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Page 1: 1. 2 Contents 5.1 Introduction 5.2 Stability Analysis 5.3 Steady-state analysis 5.4 Dynamic analysis 5.5 Controllability, reachability, observability,

1

Page 2: 1. 2 Contents 5.1 Introduction 5.2 Stability Analysis 5.3 Steady-state analysis 5.4 Dynamic analysis 5.5 Controllability, reachability, observability,

2

Contents

5.1 Introduction

5.2 Stability Analysis

5.3 Steady-state analysis

5.4 Dynamic analysis

5.5 Controllability, reachability, observability, detectability

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5.1 Introduction

To apply a CCS to industry, we must analyze

Stability (z-plane)

Steady-state analysis (steady-state error)

Dynamic analysis (transient propoty)

Analysis of CTS: s-plane

Analysis of CCS: z-plane

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5.2 Stability Analysis

5.2.1 The relationship between s-plane and z-plane

The pole and zero locations of CCS in z-plane are related to the

pole and zero locations of CTS in s-plane.

The dynamic specifications of CCS is dependable to T.

(1) The left half-s-plane:

(2) The jω of s-plane:

( ) ( 2 )

0, 1, 2,

,

sT j T T j T T j T k

T

s j

z e e e e e e

k

z e z T

1

1

Tz e

z

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5

5.2 Stability Analysis(3) The periodic strips

A point in z plane →infinite points in s planeA point in s plane → a single point in z plane

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5.2 Stability Analysis

Figure 5.2 (a) Constant attenuation lines in the s plane

(b) Corresponding loci in the z plane

(4) Some commonly used contour

a. constant-attenuation loci (σ)

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5.2 Stability Analysis

Figure 5.3 (a) Constant-frequency loci in the s plane

(b) The corresponding loci in the z plane

•Left half-s-plane, : z-plane 0-(-1) negative real axis

•Negative real axis in s-plane : z-plane 0-1 positive real axis

•Right half-s-plane, : z-plane 1-∞ positive real axis

/ 2s

sn

b. Constant frequency (ω)

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5.2 Stability Analysis

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5.2 Stability Analysis

c. Constant damping ratio (ξ)

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5.2 Stability Analysis

s

d

s

d

s

d

s

ddn

Ts

dnnn

ω

ωπz

ω

ξ

πξ(z

ωπj

ω

ω

ξ

πξ(T)jωTξω(ez

jωξωξjωξωs

2

1

2exp

21

2expexp

1

2

2

2

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5.2 Stability Analysis

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5.2 Stability Analysis

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5.2 Stability Analysis

5.2.2 Stability definition

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5.2 Stability Analysis

(1) Stability and asymptotic stability

Consider the discrete-time state-space equation (possibly nonlinear and time-varying)

x (k+1) = f (x(k), k) (4.1)

Let x0(k) and x(k) be solutions of (4.1) when the initial conditions

are x0(k0) and x(k0) respectively. Further, let ║·║denote a vector

norm.

Definitions 5.1 STABILITY

The solution x0(k) of (4.1) is stable if for a given > 0, there exist

s a (, k0)>0 such that all solutions with║x(k0) - x0(k0)║< such

that ║x(k) - x0(k)║< when k k0.

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5.2 Stability Analysis

Definitions 5.2 ASYMPTOTIC STABILITY

The solution x0(k) of (4.1) is asymptotically stable if it is stable a

nd if can be chosen such that║x(k0) - x0(k0)║< implies that ║

x(k) - x0(k)║0 when k .

For linear, time-invariant systems, stability is a property of the sy

stem and not of a special solution.

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5.2 Stability Analysis

Theorem 5.1 ASYMPTOTIC STABILITY OF LINEAR SYSTEM

x(k+1) = x (k) x(0) = a (4.2)

A discrete-time linear time-invariant system (4.2) is asymptotical

ly stable if and only if all eigenvalues of are strictly inside the

unit disc.

