[BK Control Manual] Feedback Control

Feedback Control Systems. 11

FEEDBACK CONTROL

Control is a very common concept.e.g., Human-machine interaction: Driving a car. MANUAL CONTROL.e.g., Independent machine: Room temperature control. Furnace inwinter, air conditioner in summer. Both controlled (turned on/off)by thermostat. AUTOMATIC CONTROL.

DEFINITION: Control is the process of causing a system variable toconform to some desired value, called a reference value. (e.g.,variable=temperature)

DEFINITION: Feedback is the process of measuring the controlledvariable (e.g., temperature) and using that information to influence thevalue of the controlled variable.

Some applications: Airplane autopilots that can land a plane in fog. Space telescopes with pointing accuracy of 106 degrees. Disk-drive read heads with < 1 micron accuracy. Robots for various applications. . . . (see next page for a synopsis).

FEEDBACK CONTROL 12

Typical Feedback ApplicationsCategories Specific ApplicationsEcological Wildlife management and control; control of plant chemical

wastes via monitoring lakes and rivers; air pollution abatement;water control and distribution; flood control via dams and re-sevoirs; forest growth management.

Medical Medical instrumentation for monitoring and control; artificial limbs(prosthesis).

Homeappliances

Home heating, refrigeration, and airconditioning via thermostaticcontrol; electronic sensing and control in clothes dryers; humiditycontrollers; temperature control of ovens.

Power/energy Power system control and planning; feedback instrumentation inoil recovery; optimal control of windmill blade and solar panel sur-faces; optimal power distribution via power factor control.

Transportation Control of roadway vehicle traffic flows using sensors; automaticspeed control devices on automobiles; propulsion control in railtransit systems; building elevators and escalators.

Manufacturing Sensor-equipped robots for cutting, drilling die casting, forging,welding, packaging, and assembling; chemical process control;tension control windup processes in textile mills; conveyor speedcontrol with optical pyrometer sensing in hot steel rolling mills.

Aerospace andmilitary

Missile guidance and control; automatic piloting; spacecraft con-trol; tracking systems; nuclear submarine navigation and control;fire-control systems (artillery).

(reproduced from: J.R. Rowland, Linear Control Systems: Modeling, Analysis, and Design, John Wiley & Sons, (New York: 1986), p. 6.)

FEEDBACK CONTROL 13

Rube Goldberg showed that a knowledge of feedback control waseven useful to budding cartoonists!

Rube Goldberg walks in his sleep, strolls through a cactus field in his bare feet, and screams out an idea for self-operatingnapkin: As you raise spoon of soup (A) to your mouth it pulls string (B), thereby jerking ladle (C) which throws cracker (D) pastparrot (E). Parrot jumps after cracker and perch (F) tilts, upsetting seeds (G) into pail (H). Extra weight in pail pulls cord (I),which opens and lights automatic cigar lighter (J), setting off sky-rocket (K) which causes sickle (L) to cut string (M) and allowpendulum with attached napkin to swing back and forth thereby wiping off your chin. After the meal, substitute a harmonica forthe napkin and youll be able to entertain the guests with a little music.

Simple Feedback System Example

DesiredTemp.

RoomTemp.

Heat Loss (Qout)Qin

HouseFurnaceGasValveThermo-

stat

Block Diagram of furnace-controlled room temperature controller.Identifies major components and omits details. Shows information/energy flow.

Central component = PROCESS or PLANT, one of whose variableswe want to control. e.g., Plant = ; Variable = .

FEEDBACK CONTROL 14

DISTURBANCE = some system input out of our direct control.e.g., Disturbance = . ACTUATOR = device that influences controlled variable.

e.g., Actuator = . REFERENCE SENSOR measures desired system output. OUTPUT SENSOR measures actual system output. * COMPENSATOR/ CONTROLLER = device that computes the

control effort to apply the the actuator, based on sensor readings.e.g., combines the last three functions.

A More Abstract Block Diagram:

Ref.Value Output

Disturbance

PlantReferenceSensor

OutputSensor

Compen-sator Actuator

error

Negative Feedback

An Even More Abstract Block Diagram:

Ref.Value Output

Disturbance

PlantCompen-sator

FEEDBACK CONTROL 15

The Control Problem:

Reject disturbance. Acceptable steady state errors. Acceptable transient response. Minimize sensitivity to parameter changes in the plant.

Solution Reached By:

1. Choosing output sensors.

2. Choosing actuators.

*3. Developing plant, actuator, sensor equations (models).*4. Designing compensator based on the models and design criteria.

*5. Evaluating design analytically, with simulation and prototype.

*6. Iteration!!

A First Analysis: Auto Cruise Control

Want to control speed of automobile.1. Output sensor = speedometer.2. Actuator = throttle and engine.

FEEDBACK CONTROL 16

Block diagram:

DesiredSpeed

ActualSpeed

Road Grade

Plant

Auto Body

OutputSensor

Speedometer

Sensor Noise

Compen-sator Actuator

EngineThrottle

ControlVariable

MeasuredSpeed

3. Model of system:

1. Operate system 55 MPH. Assume linear response near 55 MPH.2. Measure: 1% change in throttle 10 MPH change in speed.3. Measure: 1% change in grade 5 MPH change in speed.4. Measure: Speedometer accurate to a fraction of 1 MPH, so is

assumed exact.5. Functional block diagram:

Compen-satorr .t/ y.t/

w.t/

u.t/10

0:5

r .t/reference speed, MPH.u.t/throttle posn., %.y.t/actual speed, MPH.w.t/road grade, %.

FEEDBACK CONTROL 17

4. Design compensator/controller

First attempt = open loop controller.

Compensator

1=10r .t/ yol.t/

w.t/

u.t/10

0:5

yol.t/ D 10 .u.t/ 0:5w.t//D 10

r.t/

10 0:5w.t/

D r.t/ 5w.t/:

5. Evaluate design.

r.t/ D 55, w.t/ D 0% yol.t/ D 55. No Error. . . Good. r.t/ D 55, w.t/ D 1% yol.t/ D 50. 10% Error. . . Not Good. r.t/ D 55, w.t/ D 2% yol.t/ D 45. 20% Error. . . Not Good. Suppose you load the trunk with all your course materials and the

car becomes more sluggish.e.g., the gain 10 becomes 9

r.t/ D 55, w.t/ D 0% yol.t/ D 49:5. 10% Error. . . Not Good. r.t/ D 55, w.t/ D 1% yol.t/ D : : :

6. Iterate design. Go to step #4.

FEEDBACK CONTROL 18

4. Second attempt = Feedback controller

Compensatorz }| {100r .t/ ycl.t/

w.t/

u.t/10

0:5

Multiply your output error by 100; FEEDBACK GAIN D 100ycl.t/ D 10u.t/ 5w.t/u.t/ D 100.r.t/ ycl.t//

so; ycl.t/ D 1000r.t/ 1000ycl.t/ 5w.t/1001ycl.t/ D 1000r.t/ 5w.t/

ycl.t/ D 0:999r.t/ 0:005w.t/5. Evaluate design.

r.t/ D 55, w.t/ D 0% ycl.t/ D 54:945 r.t/ D 55, w.t/ D 1% ycl.t/ D 54:94 r.t/ D 55, w.t/ D 2% ycl.t/ D 54:935 r.t/ D 55, w.t/ D 10% ycl.t/ D 54:895 : : : 0:2% Error!

Feedback system rejects disturbances. Feedback system has steady state error.

r.t/ D 55, w.t/ D 0%, plant=9, not 10 ycl.t/ D 54:939 Feedback system less sensitive to system parameter

values.

NOTE! High feedback gain = good performance here. Not alwaystrue! e.g., Public address amplifier.

FEEDBACK CONTROL 19

4. Third attempt. Try to get rid of steady state error.

Compensator

r .t/ ycl.t/

w.t/

u.t/10

0:5

ycl.t/ D 10 .r.t/ ycl.t//

5w.t/D 10r.t/ 10ycl.t/ 5w.t/

ycl.t/ D 10r.t/ 5w.t/1C 10D

101C 10

| {z }set to 1: D1 110

r.t/

51C 10

w.t/

D r.t/ 510

w.t/:

Best results yet! (Try for yourself with D 100.)

A Brief History of Control

2nd century B.C.: Fluid level/Flowrate control. Still used to flush toilets!

SupplyFloat

FEEDBACK CONTROL 110

1624: Drebbels incubator Control temperatureusing mechanicalfeedback.

WaterEggs

Riser

Float

Metal plate

GassesFlue

MercuryAlcoholFire

Damper

1787: Thomas Meads fly-ball governor. 1788: James Watts fly-ball governor.

ButterflyTo engine

inlet

Pivot

valve

Ball

Rotation

Sleeve

Pulley from engine

Steam

Both control a rotating shaft.

1840: Airys telescope controller. and the machine (if I may soexpress myself) became perfectly wild.

Discovery of instability in a feedback control system. Analysis of system with differential equations. Beginnings of feedback control theory.


1868: Maxwell found stability criteria for second and third order(simple) systems. 1877: Routh found stability criteria for more complex (general) linear

systems.

1893: Lyapunov: stability of nonlinear systems. 1932: Nyquists graphical stability criteria (1945: Bodes simpler graphical method) 1936: PID Control. 1948: Evans root locus. 1960s: State variable (modern) control design. 1980s: Post Modern H1 Robust control. Adaptive control...?


1624

1728

1868

1877

1890

1932

1910

1927

1938

1942

1947

1948

1950

1956

1957

1960

1969

Chronological Historyof Feedback ControlDrebble, Incubator

Watt, Flyball governor

Maxwell, Flyball stability analysis

Routh, Stability

Liapunov, Nonlinear stability

Sperry, Gyroscope and autopilot

Black, Feedback electronic amplifier: Bush, Differential analyzer

Nyquist, Nyquist stability criterion

Bode, Frequency response methods

Wiener, Optimal filter design: Ziegler-Nichols PID tuning

Hurewicz, Sampled data systems; Nichols, Nichols chart

Evans, Root locus

Kochenberger, Nonlinear analysis

Pontryagin, Maximum principle

Bellman, Dynamic programming

Draper, Inertial navigation; Kalman, Optimal estimation

Hoff, Microprocessor(reproduced from: G.F. Franklin, J.D. Powell, A. Emami-Naeini, Feedback Control of Dynamic Systems, 3rd edition,Addison Wesley, (Reading, Mass: 1994), flyleaf.)

munelCross-Out


SYSTEM MODELING IN THE TIME DOMAIN

Model = Set of equations used to represent a physical system,relating output to input. Required to:

1. Understand system behavior (analysis).2. Design a controller (synthesis). Developing the model 80%90% of the effort in designing a

controller. Methods:

1. Analytic system modelingwe focus on these methods.2. Empirical system identification. (In practice, there is always an

empirical component to system modeling). No model is exact! Inaccuracies due to:

1. Unknown parameter values.2. Unmodeled dynamics (to make simpler model).

LTI Systems

This course teaches methods to control linear time invariant (LTI)systems.

None exist! But, many are close enough. In the following, suppose a system maps x.t/ 7! y.t/ as

y.t/ D T[x.t/]:

SYSTEM MODELING IN THE TIME DOMAIN 22

LINEAR: A system is linear iffT

[ax1.t/C bx2.t/] D aT[x1.t/]C bT[x2.t/]:a.k.a. superposition.

TIME INVARIANT: A system is time invariant iffy.t t0/ D T[x.t t0/] 8 t0:

(Translation: If we input a specific signal x.t/ and record the outputy.t/, then input a shifted version of the signal x.t/, the output will be ashifted version of y.t/, with the same shift.)

