MODERN CONTROL SYS-LECTURE VI.pdf

MODERN CONTROL SYSTEMS ENGINEERING

COURSE #: CS421

INSTRUCTOR: DR. RICHARD H. MGAYA

Date: October 25th, 2013

Digital Compensator Design

Techniques Employing the z-Transform

Governing difference equation algorithm:

Design problem:

Selection of the coefficients ai and bi once the sampling interval T is specified

Note: The control system designer must know whether the actuating

hardware will be able to respond to the magnitude of the control effort

generated by the algorithm in the digital processor

Modeling and digital simulations of the control system

Examination of the control effort required

Dr. Richard H. Mgaya

)()1(

)()1()()(

01

01

nkubkub

nkeakeakeaku

n

nn



General PID Direct Digital Control Algorithm

Recap:

Proportional Control

Steady-state error was necessary in order to have a steady-state output

Proportional plus Integral

Integrator reduces the steady-state error to zero

However, integrator in the loop can create instability or overshoot i.e., poor system dynamics

Proportional plus Integral plus Derivative

Steady-state error reduced to zero

Improvement in system dynamics





Time domain PID Controller

Digital Control Algorithm for PID

Approximation

Integral Trapezoidal integration

Derivative Backward difference equation


dt

deKedtKteKtu d

t

ip 0

)()(

21

1

2

2

2)(

kd

kd

pi

kdi

pk

eT

Ke

T

KK

TK

eT

KTKKuku




Compensator transfer function:

where


211 )()()()()( zzEzzEzEzzUzU

)1(1)(

)()(

2

1

21

zz

zz

z

zz

zE

zUzD

T

KTKK dip

2

T

KK

TK dp

i 2

2

T

Kd




Compensator transfer function:

The compensators numerator is quadratic

May be chosen to cancel the slow pole

Approximation algorithm: Appropriate algorithm produces causal control effort algorithm

Noncausal algorithms produces future values in the control effort

General PI

where


1)(

)()(

z

z

zE

zUzD

2

TKK ip

pi KTK

2




Example: Consider the thermal system given in the figure below. Design a PI controller to have a steady-state error of

zero to step input

Cancel the slow pole:


1

)/(

)(

)()(

z

z

zE

zUzD

1

)952.0()(

z

zzD




Let be the controller gain:

Through trial-and-error plus root locus design = 2.5 for closed-loop

damping ratio of = 0.7 and T = 0.25


Root Locus

Real Axis

Imagin

ary

Axis

-5 -4 -3 -2 -1 0 1 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

System: sys

Gain: 2.5

Pole: 0.705 + 0.221i

Damping: 0.706

Overshoot (%): 4.35

Frequency (rad/sec): 1.72




Control algorithm:


)(5.2 11 kkkk eeuu

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude



Ziegler Nichols Tuning Procedure for PID

Process for determining the gain Kp, Ki and Kd for the controller parameters , and

Experimental evaluation of the step responses of the plant to be controlled

R the slope of the response

L the plant lag to the response

Tuning strategy:

Relationship between PID parameters to the values of R and L




Ziegler Nichols Tuning Procedure for PID

PI Controller:

PID Controller:


RLK p

9.0

2

272.0

3.3

1

RLK

LK pi

RLK p

9.0

2

6.0

2

1

RLK

LK pi

RLKK pd

6.05.0

RAJAHHighlight

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RAJAHPencil

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RAJAHPencil



Example: Use Ziegler-Nichols tuning strategy to design PI controller for the thermal system with the following transfer

function same thermal system, in continuous form

Open-loop step response:


5.075.2

1

)(

)()(

2

sssU

sTsG

Step Response

Time (sec)

Am

plit

ude

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

System: sys

Time (sec): 0.47

Amplitude: 0.0763

Step Response

Time (sec)

Am

plit

ude

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

sL 47.0 37.0R



Example:

Calculations for PI parameters:

Using T = 0.25:

Control algorithm:


1

)85.0(592.5)(

z

zzD

175.547.037.0

9.09.0

RLK p

330.047.037.0

272.0272.022

RLKi

and 592.5 758.4

11 758.4592.5 kkkk eeuu0 1 2 3 4 5 6 7 8 9

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Step Response

Time (sec)

Am

plit

ude



Example:

Decreasing the gain decreases the overshoot but slows down the response


and 5.0 425.0

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude



Direct Design Method of Ragazzini

Closed-loop transfer function of a controlled system, M(z)

If the desired closed-loop transfer function M(z) and the plant transfer function G(z) are known, Then,


)()(1

)()(

)(

)()(

zDzG

zDzG

zR

zYzM

)(1

)(

)(

1)(

zM

zM

zGzD




Fundamental issues to consider:

Causality:

Current control effort depends only on the current and past inputs and the past control effort

Causal M(z), must have a zero at infinity same order as the zero of the plant, G(z), at infinity

M(z) must have as many pure delays as the plant G(z)

Zero at infinity in time domain implies that the pulse response of G(z) has delay at least one sample interval

Causal controller current effort does not dependent on the future values of the error





Stability:

Necessary conditions for stability:

Closed-loop characteristic equation

Let D(z) = c(z)/d(z) and G(z) = b(z)/a(z) the closed-loop characteristic

polynomial:

Let a common factor (z - ) be in the numerator D(z) and the denominator

of G(z). If the factor is the pole of G(z), (z - ) to be cancelled then,


0)()(1 zDzG

0)()()()( zczbzdza

0)()()()()(0)()()()()()(

zbzczdzaz

zbzczzdzaz

a




Stability:

Common factor remain a factor of the characteristic polynomial

If is outside the unit circle, then the system is unstable

Therefore, if D(z) is not to cancel the poles of G(z) then the factors of a(z) must be the factors of 1 M(z)

Similarly, if D(z) is not to cancel the zeros of G(z) such zeros must be

factors of M(z)

Constraints summary:

1 - M(z) must contain as zeros as all poles of G(z) that lies on or outside the unit circle

M(z) must contain as zeros as all zeros of G(z) that lie on or outside the unit circle





Steady-State Accuracy:

Final value theorem - steady-state error for unit step

For zero steady-state error


)](1)[()( zMzRzE

)1(1

)](1[1

1lim 11

M

zMz

zze zss

1)1( M




Steady-State Accuracy:

Final value theorem - steady-state error for ramp

LHopital rule to evaluate the limit z approaching 1


1)1(

1)](1[

11lim

2

1

1

M

KzM

z

Tzze

v

zss

vz Kdz

dMT

1

1

TKdz

dM

vz

1

1




Example: Consider the thermal plant transfer function for T = 0.25

Design a controller to have closed-loop poles at 0.4 j0.4, i.e., d =

rad/s, = 0.58, zero steady-state error for a step input and a velocity

constant of Kv = 10/T

Soln:

Closed-loop transfer function from specs


5026.048.1

)816.0(025.0)(

2

zz

zzG

21

3

3

2

2

1

10

32.08.01)(

zz

zbzbzbbzM




Example:

Causality: b0 = 0 plant has one step pure delay

Steady-state step response requirements:

Velocity constant requirements


21

21

32.08.01)(

zz

bzbzM

152.0

)1( 21

bb

M

1.01

)32.08.0(

)8.02)(()32.08.0(22

211

2

1

TKzz

zbzbbzz

dz

dM

vz

(i)




Example:

Velocity constant requirements

But b1 + b2 = 0.52 from eqn (i)

Then

Closed-loop transfer function:


21 32.08.01

628.0148.1)(

zz

zzM

1.052.0

)2.1)((52.02

211

1

bbb

dz

dM

z

628.052.0

148.1

12

1

bb

b




Example:

Controller form

Substitution

Controller :


)948.0)(1(

547.0

816.0

)952.0)(528.0(

025.0

148.1)(

zz

z

z

zzzD

)(1

)(

)(

1)(

zM

zM

zGzD

7736.06416.0132.1

)2749.03122.1027.2(92.45)(

23

23

zzz

zzzzD

Cancels Plant Dynamics Corrects Closed-loop Poles and Error Constant




Example:

Step Response





Ripple Free Design:

Plant must be stable

For finite settling time, M(z) must be finite polynomial in z-1

Transfer function from the reference input R(z) to the control effort U(z) must be of finite settling time in z-1

M(z) must be selected such that it has a factor to cancel finite zeros of G(z), i.e., no poles of


)(

)(

zG

zM

)(

)(

)(

)(

zG

zM

zR

zU




Ripple Free Design:

Example: Design a zero-ripple, finite settling time controller for the thermal plant which will yield zero steady-state error to

a step reference input

Soln:

Since G(z) has no free integrator in the form of 1 - z-1, then for zero steady-state error M(z) must have that factor


21

11

2 5026.048.11

)816.01(025.0

5026.048.1

)816.0(025.0)(

zz

zz

zz

zzG

))(1()(1 1101 zaazzM (i)




Ripple Free Design:

Example:

Plant is stable and hence for zero ripple controller M(z) must have a factor for numerator terms of G(z)

Solve a set of linear equation from i and ii


1

1

1)816.01()( zdzzM

21

1 816.01)(1 zzdzM (ii)

11

101

0

816.0

1

da

daa

a

5507.0 ,4493.0 ,1 110 daa

RAJAHHighlight




Example:

Closed-loop transfer function:

Controller

Confirming U(z)/R(z)


21 4493.05507.0)( zzzM

)4493.01)(1(

5507.0

025.0

5026.048.11)(

11

1

1

21

zz

z

z

zzzD

4493.05507.0

)5026.048.1(028.22)(

2

2

zz

zzzD

2

2 )5026.048.1(028.22

)(

)(

)(

)(

z

zz

zG

zM

zR

zU




Example:

Step response:


0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude

State-Space Representation

Solution to the State Vector Differential Equation

Consider the first-order differential equation

Laplace

Inverse

Note:


)()( tbutaxdt

dx

)()()0()( sbUsaXxssX

)()0(

)( sUas

b

as

xsX

t

taat dbuexetx0

)( )()0()(

!!21

22

k

tataate

kkat


Solution to the State Vector Differential Equation Cont

Consider the state vector differential equation

Laplace

Pre-multiply (sI A)-1

Inverse

The expression eAt is called Transition Matrix (t)

Represents the natural response of the system:


BuAxx

)()()0()( sBUsAXxssX

)()0()()( sBUxsXAsI

t

tAAt dBUexetx0

)( )()0()(

)()()0()()( 11 sBUAsIxAsIsX

1)()( AsIs


Solution to the State Vector Differential Equation Cont

Alternatively:

Thus,

The 1st term is the response to initial conditions

The 2nd term represent response to a forcing function

Characteristic equation:


t

dBUttxttx0

)())(()0()()(

!!21)(

22

k

tAtAAtt

kk

0)( AsI


Discrete Solution to the State Vector Differential Equation

Discrete solution to state equation:

General form:

Note:


)()( and )()()()()1( kTCxkTykTuTBkTxTATkx

!!21)(

22

k

TATAATTA

kk

)()()1(0

kTuBdekTxeTkx

T

AAT

BTBATA )( and )(

ATeTTA )()(

BTK

TABdeTB

k

kkT

AT

00)!1(

)(

Bk

TATAATTTB

kk

)!1(!3!21)(

1322



Example: The state-space equation for a mass damper system is

given as follows

Evaluate:

(a) Characteristic equation, its roots, n and

(b) Transition matrix (s) and (t)

(c) Transient response of state variables from initial conditions

y(0) = 0 and y`(0) = 0.1


ux

x

x

x

1

0

32

10

2

1

2

1



(a) Characteristic equation, its roots, n and

Roots: s = -1, -2

Natural frequency and damping ratio


32

1

32

10

0

0)(

s

s

s

sAsI

023)2()3()( 2 ssssAsI

06.1 ,32

/414.1 ,22

n

nn srad

2 ,1 s




(s) is given as


s

ssCofactor

s

ssMinor

1

23)(

1

23)(

A

AA

det

Adjoint 1

1)()( AsIs

)2)(1()2)(1(

1)2)(1(

2

)2)(1(

3

)(

ss

s

ss

ssss

s

s




Partial fraction of (s):

Inverse Laplace:


2

2

1

1

2

1

1

12

2

1

1

1

2

1

1

2

)(

ssss

sssss

tttt

tttt

eeee

eeeet

22

22

22

2)(



(c) Transient response of state variables from initial conditions

y(0) = 0, y`(0) = 0.1, x1(0) = 1, and x2(0) = 0

Transient response:


)0()()( xttx

0

1

22

2)(

22

22

tttt

tttt

eeee

eeeetx

)(2)(

2)(

2

2

2

1

tt

tt

eetx

eetx


Controllability

The possibility to drive the system from some arbitrary initial condition x(0) to some specified final state x(n) in n sampling

intervals

Consider the general system

State at the kth time step

Similarly,


)()()1( kBukAxkx

1

0

1 )()0()(k

i

ikk iBuAxAkx

1

0

1 )()0()(n

i

inn iBuAxAnx

)1()2()1()0()0()( 21 nBAunABBuABuAxAnx nnn


Controllability

Matrix form:

Set of n linear equation in mn unknowns

The left hand side quantities are known

Solution exists if the rank of matrix coefficient is same as the unknown

The matrix is called the Controllability matrix


)1(

)1(

)0(

)0()( 21

nu

u

u

BBABAxAnx nnn

nBBABArank nn AB 21


Controllability

Example: A linear discrete system is given as follows

Examine the controllability of the system

Controllability matrix:

Thus, uncontrollable


BABrank

)(1

1

)(

)(

4.04.0

2.01

)1(

)1(

2

1

2

1ku

kx

kx

kx

kx

8.0

8.0

1

1

4.04.0

2.01AB

118.0

18.0B

rankABrank


Observability

The possibility to determine the states and the control effort from measured output at different time intervals

Consider the general system

Can the initial state x(0) be determine from the sequence of measurements y(0), y(1), y(n-1)output

Similarly,


)()()1( kBukAxkx

)0()0()1( BuAxx

)()( kCxky

)0()0()1()1( CBuCAxCxy

1

0

1 )()0()(k

i

ikk iBuCAxCAky


Observability

In matrix form:

Output relationship: Set of nr linear equation with n unknown


)0()0()1( CAxCBuy

)0()1()0()2( 2xCACBuCBuy

1

2

)1(

)2(

)1(

)0(

nCA

CA

CA

C

ny

y

y

y

)0()()1( 11

0

2 xCAiBuACny nn

i

in


Observability

Solution exists if the rank of the observability matrix is n

Example: Determine the observability of the inertial plant with

state equation given as


n

CA

CA

CA

C

rank

n

1

2

)(2)(

)(

10

1

)1(

)1(2

2

1

2

1ku

T

T

kx

kxT

kx

kx

)(

)(10)(

2

1

kx

kxky


Observability

Observability matrix:

The system is unobsevable


110

10

rank

AC

Crank

T

T

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MODERN CONTROL SYS-LECTURE VI.pdf