Remark:

If a simple pole lie at z = 1, then the system becomes critically st

able. Also, the system becomes critically stable if a single pair of

conjugate complex poles lies on the unit circle in the z plane.

Any multiple closed-loop poles on the unit circle makes the syst

em unstable.

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5.2 Stability Analysis

Example 5.1: Consider the closed-loop control system shown in Fig. 5.9. Dete

rmine the stability of the system when K = 1, T=1. The open-loop transfer function G(s) of the system is

1( )

( 1)G s

s s

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5.2 Stability Analysis

The z transform of G(s) is

the closed-loop pulse transfer function for the system is

The characteristic equation is

which becomes

1 0.3679 0.2642( ) (1 ) ( ) /

( 0.3679)( 1)

zG z z Z G s s

z z

)(1

)(

)(

)(

zG

zG

zR

zC

0)(1 zG

02642.03679.0)1)(3679.0( zzz

06321.02 zz

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5.2 Stability Analysis

The roots of the characteristic equation are found to be

Since

the system is stable.

If K > 2.3925 , the system is stable or not?

6181.05.0,6181.05.0 11 jzjz

121 zz

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5.2 Stability Analysis

(2) Input-Output Stability

Definition 5.3 BOUNDED-INPUT BOUNDED-OUTPUT STABILITY

A linear time-invariant system is defined as bounded-input-

bounded-output (BIBO) stability if a bounded input gives a

bounded output for every initial value.

Theorem 5.2 RELATION BETWEEN STABILITY CONCEPTS

Asymptotic stability implies stability and BIBO stability.

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5.2 Stability Analysis

5.2.3 Stability test

a. Direct computation of the eigenvalues of

b. Methods based on properties of characteristic polynomials

c. The root locus method (assume the open-loop system is known)

d. The Nyquist criterion (assume the open-loop system is known)

e. Lyapunov’s method (non-linear, time-variant)

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5.2 Stability Analysis

(1) The Jury Stability Criterion Suppose the characteristic equation is

(5.4) Form the table,

0)( 110

nnn azazazA

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5.2 Stability Analysis

Theorem 5.3 JURY’S STABILITY TEST

If a0 > 0, then (4.4) has all roots inside the unit disc if and only if

all a0k, k = 0,1,…,n-1 are positive. If no a0

k is zero, then the

number of negative a0k is equal to the number of roots outside

the unit disc.

Remark

If all a0k are positive for k = 1,2,…,n-1, then the condition a0

0 > 0

can be shown to be equivalent to the conditions

A(1)>0

(-1)nA(-1)>0

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5.2 Stability Analysis

Example 5.2:

| |

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5.2 Stability Analysis

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5.2 Stability Analysis

(2) Routh stability criterion after bilinear transformation

The bilinear transformation is defined by 1 1

or 1 1

w zz w

w z

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27

5.2 Stability Analysis

1( )

1

zw

z

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28

5.2 Stability Analysis

First substitute (1+w)/(1-w) for z in the characteristic equation

as follows

Then, clearing the fractions by multiplying both sides of this last

equation by (1-w)n, we obtain

Once we transform P(z)= 0 into Q(w)= 0, it is possible to apply Routh

stability criterion in the same manner as in the continuous-time systems.

0)( 11

10

nnnn azazazazP

1

0 1 1

1 1 1........ 0

1 1 1

n n

n n

w w wa a a a

w w w

0.......)( 11

10

nnnn bzbwbwbwQ

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29

5.2 Stability Analysis

Example 5.3 Discussing the system’s stability (shown as the following Figure)

by applying Routh stability criterion.

