mcmaster - intro to linear control systems

EE 3CL4, L 11 / 21

Tim Davidson

Why are youhere?

What is acontrolsystem?

What tools willwe use?

Administrativedetails

Parting shot

EE3CL4:Introduction to Linear Control Systems

Lecture 1

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 12 / 21

Tim Davidson

Why are youhere?




Parting shot

Outline

1 Why are you here?

2 What is a control system?

3 What tools will we use?

4 Administrative details

5 Parting shot

EE 3CL4, L 14 / 21

Tim Davidson

Why are youhere?




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Why are you here?

You might be interested in:

Designing guidance systems for: the next Mars Rover, a space craft or a UAV

EE 3CL4, L 15 / 21

Tim Davidson

Why are youhere?




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Why are you here?


Designing industrial or biomedical robots

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Tim Davidson

Why are youhere?




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Why are you here?You might be interested in:

Designing humanoid robots

EE 3CL4, L 17 / 21

Tim Davidson

Why are youhere?




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Why are you here?You might be interested in: Designing control systems for hybrid or electric cars

Designing disk drives

Developing technologies for enhanced powerdistribution (Smart Grid)

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Tim Davidson

Why are youhere?




Parting shot

Why are you here?


Designing an automated insulin delivery system

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Tim Davidson

Why are youhere?




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Why are you here?


Designing and analyzing financial systems

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Tim Davidson

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What is a feedback controlsystem?

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What is a feedback controlsystem?

Example: You driving a car

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Tim Davidson

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What tools will we use?

Newtonian mechanics, linear and rotational(Phys 1D03)

Basic electromagnetism (Phys 1E03, EE 2CJ4) Electric circuit analysis (EE 2CI5, EE 2CJ4, EE 2EI5) Step response of first and second order systems

(Math 2P04/2Z03, EE 2CI5, EE 2CJ4) Laplace transforms

(Math 2P04/2Z03, EE 2CJ4, EE 3TP4) Transfer functions (EE 2CJ4, EE 3TP4) Bode diagrams (EE 2CJ4) Structured problem solving methods

(EE 2CI5, EE 2CJ4, EE 3TP4, . . . )

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Tim Davidson

Why are youhere?




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Contact details

Tim DavidsonITBA310Ext. [email protected], EE3CL4 in subject line

Course web site (soon to be fully updated for 2011)http://www.ece.mcmaster.ca/davidson/EE3CL4

A formal course outline appears on the web site.Here we will focus on some key points

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Tim Davidson

Why are youhere?




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Class details

Lectures Tuesday, Thursday, Friday, 12:30pm, TSH/B128 Tentative topic schedule will appear on web site

Tutorials (starting next week) T01: Tuesday, 8:30am, T13/125 T02: Monday, 11:30am, BSB/106

Labs (tentatively starting 24 January) Four labs One every other week, ITB/154 Significant pre-lab work will be required

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Tim Davidson

Why are youhere?




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Marking scheme

Laboratory reports: 20% Midterm test: 25%

Tentatively scheduled for week starting Monday 28 Feb(first week after midterm break), 7:00pm 8:30pm

Final examination: 55%

Students must personally complete all laboratories andall laboratory reports in order to be eligible for a finalgrade

Formally deferred tests & exams may be conductedorally

Remarking requests will require documentation

On tests & exams, expect to see problems that youhave not seen before

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Some suggestions

Be active in lectures Participate in tutorials Take advantage of the labs Do half of the assigned problems under examination

conditions In exams, explain your methodology

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Tim Davidson

Why are youhere?




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Tim Davidson

What iscontrolengineering

Examples

Designprocess

Disk driveexample


Lecture 2

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 22 / 19

Tim Davidson


Examples

Designprocess

Disk driveexample

Outline

1 What is control engineering

2 Examples

3 Design process

4 Disk drive example

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Tim Davidson


Examples

Designprocess

Disk driveexample

What is control engineering

First we have a system that we want to understand Typically, that means a mathematical model

Then we use that understanding to design a secondarysystem that controls the behaviour of the first

Must obtain appropriate control, even if modelinaccurate, or subject to noise/disturbances

Mathematical model Must balance accuracy against insight generated This course: models will be linear Hence, tools available for insight: superposition,

transfer function, Laplace

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Examples

Designprocess

Disk driveexample

Model, open-loop, closed-loop

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Examples

Designprocess

Disk driveexample


What about disturbances? Mismatch (temperature, age)

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Examples

Designprocess

Disk driveexample


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Examples

Designprocess

Disk driveexample

Multivariable control

Something for fourth year!

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Examples

Designprocess

Disk driveexample

Watts flyball governor

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Tim Davidson


Examples

Designprocess

Disk driveexample

CD player speed control:Open-Loop

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Tim Davidson


Examples

Designprocess

Disk driveexample

CD player speed control:Closed-Loop

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Examples

Designprocess

Disk driveexample

Doritos

John MacGregor (Chem Eng): Visual feedback control of flavours

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Examples

Designprocess

Disk driveexample

Structured approach to design

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Examples

Designprocess

Disk driveexample

Disk drive: Intro

Spins at between 1800 and 7200 rpm,perhaps even 10,000

Head height: 100nm A key issue: Seek time

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Examples

Designprocess

Disk driveexample

Disk drive: Establish goals

Move from track a to track b with accuracy of at least1m within 50ms

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Examples

Designprocess

Disk driveexample

Disk drive: Control architecture

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Tim Davidson

ModellingphysicalsystemsTranslationalNewtonianMechanics

Rotational NewtonianMechanics

Linearization

Laplacetransforms


Lecture 3

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 32 / 20

Tim Davidson



Linearization

Laplacetransforms

Outline

1 Modelling physical systemsTranslational Newtonian MechanicsRotational Newtonian Mechanics

2 Linearization

3 Laplace transforms

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Linearization

Laplacetransforms

Differential equation models

Most of the systems that we will deal with are dynamic Differential equations provide a powerful way to

describe dynamic systems Will form the basis of our models

We saw differential equations for inductors andcapacitors in 2CI, 2CJ

What about mechanical systems?both translational and rotational

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Linearization

Laplacetransforms

Translational Spring

F (t): resultant force in direction xRecall free body diagrams and action and reaction

Spring. k : spring constant, xr : relaxed length of spring

F (t) = k([x2(t) x1(t)] xr

)

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Linearization

Laplacetransforms

Translational Damper

F (t): resultant force in direction x

Viscous damper. b: viscous friction coefficient

F (t) = b(dx2(t)

dt dx1(t)

dt

)= b

(v2(t) v1(t)

)

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Linearization

Laplacetransforms

Mass

F (t): resultant force in direction x

Mass: M

F (t) = Md2xm(t)dt2

= Mdvm(t)dt

= Mam(t)

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Linearization

Laplacetransforms

Rotational spring

T (t): resultant torque in direction

Rotational spring. k : rotational spring constant,r : rotation of relaxed spring

T (t) = k([2(t) 1(t)] r

)

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Linearization

Laplacetransforms

Rotational damper


Rotational viscous damper.b: rotational viscous friction coefficient

T (t) = b(d2(t)

dt d1(t)

dt

)= b

(2(t) 1(t)

)

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Linearization

Laplacetransforms

Rotational inertia


Rotational inertia: J

T (t) = Jd2m(t)dt2

= Jdm(t)dt

= Jm(t)

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Linearization

Laplacetransforms

Example system (translational)Horizontal. Origin for y : y = 0 when spring relaxed

F = M dv(t)dt v(t) = dy(t)dt F (t) = r(t) b dy(t)dt ky(t)

Md2y(t)dt

+ bdy(t)dt

+ ky(t) = r(t)

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Linearization

Laplacetransforms

Example, continued

Md2y(t)dt

+ bdy(t)dt

+ ky(t) = r(t)

Resembles equation for parallel RLC circuit.