KEY POINT: If a system is LTI, then it has an impulse response. Thisentirely characterizes the systems dynamics. The Laplace transformof the impulse response is the transfer function. Working with thetransfer function eliminates the need to mess around with trying tosolve complicated differential equations.

A Simple Model

Consider a resistor:i.t/

v.t/R

Ohms Law (model) says: v.t/ D i.t/ R. Is the resistor LTI? Is the model of the resistor LTI?

EXAMPLE: Consider a 1 Ohm, 2 Watt resistor.


Apply 1 Volt;1 Amp of current is predicted to flow.Power dissipated D V 2=R D 1 Watt. Now apply 10 Volts.

10 Amps of current is predicted to flow.Power dissipated D V 2=R D 100 Watts! Model will no longer be accurate. True behavior depends on input signal levelnonlinear. Model is accurate in certain range of input-signal values.

Dynamics of Mechanical Systems I (translational) Newtons Law:

XF D ma. Vector sum of forces = mass of object

times inertial acceleration.

Free-body diagrams are a tool to apply this law.EXAMPLE: Cruise control model.

Write the equations of motion for the speed and forward motion of acar assuming that the engine imparts a forward force of u.t/.

1. Assume rotational inertia of wheels is negligible.2. Assume that friction is proportional to cars speed (viscous friction).

XF D ma

u.t/ b .x.t/ D m ..x.t/or;

..

x.t/C bm

.

x.t/ D u.t/m

b.

x.t/ u.t/

x.t/

m


If the variable of interest is speed (v.t/ D .x.t/), not position,.

v.t/C bmv.t/ D u.t/

m Notice that the differential equation has output variables on the left

of D, and input variables on the right.IMPORTANT POINT: All of our models of dynamical systems will be

differential equations involving the input (e.g., u.t/) and its derivativesand the output (e.g., y.t/) and its derivatives. No other signals(intermediate variables) are allowed in our solutions.

EXAMPLE: Car suspension.Each wheel in a car suspension system has a tire, shock absorberand spring. Write the one-dimensional (vertical) equations ofmotion for the car body and wheel.

Quarter-car model

Road SurfaceInertial Reference

kw

bks

m1

m2

x.t/

y.t/

r .t/

Free-body diagram:ks.y.t/ x.t// b.

.

y.t/ .x.t//

kw.x.t/ r .t// ks.y.t/ x.t// b. .y.t/ .x.t//

m1 m2x.t/ y.t/


The force from the spring is proportional to its stretch. The force from theshock absorber is proportional to the rate-of-change of its stretch.X

F D mab

.

y.t/ .x.t/C ks .y.t/ x.t// kw .x.t/ r.t// D m1 ..x.t/

ks .y.t/ x.t// b

.

y.t/ .x.t/D m2

..

y.t/

Re-arrange:..

x.t/C bm1.

.

x.t/ .y.t//C ksm1.x.t/ y.t//C kw

m1x.t/ D kw

m1r.t/

..

y.t/C bm2.

.

y.t/ .x.t//C ksm2.y.t/ x.t// D 0

Important components for mechanical-translational systems:

x1.t/

x1.t/

x2.t/

x2.t/f .t/ D k.x1.t/ x2.t//

f .t/ D b. .x1.t/ .x2.t//

k

b

m1. Mass

2. Spring

3. Damper

Dynamics of Mechanical Systems II (rotational) Newtons Law is modified to be:

XM D J (or I)

Vector sum of moments = moment of inertia times angularacceleration. (momentDtorque).

EXAMPLE: Satellite control.Satellites require attitude control so that sensors, antennas, etc., areproperly pointed. Lets consider one axis of rotation.


.t/

d

Gas jetFc.t/

moment D Fc.t/ d; so;Fc.t/d D J

..

.t/..

.t/ D Fc.t/dJNote: Output of system .t/ integrates

torques twicedouble-integrator plant.EXAMPLE: Torsional pendulum.A torsional pendulum is used, for example, in clocks enclosed in glassdomes. A similar device is the read-write head on a hard-disk drive.

k

bJ

;

k: Springiness of suspension wire.

b: Viscous friction.

XM D J ...t/

J..

.t/ D .t/ b ..t/ k.t/..

.t/C bJ.

.t/C kJ .t/ D.t/

J

Important components for mechanical rotational systems:

1.t/

1.t/

2.t/

2.t/ .t/ D k.1.t/ 2.t//

.t/ D b. .1.t/.

2.t//

k

b

J1. Inertia

2. Spring

3. Damper


EXAMPLE: NONLINEAR Rotational Pendulum.

l

.t/; .t/

mg

Moment of intertia: J D ml2:XM D J ...t/

J..

.t/ D .t/ mgl sin..t//..

.t/C gl sin..t//| {z }Nonlinear!

D .t/ml2

If motion is small, sin ..t// .t/...

.t/C gl .t/ D.t/

ml2Linear.

This is a preview of linearization.

Summary of Developing Models for Rigid Bodies:1. Assign variables such as x.t/ and .t/ that are both necessary and

sufficient to describe and arbitrary position of the object.2. Draw a free-body diagram of each component, and indicate all forces

acting on each body and the accelerations of the center of mass withrespect to an inertial reference.

3. Apply Newtons laws:X

F D ma;X

M D J.4. Combine the equations to eliminate internal forces.5. The final form must be in terms of ONLY the input to the system and

its derivatives, and the output of the system and its derivatives.

Dynamics of Electrical Circuits

Kirchhoffs Laws: Current Law (KCL): The algebraic sum of currents entering a node

equals the algebraic sum of currents leaving the node. Voltage Law (KVL): The algebraic sum of all voltages taken around

a closed path in a circuit is zero.


Node analysis is a tool to apply these laws. (i.e., select one node asreference (e.g., ground) and assume all other voltages are unknown.Write equations for the unknowns using KCL. KVL must be used forvoltage sources.)

EXAMPLE: Bridged-Tee circuit.

R1 R2

C1

C2

vi.t/

Select reference = .

KVL at : v.t/ D vi.t/. KCL at : v.t/ v.t/

R1 v.t/ v.t/

R2 C1 .v.t/ D 0.

KCL at : v.t/ v.t/R2

C C2. .v.t/ .v.t// D 0.

v.t/ v.t/C R2C2. .v.t/ .v.t// D 0v.t/C R2C2. .v.t/ .v.t// D v.t/

v.t/v.t/C R2C2. .v.t/ .v.t//

R1

v.t/C R2C2. .v.t/ .v.t//

v.t/R2

C1

.

v.t/C R2C2. ..v.t/ ..v.t// D 0

R2(v.t/

v.t/C R2C2. .v.t/ .v.t//

R1(v.t/C R2C2. .v.t/ .v.t//

v.t/


R1 R2C1

.

v.t/C R2C2. ..v.t/ ..v.t// D 0:(

R1 R22C1C2

..

v.t/C(R22C2 C R1 R2C2 C R1 R2C1

.

v.t/C .R2/ v.t/D .R1 R22C1C2/

..

v.t/C .R22C2 C R1 R2C2/.

v.t/C .R2/v.t/

Important components for electrical systems:

1. Resistor

2. Capacitor

3. Inductor

4. Voltage source

5. Current source

6. OperationalAmplifier

v.t/

v.t/

v.t/

v.t/

i.t/

i.t/

i.t/

i.t/

vs

is

vd.t/i.t/

iC.t/ vo.t/

v.t/ D Ri.t/

i.t/ D C d v.t/dt

v.t/ D L d i.t/dt

v.t/ D vs

i.t/ D is

vd.t/ D 0i.t/ D iC.t/ D 0vo.t/ D Ao.vC.t/ v.t//

as Ao !1


EXAMPLE: Op-amp circuit.

R1

R2 C

vi.t/ vo.t/

i.t/ D vi.t/R1

vo.t/ D R2i.t/ vc.t/:

dvo.t/dtD R2di.t/dt

dvc.t/dt

D R2R1

dvi.t/dt i.t/

C

D R2R1

dvi.t/dt 1

R1Cvi.t/

R1C.

vo.t/ D R2C .vi.t/ vi.t/(as C !1, we get an inverting amplifier.)

Dynamics of Electro-Mechanical Systems

These are systems that convert energy from electrical to mechanical,or vice versa.

EXAMPLE: DC Generator.

Assume generator is driven at constant speed. Generator has field windings (input), and rotor/armature windings

(output).R f Ra Lai f .t/ ia.t/

e f .t/ L f eg.t/ Zlea.t/

| {z }Fieldcircuit

| {z }Rotorcircuit

| {z }Loadcircuit


e f .t/ D R f i f .t/C L f di f .t/dt e f .t/ is input, i f .t/ is output.

eg.t/ D Kd.t/dt K depends on generator structure.D Kgi f .t/. d.t/=dt = angular velocity = cst.

= flux, proportional to i f .t/.

eg.t/ D Raia.t/C La dia.t/dt C ea.t/. eg.t/ is input, ia.t/ is output. ea.t/ D Zlia.t/. ia.t/ is input, ea.t/ is output.

e f .t/ ea.t/i f .t/ eg.t/ ia.t/Field

circuit KgRotorcircuit Zl

This is a preview of a block diagram used to simplify our understandingof the system dynamics.EXAMPLE: DC Motor (servo-motor). Directly generates rotational motion. Indirectly generates translational motion.

ea.t/ eb.t/

ia.t/ Ra La b

J.t/; .t/

| {z }Armature

| {z }Load

Mechanical resistance of load is translated into an electricalresistance called the back e.m.f.

eb.t/ D Ke d.t/dt Ke D K, as with generator.

ea.t/ D Raia.t/C La dia.t/dt C eb.t/ .t/ D K ia.t/ K D K1


Combining these equations of motion, recall Newton:XM D J

J..

.t/ D .t/ b ..t/D K ia.t/ b

.

.t/

Assume (FOR NOW ONLY) electrical response is faster thanmechanical. La 0.

J..

.t/ D K

ea.t/ eb.t/Ra

b ..t/

J..

.t/C

b C KKeRa

| {z }

back emf indistinguishable from friction!

.

.t/ D KRa ea.t/

Dynamics of Heat Flow/ Dynamics of Fluid Flow

These two subjects will not be covered here. Refer to texts onthermodynamics or fluid-dynamics.

Transformers and Gears

Ideally, both of these devices simply scale their input value.

Transformer : N1N2D e1

e2D i2

i1

Gears : r1r2D 21D 12

r1 r2

System Identification (SYS ID) When we generate models of system dynamics, we are performing

system identifications.


When we use known properties from physics and knowledge of thesystems structure (as we have done here) we are performing whitebox system ID.

If the system is very complex, or if the physics are not wellunderstood, we need to use input/output data to generate a systemmodel: black-box system ID.

A topic for the whole course!

Linearization

We will study how to control linear systems. Linear systems are rare. We can linearize a non-linear systemthe controller designed for

linearized model will work on the true nonlinear system (but not aswell as a controller designed directly for the non-linear system.)

KEY POINT: We can convert any differential equation into a first-ordervector differential equation:

.

Ex D f .Ex; u/ I Ex D vector; u D input:If the system is linear, this will be of the form:

.

Ex D AEx C BuI A and B are constant matrices:EXAMPLE: Torsional pendulum (pg. 226)

..

.t/C bJ.

.t/C kJ .t/ D.t/

J

let"

x1.t/

x2.t/

#D".t/.

.t/

#


.

x2.t/C bJ x2.t/CkJ

x1.t/ D .t/J".

x1.t/.

x2.t/

#D"

0 1k=J b=J

#| {z }

A

"x1.t/

x2.t/

#C"

01=J

#| {z }

B

.t/:

So, our model of the torsional pendulum is linear.EXAMPLE: Rotational pendulum (pg. 227)

..

