McMaster - Intro to Linear Control Systems
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Transcript of McMaster - Intro to Linear Control Systems
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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?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
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?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
Why are you here?
You might be interested in:
Designing industrial or biomedical robots
-
EE 3CL4, L 16 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
Why are you here?You might be interested in:
Designing humanoid robots
-
EE 3CL4, L 17 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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)
-
EE 3CL4, L 18 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
Why are you here?
You might be interested in:
Designing an automated insulin delivery system
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EE 3CL4, L 19 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
Why are you here?
You might be interested in:
Designing and analyzing financial systems
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EE 3CL4, L 111 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
What is a feedback controlsystem?
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EE 3CL4, L 112 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
What is a feedback controlsystem?
Example: You driving a car
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EE 3CL4, L 114 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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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EE 3CL4, L 116 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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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EE 3CL4, L 117 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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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EE 3CL4, L 118 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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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EE 3CL4, L 119 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
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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EE 3CL4, L 121 / 21
Tim Davidson
Why are youhere?
What is acontrolsystem?
What tools willwe use?
Administrativedetails
Parting shot
Parting shot
-
EE 3CL4, L 21 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
EE3CL4:Introduction to Linear Control Systems
Lecture 2
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 22 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Outline
1 What is control engineering
2 Examples
3 Design process
4 Disk drive example
-
EE 3CL4, L 24 / 19
Tim Davidson
What iscontrolengineering
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
-
EE 3CL4, L 25 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Model, open-loop, closed-loop
-
EE 3CL4, L 26 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Model, open-loop, closed-loop
What about disturbances? Mismatch (temperature, age)
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EE 3CL4, L 27 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Model, open-loop, closed-loop
-
EE 3CL4, L 28 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Multivariable control
Something for fourth year!
-
EE 3CL4, L 210 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Watts flyball governor
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EE 3CL4, L 211 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
CD player speed control:Open-Loop
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EE 3CL4, L 212 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
CD player speed control:Closed-Loop
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EE 3CL4, L 213 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Doritos
John MacGregor (Chem Eng): Visual feedback control of flavours
-
EE 3CL4, L 215 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Structured approach to design
-
EE 3CL4, L 217 / 19
Tim Davidson
What iscontrolengineering
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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EE 3CL4, L 218 / 19
Tim Davidson
What iscontrolengineering
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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EE 3CL4, L 219 / 19
Tim Davidson
What iscontrolengineering
Examples
Designprocess
Disk driveexample
Disk drive: Control architecture
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EE 3CL4, L 31 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
EE3CL4:Introduction to Linear Control Systems
Lecture 3
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 32 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Outline
1 Modelling physical systemsTranslational Newtonian MechanicsRotational Newtonian Mechanics
2 Linearization
3 Laplace transforms
-
EE 3CL4, L 34 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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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EE 3CL4, L 35 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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
)
-
EE 3CL4, L 36 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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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EE 3CL4, L 37 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Mass
F (t): resultant force in direction x
Mass: M
F (t) = Md2xm(t)dt2
= Mdvm(t)dt
= Mam(t)
-
EE 3CL4, L 38 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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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EE 3CL4, L 39 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Rotational damper
T (t): resultant torque in direction
Rotational viscous damper.b: rotational viscous friction coefficient
T (t) = b(d2(t)
dt d1(t)
dt
)= b
(2(t) 1(t)
)
-
EE 3CL4, L 310 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Rotational inertia
T (t): resultant torque in direction
Rotational inertia: J
T (t) = Jd2m(t)dt2
= Jdm(t)dt
= Jm(t)
-
EE 3CL4, L 311 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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)
-
EE 3CL4, L 312 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Example, continued
Md2y(t)dt
+ bdy(t)dt
+ ky(t) = r(t)
Resembles equation for parallel RLC circuit.
-
EE 3CL4, L 313 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Example, continued
Stretch the spring a little and hold. Assume an under-damped system. What happens when we let it go?
-
EE 3CL4, L 315 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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
-
EE 3CL4, L 316 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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
-
EE 3CL4, L 318 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
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
-
EE 3CL4, L 319 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Laplace transform pairs
can be tabulated
-
EE 3CL4, L 320 / 20
Tim Davidson
ModellingphysicalsystemsTranslationalNewtonianMechanics
Rotational NewtonianMechanics
Linearization
Laplacetransforms
Laplace transform pairs
Recall that complex poles come in conjugate pairs.
