equivalent to O = * I I 1 C*P 1 C*P

When assumed I = 0, found O 1 1 = i.e. O = * D D 1 C*P 1 C*P

In general, I and D wont be 0, so can find O in terms of I and D by adding O when D = 0, to O when I = 0,


Let’s Put Some Numbers In

C * P OWe have, O = * I Is O = I? i.e. is = 1?

1 C*P I

Answer NO, as 1+C*P ≠ C*P, but can be close

Let’s try with different values of C and P

If C = 5 & P = 2, O/I = 10/11 = 0.909

If C= 50 & P = 2, O/I = 100/101 = 0.9901

If C = 500 & P = 2, O/I = 1000/1001 = 0.999

Thus if High Loop Gain: O I

poornot badok?

I OC

EP


What If P Changes?In practice, parameters of a model can change gradually

Resistor warms when current through it : resistance changes

Suppose P changes by 10%, from 2 to 2.2

If C = 5, O/I = 11/12 = 0.9167 0.84% change

[ When P was 2, O/I = 0.909:

% change = 100 * (0.9167-0.909) / 0.909 = 0.84]

If C = 50, O/I = 110/111 = 0.991 0.09% change

If C = 500, O/I = 1100/1101 = 0.999 0.009% change

High gain: O largely unaffected by change in P


What about effect of Disturbances

1We've got: O = * D1 C*P

Is it good enough?

If C = 5 & P = 2, O/D = 1/11 = 0.0909

If C= 50 & P = 2, O/D = 1/101 = 0.009901

If C = 500 & P = 2, O/D = 1/1001 = 0.000999

Thus, high gain implies D has little effect on O.

Thus it is advantageous for the value C*P to be high.

We want: O = 0 * D

poornot badok?


Numbers for complete model

If C = 11, P = 9, I = 2 and D = 5;

O = 99/100 * 2 + 1/100 * 5 = 2.03

If C = 500, P = 2, I = 1 and D = 4

O = 1000/1001 * 1 + 4 / 1001 ~ 1

If loop gain (CP) high, response close to desired:

O = I + 0 and not affected by changing P or by D

BUT, as shall see later, we can’t just have high loop gain.

C*P 1O = * I * D

1 C*P 1 C*P


Lecture 3 In Class Exercise

a) Find 1 minus Loop

b) Find O/I assuming D = 0

c) Find O/D assuming I = 0

d) Evaluate O if I = 10 and D = -5

e) Find O/I if P changed to 40

Suppose

C = 27

P = 37

OD

IC

EP


Feedback Not Just for Control

Note + sign in ‘summer’. Control system if A = C * P, = -1!

ForwardIgnoring D, using we get, 1-Loop

Or do by: O = A * X, X = I + * O, O = A*I + A* O

So O*(1-A*) = A*I i.e. same result!

So more General Feedback System

Control Engineers want O = I, Audio Engineers O = I * G

O A = I 1 A

OD

IA

X

F


Analysis – and when Feedback Good

e.g. if A = -5000, = 0.2, A = -1000, 1- A = 1001, ~ - A

O = -5 * I again -1/β and independent of A

Suppose A >> 1 {much greater than 1} ‘negligibly large’

O A A 1 = = I 1 A -A

e.g. if A = -5000, = -0.2, A = 1000, 1-A = -999 ~ -A

1So O = * I = 5 * I (independent of A)

Also works if A << -1, (large and negative)

A 1O * I = * I

-A

Feedback good if modulus of Loop Gain, |A|, large

{ modulus means size, ignore sign: | 5 | = 5 |-5| = 5 }


Disturbance Response

If A << -1, O = (small positive number) * D

If A >> 1, O = (small negative number) * D

So, if feedback good (|A | large), O ~ 0

O ForwardSet I = 0, then = = D 1 Loop

OD

IA

X

F


Overall ResponseSuperposition Theorem Gives complete response:

1 1 O = *I 0 *D = - * I

If feedback is good, i.e. |A| >> 1

That is O = I times -1/feedback value, is independent of A (and hence of changes in A) and unaffected by D.

If | loop gain | is not that high, get O = X*I + Y*D

hopefully, X is close to 1 and Y small,

if A changes, X changes by a smaller amount - see later.