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5.2 Stability Analysis

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5.2 Stability Analysis

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5.2 Stability Analysis

(3) For systems with order 2, Jury’s stability test is simplified to (with the restricted condition that A(z) must be a0 = 1)

A(1) > 0A(-1) > 0|A(0)| < 1

Proof: Suppose a system with order 2 is: First we will form the Jury table 1 a b b a 1 1-b2 a-ab a-ab 1-b2

(1-b2)-{a2(1-b)2/1-b2}={(1-b)(1+b)2/(1+b)}-{a2(1-b)/1+b}=(1-b)/(1+b){ (1+b)2-a2}

2( ) 0A z z az b

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33

5.2 Stability Analysis

According to the Jury stability test rules, the system is stable if

and only if

1-b2>0 |b|<1 -1<b<1 (1-b)/(1+b)>0

and (1-b)/(1+b){(1+b)2-a2}>0 {(1+b)2-a2}>0 |a/(1+b)|<1

-1<a/(1+b)<1

-1-b<a<1+b 1+b+a>0 and 1+b-a>0

We all know that

|b|<1 is equivalent to |A(0)|<1

1+b+a>0 is equivalent to A(1)>0

1+b-a>0 is equivalent to A(-1)>0

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34

5.2 Stability Analysis

Example 5.4: The system is the same as example 5.3, directly ascertain the

K’s value scale in Z-plane.

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35

5.2 Stability Analysis

Example 5.5

Determine the range of K when system is stable where

Y(s)

R(s)

_ D(z) ZOH G(s)

G(z)

T

kzDTs

sG

)(,1,1

1)(

394.20

394.2182.51)0(

382.260104.6736.2)1(

00632.0)1(

0264.0368.0)368.1368.0(0)(1)(

)368.0)(1(

264.0368.0}

)({)1()(

2

1

K

KA

KKA

KKA

KzKzzKGzA

zz

z

s

sGZzzG

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36

5.2 Stability Analysis

5.2.4 Relative Stability

Definition 5.4 AMPLITUDE MARGIN

Let the open-loop system have the pulse-transfer function H(z)

and let 0 be the smallest frequency such that

and such that is decreasing for = 0 . The amplitude or gain m

argin is then defined as

)(arg 0hieH

|)(|

10

arg himeH

A

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37

5.2 Stability Analysis

Definition 5.5 PHASE MARGIN

Let the open-loop system have the pulse-transfer function H(z)

and further let the crossover frequency c be the smallest frequ

ency such that

The phase margin marg is then defined as

1)( hi ceH

)(argarghi

mceH

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38

Exercises

(a)

(b)

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39

Exercises

(c)

(d)

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40

5.3 Steady-state analysis

5.3.1 Steady-state response to input signal

The steady-state performance of a stable control system is gene

rally judged by the steady-state error due to step, ramp, and acc

eleration inputs.

Suppose the open-loop pulse transfer function is given by the e

quation

where B(z)/A(z) contains neither a pole nor a zero at z =1. Then

the system can be classified as a type 0 system, a type 1 syste

m, or a type 2 system according to whether N = 0, N = 1, or N =

2, respectively.

)(

)(

)1(

1

zA

zB

z N

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41

5.3 Steady-state analysis

Consider the typical discrete-time control system

From the diagram we have the actuating error

Figure 5.10 Close-loop control system

)()()( tbtrte

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42

5.3 Steady-state analysis

Consider the steady-state actuating error at the sampling instants, from the final value theorem, we have

and

where

so

1

1lim ( ) lim 1 ( )k z

e kT z E z

)()(1

1)( zR

zGHzE

s

sHsGZzzGH

)()()1()( 1

)(

)(1

11lim 1

1zR

zGHze

zss

( ) ( ) ( ) ( ) ( ) ( )E z R z B z R z GH z E z

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43

5.3 Steady-state analysis

(1) Static Position Error Constant

For a unit-step input r(t) = 1(t), we have

We define the static position error constant Kp as follows

Then the steady-state actuating error in response to a unit-step i

nput can be obtained from the equation

11

1)(

zzR

)(1

1lim

1

1

)(1

11lim

111

1 zGHzzGHze

zzss

)(lim1

zGHKz

p

pss Ke

1

1

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44

5.3 Steady-state analysis

If Kp = , which requires that GH(z) have at least one pole at z = 1, the steady-state actuating error in response to a unit-step input will become zero.