-
EE 3CL4, L 41 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
EE3CL4:Introduction to Linear Control Systems
Lecture 4
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 42 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Outline
1 Laplace transforms
2 Laplace transforms in action
-
EE 3CL4, L 44 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 45 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 47 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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)
-
EE 3CL4, L 48 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 49 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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)
-
EE 3CL4, L 410 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 411 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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!
-
EE 3CL4, L 412 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Critically damped case
s1 = s2 = n y(t) = c1ent + c2tent
y(0) = y0 = c1 = y0 dy(t)dt
t=0
= 0 = c1n + c2 = 0
-
EE 3CL4, L 413 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Underdamped case
s1, s2 = n jn
1 2 Therefore, |si | = n: poles lies on a circle Angle to negative real axis is cos1().
-
EE 3CL4, L 414 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 415 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
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
-
EE 3CL4, L 416 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 3
-
EE 3CL4, L 417 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 2.75
-
EE 3CL4, L 418 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 2.5
-
EE 3CL4, L 419 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 2.25
-
EE 3CL4, L 420 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 2
-
EE 3CL4, L 421 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 1.95
-
EE 3CL4, L 422 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 1.75
-
EE 3CL4, L 423 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 1.5
-
EE 3CL4, L 424 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 1.25
-
EE 3CL4, L 425 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 1
-
EE 3CL4, L 426 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 0.75
-
EE 3CL4, L 427 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 0.5
-
EE 3CL4, L 428 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 0.25
-
EE 3CL4, L 429 / 29
Tim Davidson
Laplacetransforms
Laplacetransforms inaction
Poles and transient response,b = 0
-
EE 3CL4, L 51 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
EE3CL4:Introduction to Linear Control Systems
Lecture 5
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 52 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Outline
1 Transfer function
2 Step response
3 Transfer function of DC motor
-
EE 3CL4, L 54 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Transfer function
Definition: Laplace transform of output over Laplacetransform of input when initial conditions are zero
Recall the mass-spring-damper system,
-
EE 3CL4, L 55 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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
-
EE 3CL4, L 57 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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.
-
EE 3CL4, L 58 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step response, b = 3
-
EE 3CL4, L 59 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 2.75
-
EE 3CL4, L 510 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 2.5
-
EE 3CL4, L 511 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 2.25
-
EE 3CL4, L 512 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 2
-
EE 3CL4, L 513 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 1.95
-
EE 3CL4, L 514 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 1.75
-
EE 3CL4, L 515 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 1.5
-
EE 3CL4, L 516 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 1.25
-
EE 3CL4, L 517 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 1
-
EE 3CL4, L 518 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 0.75
-
EE 3CL4, L 519 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 0.5
-
EE 3CL4, L 520 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 0.25
-
EE 3CL4, L 521 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Poles and step resp., b = 0
-
EE 3CL4, L 523 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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
-
EE 3CL4, L 524 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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?
-
EE 3CL4, L 525 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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
-
EE 3CL4, L 526 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Towards transfer function
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
-
EE 3CL4, L 527 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Towards transfer function
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
-
EE 3CL4, L 528 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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
-
EE 3CL4, L 529 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
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 :(
-
EE 3CL4, L 530 / 30
Tim Davidson
Transferfunction
Step response
Transferfunction of DCmotor
Second-order approximation
G(s) =(s)Va(s)
=Km
s[(Ra + sLa)(Js + b) + KbKm
] 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
-
EE 3CL4, L 61 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
EE3CL4:Introduction to Linear Control Systems
Lecture 6
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 62 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
Outline
1 Block diagram modelsExample: Loop transfer function
2 Block diagram transformations
-
EE 3CL4, L 64 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
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
-
EE 3CL4, L 65 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
Simple example
Y1(s) = G11(s)R1(s) +G12(s)R2(s) Y2(s) = G21(s)R1(s) +G22(s)R2(s)
-
EE 3CL4, L 66 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
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)?
