Download - Student Lecture Notes Part-I 476 CSULB

Transcript
Page 1: Student Lecture Notes Part-I 476 CSULB

1

Lecture 1

Introduction

1. Control:

2. Regulation:

3. Tracking:

EX. TANK LEVEL CONTROL (REGULATION)

l(t)

lR(t)

Tank Level

Leak

RainEvaporation

Control Valve

• Controlled variable:

• Manipulated variable:

• Setpoint:

Page 2: Student Lecture Notes Part-I 476 CSULB

2

EX. MISSILE DEFENSE SYSTEM (TRACKING)

θR(t)

θ(t)

gear/platform

• Controlled variable:

• Manipulated variable:

• Tracking error:

4. Block diagram:

ControlSystem

Motor andPlatform Dynamics

θR(t)

input

v(t)

control

θ(t)

output

5. Feedback:

6. Closed-loop system:

7. Open-loop system:

Page 3: Student Lecture Notes Part-I 476 CSULB

3

EX. TYPICAL WASHING MACHINE CONTROLS (OPEN-LOOP)

Preset washingmachine controls

(Normal, Delicate, etc.)

Washing machineDynamics

dR(t) d(t)

Desired amountof dirt in laundry

u1(t)

un(t)

::

Various machinecontrol signals Actual amount

of dirt in laundry

open-loop: no means of measuringhow well machine is performing ...

In this course, we learn to design model-based control systems for various mechanical/electrical

systems.

8. Disturbance:

EX. DISTURBANCE – WIND FORCES ON OUR MISSILE DEFENSE SYSTEM

ControlSystem

DC MotorDynamics

PlatformDynamics

yr +_e u τ + +

τd

y

motor torque(actuation)

wind torque(disturbance)

9. Sensor (measurement) dynamics:

10. Noise:

Page 4: Student Lecture Notes Part-I 476 CSULB

4

What is a Control System? (Draw a block diagram to represent a control

system)

Page 5: Student Lecture Notes Part-I 476 CSULB

5

Lecture 2

Topics covered in this lecture:

1. System modeling in ODE, TF, SS forms and conversion among them

2. Simplification of block diagrams

Dynamic System Models

1. System Representations – four totally equivalent models of linear systems

• Ordinary Differential Equations (ODE)

• Transfer Functions (TF) – classical control representation

Derived using Laplace transform with zero initial conditions

A ratio of output(s)/input(s)

• State Space (SS) – modern control representation

BuAxx +=&

x – n dimensional state vector

A – n×n dimensional state matrix

u – m dimensional input vector

B – n×m dimensional input matrix

DuCxy +=

y – r dimensional output vector

C – r×n dimensional output matrix

D – r×m dimensional direct transmission matrix

• Block Diagram (BD) – a graphical representation in the s domain

2. Modeling Mechanical Systems – Newton’s Second Law

• Translating

• Rotating

• Elements:

EX. A SIMPLE SYSTEM: CRUISE CONTROL

Assumption: 1) No rotational inertia of the wheel

2) The friction force is proportional to the car’s speed

Find: 1) EOM

2) TF (with velocity as an output)

3) SS (with displacement as an output)

Page 6: Student Lecture Notes Part-I 476 CSULB

6

Solution:

3. Modeling Electrical Systems

• Kirchoff’s Voltage Law 0=∑ iV (circuit loop)

• Kirchoff’s Current Law 0=∑ ji (current node)

Page 7: Student Lecture Notes Part-I 476 CSULB

7

• Elements:

4. Modeling Electromechanical Systems

EX. MODELING A DC MOTOR

Find: 1) ODE 2) SS (with angular velocity as an output) 3) TF

Solution:

Page 8: Student Lecture Notes Part-I 476 CSULB

8

Block Diagram

a) Recall from the first lecture: a functional schematic showing the major components and

interconnections of a control system.

b) If the blocks contain TFs, a BD is another totally equivalent linear system representation

(BDs are s-domain representations)

c) BDs are especially useful system representations

• They are graphical (intuitive)

• They are algebraic (due to ‘s’)

• Model simulation programs (Simulink incorporate them directly)

d) Block Diagram Algebra:

Converting from BDs to TFs is not quite as easy, but a handful of rules (block diagram

algebra) make it straightforward.

