Chapter 14 Intelligent Robot Lab z-Transform and Stability...

Intelligent Robot Lab

Pusan National UniversityIntelligent Robot Lab

Chapter 14z- Transform and Stability Analysis

Pusan National UniversityIntelligent Robot Laboratory

http://robotics.pusan.ac.kr


Intelligent Robot Lab.

v Introduction v The z-Transformv Relationship Between the s-Plane and the z-Planev The Inverse z-Transformv Theorem of the z-Transformv z-Transform Solution of Difference Equationv Stability

Table of Contents




v Digital control involves systems whose control is updated at discrete time instants.

v Discrete-time models provide mathematical relations between the system variables at these time instants.

v In this chapter, we develop the mathematical properties of discrete-time models.

v In principle, the Laplace transform can be used to model digital control systems. However, the typical Laplace transform expressions of systems involving digital or sampled signals all contain exponential terms in the form of .

v This inevitably would make the manipulation of the transform expressions in the Laplace domain unduly difficult.

v So, we may regard this as a motivation for the introduction of the z-transformation.

Introduction

Tse




v The stability and transient response of analog systems depend upon gain and component values; sampled data system’s stability and transient response depend upon the sampling rate.

v Goal is to develop a transform that contains the information of sampling from which sampled data systems can be modeled with transfer functions, analyzed, and designed with the ease and insight we enjoyed with the Laplace transform.

The z-Transform




v Definition of the z-Transform§ Equation (14.1) is the ideal sampled waveform.

§ Taking the Laplace transform of this sampled time waveform, we obtain

§ Now, letting , Eq. (14.2) can be written as

§ Equation (14.3) defines the z-transform. That is, an can be transformed to , or an can be transformed to . Alternately, we can write

The z-Transform

(14.1)*( ) ( ) ( )k

f t f kT t kTd¥

=-¥

= -å

*( ) ( ) kTsk

F s f kT e¥

-

=-¥

= å (14.2)

0( ) ( ) k

kF z f kT z

¥-

=

=å (14.3))(zF

)(zF)(kTf )(kTf

( ) ( )f kT F zÛ (14.4)

Tsz e=




v Ex14.1 z-transform of a time function§ Find the z-transform of a sampled unit ramp.§ Sol) For a unit ramp, . Hence the ideal sampled step can be written

from Eq. (14.1) as

§ Taking the Laplace transform, we obtain

§ Converting to the z-transform by letting , we have

The z-Transform

( )f kT kT=

*

0( ) ( )

kf t kT t kTd

¥

=

= -å

*

0( ) kTs

kF s kTe

¥-

=

=å

(14.5)

(14.6)

kTs ke z- -=

( )1 2 30 0

( ) 2 3k kk k

F z kTz T kz T z z z¥ ¥

- - - - -

= =

= = = + + +å å L (14.7)




§ Equation (14.7) can be converted to a closed form by forming the series for zF(z) and subtracting F(z). Multiplying Eq. (14.7) by z, we get

§ Subtracting Eq. (14.7) from Eq. (14.8), we obtain

§ But,

§ It can be verified by performing the indicated division. Substituting Eq.(14.10) into (14.9) and solving for F(z) yields

as the z-transform of f(kT)=kT.

The z-Transform

(14.8)

(14.9)

1 21

1 11

z zz

- -- = + + +-

L

( )1 2( ) ( ) ( 1) ( ) 1zF z F z z F z T z z- -- = - = + + +L

( )1 2( ) 1 2 3zF z T z z- -= + + +L

(14.10)

(14.11)2( )

( 1)zF z T

z=

-




v Relationship between the Laplace transform and the z-transform§ The Laplace transform of

§ The z-Transform of

§ In general,

§ The time equivalent of Eq.(14.13) is

§ A sampled signal should revert to the continuous-data signal as the sampling period approaches zero? We shall discuss this problem in chapter 15.

Relationship Between The s-Plane and The z-Plane

(14.12)

(14.14)

( )f t

0( ) [ ( )] ( ) stF s f t f t e dt

¥ -= = òL

( )f t

0( ) { ( )} ( ) k

kF z z f t f kT z

¥-

=

= =å

0lim ( ) ( )TF z F s

®¹

0lim ( ) ( )T

f t f t*®

¹

(14.13)




v Mapping of the primary strip§ In figure 14.1(a), the left half of the s-plane is mapped into a unit circle, centered

at the origin in the z-plane. Since


Figure 14.1 Mapping of the primary strip in the left half of the s-plane into the z-plane by the z-transform

( ) 2ss jm T Ts j m Tse e e e zw p+ = = =




v The constant-damping locus§ The constant-damping locus in the z-plane are described by

(where is real constant)


Figure 14.2 Constant-damping locus in the s- and z-planes

1TTsz e e s= = (14.15)1s




v The constant-frequency locus§ Frequency in the s-domain, the constant-frequency locus in the s-plane is a

horizontal line, .§ The corresponding z-transform is


Figure 14.3Constant-frequency locus in the s- and z-planes

1w w=1s jw=

1j TTsz e e w= = (14.16)




v The constant-damping-ratio locus§ The constant-damping ratio locus in the s-plane is described by

where as shown in figure 14.4(a).§ The z-transform relation is§ Figure 14.4(b) shows the constant-damping ratio locus( )


Figure 14.4Constant-frequency locus in the s- and z-planes

(14.17)tans jw b w= - +1sinb z-=

2 tan /( tan ) 2 /sTs T j sz e e epw b ww b w pw w-- += = = Ð

0.5z = 30b = o




v The constant-damping-ratio locus§ For practical purpose only the portion of the constant-damping-ratio locus that

corresponds to the frequency rage from to .