A 1O = I D

1 A 1 A


Example – A = 100, = 0.21 1-A = 1-21 = -20

If I = 2, and D = 0, O = -5*2 = -10

Check: F = -2.1, so X = 2+-2.1 = -0.1; O = 100*-0.1+0 = -10If I = 0 and D = 1, O = -0.05 * 1 = -0.05.

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

O A 100 = = = 5I 1-A 20

O 1 1 = = = 0.05D 1-A 20

OD

IA

X

F

Real System – in Electronics


Operational Amplifier

VOA

Vi-

Vi+

Potential Divider

2 o

1 2

R= VR R

21 2

RR + R

VO Vm

Vo = A * ( Vi+ - Vi

- )

A very big, ~105,


Combine asIn use, put feedback round op-amp :Vi

- found by potential divider = VI

Suppose R1 = 9kΩ and R2 = 1kΩAssume A very big, 105

oI

V A = V 1 A*0.1

A = 10A*0.1

Block Diagram for complete analysis

VOVI A

Vi-

0.1

o oV *1k V = 9k + 1k 10

1k = = 0.19k + 1k

5

o4I

V 10 In fact = = 9.999V 1 10


SummaryWe have analysed simple feedback systems :

used forward over one minus loop for closed loop TF

We have seen benefit of high loop gain

We have seen practical (op-amp) circuits

We will extend the analysis next week

And consider positive and negative feedback

Before next week do following Exercise on Blackboard



Suppose A = 200, β = -0.2

a) Find O/I if D = 0

b) Find O/D if I = 0

c) Find %change in O/I if

A changes by 10% to 220

a) O/I = 200/(1+40) = 4.878

b) O/D = 1/41 = 0.024

c) O/I = 220/1+44 = 4.889

%ch = 100 * (4.889 – 4.878) / 4.878 = 0.226%

OD

IA

X

F


4 : More Feedback Systems AnalysisWe extend our analysis; show how computers can be used;

and tackle the question of positive and negative feedback.

For this feedback control system: use forward/1-loopO C*PIf D = 0, = I 1+C*P

O 1If I = 0, = D 1+C*P

C * P 1Overall, O = * I + * D1 + C*P 1 + C*P

If C*P big, O ~ 1*I + 0*D - what control engineers want

OD

IC

EP

If more than one loop ...


Better control if feedback position and velocity :

TF = Forward over 1 minus sum of loops (-CP and –PV)O C*P C*PIf D = 0, = = I 1--C*P--P*V 1+C*P+P*V

O 1If I = 0, = D 1+C*P+P*V

Ok in this configuration – will return to multi loops later

OD

I CE P

V


Analysis Using MATLABMATLAB is an interactive package that you will use

It processes matrices, which have one or many numbers

You can type commands at the prompt, to assign values to variables, and/or to perform calculations

You can call functions provided by MATLAB and write your own functions, in a language like C

You can plot graphs

You can define a suitable graphical interface …

NB 35 is a number

[35 8 1 -8] 4 column vector (matrix)

[35, 67; 89, 88; 3, -1*5] 3 row 2 column matrix


Using MATLAB on Control Systems>> C = 11; P = 9; I = 2; D = 5;

>> O = I * (C * P) / (1 + C * P)

O =

1.9800

>> O = D / (1 + C * P)

O =

0.0500

>> O = I * (C * P) / (1 + C * P) + D / (1 + C * P)

O =

2.0300

‘;’ so no output shown

Do calc, assign to O show result as no ‘;’

{ could extend easily to cope with V }


Or to see effect of change in P>> C = 50; P = 2; Pnew = P * (1.1);>> OoverI = C * P / (1 + C * P)OoverI =

0.9901>> OoverD = 1 / (1 + C * P)OoverD =

0.0099>> OoverInew = C * Pnew / (1 + C * Pnew)OoverInew =

0.9910>> PCchange = 100 * (OoverInew - OoverI) / OoverIPCchange =

0.0901

Set Values

Calc O/I

Calc O/D

Calc O/I new P

Calc % change in O/I


Can Write MATLAB Functionfunction FBSys (C, P, pc); % store in file fbsys.m% FBSYS (C, P, pc)% Analyse feedback system with gains C and P% noting effect when P change by pc%, % Dr Richard Mitchell 12/7/05den = (1 + C * P); % calculate valuesOoverI = C * P / den;OoverD = 1 / den;Pa = P + P * pc/100; % find new POaoverI = C * Pa / (1 + C * Pa); % and now output them[OoverI, OoverD, OaoverI, 100*(OaoverI-OoverI)/OoverI]>> fbsys(50, 2, 10)