For type 0 system:

For type I system:

For type II system:

pss Ke

1

1

0sse

0sse

)(')1(

1lim

11zGH

zK

izp

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45

5.3 Steady-state analysis

(2) Static Velocity Error Constant

For a unit-ramp input r(t) = t, we have

We define the static velocity error constant Kv as follows

21

1

)1()(

z

TzzR

)()1(

lim)1()(1

11lim

1121

11

1 zGHz

T

z

Tz

zGHze

zzss

T

zGHzK

zv

)()1(lim

1

1

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46

5.3 Steady-state analysis

Then the steady-state actuating error in response to a unit-ramp input c

an be given by

If Kv = , then the steady-state actuating error in response to a unit-ra

mp input is zero. This requires GH(z) to possess a double pole z = 1.

For type 0 system:

For type I system:

For type II system:

vss Ke

1

izv

zT

zGHzK

)1(

)(')1(lim

1

1

1

vss Ke

1

sse

0sse

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47

5.3 Steady-state analysis

(3) Static Acceleration Error Constant For a unit-acceleration input r(t)= t2/2, we have

We define the static acceleration error constant Ka as follows

31

112

)1(2

)1()(

z

zzTzR

)()1(

lim)1(2

)1(

)(1

11lim

21

2

131

1121

1 zGHz

T

z

zzT

zGHze

zzss

2

21

1

)()1(lim

T

zGHzK

za

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48

5.3 Steady-state analysis

Then the steady-state actuating error in response to a unit-acceleration input can be obtained from the equation

The steady-state actuating error in response to a unit-acceleration input become zero if Ka = . This requires GH(z) to possess a triple pole z=1.

For type 0 system:

For type I system:

For type II system:

sse

iza

zT

zGHzK

)1(

)(')1(lim

12

21

1

ass Ke

1

sse

ass Ke

1

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49

5.3 Steady-state analysis

Table 5.1 system types and the corresponding steady-state error

in response to special input

System Step-input Ramp-input Acceleration-input

Type 0 1/(1+Kp)

Type 1 0 1/Kv

Type 2 0 0 1/Ka

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50

5.3 Steady-state analysis

Remark:

1. We should test the stability of the system before we compute the

steady-state error of the system.

2. For non-unit input signal, static error constants are not changed,

but steady-state error will be different according to the coefficien

ts of the input signal.

3. For multiple input signals, the steady-state error can be compute

d by superposing multiple steady-state errors.

For example, if r(t) = a + bt, then

vpss K

b

K

ae

1

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51

5.3 Steady-state analysis

4. For a different closed-loop configuration, it is noted that if the

closed-loop discrete-time control has a closed-loop pulse

transfer function, then the static error constants can be

determined by an analysis similar to the one just presented. If

the closed-loop discrete-time control system does not have a

closed-loop pulse transfer function, however, the static error

constants cannot be defined, because the input signal cannot

be separated from the system dynamics.

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5.3 Steady-state analysis

Example 5.6

where

a. T = 0.5, k = 6 and 10, compute the ess respectively;

b. T = 0.5, determine the range of the k for satisfying ess 0.05.

Y(s)

R(s) _ D(z) ZOH G(s)

G(z)

Y(s)

1( ) , ( ) , ( ) 1( ) 0.5

1 1

kzG s D z r t t t

s z

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53

5.3 Steady-state analysis

Solution:

Judge the stability of the system:

1 ( ) 1 (1 )( ) (1 ) { } , ( ) ( ) ( )

( 1)( )

T T

T T

G s e k e zG z z Z Q z D z G z

s z e z z e

0)1()1(0)(1)( 2 zekezezzQzA TTT

(1) (1 ) 0 0 (0 1)T TA k e k e

T

T

e

ekA

1

)1(20)1(

1)0( A

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54

5.3 Steady-state analysis

a. when T = 0.5, k should satisfy 0 < k < 8.15, so we should only

need to compute ess for k = 6.