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Linearization

Laplacetransforms

Example, continued

Stretch the spring a little and hold. Assume an under-damped system. What happens when we let it go?

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Linearization

Laplacetransforms

Taylors series

Nature does not have many linear systems However, many systems behave approximately linearly

in the neighbourhood of a given point Apply first-order Taylors Series at a given point Obtain a locally linear model

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Linearization

Laplacetransforms

Pendulum example

Torque due to gravity: T = MgL sin Linearize around = 0. At that point, T = 0 Linearized model

T MgL d sin d

=0

= MgL

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Linearization

Laplacetransforms

Laplace transform Once we have a linearized differential equation we can

take Laplace Transforms to obtain the transfer function

We will consider the one-sided Laplace transform, forsignals that are zero to the left of the origin.

F (s) =

0f (t)est dt

What does mean? limT T .

Does this limit exist? If |f (t)| < Met , then exists for all Re(s) > .

Includes all physically realizable signals

Note: When multiplying transfer function by Laplace of input, outputis only valid for values of s in intersection of regions of convergence

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Linearization

Laplacetransforms

Laplace transform pairs

can be tabulated

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Linearization

Laplacetransforms

Laplace transform pairs

Recall that complex poles come in conjugate pairs.

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Laplacetransforms

Laplacetransforms inaction


Lecture 4

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 42 / 29

Tim Davidson

Laplacetransforms


Outline

1 Laplace transforms

2 Laplace transforms in action

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Laplacetransforms


Laplace transform

We will consider the one-sided Laplace transform, forsignals that are zero to the left of the origin.

F (s) =

0f (t)est dt

Key properties

df (t)dt

sF (s) f (0)

t

f (x)dx F (s)s

+1s

0

f (x)dx

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Tim Davidson

Laplacetransforms


Final value theorem

If F (s) has all its poles in the left half plane, except,perhaps, for a single pole at the origin,

then we can compute the final value of f (t), namelylimt f (t) without inverting the Laplace Transform.

In particular,limt

f (t) = lims0

sF (s)

Common application: Steady state value of stepresponse

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Laplacetransforms


Mass-spring-damper system

Horizontal (no gravity) Set origin of y where spring is relaxed F = M dv(t)dt v(t) = dy(t)dt F (t) = r(t) b dy(t)dt ky(t)

Md2y(t)dt

+ bdy(t)dt

+ ky(t) = r(t)

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Laplacetransforms


MSD system

Md2y(t)dt

+ bdy(t)dt

+ ky(t) = r(t)

Consider t 0 and take Laplace transform

M(s2Y (s)sy(0) dy(t)

dt

t=0

)+b(sY (s)y(0))+kY (s) = R(s)

Hence

Y (s) =1/M

s2 + (b/M)s + k/MR(s)

+(s + b/M)

s2 + (b/M)s + k/My(0)

+1

s2 + (b/M)s + k/Mdy(t)dt

t=0

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Laplacetransforms


Response to static init. cond.

Spring stretched to a point y0, held, then let go at time t = 0

Hence, r(t) = 0 and dy(t)dtt=0

= 0

Hence,

Y (s) =(s + b/M)

s2 + (b/M)s + k/My0

What can we learn about this response without having toinvert Y (s)

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Tim Davidson

Laplacetransforms


Standard form

Y (s) =(s + b/M)

s2 + (b/M)s + k/My0

=(s + 2n)

s2 + 2ns + 2ny0

where n =k/M and = b

2kM

Poles: s1, s2 = n n2 1

> 1 (equiv. b > 2kM): distinct real roots, overdamped = 1 (equiv. b = 2kM): equal real roots, critically damped < 1 (equiv. b < 2kM): complex conj. roots, underdamped

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Laplacetransforms


Overdamped case

s1, s2 = n n2 1

Overdamped response: > 1 (equiv. b > 2kM)

y(t) = c1es1t + c2es2t y(0) = y0 = c1 + c2 = y0 dy(t)dt

t=0

= 0 = s1c1 + s2c2 = 0

What does this look like when strongly overdamped s2 is large and negative, s1 is small and negative Hence es2t decays much faster than es1t Also, c2 = c1s1/s2. Hence, small Hence y(t) c1es1t Looks like a first order system!

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Tim Davidson

Laplacetransforms


Critically damped case

s1 = s2 = n y(t) = c1ent + c2tent

y(0) = y0 = c1 = y0 dy(t)dt

t=0

= 0 = c1n + c2 = 0

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Tim Davidson

Laplacetransforms


Underdamped case

s1, s2 = n jn

1 2 Therefore, |si | = n: poles lies on a circle Angle to negative real axis is cos1().

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Tim Davidson

Laplacetransforms


Underdamped case

Define = n, d = n

1 2. Response is:y(t) = c1et cos(d t) + c2et sin(d t)

= Aet cos(d t + )

Homework: Relate A and to c1 and c2. Homework: Write the initial conditions y(0) = y0 and

dy(t)dt

t=0

= 0 in terms of c1 and c2, and in terms of A and

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Laplacetransforms


Numerical examples

Y (s) = (s+2n)s2+2ns+2n y0, where n =k/M, = b

2kM

Poles: s1, s2 = n n2 1

> 1: overdamped; < 1: underdamped

Consider the case of M = 1, k = 1. Hence, n = 1, b = 3 0. Hence, = 1.5 0 Initial conds: y0 = 1, dy(t)dt

t=0

= 0

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Tim Davidson

Laplacetransforms


Poles and transient response,b = 3

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Laplacetransforms


Poles and transient response,b = 2.75

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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Laplacetransforms



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Tim Davidson

Transferfunction

Step response

Transferfunction of DCmotor


Lecture 5

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 52 / 30

Tim Davidson

Transferfunction

Step response


Outline

1 Transfer function

2 Step response

3 Transfer function of DC motor

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Tim Davidson

Transferfunction

Step response


Transfer function

Definition: Laplace transform of output over Laplacetransform of input when initial conditions are zero

Recall the mass-spring-damper system,

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Tim Davidson

Transferfunction

Step response


Transfer function, MSD system

For the mass-spring-damper system,

Y (s) =1/M

s2 + (b/M)s + k/MR(s)

+(s + b/M)

s2 + (b/M)s + k/My(0)

+1

s2 + (b/M)s + k/Mdy(t)dt

t=0

Therefore, transfer function is:

1/Ms2 + (b/M)s + k/M

=1

Ms2 + bs + k

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Tim Davidson

Transferfunction

Step response


Step response

Recall that u(t) 1s Therefore, for transfer function G(s), the step response

is:L 1

{G(s)s

} For the mass-spring-damper system, step response is

L 1{ 1s(Ms2 + bs + k)

}

What is the final position for a step input?Recall final value theorem. Final position is 1/k .

Consider again the case of M = k = 1, b = 3 0.n = 1, = 1.5 0.

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Tim Davidson

Transferfunction

Step response


Poles and step response, b = 3

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Tim Davidson

Transferfunction

Step response


Poles and step resp., b = 2.75

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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response


Poles and step resp., b = 2

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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response



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Transferfunction

Step response


A DC motor

We will consider linearized model for each component Flux in the air gap: (t) = Kf if (t) (Magnetic cct, 2CJ4) Torque: Tm(t) = K1(t)ia(t) = K1Kf if (t)ia(t). Is that linear? Only if one of if (t) or ia(t) is constant We will consider armature control: if (t) constant

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Tim Davidson

Transferfunction

Step response


Armature controlled DC motor

if (t) will be constant (to set up magnetic field), if (t) = If Torque: Tm(t) = K1Kf If ia(t) = Kmia(t) Will control motor using armature voltage Va(t) What is the transfer function from Va(s) to angular

position (s)? Origin?