0.9901 0.0099 0.9910 0.0901


More General Feedback System

If D = 0, If I = 0,

Overall

If A*β big, O ~ -I/β + 0*D - what other engineers want

If A changes and A*β still big, O unaffected

O A = I 1 A

O 1 = D 1 A

A 1O = I D

1 A 1 A

OD

IA

X

F


MATLAB Analysis of Aβ System>>A=200; b=-0.1;>> OoverI=A/(1-A*b)OoverI =

9.5238>> OoverD=1/(1-A*b)OoverD =

0.0476>> Anew = A*1.1; OoverInew=Anew/(1-Anew*b)OoverInew =

9.5652>> PCchange=100*(OoverInew-OoverI)/OoverIPCchange=

0.4348

Set Values

Calc O/I

Calc O/D

Calc O/I new P

Calc % change in O/I


Lecture 4 In Class ExerciseWrite MatLab function to calculate transfer functions so as to find and print percentage change in closed loop gain when A changes by pc%

function FBSys (A, b, pc);% FBSYS (A, b, pc)% For system A in forward, b in feedback paths% find change in gain when A changes by pc%

OoverI = A/(1-A*b);Anew = A + A*pc/100; OoverInew = Anew/(1-Anew*b)PCchange = 100*(OoverInew-OoverI)/OoverI


Positive/Negative FeedbackWe 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 is

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) if |closed loop gain| < |open loop gain|

We need to quantify this. NB | x | means size : ignore sign


Negative Feedback (Harold Black 1930s)Forward Path (Open Loop) Gain = AFeedback Path Gain = Loop Gain = A

= < 1- 1- A A AA A

Closed Loop Gain =

1AA

Negative Feedback | closed loop gain | < | open loop gain |

OD

IA

X

F

1i.e. 1 or 1 1- 1-

AA

eg A = 10, = 0.05

A = 10, = 0.5

A = 10, = -0.5

|1 - A | = | 1–0.5 | = 0.5 Pos|1 - A | = | 1 – 5 | = 4 Neg|1 - A | = | 1 – -5 | = 6 Neg


Negative Feedback & DisturbancesNegative feedback reduces effect of disturbances

OOpen Loop: = 1; no reduction in effect of DD

O 1Closed Loop : = ; reduction if |1-A | 1D 1-A

O 1 1 1A = 10 and = -2: = = = < 1D (1-A ) 1 20 21

Examples

D must change by 21 if O is to change by 1.

A = 10 and = 0.02: O 1 1 = = = 1.25D (1-A ) 1 0.2


… and Changes in Parameters

A(1 )-ARelative change in A: open loop = = A

It can be shown (see next slide: closed loop =

if you are interested) that 1-A

Negative Feedback reduces effects of changes in paras

Let A change by a small proportion: call it : A := A(1 + )

Feedback reduces the effect of change in A if |1-A | > 1.

Let A = 10, δ = 0.1 (A change by 10% to 11) and = 1;

Open loop change = 0.1; Closed Loop = 0.1/-9 = -0.0111

Check: when A = 10; O/I = 10/-9 = -1.1111

when A = 11; O/I = 11/-10 = -1.1; diff is -0.0111


Interest Only: A → A(1 + )A(1 )Closed Loop Gain =

1-A(1 )

A(1 ) A1 1-A1-A(1 ) 1-A = 1A 1-A(1 )