.t/C gl sin..t// D.t/

ml2

let"

x1.t/

x2.t/

#D".t/.

.t/

#"

.

x1.t/.

x2.t/

#D24 x2.t/g

lsin.x1.t//

35C

24 01

ml2

35 .t/:

Not linear because we cannot make a constant A matrix.

Small Signal Linearization

Uses a Taylor-series expansion of the differential equation aroundsome operating condition. (Equilibrium value where.

x0 D 0 D f .x0; u0/).

let x D x0 C x x0 D operating stateu D u0 C u u0 D nominal control value:.

x D f .x; u/: Taylor-series expansion:

.

x D .x0 C .x f .x0; u0/C Ax C Bu plus higher-order terms


Subtract out equillibrium (nominal) solution;

.

x D Ax C Bu;which is linear. This is exactly how we linearized the rotationalpendulum before, with 0 D 0I 0 D 0:

sin./ D 3

3!C

5

5!

:Feedback Linearization (computed torque) For rotational pendulum, ml2 ...t/C mgl sin..t// D .t/.

COMPUTE: .t/ D mgl sin..t//C u.t/. THEN: ml2

..

.t/ D u.t/, no matter how large .t/ becomes! Sometimes used in robotics and airplane flight control, but very

computationally intensive.

Analogous Systems

The linearized differential equations of many very different physicalsystems appear identical.

One would suppose they behave in similar ways (dynamic response)and can be controlled with similar controllers.


Mechanical Translational m ..x.t/C b .x.t/C kx.t/ D u.t/Mechanical Rotational J

..

.t/C b ..t/C k.t/ D .t/Satellite J

..

.t/ D f .t/ dDC Motor (for La D 0) J

..

.t/C

b C kkeRa

.

.t/ D kRa ea.t/Generator .La L f /

..

ea.t/C .L f .Ra C Rl/C La R f / .ea.t/CR f .Ra C Rl/ D .kg Rl/e f .t/

These are all of the forma2

..

x.t/C a1 .x.t/C a0x.t/ D b2 ..u.t/C b1 .u.t/C b0u.t/which is called a second-order form.

Therefore, we have seen very specific examples of a very generalclass of system. If we learn how to control the general class, we canapply this knowledge to specific systems.


DYNAMIC RESPONSE

We can now model dynamic systems with differential equations. What do these equations mean? How does this system respond to certain inputs? If we add dynamics (a controller) how will the system respond? How SHOULD the system respond? (specifications)

KEY, ESSENTIAL, VITAL, TOOL: Laplace Transform

Some Important Input Signals

Several signals recur throughout this course. The unit step function:

1.t/ D(

1; t 0I0; otherwise:

t

1.t/

The unit ramp function:

r.t/ D(

t; t 0I0; otherwise:

t

r .t/

The unit parabola function:

p.t/ D8 0. h.t/ D ekt1.t/. Response of this system to general input:

y.t/ DZ 11

h. /u.t / d

DZ 11

ek1. /u.t / d

DZ 1

0eku.t / d:

Transfer Function

Response to impulse = impulse response h.t/. Response to general input = messy convolution: h.t/ u.t/.

Response to Sinusoid? Cosinusoid?

A cos.!t/ D A2(e j!t C e j!t

Break it Down: Response to Exponential?

Let u.t/ D est; where s is complex.y.t/ D

Z 11

h. /u.t / d

DZ 11

h. /es.t/ d

DZ 11

h. /estes d

DYNAMIC RESPONSE 35

D estZ 11

h. /es d| {z }Transfer function; H.s/

D est H.s/: An input of the form est decouples the convolution into two

independent parts: a part depending on est and a part depending onh.t/. [Complex exponentials are eigenfunctions of all LTI systems.]

EXAMPLE:.

y.t/C ky.t/ D u.t/ D est :but ; y.t/ D H.s/est; .y.t/ D s H.s/est;s H.s/est C k H.s/est D est

H.s/ D 1s C k .I never integrated!/

y.t/ D est

s C k :Response to Cosinusoid (revisited)Let sD j! u.t/De j!t y.t/DH. j!/e j!t

sD j! u.t/De j!t y.t/DH. j!/e j!tu.t/DA cos.!t/ y.t/DA

2H. j!/e j!t C H. j!/e j!t

Now; H. j!/ 4D Me jH. j!/ D Me j .can be shown for h.t/ real/

y.t/ D AM2e j .!tC/ C e j .!tC/

D AM cos.!t C /: The response of an LTI system to a sinusoid is a sinusoid! (of the

same frequency).

DYNAMIC RESPONSE 36

EXAMPLE: Frequency response of our first order system:

H.s/ D 1s C k

H. j!/ D 1j! C kM D jH. j!/j D 1p

!2 C k2 D 6 H. j!/ D tan1

!k

y.t/ D Ap

!2 C k2 cos!t tan1

!k

:

Can we use these results to simplify convolution and get an easierway to understand dynamic response?

The Laplace L Transform

We have seen that if a system has an impulse response h.t/, we cancompute a transfer function H.s/,

H.s/ DZ 11

h.t/est dt:

Since we deal with causal systems (possibly with an impulse att D 0), we can integrate from 0 instead of negative infinity.

H.s/ DZ 1

0h.t/est dt:

This is called the one-sided (uni-lateral) Laplace transform of h.t/.

DYNAMIC RESPONSE 37

Laplace Transforms of Common SignalsName Time function, f .t/ Laplace tx., F.s/Unit impulse .t/ 1

Unit step 1.t/ 1s

Unit ramp t 1.t/ 1s2

nth order ramp tn 1.t/ n!snC1

Sine sin.bt/1.t/ bs2 C b2

Cosine cos.bt/1.t/ ss2 C b2

Damped sine eat sin.bt/1.t/ b.s C a/2 C b2

Damped cosine eat cos.bt/1.t/ s C a.s C a/2 C b2

Diverging sine t sin.bt/1.t/ 2bs.s2 C b2/2

Diverging cosine t cos.bt/1.t/ s2 b2

.s2 C b2/2

Properties of the Laplace Transform

Superposition: L fa f1.t/C b f2.t/g D aF1.s/C bF2.s/. Time delay: L f f .t /g D esF.s/. Time Scaling: L f f .at/g D 1jajF

sa

.

(useful if original equations are expressed poorly in time scale. e.g.,measuring disk-drive seek speed in hours). Differentiation:

L

n.

f .t/oD s F.s/ f .0/

L

n..

f .t/oD s2F.s/ s f .0/ .f .0/

L

f .m/.t/} D sm F.s/ sm1 f .0/ : : : f .m1/.0/:

DYNAMIC RESPONSE 38

Integration: LZ t

0f . / d

D 1

sF.s/.

Convolution: Recall that y.t/ D h.t/ u.t/Y .s/ D L fy.t/g D L fh.t/ u.t/gD L

Z tD0

h. /u.t / d

DZ 1

tD0

Z tD0

h. /u.t / d est dt

DZ 1D0

Z 1tD

h. /u.t / est dt d:

t

D t

Region ofintegration

Multiply by eses

Y .s/ DZ 1D0

h. /esZ 1

tDu.t /es.t/ dt d:

Let t 0 D t :Y .s/ D

Z 1D0

h. /es dZ 1

t 0D0u.t 0/est

0 dt 0

Y .s/ D H.s/U .s/: The Laplace transform unwraps convolution for general input

signals. Makes system easy to analyze.

The Inverse Laplace Transform

The inverse Laplace Transform converts F.s/! f .t/. Once we get an intuitive feel for F.s/, we wont need to do this often.

DYNAMIC RESPONSE 39

The main tool for ILT is partial-fraction-expansion.Assume : F.s/ D b0s

m C b1sm1 C C bmsn C aasn1 C C an

D kQm

iD1.s zi/QniD1.s pi/

.zeros/

.poles/

D c1s p1 C

c2

s p2 C Ccn

s pn if fpigdistinct:

so; .s p1/F.s/ D c1 C c2.s p1/s p2 C C

cn.s p1/s pn

let s D p1 : c1 D .s p1/F.s/jsDp1ci D .s pi/F.s/jsDpi

f .t/ DnX

iD1cie

pi t1.t/ since Lekt1.t/

D 1s k :

EXAMPLE: F.s/ D 5s2 C 3s C 2 D

5.s C 1/.s C 2/.

c1 D .s C 1/F.s/sD1D 5

s C 2sD1D 5

c2 D .s C 2/F.s/sD2D 5

s C 1sD2D 5

f .t/ D .5et 5e2t/1.t/: If F.s/ has repeated roots, we must modify the procedure. e.g.,

repeated three times:

F.s/ D k.s p1/3.s p2/

D c1;1s p1 C

c1;2

.s p1/2 Cc1;3

.s p1/3 Cc2

s p2 C

c1;3 D .s p1/3F.s/sDp1

DYNAMIC RESPONSE 310

c1;2 D

dds(.s p1/3F.s/

sDp1

c1;1 D 12

d2

ds2(.s p1/3F.s/

sDp1

cx;ki D 1i!

di

dsi(.s pi/k F.s/

sDpi

:

EXAMPLE: Find ILT of s C 3.s C 1/.s C 2/2 .

ans: f .t/ D .2et 2e2t te2t|{z}from repeated root.

/1.t/:

TEDIOUS. Use Matlab. e.g., F.s/ D 5

s2 C 3s C 2.

Example 1. Example 2.

>> Fnum = [0 0 5]; >> Fnum = [0 0 1 3];

>> Fden = [1 3 2]; >> Fden = conv([1 1],conv([1 2],[1 2]));

[r,p,k] = residue(Fnum,Fden); [r,p,k] = residue(Fnum,Fden);

r = -5 r = -2

5 -1

p = -2 2

-1 p = -2

k = [] -2

-1

k = []

When you use residue and get repeated roots, BE SURE to typehelp residue to correctly interpret the result.

DYNAMIC RESPONSE 311Using the Laplace Transform to Solve Problems

We can use the Laplace transform to solve both homogeneous andforced differential equations.

EXAMPLE:..

y.t/C y.t/ D 0; y.0/ D ; .y.0/ D :s2Y .s/ s C Y .s/ D 0

Y .s/.s2 C 1/ D s C Y .s/ D s C

s2 C 1D s

s2 C 1 C

s2 C 1:From tables, y.t/ D [ cos.t/C sin.t/]1.t/. If initial conditions are zero, things are very simple.

EXAMPLE:..

y.t/C5 .y.t/C4y.t/ D u.t/; y.0/ D 0; .y.0/ D 0; u.t/ D 2e2t1.t/:s2Y .s/C 5sY .s/C 4Y .s/ D 2

s C 2Y .s/ D 2

.s C 2/.s C 1/.s C 4/D 1

s C 2 C2=3

s C 1 C1=3

s C 4:

From tables, y.t/ De2t C 2

3et C 1

3e4t

1.t/.

Time Response vs. Pole Locations: 1st-Order Pole(s) (stable) Pole = root of denominator of H.s/ D b.s/=a.s/. Poles qualitatively determine the behavior of the system.


Zeros quantify this relationship.EXAMPLE: H.s/ D 1

s C h.t/ D e t1.t/:

If > 0, pole is at s < 0, STABLE i.e., impulse response decays, andany bounded input produces bounded output.

If < 0, pole is at s > 0, UNSTABLE. is time constant factor: D 1= .