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Tim Davidson

Transferfunction

Step response


Towards transfer function

Tm(t) = Kmia(t) Tm(s) = KmIa(s) KVL: Va(s) = (Ra + sLa)Ia(s) + Vb(s) Vb(s) is back-emf voltage, due to Faradays Law Vb(s) = Kb(s), where (s) = s(s) is rot. velocity Remember: transfer function implies zero init. conds

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Transferfunction

Step response



Torque on load: TL(s) = Tm(s) Td(s) Td(s): disturbance. Often small, unknown (e.g., wind) Load torque to angle (Newton plus friction):

TL(s) = Js2(s) + bs(s)

Now put it all together

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Transferfunction

Step response



Tm(s) = KmIa(s) = Km(Va(s)Vb(s)

Ra+sLa

) Vb(s) = Kb(s) TL(s) = Tm(s) Td(s) TL(s) = Js2(s) + bs(s) = Js(s) + b(s) Hence (s) = TL(s)Js+b (s) = (s)/s

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Transferfunction

Step response


Block diagram

Tm(s) = KmIa(s) = Km(Va(s)Vb(s)

Ra+sLa

) Vb(s) = Kb(s) TL(s) = Tm(s) Td(s) TL(s) = Js2(s) + bs(s) = Js(s) + b(s) Hence (s) = TL(s)Js+b (s) = (s)/s

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Transferfunction

Step response


Transfer function

Set Td(s) = 0 and solve (you MUST do this yourself)

G(s) =(s)Va(s)

=Km

s[(Ra + sLa)(Js + b) + KbKm

]=

Kms(s2 + 2ns + 2n)

Third order :(

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Transferfunction

Step response


Second-order approximation

G(s) =(s)Va(s)

=Km


] Often armature time constant, a = La/Ra, is negligible Hence (you MUST derive this yourself)

G(s) Kms[Ra(Js + b) + KbKm

] = Km/(Rab + KbKm)s(1s + 1)

where 1 = RaJ/(Rab + KbKm) Using a power balance can show that Kb = Km

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Block diagrammodelsExample: Looptransfer function

Block diagramtransforma-tions


Lecture 6

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 62 / 10

Tim Davidson



Outline

1 Block diagram modelsExample: Loop transfer function

2 Block diagram transformations

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Bock diagram models

A convenient way to represent a transfer function is viaa block diagram

In this case, U(s) = Gc(s)R(s) and Y (s) = G(s)U(s) Hence, Y (s) = G(s)Gc(s)R(s) Consistent with the engineering procedure of breakingthings up into little bits, studying the little bits, and thenput them together

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Simple example

Y1(s) = G11(s)R1(s) +G12(s)R2(s) Y2(s) = G21(s)R1(s) +G22(s)R2(s)

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Example: Loop transfer function

Ea(s) = R(s) B(s) = R(s) H(s)Y (s) Y (s) = G(s)U(s) = G(s)Ga(s)Z (s) Y (s) = G(s)Ga(s)Gc(s)Ea(s) Y (s) = G(s)Ga(s)Gc(s)

(R(s) H(s)Y (s)

)Y (s)R(s)

=G(s)Ga(s)Gc(s)

1+G(s)Ga(s)Gc(s)H(s)

Each transfer function is a ratio of polynomials in s What is Ea(s)/R(s)?

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Block diagram transformations

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Using block diagramtransformations

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Using block diagramtransformations

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Transferfunction ofarmaturecontrolled DCmotor

Application todisk drive readsystem

Our firstcontrol systemdesign

Characteristicsof feedback


Lecture 7

Tim Davidson

McMaster University

Winter 2011

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Tim Davidson





Outline

1 Transfer function of armature controlled DC motor

2 Application to disk drive read system

3 Our first control system design

4 Characteristics of feedback control systems, Ch 4

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Last week

We constructed a block diagram for armature-controlledDC motor

if (t) constant; motor controlled using va(t) Determined the transfer function (s)Va(s)

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Transfer function

G(s) =(s)Va(s)

=Km


]=

Kms(s2 + 2ns + 2n)

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Model for a disk drive readsystem

Uses a permanent magnet DC motor Can be modelled using arm. contr. model with Kb = 0 Hence, motor transfer function:

G(s) =(s)Va(s)

=Km

s(Ra + sLa)(Js + b)

Assume for now that the arm is stiff

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Typical values

G(s) =(s)Va(s)

=Km

s(Ra + sLa)(Js + b)

G(s) =5000

s(s + 20)(s + 1000)

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Time constants

Initial model

G(s) =5000

s(s + 20)(s + 1000)

Motor time constant = 1/20 = 50ms Armature time constant = 1/1000 = 1ms Hence

G(s) 5s(s + 20)

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A simple feedback controller

Now that we have a model, how to control?

Here, Y (s) = (s).

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Simplified block diagram

What is the transfer function from command toposition? Derive this yourself

Y (s)R(s)

=KaG(s)

1 + KaG(s)

For second-order G(s), and a gain of Ka = 40,

Y (s) =200

s2 + 20s + 200R(s)

What is the response for R(s) = 0.1/s? Does it meet our design criteria?

Within 1m within 50ms?

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Tim Davidson





Step responseResponse to r(t) = 0.1u(t)

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Tim Davidson





Closed and open loop

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The error signal

Error signal E(s) = R(s) Y (s)For the case where H(s) = 1 (derive this for yourself):

E(s) =1

1 +Gc(s)G(s)R(s)

G(s)1 +Gc(s)G(s)

Td(s)

+Gc(s)G(s)

1 +Gc(s)G(s)N(s)

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Tim Davidson





Loop gain

Again, set H(s) = 1

Define loop gain: L(s) = Gc(s)G(s)

E(s) =1

1 + L(s)R(s) G(s)

1 + L(s)Td(s) +

L(s)1 + L(s)

N(s)

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Tim Davidson





Sensitivities

Again, set H(s) = 1

Define sensitivity: S(s) =L(s)

1 + L(s)

Define complementary sensitivity: C(s) =1

1 + L(s)

E(s) = S(s)R(s) S(s)G(s)Td(s) + C(s)N(s)Note that S(s) + C(s) = 1.

Trading S(s) against C(s): a key to the art of control design

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Tim Davidson



Lecture 8

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 82 / 11

Tim Davidson


Outline

1 Characteristics of feedback control systems

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Tim Davidson


The closed loop

For the whole of this lecture we will consider unity feedback:

H(s) = 1

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Tim Davidson


The output signal

Under standing assumption of H(s) = 1, what is Y (s)?

Y (s) =Gc(s)G(s)

1 + Gc(s)G(s)R(s)

+G(s)

1 + Gc(s)G(s)Td(s)

Gc(s)G(s)1 + Gc(s)G(s)

N(s)

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Tim Davidson


The error signal

Define the error signal: E(s) = R(s) Y (s)For H(s) = 1, what is E(s)?

E(s) =1

1 + Gc(s)G(s)R(s)

G(s)1 + Gc(s)G(s)

Td (s)

+Gc(s)G(s)

1 + Gc(s)G(s)N(s)

Note: For H(s) = 1, contr. input Ea(s) = R(s)(Y (s) + N(s)

)

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Tim Davidson


Loop gain

Define loop gain: L(s) = Gc(s)G(s)

E(s) =1

1 + L(s)R(s) G(s)

1 + L(s)Td(s) +

L(s)1 + L(s)

N(s)

What do we want?