-
EE 3CL4, L 68 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
Block diagram transformations
-
EE 3CL4, L 69 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
Using block diagramtransformations
-
EE 3CL4, L 610 / 10
Tim Davidson
Block diagrammodelsExample: Looptransfer function
Block diagramtransforma-tions
Using block diagramtransformations
-
EE 3CL4, L 71 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
EE3CL4:Introduction to Linear Control Systems
Lecture 7
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 72 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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
-
EE 3CL4, L 74 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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)
-
EE 3CL4, L 75 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
Transfer function
G(s) =(s)Va(s)
=Km
s[(Ra + sLa)(Js + b) + KbKm
]=
Kms(s2 + 2ns + 2n)
-
EE 3CL4, L 77 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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
-
EE 3CL4, L 78 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
Typical values
G(s) =(s)Va(s)
=Km
s(Ra + sLa)(Js + b)
G(s) =5000
s(s + 20)(s + 1000)
-
EE 3CL4, L 79 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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)
-
EE 3CL4, L 711 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
A simple feedback controller
Now that we have a model, how to control?
Here, Y (s) = (s).
-
EE 3CL4, L 712 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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?
-
EE 3CL4, L 713 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
Step responseResponse to r(t) = 0.1u(t)
-
EE 3CL4, L 715 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
Closed and open loop
-
EE 3CL4, L 716 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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)
-
EE 3CL4, L 717 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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)
-
EE 3CL4, L 718 / 18
Tim Davidson
Transferfunction ofarmaturecontrolled DCmotor
Application todisk drive readsystem
Our firstcontrol systemdesign
Characteristicsof feedback
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
-
EE 3CL4, L 81 / 11
Tim Davidson
Characteristicsof feedback
EE3CL4:Introduction to Linear Control Systems
Lecture 8
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 82 / 11
Tim Davidson
Characteristicsof feedback
Outline
1 Characteristics of feedback control systems
-
EE 3CL4, L 84 / 11
Tim Davidson
Characteristicsof feedback
The closed loop
For the whole of this lecture we will consider unity feedback:
H(s) = 1
-
EE 3CL4, L 85 / 11
Tim Davidson
Characteristicsof feedback
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)
-
EE 3CL4, L 86 / 11
Tim Davidson
Characteristicsof feedback
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)
)
-
EE 3CL4, L 87 / 11
Tim Davidson
Characteristicsof feedback
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?
-
EE 3CL4, L 88 / 11
Tim Davidson
Characteristicsof feedback
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
-
EE 3CL4, L 89 / 11
Tim Davidson
Characteristicsof feedback
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!
-
EE 3CL4, L 810 / 11
Tim Davidson
Characteristicsof feedback
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
-
EE 3CL4, L 811 / 11
Tim Davidson
Characteristicsof feedback
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
-
EE 3CL4, L 91 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
EE3CL4:Introduction to Linear Control Systems
Lecture 9(Given as the 10th lecture in Winter 2011)
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 92 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
Outline
1 More advantages of feedback
2 Price of feedback
3 Example: English Channel boring machines
4 Example: Disk drive read system
-
EE 3CL4, L 94 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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
-
EE 3CL4, L 95 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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)
-
EE 3CL4, L 96 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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
-
EE 3CL4, L 97 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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
-
EE 3CL4, L 99 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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
-
EE 3CL4, L 911 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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.
-
EE 3CL4, L 912 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
English Channel boringmachine
K = 100
-
EE 3CL4, L 913 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
English Channel boringmachine
K = 20
-
EE 3CL4, L 914 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
English Channel boringmachine
-
EE 3CL4, L 916 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
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.