Rule 1 Blocks in Series

Equivalent TF:

EX. BLOCK DIAGRAM FOR OPEN LOOP CONTROL SYSTEM

Find: the open loop TF

Solution:

Rule 2 Blocks in Parallel

Equivalent TF:

Rule 3 Negative Feedback

_

+ E(s) Y(s) Yr(s) G(s)

Y(s) U(s)

G1(s)

G2(s)

Y(s) Yr(s) s

KKC(s) i

p += 1

1

+sτ

Y(s) U(s) G1(s) G2(s)

Page 9: Student Lecture Notes Part-I 476 CSULB

9

Rule 4 Feedback with Sensor Dynamics (Negative Feedback)

EX. TRANSFER FUNCTION FROM A SIMPLE BLOCK DIAGRAM

+ +

_

+ Y R

s

4

2

s

1

_

+ E(s) Y(s) Yr(s)

G(s)

H(s)

Page 10: Student Lecture Notes Part-I 476 CSULB

10

EX. SIMPLIFY INTERCONNECTED BD LOOPS

Solution:

+ +

_

+ Y R

G(s)

H2(s)

H1(s)

Page 11: Student Lecture Notes Part-I 476 CSULB

1

Lecture 3-4

Topics covered in this lecture:

1. Concepts of poles, zeros and stability

2. Transient responses of first-order and second-order systems

Poles and Zeros

Recall that any linear system can be expressed as a transfer function, which is an output/input

ratio of polynomials in s:

011

1

011

1

)(

)()(

asasasa

bsbsbsb

sa

sb

input

outputsH

nn

nn

mm

mm

++++

++++===

−−

−−

L

L=

Where n ≥ m (for a proper TF), and ai and bj are constant coefficients.

a). Zeros:

b). Poles:

c). Stability of Linear Control System

A Linear Time Invariant (LTI) system is stable iff

EX. EVALUATE STABILITY

Given the system’s TF as

)2)(5.0(

5.0)(

+−

+=

sss

ssG

What are the sytem’s zeros and poles? Is the system stable? Why or why not?

EX. EVALUATE STABILITY

Given: A feedback control system with a proportional controller as shown

Find the closed-loop system’s poles and zeros, and determine the closed-loop system’s stability

for K =10, K=5 and K=1.5, respectively.

Solution:

_ +

R(s) K

sss

ssG

−+

+=

23 5.1

5.0)( Y(s)

Page 12: Student Lecture Notes Part-I 476 CSULB

2

%MATLAB commands

>> K = 5 %input the K value; try K=5, 10, or 15 >>num = K*[1 0.5] %define the numerator polynomial of the TF

>>den = [1 1.5 K-1 0.5*K] %define the denominator polynomial of the TF

>>sysTF = tf(num, den) %define system’s TF

>>roots(den) %find poles

>>pole(sysTF) %find poles

>>roots(num) %find zeros

>>zero(sysTF) %find zeros

>>pzmap(sysTF) %display the poles and zeros on s-plane

>>step(sysTF) %find system’s response for a unit step input

Analysis of Linear Systems

– Given a linear system model, how does this model respond with time for an input?

How to find a system’s response?

1). Solve ODE analytically

2). Use Inverse Laplace Transform (ILT) (by using Laplace Transforms Tables)

3). Use Numerical Integration, e.g. MATLAB simulink

EX. FINDING TIME RESPONSE USING INVERSE LAPLACE TRANSFORM

Given: the first-order linear model

ODE: )(5 tuyy =+& 5

1

)(

)( :

+=

ssU

sYTF

for the input u(t)= ustep(t), with zero initial condition y(0) =0.

Solution:

0 5 10 15-1

-0.5

0

0.5

1

1.5

2

2.5

3Step Response

Time (sec)

y

Kp=1.5

Kp=5

Kp=10

Page 13: Student Lecture Notes Part-I 476 CSULB

3

%MATLAB commands

>> num = [1];

>> den =[1 5];

>> sysTF = tf(num,den);

>> step(sysTF,5)

>> hold

>> t=[0:0.1:5];

>> y = 0.2- 0.2*exp(-5*t);

>> plot(t,y,'-o')

Use Matlab/simulink: download the simulink model from \Beachboard\content\matlab folder

Transient Responses of Linear Systems 1. Introduction

We can analyze the transient response (overshoot, settling time, etc.) by solving for y(t), but it

is much easier to understand how transient response relates to pole locations.