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Tim Davidson


A taste of loop shaping

E(s) =1

1 + L(s)R(s) G(s)

1 + L(s)Td(s) +

L(s)1 + L(s)

N(s)

What do we want? Good tracking: E(s) does depend only weakly on R(s)

= L(s) large where R(s) large Good disturbance rejection:

= L(s) large where Td(s) large Good noise suppression:

= L(s) small where N(s) large

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Tim Davidson


A taste of loop shapingPossibly easier to understand in pure freq. domain, s = j

Recall that L(s) = Gc(s)G(s),G(s): fixed; Gc(s): controller to be designed

Good tracking: = L(s) large where R(s) large|L(j)| large in the important frequency bands of r(t)

Good dist. rejection: = L(s) large where Td (s) large|L(j)| large in the important frequency bands of td (t)

Good noise suppr.: = L(s) small where N(s) large|L(j)| small in the important frequency bands of n(t)

Typically, L(j) is a low-pass function

How big should |L(j)| be? Any other constraints? Stability!

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Tim Davidson


Sensitivities

Define sensitivity: S(s) =1

1 + L(s)

Define complementary sensitivity: C(s) =L(s)

1 + L(s)

E(s) = S(s)R(s) S(s)G(s)Td(s) + C(s)N(s)

Note that S(s) + C(s) = 1.Trading S(s) against C(s), with stability

is the essence of the art of control design

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Tim Davidson


Model sensitivities

In practice we rarely model G(s) exactly, and it may age How does T (s) = Y (s)R(s) change as G(s) changes? G(s) + G(s). limG(s)0 T (s)/T (s)G(s)/G(s) =

lnT (s) lnG(s)

For an open loop system this sensitivity =1 For the closed loop system, with H(s) = 1,S(s) = 11+Gc(s)G(s) , as earlier

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Tim Davidson

Moreadvantages offeedback

Price offeedback

Example:EnglishChannelboringmachines

Example: Diskdrive readsystem


Lecture 9(Given as the 10th lecture in Winter 2011)

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 92 / 18

Tim Davidson


Price offeedback



Outline

1 More advantages of feedback

2 Price of feedback

3 Example: English Channel boring machines

4 Example: Disk drive read system

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Tim Davidson


Price offeedback



Advantages of feedback so far

reduced sensitivity to disturbances reduced sensitivity to model variation

can also manipulate the transient response,to some degree (next week)

can also control steady-state response to certaininputs, without tight calibration

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Tim Davidson


Price offeedback



Steady-state error

Recall standing assumption of H(s) = 1

Consider R(s) only; set Td(s) and N(s) to zero

E(s) = R(s) Y (s) = 11 +Gc(s)G(s)

R(s)

So what is the steady state error?

If E(s) = 11+Gc(s)G(s) R(s) has no poles in the close right halfplane, except, perhaps for a simple pole at the origin,

limt

e(t) = lims0

sE(s)

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Tim Davidson


Price offeedback



Steady-state error for step

Consider the case where r(t) = u(t). = R(s) = 1/s. E(s) = 1

1 +Gc(s)G(s)1s

lims0 sE(s) = lims01

1 +Gc(s)G(s)

Hence, steady-state error is 11 +Gc(0)G(0)

How to make this small? Large loop gain at DC

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Tim Davidson


Price offeedback



Steady-state error for ramp

Consider r(t) = tu(t). = R(s) = 1/s2. E(s) = 11+Gc(s)G(s)

1s2

lims0 sE(s) = lims0 11+Gc(s)G(s)1s

How to make this finite? Let Gc(s) = nc(s)dc(s) , G(s) =

n(s)d(s)

sE(s) = dc(s)d(s)dc(s)d(s)+nc(s)n(s)1s

For finite SS error, dc(s)d(s) must contain a factor s i.e., either Gc(s) or G(s) must have pole at origin

integration

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Tim Davidson


Price offeedback



Price of feedback

more components than open loop

less gain than open loopGc(s)G(s)

1+Gc(s)G(s) instead of Gc(s)G(s)

potential for instability

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Tim Davidson


Price offeedback



English Channel boringmachines

Y (s) =K + 11s

s2 + 12s + KR(s) +

1s2 + 12s + K

Td(s)

Lets consider step and step disturbance responses for twovalues of K , 100 and 20.

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Tim Davidson


Price offeedback



English Channel boringmachine

K = 100

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Tim Davidson


Price offeedback




K = 20

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Tim Davidson


Price offeedback




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Tim Davidson


Price offeedback



Disk drive read system

Y (s) =5000Ka

s3 + 1020s2 + 20000s + 5000KaR(s)

+s + 1000

s3 + 1020s2 + 20000s + 5000KaTd(s)

with Td(s) = 0, compare step resps with Ka = 10 and 80.

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Tim Davidson


Price offeedback



Step responses for Ka = 10, 80

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Tim Davidson


Price offeedback



Disturb. step resp. for Ka = 80

To reduce this response need larger Ka

but larger Ka will result in more oscilliatory response

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Tim Davidson

Performanceof feedbackcontrolsystems

Performanceofsecond-ordersystems


Lecture 10(Given as the 9th lecture in Winter 2011)

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 102 / 19

Tim Davidson



Outline

1 Performance of feedback control systems

2 Performance of second-order systems

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Tim Davidson



Basic performance criteria

Stability (next week) Steady-state response to chosen inputs Transient response to chosen inputs Compromises: the art of design

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Tim Davidson



Typical test signals

Step, ramp, parabolic

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Tim Davidson



A second-order systemNow examine, in detail, a particular class of second ordersystems

Y (s) =G(s)

1 +G(s)R(s) =

2ns2 + 2ns + 2n

R(s)

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Tim Davidson



Step response What is the step response? Set R(s) = 1/s; take inverse Laplace transform of Y (s) Y (s) = 2n

s(s2+2ns+2n

) For the case of 0 < < 1,

y(t) = 1 1ent sin(nt + )

where =

1 2 and = cos1 . Recall pole positions of G(s)1+G(s) (ignore the zero):

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Tim Davidson



Typical step responses, fixed n

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Tim Davidson



Typical step responses, fixed

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Tim Davidson



Key parameters of(under-damped) step response

With =

1 2 and = cos1 ,


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Tim Davidson



Peak time and peak value


Peak time: first time dy(t)/dt = 0 Can show that this corresponds to nTp = pi Hence, Tp =

pi

n

1 2 Hence, peak value, Mpt = 1 + e

(pi/

12)

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Tim Davidson



Percentage overshoot

Let fv denote the final value of the step response.

Percentage overshoot defined as:

P.O. = 100Mpt fv

fv

In our example, fv = 1, and hence

P.O. = 100e(pi/

12)

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Tim Davidson



Overshoot vs Peak TimeThis is one of the classic trade-offs in control

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Tim Davidson



Steady-state error, ess

In general this is not zero.