1-A1 1-A 1-A(1 ) =

1-A(1 )1 A A 1 A A = =

1-(A A ) 1-A(1 )

new - oldRelative change = old

As 1, this approximates to , as stated earlier1-A

Can also do by differentiating closed loop TF w.r.t A


Effect of Changing A : 50 to 55

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%

β = a) -1/50 b) -1/10 c) 1/100 d) 3/100 e) 1/10

β 1-Aβ % diff

a -1/50 1+1=2b -1/10 1+5=6c 1/100 1-0.5=0.5d 3/100 1-1.5=-0.5e 1/10 1-5=-4

O A = I 1-A

O 1 = D 1-A

O (A = 55)I

1/2=0.5 50/2=25 55/2.1= 26 +4%

1/6=0.17 50/6=8.3 55/6.5= 8.5 +1.5%

1/0.5=2 50/0.5=100 55/0.45= 122 +22%

1/-0.5=-2 50/-0.5=-100 55/-0.65= -84.6 -15%

1/-4=-0.25 5 -/-4=-12.5 55/-4.5= -12.2 -2%


Graphs for b) : A = 50, β = -0.1

O 50 O 1Here = and = , soI 6 D 6

It can be useful to plot graphs of O vs I and O vs D

As straight lines … associated system is said to be linearGradient of O/D < 1 : system has negative feedback

Both straight lines thru 0,0O/I thruO = 50, I = 6

O/D thruO = 1, I = 6


SummaryWe have analysed feedback system : A and CP

We have briefly noted multiple loops

We have an appropriate definition of negative feedback

We have shown that can plot graphs of O vs I and O vs D

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


Lecture 4 After Class Exercise a)Here, I = 2, A = 10.Find O and state whether positive feedback if:

a) = -1 b) = +1c) = -0.02 d) = +0.02

d) O = 2 * 10 / (1-0.2) = 25 +ve fb

c) O = 2 * 10 / (1+0.2) = 16.667 -ve fb

b) O = 2 * 10 / (1-10) = -2.222 -ve fb

a) O = 2 * 10 / (1+10) = 1.818 -ve fb

OIA

X

F


5 : Feedback Systems and Limits

Recall last week’s system with A = 50, β = -0.1

1 - A β = 1+5 = 6

O 50 O 1 = and = I 6 D 6

We consider effects on feedback systems of (practical) limits –system signals cannot be infinite.

Straight line graphs –system is linear

|1 - A β| > 1 : -ve fb

I OD

A

X

F


LimitsO/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

The output has limits; and we can incorporate them.

If -L ≤ In ≤ L, Out = In;

if In < -L, Out = -L; if In > L, Out = L

Below are shown limits graphically and as a block diagram

In

Out+L

-L

Limit Block

In Out+L


Limits in Feedback Systems

Let A = 50; = -0.1; +L = +25 and -L = -25.

Consider O vs I assuming D = 0.

When no limiting, limit box transfer function = 1.

O 50 50 50Thus = = = 8.333 i.e. O (= Z = Y) = * II 1--5 6 6

25 50So when just at limit, Y = Z = O = 25, = or I = 3 I 6

OD

IA

X

F-L

+LY Z


What happens when pass Limit?

If I increases, X and Y increase, but Z and O stay at 25

A = 50

= -0.1

L = 25

At limit

I = 3, O = 25

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

OD

IA

X

F-L

+LY Z


General Case – and other example1-A 1-AIn general A system; I just limit at L & L

A A

1--50*0.1In above example, I limit at 2550

6= 25 = 3

50

Suppose A = 100; β = -0.1 +L = 10 and –L = -10

System linear (as before) when no limit: -10 ≤ Y ≤ 10

O A 100 = = 9.09; O vs I straight lineI 1 A 11

Values of I where just limit are1--10 10 * = 1.1100


O vs D : a different response

Do similarly for when Y = Z = -L : or use symmetry

Strategy, find D when just limit: Y = Z = +L: { O/D = 1/1-Aβ }

Then L D = O; D = 1 - A *O so L + O - AβO = O

L L Hence L = AβO or O = : D = 1 - A A A

OD

IA

X

F-L

+LY Z


So, when A = 100; = -0.1

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.

When just about to limit (remember A = 100*-0.1 = -10)

L 10 O = = = -1; D = 1-A * 1 = -11; A -10

O 1 1If -10 Y 10, response as usual: = = D 1 A 11

When just about to limit (remember A = 100*-0.1 = -10)

L 10 O = = = -1; D = 1-A * 1 = -11; A -10


Lecture 5 In Class ExerciseA = 1000 = -1/5+L = 20-L = -20

Calc O & D at L = -20 to label the graph

O = L/(A) = +20/(1000 * -0.2)= -20/200 = -0.1

D = O*(1-A )= -0.1 * 201 = -20.1

Click to Run Limits Prog

OD

IA

X

F-L

+LY Z


Limits and Hole in the Ozone LayerThe ozone layer is a feedback system

Must be (according to Gaia) so correct amount of u.v. getting to Earth’s surface:

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


What of Limits & Positive Loop Gain

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.