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1impulse([0 1],[1 1]);

Time (sec )t D

1e

e t

h.t/

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1step([0 1],[1 1]);

Time (sec )t D

y.t/

KK .1 et= /

System response. K D dc gain

Response to initial condition! 0.

Time Response vs. Pole Locations: 2nd-Order Poles (stable)

H.s/ D b0s2 C a1s C a2 D

K!2ns2 C 2!ns C !2n

.standard form/:

D damping ratio:!n D natural frequency or undamped frequency:

h.t/ D !np1 2e

t .sin.!dt// 1.t/;

where; D !n;!d D !n

p1 2 D damped frequency:


D sin1. /=.s/


=.s/


tr = Rise time = time to reach vicinity of new set point. ts = Settling time = time for transients to decay (to 5%, 2%, 1%). Mp = Percent overshoot. tp = Time to peak.

Rise Time

All step responses rise in roughly the same amount of time (seepg. 3313.) Take D 0:5 to be average. time from 0:1 to 0:9 is approx !ntr D 1:8:

tr 1:8!n:

We could make this more accurate, but note: Only valid for 2nd-order systems with no zeros. Use this as approximate design rule of thumb and iterate design

until spec. is met.

Peak Time and Overshoot

Step response can be found from ILT of H.s/=s:y.t/ D 1 e t

cos.!dt/C

!dsin.!dt/

;

!d D !np

1 2; D !n: Peak occurs when .y.t/ D 0

.

y.t/ D e t

cos.!dt/C !d

sin.!dt/ e t .!d sin.!dt/C cos.!dt//

D e t 2

!dsin.!dt/C !d sin.!dt/

D 0:


So,!dtp D ;

tp D !dD !np

1 2 :

Mp D e=p

1 2 100. (common values: Mp D 16% for D 0:5; Mp D 5% for D 0:7). 0 0.2 0.4 0.6 0.8 1.00

102030405060708090

100

Mp,

%

Settling Time

Determined mostly by decaying exponentiale!n ts D : : : D 0:01; 0:02; or 0:05

EXAMPLE:

D 0:01e!n ts D 0:01!n ts D 4:6

ts D 4:6!nD 4:6

ts0:01 ts D 4:6=0:02 ts D 3:9=0:05 ts D 3:0=

Design Synthesis

Specifications on tr , ts, Mp determine pole locations. !n 1:8=tr . fn.Mp/: (read off of versus Mp graph on page 3316) 4:6=ts: (for examplesettling to 1%)


=.s/


Mp D 0; D 1: From transfer fn: 2 D 2!n !n D 1: !2n D 12 D 0:5Ka; Ka D 2:0 Note: ts D 4:6 seconds. We will need a better controller than this for a

pen plotter!

Time Response vs. Pole Locations: Higher Order Systems

We have looked at first-order and second-order systems withoutzeros, and with unity gain.

Non-unity gain

If we multiply by K , the dc gain is K . tr , ts, Mp, tp are not affected.Add a zero to a second-order system

H1.s/ =2

.s C 1/.s C 2/ H2.s/ =2.s C 1:1/

1:1.s C 1/.s C 2/=

2s C 1

2s C 2 =

21:1

0:1

s C 1 C0:9

s C 2

=

0:18s C 1 C

1:64s C 2

Same dc gain (at s D 0). Coefficient of .s C 1/ pole GREATLY reduced. General conclusion: a zero near a pole tends to cancel the effect of

that pole. How about transient response?

H.s/ D .s=!n/C 1.s=!n/2 C 2 s=!n C 1:


Zero at s D : Poles at


0 2 4 6 8 100.5

0

0.5

1

1.5

2

H .s/

Ho.s/

Hd.s/

Time (sec)

y.t/

2nd-order min-phase step resp.

0 2 4 6 8 101.5

1

0.5

0

0.5

1

1.5

Ho.s/

H .s/

Hd.s/

Time (sec)

y.t/

2nd-order nonmin-phase step resp.

Add a pole to a second order system

H.s/ D 1.s=!n C 1/[.s=!n/2 C 2 s=!n C 1]:

Original poles at


Extra zero in RHP depresses overshoot, and may cause stepresponse to start in wrong direction. DELAY .

Extra pole in LHP increases rise-time if extra pole is within a factor of 4 from the real part of complex poles.

UnstableregionDominantInsignificant

=.s/


Important tools: block diagram manipulation and Masons rule.

Block Diagram Manipulation

We have already seen block diagrams (see pg. 113). Shows information/energy flow in a system. When used with Laplace transforms, can simplify complex system

dynamics.

Four BASIC configurations:H .s/ Y .s/U.s/ Y .s/ D H .s/U.s/

H1.s/ H2.s/ Y .s/U.s/ Y .s/ D [H1.s/H2.s/] U.s/

H1.s/

H2.s/

Y .s/U.s/ Y .s/ D [H1.s/C H2.s/] U.s/

H1.s/

H2.s/

Y .s/R.s/U1.s/

Y2.s/ U2.s/

U1.s/ D R.s/ Y2.s/Y2.s/ D H2.s/H1.s/U1.s/

so; U1.s/ D R.s/ H2.s/H1.s/U1.s/D R.s/

1C H2.s/H1.s/Y .s/ D H1.s/U1.s/

D H1.s/1C H2.s/H1.s/R.s/

Alternate representation

DYNAMIC RESPONSE 323H .s/

H1.s/

H1.s/

H1.s/ H2.s/

H2.s/

H2.s/Y .s/

Y .s/

Y .s/

Y .s/

U.s/

U.s/

U.s/

R.s/

EXAMPLE: Recall dc generator dynamics from page 2210R f Ra Lai f .t/ ia.t/

e f .t/ L f eg.t/ Zlea.t/

| {z }Fieldcircuit

| {z }Rotorcircuit

| {z }Loadcircuit

e f .t/ ea.t/i f .t/ eg.t/ ia.t/Field

circuit KgRotorcircuit Zl

Compute the transfer functions of the four blocks.

e f D R f i f C L f ddt i fE f .s/ D R f I f .s/C L f s I f .s/I f .s/E f .s/

D 1R f C L f s :

eg.t/ D Kgi f .t/Eg.s/ D Kg I f .s/Eg.s/I f .s/

D Kg:


ea.t/ D ia.t/ZlEa.s/ D Zl Ia.s/Ea.s/Ia.s/

D Zl:

eg.t/ D Raia.t/C La ddt ia.t/C ea.t/Eg.s/ D Ra Ia.s/C Las Ia.s/C Ea.s/

D .Ra C Las C Zl/ Ia.s/Ia.s/Eg.s/

D 1Las C Ra C Zl :

Put everything together.Ea.s/E f .s/

D Ea.s/Ia.s/

Ia.s/Eg.s/

Eg.s/I f .s/

I f .s/E f .s/

D Kg Zl(L f s C R f

.Las C Ra C Zl/:

Block Diagram Algebra

()

()

()

H .s/

H .s/

H .s/H .s/

H .s/

1H .s/

H1.s/H1.s/ H2.s/

H2.s/

1H2.s/

Y .s/Y .s/

Y .s/Y .s/

Y1.s/Y1.s/

Y2.s/Y2.s/

U.s/U.s/

U1.s/U1.s/

U2.s/U2.s/

R.s/R.s/

Unity Feedback


EXAMPLE: Simplify:

H1.s/ H2.s/

H3.s/

H4.s/

H5.s/

H6.s/

Y .s/R.s/

H1.s/1 H1.s/H3.s/ H2.s/

H4.s/

H5.s/

H6.s/

Y .s/R.s/

H1.s/1 H1.s/H3.s/ H2.s/

H4.s/

H5.s/

H6.s/H2.s/

Y .s/R.s/

| {z }H1.s/H2.s/

1H1.s/H3.s/

1CH1.s/H2.s/H4.s/1H1.s/H3.s/

| {z }

H5.s/CH6.s/H2.s/

H1.s/H2.s/H5.s/C H1.s/H6.s/1 H1.s/H3.s/C H1.s/H2.s/H4.s/R.s/ Y .s/

DYNAMIC RESPONSE 326Masons Rule

If you dont care to simplify a block diagram using block diagrammanipulation, you can use Masons Rule.

Node = Common input to several blocks, or output of summingjunction.

H1.s/

H2.s/

Y .s/U.s/

Path = Sequence of connected blocks, from one variable to another,in the direction of signal flow, without including any variable more thanonce.

Forward Path = Path from input to output. Loop = Path from node back to itself. Path Gain = Product of all transfer functions in a path. Loop Gain = Product of all transfer functions in a loop. Nontouching = Two loops are nontouching if they have no nodes in

common.

G1 G2 G3 G4 G5

G6

H1 H2YR

2 Loops: , Nontouching. 2 Forward paths: Touches both loops.

Does not touch first loop, but touches second loop.


Masons RuleT D 1

1

pXkD1

Mk1k D 11

(M111 C M212 C C Mp1p

1 D 1 .sum of ALL .touching or not/ individual loop gains/C.sum of the products of loop gains of all possiblecombinations of nontouching loops taken two at a time/.sum of the products of loop gains of all possiblecombinations of nontouching loops taken three at a time/C

Mk D Path gain of kth forward path1k D Value of 1 for that part of the flow graph not touching the kth

forward path:

Key to using rule = organization! Above example:

Find the loop- and the forward path-gains. Loops: L1 : ; Gain=H1G2. L2 : ; Gain=H2G4.

Forward Paths: M1 : ; Gain=G1G2G3G4G5 M2 : ; Gain=G6G4G5.

1 D 1 .L1 C L2/C L1L2 D 1C H1G2 C H2G4 C H1 H2G2G4. 1k is easiest determined by redrawing flowgraph without path k.


G1 G2 G3

G6

H1

H1

H2

H2

Path 1 removed. No loops.11 D 1 0

Path 2 removed. One loop withgain G2 H1. 12 D 1 .G2 H1/

T D M111 C M2121

D G1G2G3G4G5 C G6G4G5.1C H1G2/1C H1G2 C H2G4 C H1 H2G2G4 :

Masons rule may be shorter that block-diagram manipulation. Use the method with which you are most comfortable.


BASIC PROPERTIES OF FEEDBACK

Two basic types of control systems:OPEN LOOP:

Ctrlrr .t/ y.t/

Disturbance

Plant

CLOSED LOOP:

Ctrlrr .t/ y.t/

Disturbance

Plant

Sensor

We will compare these systems in a number of ways: disturbancerejection, sensitivity, dynamic tracking, steady state error and stability.

DC Motor Speed Control

Recall equations of motion for a dc motor but add loadtorque.

ea.t/ eb.t/

ia.t/ Ra La b

J.t/; .t/

| {z }Armature

| {z }Load

l , load

Assume that we are trying to control motor speed:

BASIC PROPERTIES OF FEEDBACK 42

J..

C b .ke

.

C La diadt C RaiaDD

k ia C lea

9=; let output y D

.

;

disturbance w 4D l:

J.

y C bykey C La diadt C Raia

DD

k ia C wea

9=; s JY .s/C bY .s/ D k Ia.s/CW .s/keY .s/C sLa Ia.s/C Ra Ia.s/ D Ea.s/

.J Lass C bLas C J Ras C bRa C kke/Y .s/ D kEa.s/C .Ra C Las/W .s/ This can be re-written as (magic happens)

.1s C 1/.2s C 1/Y .s/ D AEa.s/C BW .s/1 Ra J=.kke/ D mechanical time constant2 La=Ra D electrical time constantA D k=.bRa C kke/B D Ra=.bRa C kke/

So,Y .s/ D A

.1s C 1/.2s C 1/Ea.s/CB

.1s C 1/.2s C 1/W .s/: If ea.t/ D ea 1.t/ (constant) and w.t/ D w 1.t/ (constant), What is

steady state output?