However, for our second-order system,


Hence ess = 0

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Tim Davidson



Settling time


How long does it take to get within x% of final value? Approx. when ent < x/100 When x = 2, that corresponds to nTs 4;

that is, 4 time constants

In that case, Ts =4n

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Tim Davidson



Rise time (under-damped)


How long to get to the target (for first time)? Tr , the smallest t such that y(t) = 1

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10%90% Rise time

What is Tr in over-damped case? Hence, typically use Tr1, the 10%90% rise time

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Tim Davidson



10%90% Rise time

Difficult to get an accurate formula Linear approx. for 0.3 0.8 (under-damped),

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Tim Davidson

Step responseof a class ofsecond-ordersystems(review)

A taste ofpole-placementdesign

Transientperformance,poles andzeros

Summary andplan


Lecture 11(Given as the 12th lecture in 2011)

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 112 / 19

Tim Davidson




Summary andplan

Outline

1 Step response of a class of second-order systems(review)

2 A taste of pole-placement design

3 Transient performance, poles and zeros

4 Summary and plan

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Tim Davidson




Summary andplan

Class of systems

We considered the step response of an (under-damped)second order system

Y (s) = T (s)R(s) =2n

s2 + 2ns + 2nR(s)

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Tim Davidson




Summary andplan

Poles of T (s)T (s) =

2ns2 + 2ns + 2n

Where are the poles?

s1, s2 = n jn

1 2= n cos() jn sin()

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Tim Davidson




Summary andplan

Step response

Y (s) = T (s)1s

=2n

s2 + 2ns + 2n

1s

= y(t) = 1 1ent sin(nt + )

where =

1 2 and = cos1 .

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Tim Davidson




Summary andplan

Properties of step response

Settling time, Ts,: 2% settling time 4n

Percentage overshoot: P.O. = 100e(pi/

12)

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Tim Davidson




Summary andplan

Design problem

For what values of K and p is the settling time 4 secs, and the percentage overshoot 4.3%?

T (s) =G(s)

1 +G(s)=

Ks2 + ps + K

=2n

s2 + 2ns + 2n,

where n =K and = p/(2

K )

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Tim Davidson




Summary andplan

Pole positions

Ts 4n

P.O. = 100e(pi/

12)

For Ts 4, n 1 For P.O. 4.3%, 1/2

Where should we put the poles of T (s)?

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Tim Davidson




Summary andplan

Pole positions

n 1 1/

2

s1, s2 = n jn

1 2 = n cos() jn sin()where = cos1().

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Tim Davidson




Summary andplan

Final design constraints

n 1 1/

2

p 2 p

2K

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Tim Davidson




Summary andplan

Caveat

Our work on transient response has been for systemswith

T (s) =2n

s2 + 2ns + 2n

What about other systems?

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Tim Davidson




Summary andplan

Poles, zeros and transientresponse

Consider a general transfer function T (s) = Y (s)R(s) Step response: Y (s) = T (s)1s Consider case with DC gain = 1; no repeated poles Partial fraction expansion

Y (s) =1s+i

Ais + i

+k

Bks + Cks2 + 2ks + (2k +

2k )

Step response

y(t) = 1 +i

Aiei t +k

Dkek t sin(k t + k )

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Tim Davidson




Summary andplan

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Tim Davidson




Summary andplan

Summary: Desirable properties

With H(s) = 1, E(s) = R(s) Y (s), L(s) = Gc(s)G(s),

E(s) =1

1 + L(s)R(s) G(s)

1 + L(s)Td (s) +

L(s)1 + L(s)

N(s)

Stability Good tracking in the steady state Good tracking in the transient Good disturbance rejection (good regulation) Good noise suppression Robustness to model mismatch

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Tim Davidson




Summary andplan

Plan: Analysis and designtechniques

Rest of course: about developing analysis and designtechniques to address these goals

Routh-Hurwitz: Enables us to determine stability without having to find

the poles of the denominator of a transfer function

Root locus Enables us to show how the poles move as a single

design parameter (such as an amplifier gain) changes

Bode diagrams There is often enough information in the Bode diagram

of the plant/process to construct a highly effectivedesign technique

Nyquist diagram More advanced analysis of the frequency response that

enables stability to be assessed even for complicatedsystems

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Tim Davidson

Steady-stateerror


Lecture 12(Given as the 11th lecture in 2011)

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 122 / 15

Tim Davidson

Steady-stateerror

Outline

1 Steady-state error

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Tim Davidson

Steady-stateerror

Steady-state error

E(s) = R(s) Y (s) = 11 +Gc(s)G(s)

R(s)

If the the conditions are satisfied, the final value theoremgives steady-state tracking error:

ess = limt

e(t) = lims0

s1

1 +Gc(s)G(s)R(s)

One of the fundamental reasons for using feedback, despitethe cost of the extra components, is to reduce this error.

We will examine this error for the step, ramp and parabolicinputs

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Tim Davidson

Steady-stateerror

Step input

ess = limt

e(t) = lims0

s1

1 +Gc(s)G(s)R(s)

Step input: R(s) = As ess = lims0 sA/s1+Gc(s)G(s) =

A1+lims0 Gc(s)G(s)

Now lets examine Gc(s)G(s). Factorize num., den.

Gc(s)G(s) =KM

i=1(s + zi)

sNQ

k=1(s + pk )

where zi 6= 0 and pk 6= 0. Limit as s 0 depends strongly on N. If N > 0, lims0 Gc(s)G(s) and ess = 0 If N = 0,

ess =A

1 +Gc(0)G(0)

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Tim Davidson

Steady-stateerror

System types

Since N plays such a key role,it has been given a name

It is called the type number

Hence, for systems of type N > 1,ess for a step input is zero

For systems of type 0, ess = A1+Gc(0)G(0)

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Tim Davidson

Steady-stateerror

Position error constant

For type-0 systems, ess = A1+Gc(0)G(0) Sometimes written as ess = A1+Kp

where Kp is the position error constant

Recall Gc(s)G(s) = KQM

i=1(s+zi )sNQQ

k=1(s+pk )

Therefore, for a type-0 system

Kp = lims0

Gc(s)G(s) =KM

i=1(zi)Qk=1(pk )

Note that this can be computed from positions of thenon-zero poles and zeros

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Tim Davidson

Steady-stateerror

Ramp input

The ramp input, which represents a step change invelocity is r(t) = At .

Therefore R(s) = As2 Assuming conditions of final value theorem are

satisfied,

ess = lims0

s(A/s2)1 +Gc(s)G(s)

= lims0

As + sGc(s)G(s)

= lims0

AsGc(s)G(s)

Again, type number will play a key role.

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Tim Davidson

Steady-stateerror

Velocity error constant

For a ramp input ess = lims0 AsGc(s)G(s) Recall Gc(s)G(s) = K

QMi=1(s+zi )

sNQQ

k=1(s+pk )

For type-0 systems, Gc(s)G(s) has no poles at origin.Hence, ess

For type-1 systems, Gc(s)G(s) has one pole at the origin.Hence, ess = AKv , where Kv =

KQ

i ziQk pk

Note Kv can be computed from non-zero poles and zeros Suggests formal definition of velocity error constant

Kv = lims0

sGc(s)G(s)

For type-N systems with N 2, for a ramp input ess = 0

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Tim Davidson

Steady-stateerror

Parabolic input

The parabolic input, which represents a step change inacceleration is r(t) = At2/2.

Therefore R(s) = As3 Assuming conditions of final value theorem are

satisfied,

ess = lims0

s(A/s3)1 +Gc(s)G(s)

= lims0

As2Gc(s)G(s)

Again, type number will play a key role.

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Tim Davidson

Steady-stateerror

Acceleration error constant

For a parabolic input ess = lims0 As2Gc(s)G(s) Recall Gc(s)G(s) = K

QMi=1(s+zi )

sNQQ

k=1(s+pk )

For type-0 and type-1 systems, Gc(s)G(s) has at most onepole at origin. Hence, ess

For type-2 systems, Gc(s)G(s) has two poles at the origin.Hence, ess = AKa , where Ka =

KQ

i ziQk pk

Again, Ka can be computed from non-zero poles and zeros Suggests formal definition of acceleration error constant

Ka = lims0

s2Gc(s)G(s)

For type-N systems with N 3, for a parabolic input ess = 0

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Tim Davidson

Steady-stateerror

Summary of steady-state errors

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Tim Davidson

Steady-stateerror

Robot steering system

Lets examine a proportional controller:

Gc(s) = K1

Gc(s)G(s) = K1K/(s + 1) Hence, type-0 system. Hence, for a step input,

ess =A

1 + Kp

where Kp = K1K .

