linear control systems

LINEAR CONTROLLINEAR CONTROL SYSTEMS SYSTEMS

Ali Karimpour

Assistant Professor

Ferdowsi University of Mashhad

2

lecture 5

Ali Karimpour Oct 2009

Lecture 5

Nonlinear systems

Topics to be covered include:

Nonlinear systems

Linearization of nonlinear systems

More study on transfer function representation

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lecture 5


Nonlinear systems.

Although almost every real system includes nonlinear features, many systems can be reasonably described, at least within certain operating ranges, by linear models.

سیستمهای غیر خطی

گرچه تقریبا تمام سیستمهای واقعی دارای رفتار غیر خطی هستند، بسیاری از سیستمها را می توان حداقل

در یک رنج کاری خاص خطی نمود.

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lecture 5


Nonlinear systems.سیستمهای غیر

خطی

Say that {xQ(t), uQ(t), yQ(t)} is a given set of trajectories that satisfy the above equations, so we have

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Linearized system سیستم خطی

شده

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Example 1

Consider a continuous time system with true model given by

Assume that the input u(t) fluctuates around u = 2. Find an operating point with uQ = 2 and a linearized model around it.

9

16

3

202

2

QQQ xxu

Operating point:

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lecture 5


Example 1 (Continue)

8

3

2

1| ,

Q

uuxxxx

fA

QQ

3

4

3

2| ,

Quuxx uu

fB

QQ

Operating point:

uxx 3

4

8

3

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Simulation شبیه سازی

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Example 2 (Pendulum)

22

2

,sin mlJmgluldt

dJ

sin

2

2

l

g

ml

u

dt

d 21 , xx

Find the linear model around equilibrium point.

θ l

u

mg12

21

sin xl

g

ml

ux

xx

pointoperatingis

021 QQQ uxx

u

mlx

x

l

gx

x

1

0

0

10

2

1

2

1

Remark: There is also another equilibrium point at x2=u=0, x1=π

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Example 3 Suppose electromagnetic force is i2/y and find linearzed model around y=y0

dt

diLiRte

y

tiMg

dt

ydM

1

2

2

2

)(

)(

ixyxyx 321 ,,

L

tex

L

Rx

x

x

Mgxxx

)(13

13

1

23

221

01 yx Q

M

y

01 MgyReQ 02 Qx 0Q3 i Mgyx Q

Equilibrium point:

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lecture 5


Example 3 Suppose electromagnetic force is i2/y and find linearzed model around y=y0

L

tex

L

Rx

x

x

Mgxxx

)(13

13

1

23

221

0100321 ,,0,,,, MgyRMgyyexxx QQQQ M

y Equilibrium point:

L

tex

L

Rx

xMy

gx

y

gx

xx

)(

2

31

3

30

10

2

21

)(

1

0

0

00

20

010

3

2

1

1

003

2

1

te

L

x

x

x

L

R

My

g

y

g

x

x

x

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Example 4 Consider the following nonlinear system. Suppose u(t)=0 and initial condition is x10=x20=1. Find the linearized system around

response of system.

)()()(

)(

1)(

12

22

1

txtutx

txtx

1)(0)(.0)( 212 atxtxtx

)(1

0

00

20

2

1

2

1 tutx

x

x

x

1)(1)( 11 tbttxtx

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Linear model for time delay

se

........62

1

11

33

22

ssse s

22

21

1

ss

Pade approximations

s

21

21

se

s

s

e

e

2

2

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lecture 5


Example 5(Inverted pendulum)

Figure 3.5: Inverted pendulum

In Figure 3.5, we have used the following notation:

y(t) - distance from some reference point(t) - angle of pendulumM - mass of cartm - mass of pendulum (assumed concentrated at tip)l - length of pendulumf(t) - forces applied to pendulum

پاندول معکوس5مثال

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Example of an Inverted Pendulum

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Nonlinear model

Application of Newtonian physics to this system leads to the following model:

where m = (M/m)

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Linear model

This is a linear state space model in which A, B and C are:

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TF model properties خواص مدل تابع انتقال

1- It is available just for linear time invariant systems. (LTI)

2- It is derived by zero initial condition.

3- It just shows the relation between input and output so it may lose some information.

4- It is just dependent on the structure of system and it is independent to the input value and type.

5- It can be used to show delay systems but SS can not.

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lecture 5


Poles of Transfer functionقطب تابع انتقال

Pole: s=p is a value of s that the absolute value of TF is infinite.

How to find it?

)(

)()(

sd

snsG Transfer function

Let d(s)=0 and find the roots

Why is it important? It shows the behavior of the system.Why?????

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lecture 5


Example 1

65

6)(

2

ss

ssG

3

1

2

21

)65(

6)(ofResponseStep

2

ssssss

ssc

S2+5s+6=0 p1= -2, p2=-3

Transfer function

roots([1 5 6])

)(1)(2)(1)( 32 tuetuetutc tt

p1= -2, p2=-3

Poles and their physical meaning

قطبها و خواص فیریکی آنها

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lecture 5


Example 1 (Continue)

65

6)(

2

ss

ssGTransfer function

step([1 6],[1 5 6])

0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plitu

de

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lecture 5


Zeros of Transfer functionصفر تابع انتقال

Zero: s=z is a value of s that the absolute value of TF is zero.

Who to find it?

)(

)()(

sd

snsG Transfer function

Let n(s)=0 and find the roots

Why is it important? It also shows the behavior of system.Why?????

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lecture 5


Example 2

65

610)(

65

6)(

2221

ss

ssG

ss

ssG

)(121)(3

1

2

21

)65(

6)( 32

121 tueetcssssss

ssc tt

Transfer functions

z1= -6 z2=-0.6

)(871)(3

8

2

71

)65(

610)( 32

222 tueetcssssss

ssc tt

Zeros and their physical meaning

صفرها و خواص فیریکی آنها

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lecture 5


Example 2

Transfer function65

610)(

65

6)(

2221

ss

ssG

ss

ssG

step([1 6],[1 5 6]) ;hold on;step([10 6],[1 5 6])

Zeros and their physical meaning


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lecture 5


Example 3

uuyyy 65

Zeros and its physical meaning

Transfer function of the system is:

65

1

)(

)(2

ss

s

su

syIt has a zero at z=-1

What does it mean? u=e-1t and suitable y0 and y’0 leads to y(t)=0


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lecture 5


Exercises

333

4)(

23

2

ssS

ssG5-1 Find the poles and zeros of following system.

5-2 Is it possible to apply a nonzero input to the following system for t>0, but the output be zero for t>0? Show it. uuyyy 265

5-3 Find the step response of following system.

333

4)(

23

2

sss

ssG

5-4 Find the step response of following system for a=1,3,6 and 9.

22

2)(

2

ss

assG

5-5 Find the linear model of system in example 2 around x2=u=0, x1=π

linear control systems

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