We will now see different response when loop gain > 0 …

OI100

0.1

X

F

-10+10Y


Limits and Positive Loop Gain

This change from +10 to -10 is represented graphically as

If I reduced now no change.

But if I increased to 0

X = -1, O = -10 still

Even if I upped to +0.9:

X = -0.1, O = -10, still, but..

I

O

-0.9

OI100

0.1

X

F

-10+10Y


O stays at 10 until I again < -0.9.

O depends on I and on O!

Hysteresis: figure to right

How to Flip Back

Thus, when I exceeds 0.9, O flips back to + 10

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

I

O

0.9

I

O

-0.9 0.9


How To Find I Where O Flips

O 100 100 9 = = - ; or I = - * O = -0.9I 1-0.1*100 9 100

If I ≥ -0.9, O stays at 10, but if I < -0.9, O will flip to -10

By symmetry, I must exceed +0.9 for O to flip to +10

Could be found by same analysis.

Again find I when system just limits,

ie Y = O = 10

1-AIn general, flip when I = L * A

OI100

0.1

X

F

-10+10Y


MATLAB Code to Plot Limitsfunction fblims (A, b, lim, whatplot); % plot limit systemOmAb = 1 - A*b; % 1 minus A * bIlim = lim * OmAb / A; % value of I when just limitingif A*b > 0 % +ve loop gain – plot hysteresis

plot([-Ilim-5, Ilim, Ilim, Ilim+5, -Ilim, -Ilim, -Ilim-5], ...[-lim, -lim, lim, lim, lim, -lim, -lim]);

elseif whatplot == 0 % -ve loop gain – plot O/Iplot ([-Ilim-5, -Ilim, Ilim, Ilim+5], [-lim, -lim, lim, lim]);

else % -ve loop, plot O/DOlim = lim / (A*b); % O when just limitsDlim = Olim * OmAb; % D when just limitsplot([-Dlim-5, -Dlim, Dlim, Dlim+5], [-Olim-5, -Olim, Olim, Olim + 5]);

end


Example Output + General SummaryGraphs A = 100, β = ± 0.1, lim = 10

-5 0 5-10-505

10

I

O

-10 0 10-5

0

5

DO

-5 0 5-10-50510

I

O

1-ALA

LA

1-A-LA

Aβ < 0 Aβ > 0

1-ALA


‘hysteresis’ loop input,

I = -C * ∫ O

I changes at rate -CO

As O > 0, I is integral of –ve constant, so I decreases

At time t, I = Flip - C*Lim*t

At t = 2*Flip / C*Lim, I = -Flip, O := -Lim

Input to ∫ C*Lim (>0), I increases constantly

At t = 4*Flip / C*Lim, I = +Flip, O := +Lim

I

Ot = 0, I = +Flip, O = +Lim

4*Flip / C*Lim

Application: Square Wave Generator

OI100

0.1

X-10

+10Y-C∫

To introduce dynamic systems – see simple use of integratorNB if input to integrator is constant, output changes at constant rate


Summary

We have seen effect of limits on feedback systems

If negative loop gain, get different responses O/I O/D

For positive loop gain, get hysteresis

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



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?

OD

IA

X

F

-L+LY Z


Answer : L5 After Class ExerciseO = -50. First find value of I such that O just not flip

So find case when O = Y = -50;

I flip = -Lim * (1 – A ) / A

= -50 * (1 – 100*0.2) / 100 = -50 * -19 / 100 = 9.5

Thus O will change to +50 when I exceeds 9.5

When O = Y = 50,

I flip = Lim * (1 – A ) / A

= 50 * (1 – 100*0.2) / 100 = 50 * -19 / 100 = -9.5

Thus O will change back to -50 when I is less than -9.5.

[Could have said -9.5 by symmetry argument]

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Cybernetics and Circuits Feedback