Recall Laplace-transform final value theorem:

If a signal has a constant final value, it may be found asyss D lim

s!0sY .s/:

So,Ea.s/ D ea

s; W .s/ D w

s; yss D Aea C Bw:

This is the response of the open-loop system (without a controller).


Ctrlrr .t/ y.t/

Dist. w.t/

A.1s C 1/.2s C 1/

BA

ea.t/

Motor

Lets make a simple controller for the open-loop system.ea.t/ D Kolr.t/; .gain of Kol/

Choose Kol so that there is no steady-state error when w D 0.yss D AKolrss C Bwss) Kol D 1=A:

Is closed-loop any better?

Ctrlrr .t/ y.t/

Dist. w.t/

A.1s C 1/.2s C 1/

BA

ea.t/

Tachometer

1

Lets make a similar controller for the closed-loop system (with thepossibility of a different value of K .)

ea.t/ D Kcl.r.t/ y.t// The transfer function for the closed-loop system is:

Y .s/ D AKcl.1s C 1/.2s C 1/ .R.s/ Y .s//C

B.1s C 1/.2s C 1/W .s/

D AKcl.1s C 1/.2s C 1/C AKcl R.s/C

B.1s C 1/.2s C 1/C AKcl W .s/

yss D AKcl1C AKcl rss; assuming w D 0:


If AKcl 1; yss rss. Open-loop with load:

yss D AKolrss C Bwss D rss C Bwssy D Bwss:

Closed-loop with load:yss D AKcl1C AKcl rss C

B1C AKclwss

y B1C AKclwss:

which is much better than open-loop since AKcl 1.ADVANTAGE OF FEEDBACK: Better disturbance rejection (by factor of1C AKcl).

Sensitivity

The steady-state gain of the open-loop system is: 1.0 How does this change if the motor constant A changes?

A! A C AGol C Gol D Kol.A C A/

D 1A.A C A/

D 1C AA|{z}

gain error

:

In relative terms: GolGolD A

AD 1:0|{z}sensitivity

AA:

Therefore, a 10% change in A 10% change in gain. Sensitivity=1.0.


Steady-state gain of closed-loop system is: AKcl1C AKcl :

Gcl C Gcl D .A C A/Kcl1C .A C A/Kcl : From calculus (law of total differential)

Gcl D dGcld A Aor

GclGclD

A

GcldGcld A

| {z }

sensitivity SGclA

AA

SGclA DA

AKcl=.1C AKcl/.1C AKcl/Kcl Kcl.AKcl/

.1C AKcl/2

D 11C AKcl :

ADVANTAGE OF FEEDBACK: Lower sensitivity to modeling error (by afactor of 1C AKcl)

Dynamic Tracking

Steady-state response of closed-loop better than open-loop: Betterdisturbance rejection, better (lower) sensitivity. What about transient response? Open-loop system: Poles at roots of .1s C 1/.2s C 1/

s D 1=1; s D 1=2: Closed-loop system: Poles at roots of .1s C 1/.2s C 1/C AKcl:

s D .1 C 2/p.1 C 2/2 412.1C AKcl/

212:


FEEDBACK MOVES POLES System may have faster/slower response System may be more/less damped System may become unstable!!! Often a high gain Kcl results in instability. We need design tools to

help us design the dynamic response of the closed-loop system. For this dc motor example, we can get step responses of the

following form:

0 2 4 6 8 100

0.5

1

1.5

High Kcl

Low Kcl

Time (sec)

y.t/

PID Control

General control setup:

D.s/R.s/ Y .s/

Disturbance

G.s/

Need to design controller D.s/. One option is PID (Proportional Integral Derivative) control design.


Extremely popular. 90C% of all controllers are PID. Doesnt mean that they are great, just popular.

We just saw proportional control where u.t/ D K e.t/, or D.s/ D K . Proportional control allows non-zero steady state error.

Increases speed of response. Larger transient overshoot.

Integral control eliminates steady state error. D.s/ D K Is:

Dynamic response gets worse.

Derivative control damps dynamic response. D.s/ D K Ds.Proportional Control

u.t/ D K .r.t/ y.t// D K e.t/ : : : D.s/ D K : May have steady state error, may not be able to completely reject a

constant disturbance.

EXAMPLE: Determine behavior of closed loop poles for the dc motor.

Y .s/R.s/

D AK.1s C 1/.2s C 1/C AK :

Poles are roots of .1s C 1/.2s C 1/C AK : Without feedback, K ! 0.

s1 D 1=1; s2 D 1=2: With feedback,

s1; s2 D .1 C 2/p.1 C 2/2 412.1C AK /

212:


K D 0K D 0 11

12

K D .1 2/2

412 A

1 C 2212


12..

y.t/C .1 C 2/.

y.t/C y.t/ D A

KTI

Z t0.r. / y. // d

C Bw.t/:

Differentiate,

12...

y.t/C .1 C 2/..

y.t/C .y.t/ D AKTI.r.t/ y.t//C B .w.t/

12...

y.t/C .1 C 2/..

y.t/C .y.t/C AKTI

y.t/ D AKTI

r.t/C B .w.t/:

If r.t/ D cst, w.t/ D cst, .w.t/ D 0,AKTI

yss D AKTI rss no error: Steady-state improves Dynamic response degrades.

Very oscillatory. Possibly unstable.

Can be improved by adding propor-tional term to integral term.

K D 0K D 0 11

12


TD = Derivative time. PURE DERIVATIVE CONTROL IMPRACTICAL SINCE DERIVATIVE

MAGNIFIES SENSOR NOISE! Practical version = Lead control, which we will study later. Stabilizes system. Does nothing to reduce constant error! If .e.t/ D 0, then u.t/ D 0. Motor Control: Poles at roots of 12s2 C .1 C 2 C AK TD/s C 1 D 0.

TD enters term, can make damping better.

PD = Proportional plus derivative controlD.s/ D K .1C TDs/

Root locus for dc motor, PD control.

11

12

1TD


For speed control problem,u.t/ D K

.r.t/ y.t//C 1

TI

Z t0.r. / y. // d C TD. .r.t/

.

y.t//:

(math happens). Solve for poles12TI s3 C TI ..1 C 2/C AK TD/s2 C TI .1C AK /s C AK D 0

s3 C1 C 2 C AK TD

12

s2 C

1C AK12

s C AK

12TID 0:

Three coefficients, three parameters. We can put poles anywhere!Complete control of dynamics in this case. Entire transfer functions are:

Y .s/W .s/

D TI BsTI12s3 C TI .1 C 2/s2 C TI .1C AK /s C AK :

Y .s/R.s/

D AK .TI s C 1/TI12s3 C TI .1 C 2/s2 C TI .1C AK /s C AK :

We can plot responses in MATLAB:num = [TI*B 0];den = [TI*TAU1*TAU2 TI*(TAU1+TAU2) TI*(1+A*K) A*K];step(num, den)

0 0.01 0.02 0.03 0.04 0.050

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time (sec)

y.t/

(rad/

sec) P

PI

PID

Step Reference Response

0 0.01 0.02 0.03 0.04 0.050

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time (sec)

y.t/

(rad/

sec) P

PIPID

Step Disturbance Response

Ziegler - Nichols Tuning of PID Controllers

Rules of Thumb for selecting K , TI , TD.


Not optimal in any sensejust provide good performance.METHOD I: If system has step response like this,

Slope, A=A

d

Y .s/U .s/

D Aeds

s C 1;(first-order system plus delay)

We can easily identify A, d, from this step response. Dont need complex model! Tuning criteria: Ripple in impulse response decays to 25% of its value

in one period of ripple

Period1

0:25

RESULTING TUNING RULES:P PI PID

K D Ad

K D 0:9Ad

TI D d0:3

K D 1:2Ad

TI D 2dTD D 0:5d

METHOD II: Configure system as

Kur .t/ y.t/Plant

Turn up gain Ku until system produces oscillations (on stabilityboundary) Ku = ultimate gain.


Period, Pu1 RESULTING TUNING RULES:

P PI PID

K D 0:5KuK D 0:45KuTI D 11:2 Pu

K D 0:6KuTI D 0:5PuTD D Pu8

Practical Problem: Integrator Overload

Integrator in PI or PID control can cause problems. For example, suppose there is saturation in the actuator.

Error will not decrease. Integrator will integrate a constant error and its value will blow up.

Solution = integrator anti-windup. Turn off integration when actuatorsaturates.

K

KaK

TI s

e.t/ u.t/ umin

umax

Doing this is NECESSARY in any practical implementation. Omission leads to bad response, instability.


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time (sec)

Step Response

With antiwindup

Without antiwindup

y.t/

0 2 4 6 8 100.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

Time (sec)

Control Effort

With antiwindup

Without antiwindup

u.t/

Steady-State Error

System error is any difference between r.t/ and y.t/. Two sources:

1. Imprecise tracking of r.t/.2. Disturbance affecting the system output.

Steady-State Error (w.r.t. Reference Input) We have already seen examples of CL systems that have some

tracking error (proportional ctrl) or not (integral ctrl) to a step input.We will formalize this concept here.

Start with very general control structure:T .s/r .t/ y.t/

The closed-loop transfer function for the whole system,T .s/ D Y .s/

R.s/:

The error isE.s/ D R.s/ Y .s/


D R.s/ T .s/R.s/D [1 T .s/] R.s/:

Assume conditions of final value theorem are satisfied (i.e., T .s/ isstable)

ess D limt!1 e.t/ D lims!0 s[1 T .s/]R.s/:

We use test inputs of the type: r.t/ D tk

k!1.t/; R.s/ D 1

skC1.

t

1.t/

t

t 1.t/

t

t2

2 1.t/

As k increases, tracking is progressively harder.

ess D lims!0

1 T .s/sk

D

8>:

0; type > K Iconstant; type D K I1; type < K :

If system type = 0, constant steady-state error for step input, infinites.s. error for ramp or parabolic input.

If system type = 1, no steady-state error for step input, constant s.s.error for ramp input, infinite s.s. error for parabolic input.

If system type = 2, no steady-state error for step or ramp inputs,constant s.s. error for parabolic inputs.

And so forth, for higher-order system types. Ramp responses of different system types:


t

tType 0 System

t

tType 1 System

t

tType 2 System

DANGER: Higher order sounds better but they are harder to stabilizeand design. Transient response may be poor.

UNITY FEEDBACK: SPECIAL CASE

NOTE: The following method is a special case of the above generalmethod. Be careful to use the apropriate method for the problem athand!

Unity-feedback is when the control system looks like:Gol.s/r .t/ y.t/

There are some important simplifications:T .s/ D Gol.s/

1C Gol.s/1 T .s/ D

1C Gol.s/1C Gol.s/

Gol.s/

1C Gol.s/D 1

1C Gol.s/:So,

E.s/ D 11C Gol.s/R.s/:


For the test inputs R.s/ D 1skC1

ess D lims!0

s E.s/

D lims!0

1[1C Gol.s/]sk

For type 0,ess D lim

s!01

1C Gol.s/ D1

1C K p ; K p D lims!0 Gol.s/:


s!01

1C Gol.s/1sD lim

s!01

sGol.s/D 1

Kv; Kv D lim

s!0sGol.s/:


s!01

1C Gol.s/1s2D lim

s!01

s2Gol.s/D 1

Ka; Ka D lim

s!0s2Gol.s/:

These formulas only meaningful for unity-feedback!K p D lim

s!0Gol.s/: position error constant

Kv D lims!0

sGol.s/: velocity error constantKa D lim

s!0s2Gol.s/: acceleration error constant

Steady-state tracking errors ess for unity-feedback case ONLY.Sys. Type Step Input Ramp Input Parabola Input

Type 0 11C K p 1 1

Type 1 0 1Kv

1Type 2 0 0 1

Ka

BASIC PROPERTIES OF FEEDBACK 418EXAMPLES:

(1) Consider Gol.s/ D s C 1.s C 2/.s C 3/; System type?