2. First-Order System

Standard form of a first-order system:

Unit Step Response of First-Order Systems

Since the Laplace transform of the unit-

step function iss

1, we obtain

=)(sY

Express Y(s) into partial fractions gives

=)(sY

Taking the inverse Laplace transform,

we obtain

=)(ty

Page 14: Student Lecture Notes Part-I 476 CSULB

4

• Transient response of first-order systems – think in terms of the s-plane

EX. SYSTEM WITH TWO SIMPLE REAL POLES

Given: A system model )10)(1(

1

)(

)(

++=

sssU

sY

Find: The time response of this system to a unit step input.

Solution:

%Matlab t=[0:0.1:4]; y = 0.1-0.111*exp(-

t)+0.0111*exp(-10*t); y2 = 0.1-0.111*exp(-t); figure plot(t,y) hold plot(t,y2,'--')

y3 = exp(-t); y4 = exp(-10*t) figure plot(t,y3) hold plot(t,y4,'--')

0 0.5 1 1.5 2 2.5 3 3.5 4-0.02

0

0.02

0.04

0.06

0.08

0.1

time (s)

Response

y=0.1-0.111*e(-t)

+0.0111*e(-10t)

y2=0.1-0.111*e(-t)

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time (s)

Res

po

nse

e(-t)

e(-10t)

Page 15: Student Lecture Notes Part-I 476 CSULB

5

3. Second-Order System

Standard form of a second-order system:

(a)

(b)

EX. Given a second-order linear system model )()(500)(50)(5 tFtxtxtx =++ &&&

Find: The natural frequency and damping ratio for this system.

Solution:

(c) Second order systems have two poles, which can be computed using quadratic equation.

Case 1: Overdamped Response: ζ >1.0

If 012 >−ζ , then 12 −ζ is a real number, and the system doesn’t have complex

conjugate poles. It has two simple, real poles.

EX. STEP RESPONSE OF AN OVERDAMPED SECOND-ORDER SYSTEM

22

2

2)(

)()(

nn

n

ssU

sYsG

ωζω

ω

++==

)1)(1( 22

2

−−+−++=

ζωζωζωζω

ω

nnnn

n

ss

)1)(1(

1)(

22

2

−−+−++=

ζωζωζωζω

ω

nnnn

n

ssssY

−+=

−−

212

21

121)(

s

e

s

ety

tstsn

ζ

ω

Where ns ωζζ )1( 21 −+= and ns ωζζ )1( 2

2 −−= .

The response includes two decaying exponential terms.

Page 16: Student Lecture Notes Part-I 476 CSULB

6

Case 2: Critically Damped System: ζ =1.0

If 012 =−ζ , then the system poles are repeated and np ω−=2,1 .

The unit step response is )1(1)( tety n

tn ωω +−= −

Case 3: Underdamped Response: ζ <1.0

EX. STEP RESPONSE OF AN UNDERDAMPED SECOND-ORDER SYSTEM

22

2

2

1)(

nn

n

sssY

ωζω

ω

++=

2222)()(

1

dn

n

dn

n

ss

s

s ωζω

ζω

ωζω

ζω

++−

++

+−=

Taking inverse Laplace gives

−+

−−= −

ζ

ζω

ζ

ζω 21

2

1tansin

11)( t

ety d

tn

, ζ <1.0

By setting the natural frequency nω =constant, use Matlab command plot to plot

the step responses for a standard 2nd

order system for different damping ratio

values ζ =[0:0.1:1 2].

Use Matlab/simulink: download the simulink model from \Beachboard\content\matlab folder

Page 17: Student Lecture Notes Part-I 476 CSULB

7

(d) Responses can be easily predicted by viewing pole locations in the s-plane.

In summary: the stability and transient response characteristics are completely determined by pole

locations.

Page 18: Student Lecture Notes Part-I 476 CSULB

1

Lecture 5

Topics covered in this lecture:

1. Relate time-domain specifications (rise time, settling time, overshoot, etc.) to pole locations

in s-domain;

Time-Domain Specifications

1. Rise Time rt :

2. Settling Time st :

3. Overshoot pM :

4. Peak Time pt :

Note: Second-Order System 22

2

2)(

)(

nn

n

sssU

sY

ωζω

ω

++=

Step Response

−+

−=−

ζ

ζω

ζ

ζω 21

2

1tansin

11)( t

ety d

tn

+−=

−tte d

d

dt

ωω

σω

σ sincos1

Steady State Value: yss = ?