b.

with the stability condition 0 < k < 8.15, we conclude that

1 1

(1 )lim ( ) lim

( 1)( )

T

p Tz z

k e zK Q z

z z e

1 1

1 1

1 1

1 11

(1 ) ( ) (1 ) (1 )lim lim

( 1)( )

(1 ) (1 )lim

(1 )(1 )

T

v Tz z

T

Tz

z Q z z k e zK

T T z z e

z k e z k

T Tz z e

042.02

5.0

1

1

k

T

KKe

vpss

505.02

kk

Tess

15.85 k

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55

5.3 Steady-state analysis

Table 5.2 Static error constants for typical closed-loop configurations of discrete-time control systems

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5.3 Steady-state analysis

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5.3 Steady-state analysis

5.3.2 Steady-state response to disturbances

Let us assume that the reference input is zero, or R(z) = 0 in the

system shown in Fig. 5.11(a), but the system is subjected to dist

urbance N(z). For this case the block diagram of the system can

be redrawn as shown in Figure 5-11(b).

Then the response C(z) to disturbance N(z) can be found from t

he closed-loop transfer function

If |GD(z)G(z)|>>1, then we find that

)()(1

)(

)(

)(

zGzG

zG

zN

zC

D

)(

1

)(

)(

zGzN

zC

D

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58

5.3 Steady-state analysis

Figure 5.11 (a) Digital closed-loop control system subjected to reference input and

disturbance input; (b) modified block diagram where the disturbance input is

considered the input to the system

C(z) N(z) _ G(z)

GD(z)

C(z) R(z) = 0

_ GD(z) G(z)

N(z)

(a)

(b)

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59

5.3 Steady-state analysis

Since the system error is

We find the error E(z) due to the disturbance N(z) to be

Thus, the larger the gain of GD(z) is, the smaller the error E(z). If

GD(z) includes an integrator [which means that GD(z) has a pole

at z = 1], then the steady-state error due to a constant disturban

ce is zero.

)()()()( zCzCzRzE

)()(

1)( zN

zGzE

D

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60

5.3 Steady-state analysis

where ĜD(z) does not involve any zeros at z = 1.

1

1 1

1 1

( )1

111

1

11

( )lim 1 ( ) lim 1

( )

1lim 1

( )1

1lim 0

ˆ ( )

ss z zD

NN z

z

zD

zD

N ze z E z z

G z

Nz

G zz

z N

G z z

1

1

1

)(ˆ

1

)(ˆ)(

z

zzG

z

zGzG DD

D

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61

5.3 Steady-state analysis

Remark:

1. If a linear system is subjected to both the reference input and a d

isturbance input, then the resulting error is the sum of the errors

due to the reference input and the disturbance input. The total e

rror must be kept within acceptable limits.

(1) Suppose N(z)=0, essr can be obtained using the above method.

(2) Suppose R(z)=0, essn can be obtained using the above method.

(3) So at last ess=essn+essr.

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5.3 Steady-state analysis

2. The point where the disturbance enters the system is very import

ant in adjusting the gain of GD(z)G(z). For example, consider the

system shown in Fig. 5.12(a) and (b), their closed-loop pulse tra

nsfer function for the disturbance are

)()(1

1

)(

)(

)(

)(

zGzGzN

zE

zN

zC

D

)()(1

)()(

)(

)(

)(

)(

zGzG

zGzG

zN

zE

zN

zC

D

D

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63

5.3 Steady-state analysis

Figure 5.12 (a) Digital closed-loop control system subjected to reference input

and disturbance input; (b) digital closed-loop control system where the

disturbance enters the feedback loop.

C(z) R(z) = 0

_ GD(z) G(z)

N(z)

(a)

(b)

C(z) R(z) = 0

_ GD(z) G(z)

N(z)

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64

5.4 Dynamic analysis

Considering the Fig. 5.9

The z transform of the output of the system is

where (z) is the z pulse transfer function of the closed-loop control

system.