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Tim Davidson

Steady-stateerror

Robot steering system

Lets examine a proportional-plus-integral controller:

Gc(s) = K1 +K2s

=K1s + K2

s

When K2 6= 0, Gs(s)G(s) = K (K1s+K2)s(s+1) Hence, type-1 system. Hence, for a step input, ess = 0 For ramp input,

ess =AKv,

where Kv = lims0 sGc(s)G(s) = KK2

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Tim Davidson

Steady-stateerror

Typical response

to a sawtooth input

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Tim Davidson

StabilityCondition in terms ofpoles

Condition in terms ofdenominatorcoefficients EE3CL4:

Introduction to Linear Control SystemsLecture 13

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 132 / 12

Tim Davidson


Condition in terms ofdenominatorcoefficients

Outline

1 StabilityCondition in terms of polesCondition in terms of denominator coefficients

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Tim Davidson



Stability

A systems is said to be stable if all bounded inputs r(t) giverise to bounded outputs y(t)

Counterexamples Albert Collins, Jeff Beck (Yardbirds),

Pete Townshend (The Who), Jimi Hendrix,Tom Morello (Rage Against the Machine),Kurt Cobain (Nirvana)

Tacoma Narrows

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Tim Davidson



Conditions for stability

y(t) =

g()r(t )d

Let r(t) be such that |r(t)| r

|y(t)| =

g()r(t )d

g()r(t )d r

g()dUsing this: system G(s) is stable iff

g()d is finite

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Tim Davidson



Condition in terms of poles?We want

g()d to be finite

Can we determine this from G(s)?

We can write a general rational transfer function in the form

G(s) =K

i(s + zi)sN

k (s + k )

m(s2 + 2ms + (2m + 2m))

Poles: 0, k , m jmAssuming N = 0 and no repeated roots, the impulse response is

g(t) =k

Akek t +m

Bmem t sin(mt + m)

Stability requires |g(t)|dt to be bounded;

that requires k > 0, m > 0In fact, system is stable iff poles have negative real parts

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Tim Davidson



Marginal stability

Consider G(s) = 1/s, simple pole at origin y(t) =

r(t)dt

if r(t) = cos(t), which is bounded,then y(t) = sin(t). Bounded

If r(t) = u(t), which is bounded,then y(t) = t . Not bounded

Consider G(s) = 1/(s2 + 1), simple poles at s = j1 Unit step response: u(t) cos(t). Bounded What if r(t) is a sinusoid of frequency 1/(2pi) rad/sec?

Not bounded

If G(s) has a pole with positive real part,or a repeated pole on j-axisoutput is always unbounded

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Tim Davidson



Routh-Hurwitz condition

We have seen how to determine stability from the poles.

Much easier than having to determine impulse response

Can we determine stability without having to determine thepoles?

Yes! Routh-Hurwitz condition

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Tim Davidson




Let G(s) = p(s)q(s) , where

q(s) = ansn + an1sn1 + . . . a1s + a0= an(s r1)(s r2) . . . (s rn)

where ri are the roots of q(s) = 0.

By multiplying out, q(s) = 0 can be written as

q(s) = ansn an(r1 + r2 + + rn)sn1+ an(r1r2 + r2r3 + . . . )sn2

an(r1r2r3 + r1r2r4 + . . . )sn3+ + (1)nan(r1r2r3 . . . rn) = 0

If all ri are real and in left half plane, what is sign of coeffs of sk?the same!

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Tim Davidson




That observation leads to a necessary condition.

Hence, not that useful for design

A more sophisticated analysis leads to the Routh-Hurwitzcondition, which is necessary and sufficient

Hence, can be quite useful for design

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Tim Davidson



R-H cond: A first lookConsider

ansn + an1sn1 + an2sn1 + + a1s + a0 = 0

Construct a table of the form

sn an an2 an4 . . .sn1 an1 an3 an5 . . .sn2 bn1 bn3 bn5 . . .sn3 cn1 cn3 cn5 . . .

......

...... . . .

s0 hn1

where

bn1 =an1an2 anan3

an1=1an1

an an2an1 an3

bn3 =1an1

an an4an1 an5 cn1 = 1bn1

an1 an3bn1 bn3

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Tim Davidson



R-H cond: A first lookConsider

ansn + an1sn1 + an2sn1 + + a1s + a0 = 0Construct a table of the form

sn an an2 an4 . . .sn1 an1 an3 an5 . . .sn2 bn1 bn3 bn5 . . .sn3 cn1 cn3 cn5 . . .

......

...... . . .

s0 hn1

Loosely speaking:

Number of roots in the right half plane is equal to the numberof sign changes in the first column of the table

Stability iff no sign changes in the first columnMore sophisticated statement in next lecture

EE 3CL4, L 141 / 19

Tim Davidson

Routh HurwitzconditionDisk drive readcontrol


Lecture 14

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 142 / 19

Tim Davidson


Outline

1 Routh Hurwitz conditionDisk drive read control

EE 3CL4, L 144 / 19

Tim Davidson


Stability

Let G(s) = p(s)q(s) , where

q(s) = ansn + an1sn1 + . . . a1s + a0

System is stable iff all poles of G(s) have negative real parts

Recall, poles are solutions to q(s) = 0

Can we find a necessary and sufficient condition that dependsonly on ak so that we dont have to solve q(s) = 0?

EE 3CL4, L 145 / 19

Tim Davidson


Routh-Hurwitz condition1 Consider, with an > 0

ansn + an1sn1 + an2sn2 + . . . a1s + a0 = 0

2 Construct a table of the form

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 bn5 . . .Row n 3 cn1 cn3 cn5 . . ....

......

... . . .Row 0 hn1

Procedure provided on the following slides

3 Count the sign changes in the first column

4 That is the number of roots in the right half plane

Stability (poles in LHP) iff ak > 0 and all terms in first col. > 0

EE 3CL4, L 146 / 19

Tim Davidson


Constructing RH table

ansn + an1sn1 + an2sn2 + . . . a1s + a0 = 0

Step 2.1: Arrange coefficients of q(s) in first two rows

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .

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Tim Davidson


Interlude

Determinant of a 2 2 matrix: a bc d = ad cb

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Tim Davidson



Step 2.2: Construct 3rd row using determinants of 2 2matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1

bn1 =1an1

an an2an1 an3

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Tim Davidson



Step 2.2, cont: Construct 3rd row using determinants of2 2 matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 . . .

bn3 =1an1

an an4an1 an5

EE 3CL4, L 1410 / 19

Tim Davidson



Step 2.3: Construct 4th row using determinants of 2 2matrices constructed from rows above

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 . . .Row n 3 cn1 . . .

cn1 =1bn1

an1 an3bn1 bn3

Step 2.4: Continue in this pattern.

