modern control sys-lecture ii.pdf

8/15/2019 MODERN CONTROL SYS-LECTURE II.pdf

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MODERN CONTROL

SYSTEMS ENGINEERING

COURSE : CS421

INSTRUCTOR:

DR. RICHARD H. MGAYA

Date: October 25th, 2013


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Linear Difference Equation and Z-Transform

Linear difference equations and the z-transform

Techniques used in design and analysis of a digital control system.

• Example of systems that use digital computers:

Radar antenna positioning system

Airplane autopilot control

Chemical process control

Machine tool control

Goal : Representing a digital computer as a transfer function

similar to other subsystems.

Note: Control system design using analog, i.e., continuous time

devices, are still used and are valid

Dr. Richard H. Mgaya


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Difference Equation

• Sequence of real values associated with temporal index k ,{k = 0, 1, 2, …}

• Digital control concerns in generating a sequence u(k ), i.e., controleffort, given a sequence y(k ), i.e., sampled data measurements

sequence The k th control effort is defined in terms of k th measurement or sample

Assumption : u(k ) is a linear combination of measurements and past control

efforts

Where: ai and bi are independent of k , i.e., time invariant

Task : Selection of ai and bi such that the control signal has the dynamic properties

if the desired controlDr. Richard H. Mgaya

)](,),2(),1(),(,),1(),([)( nk uk uk unk yk yk y f k u

)()1()()()1()( 0101 mk yak yak yank ubk ubk u mmn


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Z-Transform of a Sequence

• Long-hand form:

Example: Sequence of finite series


21 )2()1()0()()( z f z f f k f Z z F

0

)()()(k

k z k f k f Z z F

0

** )()()]([)(k

kTsekT f kT f Z t f L s F

0 )()( k k

z kT f z F

Let z = eTs

k z kT f z T f z T f f z F )()2()()0()( 21


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Unit Step Sequence

• z-transform

• Geometric series convergence


0 1

0 0)(

k

k kT u

0

)()]([)(k

k z kT f kT u Z z U

12 1

1

1

111

z z

z

z z


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Unit Ramp Sequence

• z-transform

Multiply by z both sides

Subtract eqn. i and ii


kT kT f )(

00

)()]([)(k

k

k

k kz T z kT f kT f Z z F

)32(321

z z z T

)4321()( 321 z z z T z zF

)1()()1()()(321

z z z T z F z z F z zF

21

11

1

1 z z

z

11

1)()1(

1

z

Tz

z T z F z

2)1()(

z

Tz z F

But

….ii

….i


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Exponential Function Sequence

• z-transform

• Geometric series convergence


0

0 0)(

k e

k kT f

akT

0

1)(][)(k

k aT akT z ee Z z F

aT e z

z


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General Exponential function Sequence

•

z-transform


k r k f )(

0

1 )(][)(k

k k z r f r Z z F

r r z

z z F

zfor)(


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Discrete Impulse Function Sequence

• z-transform

Delayed impulse function sequence

• Z-transform


0 1]1[)]([)( k k

z Z k Z z F

0 0

0 1)(

k

k k f

01

0)(

nk

nk k f

n Z nk Z z F )]([)(


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Solving Linear Difference Equation with Z -Transform

• Delay and advance theorem

Theorem 1:

Theorem 2:

Note: s is associated with differentiation of a differential equation

z is associated with the shifting of the difference equation

Example: Consider a homogenous first-order differential equation

• Initial value x(0) = 1


)()( z F z nk f Z n

0)(8.0)1(

k xk x

)1()0()()( n zf f z z F z nk f Z nn

0)(8.0)0()( z X zx z zX

8.0

)(

z

z z X k k x )8.0()(


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Pulse Transfer Function

• Consider the following block diagram

• U *( s) – Sample input to G( s)

• X 0( s) – Continuous output

• X 0*( s) – Sampled output

• The figure is the pulse transfer function where U *( s) = U ( z )and X 0( z ) = X 0

*( s)


)()(

)(0 z G z U

z X


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Pulse Transfer Function

• Cascaded blocks

• Note: G1G2( z ) ≠ G1( z )G2( z )


)()()(

)(21

0

z G z G z U

z X

)()()(

21

0

sGG Z z U

z X

)()(

)(21

0 z GG z U

z X


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Inverse Z-Transform

Long Division

Example: Find inverse sequence of the function:

• Multiply by z -2 in the numerator and denominator


1)0( f

43)(

2

2

z z

z z z F

21

1

431

1)(

z z

z z F

21

121

841

1431

z z

z z z 21

431

z z 21 44 z z 321 16124 z z z 32 168

z z 432 32248 z z z 43

328 z z

Division Sequence

4)1( f 8)2( f


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Inverse Z-Transform

Partial Fraction Expansion

• Consider the following function:

• The limit of β implies the poles of the quadratic term arecomplex

• Partial fraction expansion:

• The 1st and 2nd terms are the z-transform of

Rk cosΩT and Rk sinΩT respectively, Q( z ) is the reminder


Tcos

)2)((

)()(

2

22

T e R

R z R z z P

z N z F

)(2

sin

2

)(

)( 2222

2

z Q R z R z

T BzR

R z R z

zR z A

z F


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Inverse Z-Transform

• Correct form of the partial fraction expansion:


787.4376.0

8.1

64.013.1

18.0)(

8.0

2

z z z

z

z

z z F C

8.064.013.1

)7079.0)(8.0(

64.013.1

))7063.0)(8.0(()(

22

2

z

Cz

z z

Bz

z z

z z A z F

8.064.013.1

)5663.0(

64.013.1

)565.0(22

2

z

Cz

z z

z B

z z

z z A


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Pole location on the z -Plane

Relationship Between Pole location and the nature of the

temporal sequence

• Consider the following function

• Partial fraction expansion:

• In polar notation the complex poles are written as follows:


rootscomplexhasrdenominato

)()(

2 cbz z

z N z F

conjugatecomplexdenotes-*

)(*

*

p z

A

p z

Az z F

j j e pe p *


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Pole location on the z -Plane

Relationship Between Pole location and the nature of the

temporal sequence

• Pole location: z -plane

• Nature of the sequence

D Ri h d H M

modern control sys-lecture ii.pdf

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