1)()()()(

z

zzzRzzY

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65

5.4 Dynamic analysis

n

ii

m

jj

n

m

ni

mj

n

m

nn

nn

mmm

m

pz

zz

a

b

pzpzpzpz

zzzzzzzz

a

b

azazaza

bzbzbzb

zq

zpz

1

1

21

21

011

1

0111

)(

)(

))...().....()((

))...().....()((

...

...

)(

)()(

where n > m. When there is no multiple roots of the characteristic

equation, Y(z) can be rewritten as

1

1

( )( )

( ) 1 1

( )( )( )where

( ) ( )( 1)i

ni

i i

ii

z z p

B zp z z AzY z

q z z z z p

p z z pp zA B

q z q z z

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66

5.4 Dynamic analysis

(1) pi is the positive real number

The transient response for the poles to be positive real number is

Let

then the transient response component of the system is given by

Since |pi| < 1, and therefore < 0. The transient response is the

exponential attenuation curve and is monotonic. If |pi| is smaller, and the

|| is larger, so that the result is that the pole is nearer to the original and

the transient response is faster.

-11

1( ) ( [ ] )

1-

nk k

i ii

y kT B p Z aaz

ipTln

1

kTieB

1

( )n

i

i i

B zY z

z p

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67

5.4 Dynamic analysis

(2) pi is the negative real number

The transient response for the poles to be negative real number is

The transient response is the exponential positive and negative

attenuation curve respectively. So the negative roots correspond to high-

frequency oscillation with frequency ωs/2, which is also called ringing.

Ringing period is 2T, frequency is 1/2T, radian frequency is 2π/2T= ωs/2.

1

( )n

ki i

i

y kT B p

1

( )n

i

i i

B zY z

z p

where

k

i iki i k

i i

B p k is evenB p

B p k is odd

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68

5.4 Dynamic analysis

(3) pi is equal to zero

For single pole case, it is a one step attenuation. When pi is zer

o, the transient response is the fastest, which is also called dea

dbeat control.

)()( kBkTy ii 1

( )n

i

i i

B zY z

z p

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69

5.4 Dynamic analysis

(4) Multiple poles case: suppose there are m poles p

( remark: )

For |p|<1 p 0,

12211

1 )(

)(

)(

)(

)1(

)1(

1

1

)(

)()(

mm pz

zza

pz

zza

q

p

zzq

zpzY

122

111)( kmkm pkapkakTy

,...2,1,1

}{1

11

k

pz

zpZ k

- 21

11

)1(}{

pz

zkpZ k

1 12 1

1 3

(1 ){ }

(1 )k z pz

Z k ppz

,...3,2,1,0)

1(

lim1

i

p

ka

k

imi

k

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70

5.4 Dynamic analysis

For 0<p<1 , the response is monotonic exponential attenuation

curve.

For -1<p<0 , the response is a high-frequency oscillation attenu

ation curve with frequency s/2.

For p=0, e.g., two poles system,

M multiple poles imply a m steps attenuation.

)()1()( 21221 kakakTyaz

za

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71

5.4 Dynamic analysis

(5) The poles are a pair conjugate complex number

where

and the similar transient response is

ii pz

ii

pz

ii zzq

pzzp

zzq

pzzpa

)1)((

))((arg

)1)((

))((

ii

i

jii

jii

jiii

eppepp

eaaa

||,||

,

1

1

iii

k

ii

jkk

ij

ijkk

ij

i

k

ii

k

iiii

kpa

epeaepea

papakTykTy

iiii

0cos2

)()( 111

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72

5.4 Dynamic analysis

The transient response is periodic damping oscillation form. The

more closed to the origin the pole is, the faster attenuation is.

The larger i is (T: 0 /2 ), the more intensely oscillation

does (oscillation frequency 0 s/4 s/2).

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73

5.4 Dynamic analysis

Summary

1. System with poles in the unit circle corresponds to attenuation

curve (stable).