Gol.0/ D 12 3 D16:

Therefore, typeD 0, ess to unit stepD 11C 1=6 D67:

(2) Consider Gol.s/ D .s C 1/.s C 10/.s 5/.s2 C 3s/.s4C s2 C 1/ ; System type?

Gol.0/ D 1 10 .5/0 1 D 1 Type > 0:

sGol.s/ D .s C 1/.s C 10/.s 5/.s C 3/.s4 C s2 C 1/

sGol.s/jsD0 D 1 10 .5/3 1 D50

3:

Therefore, typeD 1, ess to unit rampD 350 :

(3) Consider Gol.s/ D s2 C 2s C 1

s4 C 3s3 C 2s2 ; System type?.

Gol.0/ D 10 D 1 Type > 0

sGol.s/ D s2 C 2s C 1

s3 C 3s2 C 2ssGol.s/jsD0 D 10 D 1 Type > 1

s2Gol.s/ D s2 C 2s C 1

s2 C 3s C 2s2Gol.s/

sD0 D

12

Type D 2:


Therefore, typeD 2, ess to unit parabolaD 2.KEY POINT: Open-loop G.s/ D Gol.s/ tells us about closed-loop s.s.

response.

EXAMPLE: DC-motor example with Proportional control.

Ctrlrr .t/ y.t/A

.1s C 1/.2s C 1/

Controller: D.s/ D K .D.s/G.s/ D K A

.1s C 1/.2s C 1/; lims!0 D.s/G.s/ D K A:

So system is type 0, with s.s. error to step input of 11C K A : (This

agrees with prior results.)EXAMPLE: DC-motor example with PI control

Controller: D.s/ D K

1C 1TI s

:

D.s/G.s/ D K A CK ATI s

.1s C 1/.2s C 1/:lims!0

D.s/G.s/ D 1

lims!0

s D.s/G.s/ D K ATI:

System is type 1, with s.s. error to ramp input of TIK A

.

EXAMPLE: DC-motor with another integrator


Controller: D.s/ D K

1C 1TI sC 1

TI s2

:

D.s/G.s/ D K A CK ATI sC K ATI s2

.1s C 1/.2s C 1/lims!0

s2D.s/G.s/ D K ATI:

System is type 2, with s.s. error to parabolic input of TIK A

.

Steady State Error (w.r.t. Disturbance) Recall, system error is any difference between r.t/ and y.t/. One source of system error is disturbance. Can find system type with respect to disturbance.

Y .s/W .s/

D Tw.s/such that

Y .s/ D T .s/R.s/C Tw.s/W .s/: We do not wish the output to have ANY disturbance term in it, so the

output error due to the disturbance is equal to the output due to thedisturbance.

ess D yss D lims!0

sTw.s/W .s/:

Type 0 system (w.r.t. disturbance) has constant lims!0

Tw.s/:


Tw.s/s

:


Tw.s/s2

:


STABILITY ANALYSIS

Many classifications of stability in system analysis. For LTI systems, all are basically the same BIBO.

Stable Neutral Unstable

If input is bounded: ju.t/j < K1, then output is also boundedjy.t/j < K2. In the time domain,

y.t/ DZ 11

h. /u.t / d

jy.t/j DZ 11

h. /u.t / d

STABILITY ANALYSIS 52

Z 11jh. /j ju.t /j d

K1Z 11jh. /j d:

So if a system is BIBO stable,Z 11jh. /j d < 1.

y.t/u.t/ C Stable? Note: .h.t/ D 1.t//.

In the Laplace domain,H.s/ D Y .s/

R.s/D K

Qm1 .s zi/Qn

1.s pi/m n:

(assume poles unique)h.t/ D

nXiD1

kiepi t

Z 11jh. /j d


EXAMPLE: T .s/ D 2s2 C 3s C 2. Is T .s/ stable?

T .s/ D 2.s C 1/.s C 2/;

Roots at s D 1; s D 2: Stable!EXAMPLE: T .s/ D 10s C 24

s3 C 2s2 11s 12 . Is T .s/ stable?

T .s/ D 10.s C 2:4/.s C 1/.s 3/.s C 4/;

Roots at s D 1, s D C3, s D 4. Unstable!EXAMPLE: T .s/ D s

s2 C 1. Is T .s/ stable?

T .s/ D s.s j/.s C j/;

Roots at s D j . MARGINALLY stable. Use input = sin.t/.

Y .s/ D ss2 C 1

1s2 C 1 D

s

.s2 C 1/2y.t/ D t sin.t/ : : : unbounded:

MARGINALLY stable = unstable (bounded impulse response, butunbounded output for some inputs.)

Routh Hurwitz Stability Criterion

Factoring high-degree polynomials to find roots is tedious andnumerically not well conditioned.

Want other stability tests, and also MARGINS of stability.


.1/ Routh test

.2/ Root locus

.3/ Nyquist test

.4/ Bode stability margins

9>>>>=>>>>;

Can you tell this isan important topic?

In 1868, Maxwell found conditions on the coefficients of a transferfunction polynomial of 2nd and 3rd order to guarantee stability.

It became the subject of the 1877 Adams Prize to determineconditions for stability for higher-order polynomials.

Routh won this prize, and the method is still useful.Case 0 Consider the denominator a.s/.

2nd order:a.s/ D s2 C a1s C a0 D .s p1/.s p2/ D s2 .p1 C p2/s C p1 p2:3rd order:

a.s/ D s3 C a2s2 C a1s C a0D .s p1/.s p2/.s p3/D s2 .p1 C p2/s C p1 p2 s p3D s3 .p1 C p2 C p3/s2 C .p1 p2 C p1 p3 C p2 p3/s p1 p2 p3:

TREND: Stability none of the coefficients ai can be 0.Does this trend continue?


an1 D .1/ sum of all roots.an2 D sum of products of roots taken two at a time.an3 D .1/ sum of products of roots taken three at a time.

:::

a0 D .1/n product of all roots. Conclusions:

1. If any coefficient ai D 0, not all roots in LHP.2. If any coefficient ai < 0, at least one root in RHP.

TEST: If any coefficient ai 0, system is unstable.EXAMPLE: a.s/ D s2 C 0s C 1. From conclusion (1), not all roots in LHP. Roots at s D j . Marginally stable.

EXAMPLE: a.s/ D s3 C 2s2 11s 12. From conclusion (2), at least one root is in RHP. Roots at s D 1, s D 4, s D C3. Unstable.

EXAMPLE: a.s/ D s3 C s2 C 2s C 8: We dont know yet if this system is stable or not.

Roots at s D 2, s D 12 jp

152

. Unstable, but how do we find outwithout factoring?


Case 1 Once we have determined that ai 0 8 i , we need to run the Routh

test.

Very mechanical, not intuitive. Proof difficult. a.s/ D sn C an1sn1 C C a1s C a0: Form Routh Array.

sn an an2 an4 sn1 an1 an3 an5 sn2 b1 b2 sn3 c1 c2 :::

s1 j1s0 k1

b1 D 1an1

an an2an1 an3

b2 D 1an1

an an4an1 an5

c1 D 1b1

an1 an3b1 b2

c2 D 1b1

an1 an5b1 b3

TEST: Number of unstable roots = number of sign changes in left column.

EXAMPLE: a.s/ D s3 C s2 C 2s C 8.

s3 1 2s2 1 8s1 6s0 8

b1 D 11

1 21 8 D .8 2/ D 6

c1 D 16

1 86 0 D 16.0C 68/ D 8

Two sign changes (1!6; 6! 8) Two roots in RHP.


EXAMPLE: a.s/ D s2 C a1s C a0.

s2 1 a0s1 a1

s0 a0

b1 D 1a1

1 a0a1 0 D 1a1 .0 a0a1/ D a0

Stable iff a1 > 0, a0 > 0.EXAMPLE: a.s/ D s3 C a2s2 C a1s C a0.

s3 1 a1s2 a2 a0

s1 a1 a0a2

s0 a0

b1 D 1a2

1 a1a2 a0 D 1a2 .a0 a1a2/D a1 a0

a2

c1 D 1b1

a2 a0b1 0 D 1b1 .a0b1/ D a0:

Stable iff a2 > 0, a0 > 0, a1 > a0a2

.

Case 2 Sometimes find an element in first column = 0 0

0(!)

Replace with as ! 0.


EXAMPLE: a.s/ D s5 C 2s4 C 2s3 C 4s2 C 11s C 10.

s5 1 2 11s4 2 4 10s3 0 6

new s3 6

s212

10

s1 6s0 10

If > 0, two sign changes.If < 0, two sign changes.

Two poles in RHP.

b1 D 12

1 22 4 D 12 .4 4/ D 0

b2 D 12

1 112 10 D 12 .10 22/ D 6

c1 D 1

2 4 6 D 1 .12 4/ 12

c2 D 1

2 10 0 D 1 .0 10/ D 10

d1 D 12

612

10

D

12.10 C 72

/ 6

e1 D 16

12

10

6 0

D16.0 60/ D 10

Case 3 Sometimes an entire row in Routh array = 0. This means that polynomial factors such that one factor has

conjugate-MIRROR-roots.=.s/


Complete Routh array by making polynomial a1.s/ from last non-zerorow in array. Poly has even orders of s only!!

a1.s/ is a factor of a.s/. It is missing some orders of s, so IS NOTSTABLE (by case 0.) We want to see if it has roots in the RHP or only on the j!-axis. Replace zero row with da1.s/

dsand continue.

EXAMPLE: a.s/ D s4 C s3 C 3s2 C 2s C 2.

s4 1 3 2s3 1 2 0s2 1 2s1 0

new s1 2s0 2

No sign changes in Routharray. So, no RHP roots.Roots of a1.s/ at s D j

p2.

Marginally stable.

b1 D 11

1 31 2 D 1

b2 D 11

1 21 0 D 2

c1 D 11

1 21 2 D 0

a1.s/ D s2 C 2da1.s/

dsD 2s

d1 D 12

1 22 0 D 2


EXAMPLE: a.s/ D s4 C 4 (Hard).

s4 1 0 4s3 0 0

new s3 4 0s2 0 4

new s2 4

s116

s0 4

Two sign changes. Therefore2 RHP poles. Other twopoles are mirrors in LHP.

a1.s/ D s4 C 4I da1.s/ds D 4s3:

b1 D 14

1 04 0 D 0

b2 D 14

1 44 0 D 4

c1 D 1

4 0 4 D 16

d1 D 16

416

0

D 4

Design Tool

Consider the system:

r .t/ y.t/Ks C 1

s.s 1/.s C 6/

T .s/ D Y .s/R.s/

D KsC1

s.s1/.sC6/1C K sC1

s.s1/.sC6/

D K .s C 1/s.s 1/.s C 6/C K .s C 1/

a.s/ D s.s 1/.s C 6/C K .s C 1/


D s3 C 5s2 C .K 6/s C K :

s3 1 K 6s2 5 K

s14K 30

5s0 K

For stability of theclosed-loop system, K > 0,and K > 30=4.

b1 D 15

1 K 65 K D 15 .K 5.K 6//D .4K 30/=5:

c1 D 1b1

5 Kb1 0 D 1b1 .b1K / D K

Step response for different values of K .