-
EE 3CL4, L 917 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
Step responses for Ka = 10, 80
-
EE 3CL4, L 918 / 18
Tim Davidson
Moreadvantages offeedback
Price offeedback
Example:EnglishChannelboringmachines
Example: Diskdrive readsystem
Disturb. step resp. for Ka = 80
To reduce this response need larger Ka
but larger Ka will result in more oscilliatory response
-
EE 3CL4, L 101 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
EE3CL4:Introduction to Linear Control Systems
Lecture 10(Given as the 9th lecture in Winter 2011)
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 102 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Outline
1 Performance of feedback control systems
2 Performance of second-order systems
-
EE 3CL4, L 104 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Basic performance criteria
Stability (next week) Steady-state response to chosen inputs Transient response to chosen inputs Compromises: the art of design
-
EE 3CL4, L 105 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Typical test signals
Step, ramp, parabolic
-
EE 3CL4, L 107 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
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)
-
EE 3CL4, L 108 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
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):
-
EE 3CL4, L 109 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Typical step responses, fixed n
-
EE 3CL4, L 1010 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Typical step responses, fixed
-
EE 3CL4, L 1011 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Key parameters of(under-damped) step response
With =
1 2 and = cos1 ,
y(t) = 1 1ent sin(nt + )
-
EE 3CL4, L 1012 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Peak time and peak value
y(t) = 1 1ent sin(nt + )
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)
-
EE 3CL4, L 1013 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
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)
-
EE 3CL4, L 1014 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Overshoot vs Peak TimeThis is one of the classic trade-offs in control
-
EE 3CL4, L 1015 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Steady-state error, ess
In general this is not zero.
However, for our second-order system,
y(t) = 1 1ent sin(nt + )
Hence ess = 0
-
EE 3CL4, L 1016 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Settling time
y(t) = 1 1ent sin(nt + )
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
-
EE 3CL4, L 1017 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
Rise time (under-damped)
y(t) = 1 1ent sin(nt + )
How long to get to the target (for first time)? Tr , the smallest t such that y(t) = 1
-
EE 3CL4, L 1018 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
10%90% Rise time
What is Tr in over-damped case? Hence, typically use Tr1, the 10%90% rise time
-
EE 3CL4, L 1019 / 19
Tim Davidson
Performanceof feedbackcontrolsystems
Performanceofsecond-ordersystems
10%90% Rise time
Difficult to get an accurate formula Linear approx. for 0.3 0.8 (under-damped),
-
EE 3CL4, L 111 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
Summary andplan
EE3CL4:Introduction to Linear Control Systems
Lecture 11(Given as the 12th lecture in 2011)
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 112 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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
-
EE 3CL4, L 114 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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)
-
EE 3CL4, L 115 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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()
-
EE 3CL4, L 116 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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 .
-
EE 3CL4, L 117 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
Summary andplan
Properties of step response
Settling time, Ts,: 2% settling time 4n
Percentage overshoot: P.O. = 100e(pi/
12)
-
EE 3CL4, L 119 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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 )
-
EE 3CL4, L 1110 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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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EE 3CL4, L 1111 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
Summary andplan
Pole positions
n 1 1/
2
s1, s2 = n jn
1 2 = n cos() jn sin()where = cos1().
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EE 3CL4, L 1112 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
Summary andplan
Final design constraints
n 1 1/
2
p 2 p
2K
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EE 3CL4, L 1113 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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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EE 3CL4, L 1115 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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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EE 3CL4, L 1116 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
Summary andplan
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EE 3CL4, L 1118 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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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EE 3CL4, L 1119 / 19
Tim Davidson
Step responseof a class ofsecond-ordersystems(review)
A taste ofpole-placementdesign
Transientperformance,poles andzeros
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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EE 3CL4, L 121 / 15
Tim Davidson
Steady-stateerror
EE3CL4:Introduction to Linear Control Systems
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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EE 3CL4, L 124 / 15
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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EE 3CL4, L 125 / 15
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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EE 3CL4, L 126 / 15
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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EE 3CL4, L 127 / 15
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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EE 3CL4, L 128 / 15
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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EE 3CL4, L 129 / 15
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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EE 3CL4, L 1210 / 15
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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EE 3CL4, L 1211 / 15
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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EE 3CL4, L 1212 / 15
Tim Davidson
Steady-stateerror
Summary of steady-state errors
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EE 3CL4, L 1213 / 15
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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EE 3CL4, L 1214 / 15
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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EE 3CL4, L 1215 / 15
Tim Davidson
Steady-stateerror
Typical response
to a sawtooth input
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EE 3CL4, L 131 / 12
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
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
Outline
1 StabilityCondition in terms of polesCondition in terms of denominator coefficients
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EE 3CL4, L 134 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 135 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 136 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 137 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 138 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 139 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
Routh-Hurwitz condition
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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EE 3CL4, L 1310 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
Routh-Hurwitz condition
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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EE 3CL4, L 1311 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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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EE 3CL4, L 1312 / 12
Tim Davidson
StabilityCondition in terms ofpoles
Condition in terms ofdenominatorcoefficients
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
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EE 3CL4, L 141 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
EE3CL4:Introduction to Linear Control Systems
Lecture 14
Tim Davidson
McMaster University
Winter 2011
-
EE 3CL4, L 142 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Outline
1 Routh Hurwitz conditionDisk drive read control
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EE 3CL4, L 144 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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?