2. Positive real pole corresponds to monotonic response Negative

real pole corresponds to high-frequency oscillation with

frequency s/2.

3. The more the pole closed to the origin is, the faster attenuation

is, when poles are on the origin, the transient response is the

fastest, which is also called deadbeat control.

4. The larger i is (T: 0 /2 ), the more intensely oscillation

does (oscillation frequency : 0 s/4 s/2).

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74

5.4 Dynamic analysis

So we have the Fig. 5.13. If

we desire to get the better

transient response (better

performance), the poles of

the closed-loop pulse-

transfer-function should

locate in the right half of unit

circle on z plane, and near

the real axis and origin.

Figure 5.13 The relationship between the transient

response and the poles location

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75

5.5 Controllability, reachability, observability, detectability1. Whether it is possible to steer a system from a given initial state t

o any other state.

2. How to determine the state of a dynamic system from

observations of inputs and outputs.

5.5.1 Controllability and Reachability

Consider the system

(5.17)

Assume that the initial state x(0) is given.

)()(

)()()1(

kCxky

kukxkx

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76

5.5 Controllability, reachability, observability, detectability The state at time n, where n is the order of the system, is given

by

(5.17)

If Wc has rank n, then it is possible to find n equations from whic

h the control signals can be found such that the initial state is tra

nsferred to the desired final state x(n).

Remark: The solution is not unique if there is more than one inp

ut signal.

1

1

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

[ ]

[ ( 1) (0)]

n n nc

nc

T T T

x n x u u n x WU

where

W

U u n u

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77

5.5 Controllability, reachability, observability, detectabilityControllability:

The system (5.17) is controllable if it is possible to find a control

sequence can be found such that the origin can be reached fro

m any initial state in finite time.

Reachability:

The system (5.17) is reachable if it is possible to find a control s

equence such that an arbitrary state can be reached from any in

itial state in finite time.

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78

5.5 Controllability, reachability, observability, detectabilityTheorem 5.4 REACHABILITY

The system in (5.17) is reachable if and only if the matrix Wc ha

s rank n.

e.unreachabl is system the11

11

)(1

1)(

10

01)1(

cW

kukxkx

e.unreachabl is system the25.05.0

5.01

)(5.0

1)(

025.0

11)1(

cW

kukxkx

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79

5.5 Controllability, reachability, observability, detectabilityRemark:

1.The matrix Wc is usually referred to as the controllability matrix

because of its analogy with continuous-time system.

2.The reachability of a system is independent of the coordinates.

TTTnc

cnnn

unuUW

UWxnuuxnx

)0(...)1(,...

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

1

1

cnn

c TWTTTTTTTW 1111 ...~~

...~~~~

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80

5.5 Controllability, reachability, observability, detectability3. Controllability and reachability are equivalent if is invertible. Otherwise,

controllability does not imply reachability.

Example: The system (5.17), where

The system is reachable because has rank 2 , that is it

has full rank. But when

We know the system is not reachable, but is controllable, because , the

origin is reached in two steps for any initial condition by using u(0) = u

(1) = 0.

0

1,

01

00

10

01cW

1

0,

01

00

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81

5.5 Controllability, reachability, observability, detectabilityExample: Determine the system is controllable or not.

(1)

)(10)(

)(368.0

632.0)(

1632.0

0368.0)1(

kxky

kukxkx

2767.0368.0

2326.0632.0

767.0

2326.0

368.0

632.0

1632.0

0368.0

rankABBrank

AB

The system is controllable.

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82

5.5 Controllability, reachability, observability, detectability(2)

)(01)(

)(8.0

1)(

116.0

10)1(

kxky

kukxkx

164.08.0

8.01

rankABBrank

The system is not controllable.

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83

5.5 Controllability, reachability, observability, detectability5.5.2 Observability and Detectability

UNOBSERVABLE STATES

The states x0≠0 is unobservable if there exists a finite k1≥n-1 such that

y(k) = 0 and u(k)=0 for 0 ≤ k ≤ k1 when x(0) = x0.