0 2 4 6 8 10 120.5

0

0.5

1

1.5

2

2.5

K =7:5K =13K =25

Ampl

itude

Time (sec.)EXAMPLE:

r .t/ y.t/K

1C 1TI s

1

.s C 1/.s C 2/


T .s/ D Y .s/R.s/

DK TI sCK

s.sC1/.sC2/1C K TI sCK

s.sC1/.sC2/

D K TI s C Ks.s C 1/.s C 2/C K TI s C K :

a.s/ D s3 C 3s2 C .2C K TI /s C K .

s3 1 2C K TIs2 3 K

s16C 3K TI K

3s0 K

b1 D 13

1 2C K TI3 K D 13 .K 3.2C K TI //

c1 D 1b1

3 Kb1 0 D 1b1 .b1K / D K

For stability of the closed-loop system, 0 < K < 61 3TI for TI 0 for TI >13

.

0 2 4 6 8 10 12

0

0.2

0.4

0.6

0.8

1

1.2

K D 3; TI D 0:5

K D 1; TI D 0:25

K D 1; TI D 1Ampl

itude

Time (sec.)


ROOT-LOCUS ANALYSIS

Recall step response: we have seen that pole locations in the systemtransfer function determine performance characteristics; such as risetime, overshoot, settling time.

We have also seen that feedback can change pole locations in thesystem transfer function and therefore performance is changed.

Suppose that we have one variable parameter in our control system.Then, we make a parametric plot of pole locations as that parameterchanges. The poles are the roots of the denominator of the transferfunction (a.k.a. the characteristic equation.) This plot is a plot of thelocus of the roots or the ROOT LOCUS PLOT (First suggested byEvans.)

VERY IMPORTANT NOTE: Root locus is a parametric plot (vs. K ) of theroots of an equation

1C K b.s/a.s/D 0:

A common control configuration is

r .t/ y.t/K G.s/

Unity feedback, proportional gain. We will generalize control configuration later.

ROOT-LOCUS ANALYSIS 62

Closed-loop transfer function

T .s/ D K G.s/1C K G.s/:

Poles = roots of 1C K G.s/ D 0. Assume plant transfer function G.s/ is rational polynomial:

G.s/ D b.s/a.s/

; such that

b.s/ D .s z1/.s z2/ .s zm/ .b.s/ is monic:/a.s/ D .s p1/.s p2/ .s pn/ n m .a.s/ is monic:/

[a.s/ may be assumed monic without loss of generality. If b.s/ is notmonic, then its gain is just absorbed as part of K in 1C K G.s/ D 0] zi are zeros of G.s/, the OPEN-LOOP transfer function. pi are poles of G.s/, the OPEN-LOOP transfer function. CLOSED-LOOP poles are roots of equation

1C K G.s/ D 0a.s/C K b.s/ D 0;

which clearly move as a function of K .

Zeros are unaffected by feedback.EXAMPLE: G.s/ D 1

s.s C 2/: Find the root locus. a.s/ D s.s C 2/; b.s/ D 1. Locus of roots: (aside, stable for all K > 0 .)

s.s C 2/C K D 0s2 C 2s C K D 0:


For this simple system we can easily find the roots.s1;2 D 2

p4 4K

2D 1p1 K :

Roots are real and negative for 0 < K < 1. Roots are complex conjugates for K > 1.

s1; K D 0s2; K D 0K D 1

=.s/

=>; K D 2

or we can locate the point on the root locus wherej


[A value of s D s1 is on the (0) complementary root locus iff1C K G.s1/ D 0 for some REAL value of K , 1 < K < 0.] So, K D 1

G.s/ Note, G.s/ is complex, so this is really two equations!

jK j D 1G.s/

(6.1)6 G.s/ D 6

1K

: (6.2)

Since K is real and positive, 6 K D 0.Therefore, 6 G.s/ D 180 l360; l D 0; 1; 2; : : : So once we know a point on the root locus, we can use the

magnitude equation 6.1 to find the gain K that produced it. We will use the angle equation 6.2 to plot the locus. i.e., the locus of

the roots = all points on s-plane where 6 G.s/ D 180 l360:NOTES:

1. We will derive techniques so that we dont need to test every point onthe s-plane!

2. The angle criteria explains why the root locus is sometimes called the180 root locus.

3. Similarly, the complementary root locus is also called the 0 rootlocus.

KEY TOOL: For any point on the s-plane6 G.s/ D

X6 .due to zeros/

X6 .due to poles/:

EXAMPLE: G.s/ D .s z1/.s p1/.s p2/:


p1p2z1

Test points1

=.s/


If the test point s1 is to the left of 1 pole and 1 zero: 6 G.s1/ D 180 .180/ D 360 D 0. 2 poles: 6 G.s1/ D 180 180 D 360 D 0. 2 zeros: 6 G.s1/ D 180 C 180 D 360 D 0.

NOT ON THE LOCUS.

GENERAL RULE #1All points on the real axis to the left of an odd number of poles and zerosare part of the root locus.

EXAMPLE: G.s/ D 1s.s C 4C 4 j/.s C 4 4 j/:

=.s/

ROOT-LOCUS ANALYSIS 67Locus Not on the Real Axis

OBSERVATION: Because we assume that system models arerational-polynomial with real coefficients, all poles must be either realor complex conjugate pairs. Therefore, THE ROOT LOCUS ISSYMMETRICAL WITH RESPECT TO THE REAL AXIS. What happens at K D 0,1?

1C K G.s/ D 1C K .s z1/.s z2/ .s zm/.s p1/.s p2/ .s pn/ D 0

.s p1/.s p2/ .s pn/C K .s z1/.s z2/ .s zm/ D 0: At K D 0 the closed loop poles equal the open-loop poles. As K approaches1

.s p1/.s p2/ .s pn/K

C .s z1/.s z2/ .s zm/ D 0Ithe n poles approach the zeros of the open-loop transfer function,INCLUDING THE n m ZEROS AT C1. Plug s D 1 into G.s/ and notice that it equals zero if m < n. The idea of1 in the complex plane is a number with infinite

magnitude and some angle. To find where the n m remaining poles go as K !1, consider that

the m finite zeros have canceled m of the poles. Looking back at theremaining n m poles (standing at C1), we have approximately

1C K 1.s /nm ;

or n m poles clustered/centered at . We need to determine , the center of the locus, and the directions

that the poles take.


Assume s1 D Re j is on the locus, R large and fixed, variable. Weuse geometry to see what must be for s1 to be on the locus.

Since all of the open-loop poles are at approximately the same place, the angle of 1C K G.s1/ is 180 if the n m angles from to s1 sumto 180.

.n m/l D 180 C l360; l D 0; 1; : : : ; n m 1or

l D 180 C 360.l 1/

n m ; l D 1; 2; : : : ; n m: So, if

n m D 1I D C180:

There is one pole going to C1along the negative real axis:

n m D 2I D 90:

There are two poles going toC1 vertically:

n m D 3I D 60; 180:

One goes left and the other twogo at plus and minus 60:

(etc)

To find the center, , reconsider our approximation b.s/a.s/ 1.s /nm

as jsj ! 1. Divide (by long division) b.s/

a.s/and 1

.s /nm and match the toptwo coefficients of s, the dominant ones.

An easier way...


Note that roots of denominator of G.s/ satisfy:sn C a1sn1 C a2sn2 C C an D .s p1/.s p2/ .s pn/

a1 D X

pi :

Note that the roots of the denominator of T .s/ aresn C a1sn1 C a2sn2 C C an C K .sm C b1sm1 C C bm/ D 0:

If n m > 1 then the .n 1/st coefficient of the closed loop systemis such that a1 D

Xri where ri are the closed-loop poles.

We know that m poles go to the zeros of G.s/, and assume theother n m are clustered at 1

.s /nm . Therefore, the asymptoticsum of roots is .n m/

Xzi :

Putting this all together,Xri D .n m/ C

Xzi D

Xpi

or

DP

pi P zi.n m/ :

GENERAL RULE #2 All poles go from their open-loop locations at K D 0 to:

The zeros of G.s/, or To C1.

Those going to C1 go along asymptotesl D 180

C 360.l 1/n m

centered at D

Ppi P zin m :

ROOT-LOCUS ANALYSIS 610EXAMPLES:

Additional Techniques

The two general rules given, plus some experience are enough tosketch root loci. Some additional rules help when there is ambiguity.(As in examples , )

ROOT-LOCUS ANALYSIS 611Departure Angles, Arrival Angles

We know asymptotically where poles go, but need to know how theystart, and how they end up there.

Importance: One of the following systems is stable for all K > 0, theother is not. Which one?

=.s/


whereN1 D Angle from Np1 to s0 90:1 D Angle from p1 to s0:2 D Angle from p2 to s0 135:

1 D 45: Can now draw departure of poles on locus. Single-pole departure rule:

dep DX

6 .zeros/X

6 .remaining poles/ 180 360l: Multiple-pole departure rule: (multiplicity q 1)

qdep DX

6 .zeros/X

6 .remaining poles/ 180 360l: Multiple-zero arrival rule: (multiplicity q 1)

q arr DX

6 .poles/X

6 .remaining zeros/C 180 360l: Note: The idea of adding 360l is to add enough angle to get the

result within 180. Also, if there is multiplicity, then l counts off thedifferent angles.

Imaginary Axis Crossings

Routh stability test can be run to find value for K D K 0 that causesmarginal stability.

Substitute K 0 and find roots of a.s/C K 0b.s/ D 0: Alternatively, substitute K 0, let s D j!0, solve for

a. j!0/C K 0b. j!0/ D 0: (Real and Imaginary parts)

ROOT-LOCUS ANALYSIS 613Points with Multiple Roots

Sometimes, branches of the locus intersect. (see , on pg.6610). Computing points of intersection can clarify ambiguous loci. Consider two poles approaching each other on the real axis:

=.s/


Some similar loci for which finding saddle points helps clarifyambiguity ...

=.s/


The magnitude equation may be used to find the value of K to get aspecific set of closed-loop poles.

That is, K D 1G.s0/

is the gain to put a pole at s0, if s0 is on the locus.Summary: Root-Locus Drawing Rules180 Locus The steps in drawing a 180 root locus follow from the basic phase definition. This is the locus of

1C K b.s/a.s/D 0; K 0

phase of b.s/

a.s/D 180

:

They are STEP 1: On the s-plane, mark poles (roots of a.s/) by an and zeros (roots of a.s/) with an .

There will be a branch of the locus departing from every pole and a branch arriving at every zero. STEP 2: Draw the locus on the real axis to the left of an odd number of real poles plus zeros. STEP 3: Draw the asymptotes, centered at and leaving at angles , where

n m D number of asymptotes:n D order of a.s/m D order of b.s/

DP

pi P zin m D

a1 C b1n m

l D 180 C .l 1/360

n m ; l D 1; 2; : : : n m:

For n m > 0, there will be a branch of the locus approaching each asymptote and departing toinfinity. For n m < 0, there will be a branch of the locus arriving from infinity along eachasymptote.

STEP 4: Compare locus departure angles from the poles and arrival angles at the zeros whereqdep D

X i

Xi 180 l360

q arr DX

i X

i C 180 l360;where q is the order of the pole or zero and l takes on q integer values so that the angles arebetween 180 . i is the angle of the line going from the i th zero to the pole or zero whoseangle of departure or arrival is being computed. Similarly, i is the angle of the line from the i thpole.

STEP 5: If further refinement is required at the stability boundary, assume s0 D j!0 andcompute the point(s) where the locus crosses the imaginary axis for positive K .