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EE 3CL4, L 145 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 146 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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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EE 3CL4, L 147 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Interlude
Determinant of a 2 2 matrix: a bc d = ad cb
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EE 3CL4, L 148 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Constructing RH table
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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EE 3CL4, L 149 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Constructing RH table
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
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EE 3CL4, L 1410 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Constructing RH table
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
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EE 3CL4, L 1411 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 1412 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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.
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EE 3CL4, L 1413 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 1414 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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)
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EE 3CL4, L 1415 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 1416 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
Disk drive read controlAdd velocity feedback (switch closed)
Using block diagram manipulation
G1(s) =5000
s + 1000Gs(s) =
1s(s + 20)
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EE 3CL4, L 1417 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 1418 / 19
Tim Davidson
Routh HurwitzconditionDisk drive readcontrol
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
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EE 3CL4, L 151 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
EE3CL4:Introduction to Linear Control Systems
Lecture 15
Tim Davidson
McMaster University
Winter 2011
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EE 3CL4, L 152 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
Outline
1 Routh Hurwitz revisionDealing with zero rows
2 Applications of Routh Hurwitz conditionTurning control of a tracked vehicle
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EE 3CL4, L 154 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 155 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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)
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EE 3CL4, L 156 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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)
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EE 3CL4, L 157 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 158 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 159 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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.
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EE 3CL4, L 1511 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 1512 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 1513 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 1514 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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
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EE 3CL4, L 1515 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
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)
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EE 3CL4, L 1516 / 16
Tim Davidson
Routh HurwitzrevisionDealing with zerorows
Applicationsof RouthHurwitzconditionTurning control of atracked vehicle
Ramp responseRamp response for case of K = 70 and a = 0.6
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EE 3CL4, L 161 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
Formal ProcedureEE3CL4:
Introduction to Linear Control SystemsLecture 16
Tim Davidson
McMaster University
Winter 2011
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EE 3CL4, L 162 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
Formal Procedure
Outline
1 The Root Locus ProcedurePreliminary examplesFormal Procedure
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EE 3CL4, L 164 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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 +?
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EE 3CL4, L 165 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
Formal Procedure
Simple example
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EE 3CL4, L 166 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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 +?
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EE 3CL4, L 167 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
Formal Procedure
Another example
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EE 3CL4, L 168 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 169 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 1610 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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?
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EE 3CL4, L 1611 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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)
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EE 3CL4, L 1612 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 1613 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
Formal Procedure
Formal ProcedureWe will first consider the case of K going from 0 to +
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EE 3CL4, L 1614 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 1615 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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.
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EE 3CL4, L 1616 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 1617 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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)
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EE 3CL4, L 1618 / 18
Tim Davidson
The RootLocusProcedurePreliminaryexamples
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
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EE 3CL4, L 171 / 15
Tim Davidson
Sketching theRoot LocusReview of Principles
Outline of steps
Review of Steps 1and 2
Step 3
Step 4
EE3CL4:Introduction to Linear Control Systems
Lecture 17
Tim Davidson
McMaster University
Winter 2011
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EE 3CL4, L 172 / 15
Tim Davidson
Sketching theRoot LocusReview of Principles
Outline of steps
Review of Steps 1and 2
Step 3
Step 4
Outline
1 Sketching the Root LocusReview of PrinciplesOutline of stepsReview of Steps 1 and 2Step 3Step 4
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EE 3CL4, L 174 / 15
Tim Davidson
Sketching theRoot LocusReview of Principles
Outline of steps
Review of Steps 1and 2
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)