Observable: The system in (5.17) is observable if there is a finite k such that knowle

dge of the input u(0), …, u(k-1) and the outputs y(0), …, y(k-1) is suffici

ent to determine the initial state of the system.

Detectability:

A system is detectable if the only unobservable states are such that dec

ay to the origin. That is, the corresponding eigenvalues are stable. The

observability matrix is independent of the coordinates in the same way

as in the controllability matrix.

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84

5.5 Controllability, reachability, observability, detectabilityTheorem 5.5 OBSERVABILITY

The system (5.17) is observable if and only if Wo has rank n.

The state x(0) is unobservable if x(0) is in the null space of Wo.

)1(

)1(

)0(

)0(

)0()1(

)0()1()1(

)0()0(

11 ny

y

y

x

C

C

C

xCny

xCCxy

Cxy

nn

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85

5.5 Controllability, reachability, observability, detectabilityExample: A system with unobservable state

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86

5.5 Controllability, reachability, observability, detectability5.5.3 Kalman’s Decomposition

The state space is partitioned into four parts :

1. reachable and observable,

2. not reachable but observable,

3. reachable and not observable,

4. neither reachable nor observable.

)(00)(

)(

0

0)(

00

000

00

)1(

21

3

1

4442

34333231

22

1211

kxCCky

kukxkx

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87

5.5 Controllability, reachability, observability, detectability The pulse-transfer operator of the system is given by

That is, the pulse-transfer operator is only determined by the rea

chable and observable part of the system.

Theorem 5.6 Kalman’s decomposition

A linear system can be partitioned into four subsystems with the

following properties (shown in Fig. 5.14):

11

111 )()( qICqH

subsystemreachablenotandobservableNotS

subsystemreachableandobservableNotS

subsystemreachablenotandobservableS

subsystemreachableandobservableS

ro

ro

ro

or

:

:

:

:

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88

5.5 Controllability, reachability, observability, detectability Further, the pulse-transfer function of the system is uniquely

determined by the subsystem that is observable and reachable.

Figure 5.14 Block diagram of the Kalman decomposition when

the system is diagonalizable

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89

Summarization

Stability Analysis

The relationship between s-plane and z-plane

-The left half-s-plane -The jw of s-plane

-Periodic strips -Constant attenuation line

-Constant frequency line -Constant damp ratio

Stability definition

-Stability -Asymptotic stability -BIBO stability

Stability test

a. The eigenvalues of

b. The properties of characteristic polynomials (Jury, Rout

h)

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90

Summarization

Steady-state analysis

Steady-state response to input signal Static Position Error Constant Static Velocity Error Constant Static Acceleration Error Constant

Steady-state response to disturbances

Dynamic analysis

pi is the negative real number

pi is the positive real number

pi is zero

The poles are a pair conjugate complex number

Controllability, reachability, observability, detectability

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91

Homework

1. Determine the range of K when system is stable where

2. Use Jury stability criterion and Modified Routh criterion to

analyze the stability of the system :

03911917745)( 23 zzzzA

Y(s)

R(s)

_ D(z) ZOH G(s)

G(z)

T

kzDTs

sG

)(,1,1

1)(

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92

Homework

3. a. T = 0.5, k = 6 and 10, compute the ess respectively;

b. T = 0.5, determine the range of the k for satisfying ess 0.05.

Y(s)

R(s) _ D(z) ZOH G(s)

G(z)

Y(s)

,5.0)(1)(,1

)(,1

1)( tttr

z

kzzD

ssG

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93

Homework

4. Determine steady-state error coefficients and steady-state error

when the input is t and t2 respectively, where k =1.

When , determine the interval of the k which sh

ould make ess 0.05.

Y(s)

R(s) _ D(z) ZOH G(s)

G(z)

Y(s)

1.0,)(,)11.0(

1)(

TkzD

sssG

tttr3

1)(1)(

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94

Homework

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95

Homework

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96

Homework

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97

Homework

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98

Homework