Caveat: If all elements of first column are non-zeroWill come back to that. Lets see some examples, first

EE 3CL4, L 1411 / 19

Tim Davidson


RH table, second-order system

q(s) = a2s2 + a1s + a0

Row 2 a2 a0Row 1 a1 0Row 0 b1

b1 =1a1

a2 a0a1 0 = a0

Therefore, second order system is stable iff all three denominatorcoefficients have the same sign

EE 3CL4, L 1412 / 19

Tim Davidson


RH table, third order system

q(s) = a3s3 + a2s2 + a1s + a0

Row 3 a3 a1Row 2 a2 a0Row 1 b1 0Row 0 c1 0

b1 =1a2

a3 a1a2 a0 c1 = 1b1

a2 a0b1 0 = a0

Therefore, if a3 > 0, necessary and sufficient condition forthird-order system to be stable is that a2 > 0, b1 > 0 and a0 > 0.

b1 > 0 is equiv. to a2a1 > a0a3, and this implies a1 > 0.

EE 3CL4, L 1413 / 19

Tim Davidson


RH table, higher order systemsConsider a general case:

sn + an1sn1 + an2sn2 + + a1s + nn = 0

Normalize to natural frequency by defining s = s/n

sn+(an1/n)sn1+(an2/2n)sn2+ +(a1/n1n )s+1 = 0

EE 3CL4, L 1414 / 19

Tim Davidson


RH table, dealing with zeros

The Routh-Hurwitz table encounters trouble when thereis a zero in the first column

The next row involves (1/0) times a determinant When some other elements in that row are not zero, we

can proceed by replacing the zero by a small positivenumber , and then taking the limit as 0 after thetable has been constructed.

When a whole row is zero, we need to be a bit moresophisticated (see next lecture)

EE 3CL4, L 1415 / 19

Tim Davidson


RH table, zero first element innon-zero row

As an example, consider

q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10

Routh table

Row 5 1 2 11Row 4 2 4 10Row 3 0 6 0Row 2 c1 10 0Row 1 d1 0 0Row 0 10 0 0

c1 =4 12

=12

d1 =6c1 10

c1 6

Two sign changes, hence unstable with two RHP poles

EE 3CL4, L 1416 / 19

Tim Davidson


Disk drive read controlAdd velocity feedback (switch closed)

Using block diagram manipulation

G1(s) =5000

s + 1000Gs(s) =

1s(s + 20)

EE 3CL4, L 1417 / 19

Tim Davidson


Closed loop

T (s) =Y (s)R(s)

=KaG1(s)G2(s)

1 + KaG1(s)G2(s)(1 + K1s)

Hence, char. eqn:s3 + 1020s2 + (20000 + 5000KaK1)s + 5000Ka = 0

EE 3CL4, L 1418 / 19

Tim Davidson


Stabilizing values of K1 and Ka

s3 + 1020s2 + (20000 + 5000KaK1)s + 5000Ka = 0

Routh table

Row 3 1 20000 + 5000KaK1Row 2 1020 5000KaRow 1 b1Row 0 5000Ka

b1 =1020(20000 + 5000KaK1) 5000Ka

1020

For stability we require b1 > 0 and Ka > 0

For example, Ka = 100 and K1 = 0.05.That pair gives a 2% settling time of 260ms

EE 3CL4, L 151 / 16

Tim Davidson

Routh HurwitzrevisionDealing with zerorows

Applicationsof RouthHurwitzconditionTurning control of atracked vehicle


Lecture 15

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 152 / 16

Tim Davidson



Outline

1 Routh Hurwitz revisionDealing with zero rows

2 Applications of Routh Hurwitz conditionTurning control of a tracked vehicle

EE 3CL4, L 154 / 16

Tim Davidson



Recall Routh Hurwitz condition

1 Determine the characteristic polynomial(denominator of transfer function), with an > 0

q(s) = ansn + an1sn1 + an2sn2 + . . . a1s + a0

2 Construct the Routh Table

Row n an an2 an4 . . .Row n 1 an1 an3 an5 . . .Row n 2 bn1 bn3 bn5 . . .Row n 3 cn1 cn3 cn5 . . ....

......

... . . .Row 0 hn1

Construction procedure reviewed on next slide

3 System is stable iff ak > 0 and all terms in first col. > 0

EE 3CL4, L 155 / 16

Tim Davidson



Construction procedure

Row k + 2 p1 a3 p5 . . .Row k + 1 q1 q3 q5 . . .Row k r1 r3 r5 . . .

To compute r3, multiply 1first element of previous row = 1q1 by determinant of 22 matrix formed in the following way:

The first column contains the first elements of the tworows above the element to be calculated

The second column contains the elements of the tworows above that lie one column to the right of theelement to be calculated

Therefore

r3 =1q1

p1 p5q1 q5 = 1q1 (p1q5 q1p5)

EE 3CL4, L 156 / 16

Tim Davidson



Zero in the first column

When there is a zero in the first column, but otherelements in the row are not zero

Replace the zero by a small positive number, say , andonce the table has been constructed, take the limit as 0. (See previous lecture)

EE 3CL4, L 157 / 16

Tim Davidson



Zero row It is possible that the Routh Hurwitz procedure can

produce a zero row

While this complicates the procedure, it yields usefulinformation for design

Zero rows occur when polynomial has either equal andopposite roots on the real axis, or a pair of complexconjugate roots on the imaginary axis.The latter is more common, and more useful in design

So how can we deal with this? Routh Hurwitz procedure provides an auxiliary

polynomial that contains the roots of interest as factors The coefficients of this polynomial appear in the row

above the zero row We replace the zero row by the coefficients of the

derivative of the auxiliary polynomial

EE 3CL4, L 158 / 16

Tim Davidson



Zero row example q(s) = s5 + 2s4 + 24s3 + 48s2 25s 50 = 0 Construct table

Row 5 1 24 25Row 4 2 48 50Row 3 0 0

Auxiliary polynomial: q(s) = 2s4 + 48s2 50This is actually a factor of q(s).Using quadratic formula, roots of q(s) are s2 = 1,25Hence roots of q(s) are s = 1,j5

dq(s)ds = 8s3 + 96s.Replace zero row by these coefficients

Row 5 1 24 25Row 4 2 48 50Row 3 8 96

EE 3CL4, L 159 / 16

Tim Davidson



Zero row example, cont Now complete the table in the usual way

Row 5 1 24 25Row 4 2 48 50Row 3 8 96Row 2 24 50Row 1 112.7 0Row 0 50

One sign change in first column.Indicates one root in right half plane.

Recall q(s) is a factor of q(s).Indeed, by polyn division q(s) = (s + 2)q(s)We have seen that roots of q(s) are 1 and j5.

Hence q(s) does indeed have one root with a positivereal part.

EE 3CL4, L 1511 / 16

Tim Davidson



Turning control of a trackedvehicle

Select K and a so that the closed-loop is stable, and the steady-state error due to a ramp is at most 24% of

the magnitude of the command

EE 3CL4, L 1512 / 16

Tim Davidson



Deal with stability first

Transfer function: T (s) = Gc(s)G(s)1+Gc(s)G(s)

Char. equation: s4 + 8s3 + 17s2 + (K + 10)s + Ka = 0

Routh table

Row 4 1 17 KaRow 3 8 K+10Row 2 b3 KaRow 1 c3Row 0 Ka

b3 =126 K

8c3 =

b3(K + 10) 8Kab3

For stability we require b3 > 0, c3 > 0 and Ka > 0

EE 3CL4, L 1513 / 16

Tim Davidson



Stability regionThese constraints can be rewritten as

K < 126Ka > 0

(K + 10)(126 K ) 64Ka > 0For positive K , last constraint becomes a < (K+10)(126K )64K Region of stable parameters

EE 3CL4, L 1514 / 16

Tim Davidson



Steady-state error to ramp

For a ramp input r(t) = At , we have ess = A/Kv , where

Kv = lims0

sGc(s)G(s) = Ka/10

Therefore, ess = 10A/(Ka)