Relate time-domain specifications to the pole-location parameters ζ and nω

Page 19: Student Lecture Notes Part-I 476 CSULB

2

Rise Time ≅rt

Overshoot and Peak Time: =pM

tp =

+−=

−ttety d

d

dt

ωω

σω

σ sincos1)(

When y(t) reaches its maximum value at peak time, its derivative will be zero.

+−−

+=

−−ttettety dd

d

ddt

d

d

dt

ωωω

σωωω

ω

σωσ

σσ cossinsincos)(&

tetetete ddt

d

d

tdd

d

td

tωωω

ω

σσωω

ω

σωσ

σσσσ sinsincoscos −−−−++−=

te dd

d

tωω

ω

σσsin

2

+=

−= 0

Note: This occurs when 0sin =tdω .

So πω =pd t � d

ptω

π=

The overshoot 21

sincos1)(ζ

ζπ

ω

σπ

πω

σπ

−−

=

+=−= eetyM

d

ppd 10 <≤ ζ

Page 20: Student Lecture Notes Part-I 476 CSULB

3

Settling Time: =st for ∆=2%; =st for ∆=5%;

Design Synthesis:

Note: (a) Qualitative guides, not precise design formulas;

(b) Providing only a starting point for the design iteration;

(c) Checking time response to verify the time specifications after control design.

Page 21: Student Lecture Notes Part-I 476 CSULB

4

EX. TRANSFORMATION OF THE SPECIFICATIONS TO THE S-PLANE (EX. 3.27)

Given: system response requirements 6.0≤rt sec, %10≤pM , and 3≤st sec

Find: (a). allowable regions in the s-plane for the poles;

(b). select a pair of poles in the design region in (a) and construct the corresponding

standard 2nd

order TF;

(c). use the MATLAB command “step()” to verify your selection.

Solution:

Page 22: Student Lecture Notes Part-I 476 CSULB

1

Lecture 6

Topics covered in this lecture:

1. Learn the concept of system type

2. Learn the relationship between system type and steady-state error

System Type

Consider a unity-feedback control system

The open-loop transfer function

)1()1)(1(

)1()1)(1()(

21 +++

+++=

sTsTsTs

sTsTsTKsG

pN

mba

L

L

System type:

EX. DETERMINE SYSTEM TYPE

Find: System Type if system’s open-loop TF is given as

(a) )3)(2(

)1(5

++

+=

sss

sG (b)

3

1)(

ssG =

Steady-state error

E(s) R(s) Y(s)

_ + G(s)

Page 23: Student Lecture Notes Part-I 476 CSULB

2

Test Signals:

EX. SYSTEM TYPE AND STEADY-STATE TRACKING ERROR

Given: the Type 1 system of the previous example

)3)(2(

)1(5

++

+=

sss

sG

Show: that its steady-state tracking errors are

(a) 0=sse for a unit step input

(b) =sse constant 0≠ for a unit ramp input

(c) ∞=sse for a unit parabolic input

Solution:

Page 24: Student Lecture Notes Part-I 476 CSULB

3

Steady-state error for a step Input

In general, the steady-state error of the system for a unit step input is

Define the Static Position Error Constant pK as

Thus, the steady-state error for the unit-step input can be summarized as follows:

Steady-state error for a ramp input

Define the Static Velocity Error Constant vK as

Thus, the steady-state error for the unit-ramp input can be summarized as follows:

Page 25: Student Lecture Notes Part-I 476 CSULB

4

Steady-state error for a parabola input

Define the Static Acceleration Error Constant aK as

Thus, the steady-state error for the unit-parabola input can be summarized as follows:

Step Input Ramp Input Parabola Input

Type 0

Type 1

Type 2

EX. STEADY-STATE ERROR AND SYSTEM TYPE WITH CONTROL

Given: A feedback control system

Determine: the steady-state error and the system type

(a) with pksC =)( (P control)

(b) with s

kksC I

p +=)( (PI control)

Solution:

(a) pksC =)(

(b) s

kksC I

p +=)(

E(s) R(s) Y(s) _

+ )15)(2(

1

++ ss C(s)

Page 26: Student Lecture Notes Part-I 476 CSULB

1

Lectures 7-8 Topics covered: 1. Discuss pros and cons of PID control 2. Learn to design PID controllers Feedback Control System PID Control (Three-Term Controller) 1. Proportional Control

kp G(s)R(s) Y(s)E(s) U(s)+

_

EX. STABILITY OF A THIRD-ORDER SYSTEM WITH P CONTROL

kp G(s)R(s) Y(s)E(s) U(s)+

_

Given: 6116

1

)3)(2)(1(

1)(

23

sssssssG ,

check the system stability for pk =1, 10, 40, 58, 61

Solution:

Page 27: Student Lecture Notes Part-I 476 CSULB

2

2. PI (Proportional Plus Integral) Control

G(s)R(s) Y(s)E(s) U(s)+

_ s

kk I

p

EX. STEADY-STATE ERROR FOR PI CONTROL: Given ))((

1)(

bsassG

3. PD (Proportional Plus Derivative) Control

G(s)R(s) Y(s)E(s) U(s)+

_skk Dp

EX. PD CONTROL OF A MARGINALLY STABLE SYSTEM

Given: 4

1)(

2

ssG

Determine: The closed-loop stability with PD control Solution:

Page 28: Student Lecture Notes Part-I 476 CSULB

3

4. PID (Proportional-Integral-Derivative) Control

G(s)R(s) Y(s)E(s) U(s)+

_sk

s

kk D

Ip

The effects of increasing each of the gains Kp, KI, KD of a PID controller can be summarized as Term Rise time Overshoot Settling time S.S. error

Kp

KI

KD

EX. P- & PD CONTROL DESIGN

Given: a second-order system )16(

4)(

sssG

1). Try to design a P-controller to satisfy the following performance requirements: Mp ≥ 10%, ts ≤ 0.4s for =2%; 2). Design a PD controller such that the response of the closed-loop system has an overshoot of less than 10% and rise time of faster than 0.1 seconds.

Solution:

Y(s) R(s) _ + )(sGskk Dp

Page 29: Student Lecture Notes Part-I 476 CSULB

4

PID Tuning Technique – Ziegler-Nichols Tuning Quarter Decay Tuning Examine a system’s open-loop unit step response.

P control: RL

k p

1

PI control: RL

k p

9.0

3.0

LTI

PID control: RL

k p

2.1 LTI 2 LTD 5.0

� � � � � �

� � � � � �

Page 30: Student Lecture Notes Part-I 476 CSULB

5

Ultimate Stability Technique

Examine a system’s closed-loop response using P controller only. Keep increasing pk until the

system goes marginally unstable. This gain is the ultimate gain uK . The period of oscillation is

the ultimate period uP .

P control up Kk 5.0

PI control up Kk 45.0 2.1u

I

PT

PID control up Kk 6.0 uI PT2

1 uD PT

8

1

Ex.

� �

Page 31: Student Lecture Notes Part-I 476 CSULB

1

Lecture 9

Topic covered in this lecture:

1. Review concept of stability

2. Determine stability using Routh’s stability criterion.

Stability – The most important and fundamental design specification of a control system

Consider a Linear Time-Invariant (LTI) system:

)(

)(

)(

)(

)(

)(

1

1

011

1

011

1

ini

imi

nn

nn

mm

mm

ps

zsK

asasasa

bsbsbsb

sa

sb

sU

sY

−Π

−Π=

++++

++++==

=

=−

−−

L

L where m ≤ n

Routh’s Stability Criterion

• An alternative to locating the roots of the CE without solving for the roots.

• Useful for determining the ranges of coefficients of polynomials for stability, especially when

the coefficients are in symbolic form.

Consider the characteristic equation of an nth-order system:

0)( 11

1 =++++= −−

nnnn

asasassa L

• A necessary (but not sufficient) condition for stability:

• A system is stable if and only if

Routh Array

Row n n

s :

Row n-1 1−n

s :

Row n-2 2−n

s :

M

Row 2 2

s :

Row 1 s :

Row 0 0

s :

The number of RHP (unstable) poles = the number of sign changes in the first column of the

Routh array.

Page 32: Student Lecture Notes Part-I 476 CSULB

2

EX. ROUTH’S TEST

Given the closed-loop C.E. of a control system as 010532)(234 =++++= sssssa

Check the stability of this system using Routh’s criterion

Solution:

Page 33: Student Lecture Notes Part-I 476 CSULB

3

EX. STABILITY VERSUS PARAMETER RANGE

Given:

The stability properties of the system are a function of the proportional feedback gain pK .

Determine: the range of pK over which the system is stable.

Solution:

0 5 10 15-1

-0.5

0

0.5

1

1.5

2

2.5

3Step Response

Time (sec)

y

Kp=1.5

Kp=5

Kp=10

_ + Y(s) R(s) pK

)2)(5.0(

5.0)(

+−

+=

sss

ssG