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CSULB 476 Lecture Notes

### Transcript of 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

RainEvaporationControl Valve

Controlled variable:

Manipulated variable:

Setpoint:

• 2

EX. MISSILE DEFENSE SYSTEM (TRACKING)

R(t)

(t)

gear/platform

Controlled variable:

Manipulated variable:

Tracking error:

4. Block diagram:

ControlSystem

Motor andPlatform Dynamics

R(t)input

v(t)control

(t)output

5. Feedback:

6. Closed-loop system:

7. Open-loop system:

• 3

EX. TYPICAL WASHING MACHINE CONTROLS (OPEN-LOOP)

Preset washingmachine controls

(Normal, Delicate, etc.)Washing machine

DynamicsdR(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 + +

d

y

motor torque(actuation)

wind torque(disturbance)

9. Sensor (measurement) dynamics:

10. Noise:

• 4

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

• 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 nn dimensional state matrix u m dimensional input vector B nm dimensional input matrix

DuCxy += y r dimensional output vector C rn dimensional output matrix D rm dimensional direct transmission matrix

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

2. Modeling Mechanical Systems Newtons 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 cars speed Find: 1) EOM

2) TF (with velocity as an output) 3) SS (with displacement as an output)

• 6

Solution:

3. Modeling Electrical Systems Kirchoffs Voltage Law 0= iV (circuit loop)

Kirchoffs Current Law 0= ji (current node)

• 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:

• 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) ip += 11+s

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

• 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)

• 10

EX. SIMPLIFY INTERCONNECTED BD LOOPS

Solution:

+ +

_

+ Y R

G(s)

H2(s)

H1(s)

• 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

bsbsbsbsa

sbinputoutput

sHn

nn

n

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 systems TF as

)2)(5.0(5.0)(

+

+=

sss

ssG

What are the sytems 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 systems poles and zeros, and determine the closed-loop systems stability for K =10, K=5 and K=1.5, respectively. Solution:

_

+ R(s) K sss

ssG

+

+= 23 5.1

5.0)( Y(s)

• 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 systems 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 systems 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 systems 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)()(

:+

=

ssUsYTF

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

• 3

%MATLAB commands >> num = ; >> 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

• 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

)()(

++=

sssUsY

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)

Respo

nse

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

pon

se

e(-t)

e(-10t)

• 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: >1.0 If 012 > , then 12 is a real number, and the system doesnt have complex conjugate poles. It has two