To obtain ess < 0.24A, we need Ka > 10/0.24 41.67,

Any (K ,a) pair in stable region with Ka > 41.67 willsatisfy design constraints

EE 3CL4, L 1515 / 16

Tim Davidson



Set of parameters with desiredperformance

For positive K , stability region is below the blue solid curve desired steady-state error region is above the red

dashed curve and below the blue solid curve Design example: (70,0.6)

EE 3CL4, L 1516 / 16

Tim Davidson



Ramp responseRamp response for case of K = 70 and a = 0.6

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Tim Davidson

The RootLocusProcedurePreliminaryexamples

Formal ProcedureEE3CL4:

Introduction to Linear Control SystemsLecture 16

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 162 / 18

Tim Davidson


Formal Procedure

Outline

1 The Root Locus ProcedurePreliminary examplesFormal Procedure

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Tim Davidson


Formal Procedure

Simple example

Transfer function T (s) = KG(s)1+KG(s)

Char. eqn: s2 + 2s + K = 0

Closed-loop poles: s1, s2 = 1

1 KWhat does this look like as K goes from 0 to +?

EE 3CL4, L 165 / 18

Tim Davidson


Formal Procedure

Simple example

EE 3CL4, L 166 / 18

Tim Davidson


Formal Procedure

Another example

Transfer function T (s) = KG(s)1+KG(s)

Consider K to be fixed

Char. eqn: s2 + as + K = 0

Closed-loop poles: s1, s2 = (aa2 4K )/2

What does this look like as a goes from 0 to +?

EE 3CL4, L 167 / 18

Tim Davidson


Formal Procedure

Another example

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Tim Davidson


Formal Procedure

What to do in the general case?

In the previous examples we exploited the simplefactorization of second order polynomials

To be truly useful, we need a more general procedure

EE 3CL4, L 169 / 18

Tim Davidson


Formal Procedure

Principles of general procedure

Transfer function T (s) = KG(s)1+KG(s) =p(s)q(s)

Closed loop poles are solutions to q(s) = 0

These are also solutions to 1 + KG(s) = 0

In polar form, |KG(s)|KG(s) = 1 + j0 = 1(180 + k360)

Therefore, for s0 to be a closed-loop pole, we must have

|KG(s0)| = 1 and KG(s0) = (180 + k360)where k is any integer

We will also keep in mind that R(s) and Y (s) correspond to realsignals. Hence, closed-loop poles are either real or occur incomplex-conjugate pairs

EE 3CL4, L 1610 / 18

Tim Davidson


Formal Procedure

In terms of poles and zerosFor s0 to be a closed-loop pole, we must have

|KG(s0)| = 1 and KG(s0) = (180 + k360)

Write G(s) = KGQM

i=1(s+zi )Qnj=1(s+pj )

, which means that the

open loop zeros are zi s; open loop poles are pi sFor s0 to be a closed-loop pole

|KKG|M

i=1 |s0 + zi |nj=1 |s0 + pj |

= 1

K + KG +Mi=1

(s0 + zi)n

j=1

(s0 + pi) = 180 + k360

Can we interpret these expressions in a geometric way?

EE 3CL4, L 1611 / 18

Tim Davidson


Formal Procedure

Vector difference Let u and v be complex numbers. Can you describe v u in geometric terms? Use the fact that v = u + (v u). That means that v u is the vector from u to v

v u = `ej. That is, |v u| is the length of the vector from u to v . (v u) is the angle of the vector from u to v

In our expressions we have terms of the forms0 + zi = s0 (zi) and s0 + pj = s0 (pj)

EE 3CL4, L 1612 / 18

Tim Davidson


Formal Procedure

Geometric interpretationMagnitude criterion:

|KKG|M

i=1 |s0 + zi |nj=1 |s0 + pj |

= 1

|KKG|M

i=1 distances from zeros of G(s) to s0nj=1 distances from poles of G(s) to s0

= 1

Phase criterion:

K + KG +Mi=1

(s0 + zi)n

j=1

(s0 + pi) = 180 + k360

K + KG +Mi=1

angles from zeros of G(s) to s0

n

j=1

angles from poles of G(s) to s0 = 180 + k360

EE 3CL4, L 1613 / 18

Tim Davidson


Formal Procedure

Formal ProcedureWe will first consider the case of K going from 0 to +

EE 3CL4, L 1614 / 18

Tim Davidson


Formal Procedure

Step 1 Write the characteristic equation as 1 + F (s) = 0 Rearrange so that the parameter of interest is

contained in the multiplier K in an exprn of the form

1 + KP(s) = 0,

where the numerator and denominator of P(s) aremonic polynomials (the coefficient of the highest powerof s is 1).

Factorize P(s) into poles and zeros, P(s) =QM

i=1(s+zi )Qnj=1(s+pj )

Hence characteristic equation is equiv. tonj=1(s + pj) + K

Mi=1(s + zi) = 0

Where does the locus start? Where are poles for K = 0? They are the poles of P(s). Mark each with an

EE 3CL4, L 1615 / 18

Tim Davidson


Formal Procedure

Step 1

nj=1

(s + pj) + KMi=1

(s + zi) = 0

Where do the poles end up? Where are poles for K ? Rewrite as (1/K )nj=1(s + pj) +Mi=1(s + zi) = 0 The zeros of P(s). Mark each with a Since M n there will often be zeros at, too

Summary: Root locus starts at poles of P(s) and ends atzeros of P(s)

Note: Often P(s) = Gc(s)G(s) and K is an amplifier gain.In that case, root locus (of the closed loop) starts at theopen loop poles and ends at the open loop zeros.

EE 3CL4, L 1616 / 18

Tim Davidson


Formal Procedure

Step 2Phase condition:

K +Mi=1

(s0 + zi)n

j=1

(s0 + pj) = 180 + k360

Recall that for K > 0, K = 0.What does this tell us when s0 is on the real axis?

Any complex conjugate pairs have no impact

EE 3CL4, L 1617 / 18

Tim Davidson


Formal Procedure

Step 2, cont.Phase condition for K > 0:M

i=1 (s0 + zi)n

j=1 (s0 + pj) = 180 + k360

Lets examine effects of poles on the real axis

For s0,1, all angles from poles to s0,1 are zero For s0,2, right pole generates an angle of 180, others zero For s0,3,

nj=1 (s0 + pj) = 360

For s0,4, n

j=1 (s0 + pj) = 540

Something similar for zeros.

Therefore: sections of real axis on the locus must lie to left of oddnumber of (real-valued) poles and (real-valued) zeros of P(s)

EE 3CL4, L 1618 / 18

Tim Davidson


Formal Procedure

ExampleP(s) = 2(s+2)s(s+4)

Step 1: Poles s = 0,4; Zeros s = 2

Step 2: Determine segments on real axis

In this case, this is enough to generate the complete rootlocus

EE 3CL4, L 171 / 15

Tim Davidson

Sketching theRoot LocusReview of Principles

Outline of steps

Review of Steps 1and 2

Step 3

Step 4


Lecture 17

Tim Davidson

McMaster University

Winter 2011

EE 3CL4, L 172 / 15

Tim Davidson


Outline of steps


Step 3

Step 4

Outline

1 Sketching the Root LocusReview of PrinciplesOutline of stepsReview of Steps 1 and 2Step 3Step 4

EE 3CL4, L 174 / 15

Tim Davidson


Outline of steps


Step 3

Step 4

Principles

We would like to know where the closed-loop poles goas a parameter of the loop (typically a controller designparameter)

mcmaster - intro to linear control systems

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