STEP 6: For the case of multiple roots, two loci come together at 180 and break away at 90.Three loci segments approach each other at angles of 120 and depart at angles rotated by 60.

ROOT-LOCUS ANALYSIS 616 STEP 7 Complete the locus, using the facts developed in the previous steps and making

reference to the illustrative loci for guidance. The loci branches start at poles and end at zeros orinfinity.

STEP 8 Select the desired point on the locus that meets the specifications (s0), then use themagnitude condition to find that the value of K associated with that point is

K D 1jb .s0/ =a .s0/j :

EXAMPLE:

r .t/ y.t/K.s C 1/.s C 2/

s.s C 4/

=.s/


Compute T .s/ to determine the characteristic equation as a functionof K .

T .s/ D KsC2

s101C K .sC2/.sC3/

.sC1/.s10/

D K .s C 2/.s C 1/.s C 1/.s 10/C K .s C 2/.s C 3/

Poles of T .s/ at.s C 1/.s 10/C K .s C 2/.s C 3/ D 0

or

1C K .s C 2/.s C 3/.s C 1/.s 10/ D 0:

=.s/


Routh Arrays2 K C 1 6K 10 K > 1s1 5K 9 K > 9

5D 1:8 stability criterion

s0 6K 10 K > 106D 1:66

When K D 9=5, the s1 row of the Routh array is zerotop rowbecomes a factor of the characteristic equation. Imag.-axis crossingswhere

95C 1

s2 C

69

5 10

D 0

14s2 C 4 D 0 : : : s D jr

27:

=.s/


D .2C 2/s3 C .18 5C 10 9/s2 C.20C 45C 12 45/s C .50 54/ D 0

D 14s2 C 32s 4 D 0

s D 32p

322 4.14/.4/28

D 32p

124828

roots at f0:118; 2:40g. Now we can complete the locus:=.s/


3) Asymptotes:n m D 2

DP

pi P zi2

D 0C 0C .a/ .1/2

D 1 a2

D f0:5; 24:5; 4gl D 180

C 360.l 1/2

D 90:4) Departure angles for two poles at s D 0.

2dep DX

6 .zeros/X

6 .remaining poles/ 180 360lD 0 0 180 360l

dep D 90 360

2l D 90:

5) Will always be stable for K > 0.6) Breakaway points:

K .s/ D s2.s C a/s C 1

dds

K .s/ D .s C 1/.3s2 C 2as/ s2.s C a/.1/

.s C 1/2D 0

D s (2s2 C .a C 3/s C 2a D 0breakaway at

(0;a C 3p.a C 3/2 4.2/.2a/

2.2/

)

saDf2;50;9g D1:25 jp7=16| {z }

not on locus; .2:04 and 24:46/; 3} and 0


=.s/


6 5 4 3 2 1 0 1 2 36

4

2

0

2

4

6

Real Axis

Imag

Axis TAKE ME TO YOUR

LEADER!

Summary: Root-Locus Drawing Rules0 Locus

We have assumed that 0 K

ROOT-LOCUS ANALYSIS 623 STEP 3: Draw the asymptotes, centered at and leaving at angles , where

n m D number of asymptotes:n D order of a.s/m D order of b.s/

DP

pi P zin m D

a1 C b1n m

l D .l 1/360

n m ; l D 1; 2; : : : n m:

For n m > 0, there will be a branch of the locus approaching each asymptote and departing toinfinity. For n m < 0, there will be a branch of the locus arriving from infinity along eachasymptote.

STEP 4: Compare locus departure angles from the poles and arrival angles at the zeros where qdep D

X i

Xi l360

q arr DX

i X

i l360;where q is the order of the pole or zero and l takes on q integer values so that the angles arebetween 180 . i is the angle of the line going from the i th zero to the pole or zero whoseangle of departure or arrival is being computed. Similarly, i is the angle of the line from the i thpole.

STEP 5: If further refinement is required at the stability boundary, assume s0 D j!0 andcompute the point(s) where the locus crosses the imaginary axis for positive K .

STEP 6: For the case of multiple roots, two loci come together at 180 and break away at 90.Three loci segments approach each other at angles of 120 and depart at angles rotated by 60.

STEP 7: Complete the locus, using the facts developed in the previous steps and makingreference to the illustrative loci for guidance. The loci branches start at poles and end at zeros orinfinity.

STEP 8: Select the desired point on the locus that meets the specifications (s0), then use themagnitude condition to find that the value of K associated with that point is

K D 1jb .s0/ =a .s0/j :

Extensions to Root Locus MethodTime Delay

A system with a time delay has the formG.s/ D edsG 0.s/

where d is the delay and G 0.s/ is the non-delayed system.


This is not in rational-polynomial form. We cannot use root locustechniques directly.

METHOD 1:

Approximate eds by

b0s C b1a0s C 1

; a polynomial.

Pad approximation.eds 1 .ds=2/

1C .ds=2/ First-order approximation

1 ds=2C .ds/2=12

1C ds=2C .ds/2=12 Second-order approximation.

11C ds Very crude.

Extremely important for digital control!!!METHOD 2:

Directly plot locus using phase condition. i.e., 6 G.s/ D .d!/C 6 G 0.s/ if s D C j!. i.e., look for places where 6 G 0.s/ D 180 C d! C 360l: Fix !, search horizontally for locus.


ROOT-LOCUS DESIGN

We have seen how to draw a root locus for a given plant dynamics. We include a variable gain K proportional control. What if desired pole locations are not on this locus? We need to modify the locus itself by adding extra

dynamicscompensator:

r .t/ y.t/K G.s/D.s/

We redraw the locus and pick K in order to put the poles where wewant them. HOW?

T .s/ D K D.s/G.s/1C K D.s/G.s/ let G

0.s/ D D.s/G.s/

D K G0.s/

1C K G 0.s/ We know how to draw this locus! Adding a compensator effectively adds dynamics to the plant. What types of (1) compensation should we use, and (2) how do we

figure out where to put the additional dynamics?

Types of Compensator Dynamics

There are 3 classical types of controllers with some importantvariations.

ROOT-LOCUS DESIGN 72

1) Proportional FeedbackD.s/ D 1: u.t/ D K e.t/T .s/ D K G.s/

1C K G.s/: Same as what we have already looked at. Controller only consists of

a gain knob. We have to take the locus as given since we have noextra dynamics to modify it. Usually a very limited approach, but a good place to start.

2) Integral FeedbackD.s/ D 1

TI su.t/ D K

TI

Z t0

e. / d

T .s/ DKTI

G.s/s

1C KTI G.s/s:

Usually used to reduce/eliminate steady-state error. i.e., if e.t/constant, u.t/ will become very large and hopefully correct the error. Penalty: Steady-state response improves, but often transient

response degrades.

Steady-State Response We apply a step (ramp, parabola) input. How well does our (closed

loop) system track this? Ultimately, we would like zero error. e.t/ D 0. (Maybe 1%2% in

reality) Recall: For a unity-feedback control system as drawn above, the

steady-state error to a unit-step input is:

ess D 11C K D.0/G.0/:


If we make D.s/ D 1TI s

, then as s ! 0, D.s/!1

ess! 11C1 D 0: So by adding the integrator into the compensator, the error has been

reduced from 11C K p to zero for systems that do not have any free

integrators.

EXAMPLE: G.s/ D 1.s C a/.s C b/; a > b > 0:

Proportional feedback, D.s/ D 1, G.0/ D 1ab

, ess D 11C Kab:

a b

=.s/


Proportional-Integral (PI) Control Now, D.s/ D 1C 1

TI sD

s C .1=TI /s

. Both a pole and a zero.

a b

=.s/

jp0j, the phase added to the open-loop transferfunction is negative: : : phase lag

Pole often placed very close to zero. e.g., p0 0:01. Zero is placed near pole. e.g., z0 0:1.


a b

=.s/


a1 and b1 are chosen to make locus go through s D s1,a1s1 C a0b1s1 C 1

G.s1/ D 1

for that point to be on the root locus.

Magnitudea1s1 C a0b1s1 C 1

jG.s1/j D 1 Phase 6

a1s1 C a0b1s1 C 1

C 6 G.s1/ D 180:

(math happens)a1 D sin./C a0jG.s1/j sin. /js1jjG.s1/j sin. /b1 D sin. C /C a0jG.s1/j sin./js1j sin. /

9>>=>>;

s1 D js1je jG.s1/ D jG.s1/je j :

EXAMPLE: G.s/ D 1s2: s1 D 2

p2e j135 D 2C 2 j .

The point s1 is chosen to achieve D 0:707 and D 0:5 sec. a0: We cannot compute a0 since 1

s2

sD0!1. So, arbitrarily choose

a0 D 2: a1: Note, D 135; D 270 because

G.s1/ D 1s2

sD2p2e j135

D 18

e j270:

a1 D sin.135/C 2.1=8/ sin.45/

.2p

2/.1=8/ sin.270/ D.1=p

2/.1C 1=4/p2=4

D 52:

b1:b1 D sin.135

/C 2.1=8/ sin.135/.2p2/ sin.270/ D

.1=p2/.1 1=4/2p2 D

316:


So, the compensator is:D.s/ D .5=2/s C 2

.3=16/s C 1:

6 5 4 3 2 1 0 14

3

2

1

0

1

2

3

4

Real Axis

Imag

Axis

Example locus passing through (-2,2)

Proportional-Integral-Derivative (PID) Control Most common type of control, even though often approximated by

Lead-Lag control.

D.s/ D K

1C 1TI sC TDs

:

K D sin. C /TI js1jjG.s1/j sin./[TI js1j C 2 cos./] TD D sin. /js1jjG.s1/j sin./ C

KTI js1j2 , where s1 D js1je

j and

G.s1/ D jG.s1/je j for both cases. TI chosen to match some design criteria. e.g., Steady-state error.

Compensator Implementation

Analog compensators commonly use op-amp circuits. See the following pages: : :


z1

K D p1z1

1

V1 V2

1p1

p1 z1p1

LEAD

z1


FREQUENCY-RESPONSE ANALYSIS

Advantages and disadvantages to root-locus design approach:ADVANTAGES:

Good indicator of transient response. Explicitly shows location of closed-loop poles. Tradeoffs are

clear.

DISADVANTAGES:

Requires transfer function of plant be known. Difficult to infer all performance values. Hard to extract steady-state response (sinusoidal inputs).

Frequency-response methods can be used to supplement root locus: Can infer performance and stability from same plot. Can use measured data when no model is available. Design process is independent of system order (# poles). Time delays handled correctly (es ). Graphical techniques (analysis/synthesis) are quite simple.

Frequency Response

We want to know how a linear system responds to sinusoidal input, insteady state.

FREQUENCY-RESPONSE ANALYSIS 82

Consider systemY .s/U .s/

D G.s/: Consider the input:

u.t/ D u0 sin.!t/U .s/ D u0!

s2 C !2 : With zero initial conditions,

Y .s/ D G.s/ u0!s2C !2 :

Do a partial-fraction expansion (assume distinct roots)Y .s/ D 1

s a1 C2

s a2 C Cn

s an C0

s C j! C0

s j!y.t/ D 1ea1t C 2ea2t C C neant| {z }

If stable; these decay to zero:C2j0j sin.!t C /

yss.t/ D 2j0j sin.!t C /; D tan1=.0/


D tan1=.G. j!//


Phase = phase of numerator phase of denominatortan1

=.num/

FREQUENCY-RESPONSE ANALYSIS 85Plot Method #2: Magnitude and Phase Plots

We can replot the data by separating the plots for magnitude andphase making two plots versus frequency.

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

Frequency, (rads/sec.)

jG.

j!/j

0 1 2 3 4 5 69

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