Figure 14.5Constant-damping-ratio locus bounded by the primary strip in the s-plane and the corresponding locus in the z-plane

0w = / 2sw w=





TABLE 14.1 Partial table of z- and s-transforms




v Non-uniqueness of the z-transform§ The correct result of the inverse z-transform of is which is equal to

only at the sampling instants.

§ The inverse z-transform of is not unique, since can be any function of t that has a value of unity at the sampling instants.

§ Figure 14.6 illustrates the simple case of taking the z-transform of a unit-step function, unit impulses with the sampling period of T s.

Inverse z-Transform

( )F z ( )f t

( )F z ( )f kT( )f t

Figure 14.6 Illustration of the nonuniqueness of the inverse z-transform




v The inverse z-transform

§ Two methods for finding the inverse z-transform (the sampled time function from its z-transform) will be described:

• Partial-fraction expansion• The power series method

§ Inverse z-Transforms via Partial-Fraction Expansion• Taking this lead and looking at Table 14.1, we find that sampled exponential time functions are

related to their z-transforms as follows:

• We thus predict that a partial-fraction expansion should be of the following form:

Inverse z-Transform

akTaT

zez e

-

-Û

-

1 2

( ) Az BzF zz z z z

= + +- -

L (14.193)

(14.18)




v Ex 14.2 Inverse z-transform via partial-fraction expansion

§ Given the function in Eq. (14.20), find the sampled time function.

§ Sol) Begin by dividing Eq.(14.20) by z and performing a partial-fraction expansion.

§ Next, multiply through by z.

Inverse z-Transform

0.5( )( 0.5)( 0.7)

zF zz z

=- -

(14.20)

( ) 0.5 2.5 2.5( 0.5)( 0.7) 0.5 0.7 0.5 0.7

F z A Bz z z z z z z

-= = + = +

- - - - - - (14.21)

0.5 2.5 2.5( )( 0.5)( 0.7) 0.5 0.7

z z zF zz z z z

-= = +

- - - -(14.22)




§ Using Table 14.1. we find the inverse z-transform of each partial fraction. Hence, the value of the time function at the sampling at the sampling instants is

§ Also, from Eqs.(14.1) and (14.23), the ideal sampled time function is

§ If we substitute k=0,1,2, and 3, we can find the first four samples of the ideal sampled time waveform. Hence,

Inverse z-Transform

( ) ( )( ) 2.5 0.5 2.5 0.7k kf kT = - +

( ) ( )*( ) ( ) ( ) [ 2.5 0.5 2.5 0.7 ] ( )k kk k

f t f kT t kT t kTd d¥ ¥

=-¥ =-¥

= - = - + -å å

(14.23)

(14.24)

*( ) 0 ( ) 0.5 ( ) 0.6 ( 2 ) 0.545 ( 3 )f t t t T t T t Td d d d= + - + - + - (14.25)




Theorem of the z-Transform

TABLE 14.2 z-transforms theorems




v Multiplication by a Constant

§ If is the z-transform of , then

where a is a constant.

§ Proof: By the z-transform definition,


( )F z ( )f t

{ ( )} { ( )} ( )z af t az f t aF z= = (14.26)

0 0{ ( )} ( ) ( ) ( )k k

k kz af t af kT z a f kT z aF z

¥ ¥- -

= =

= = =å å (14.27)




v Addition and Subtraction§ If and have z-transforms and

and

then, the z-transform of is

§ Proof: By definition,


(14.28)

(14.29)

1( )f t 2 ( )f t 1( )F z 2 ( )F z

1 1 10

( ) { ( )} ( ) kk

F z z f t f kT z¥

-

=

= =å

2 2 20

( ) { ( )} ( ) kk

F z z f t f kT z¥

-

=

= =å

1 2( ) ( )f t f t±

1 2 1 2{ ( ) ( )} ( ) ( )z f t f t F z F z± = + (14.30)

1 2 1 20

1 20 0

1 2

{ ( ) ( )} [ ( ) ( )]

( ) ( )

( ) ( )

k

k

k k

k k

z f t f t f kT f kT z

f kT z f kT z

F z F z

¥-

=

¥ ¥- -

= =

± = ±

= ±

= +

å

å å (14.31)




v Complex Translation Theorem

§ If has the z-transform of , then

where a is a constant.

§ Proof: From the z-transform definition we can write the z-transform of as

Let ; then Eq.(14.33) becomes


( )F z( )f t

(14.32)

(14.33)

{ ( )} [ ( )] ( )Tsat aTz ez e f t F s a F ze±

== ± =m

( )ate f tm

0{ ( )} ( )at akT k

kz e f t f kT e z

¥-

=

=åm m

1aTz ze±=

1 10

{ ( )} ( ) ( )

( )

at k

kaT

z e f t f kT z F z

F ze

¥-

=

±

= =

=

åm

(14.34)




v Real Translation (shifting) Theorem-Right shift(Time delay)

§ If is the z-transform of , then the z-transform of shifted to the right by nT, where n is a positive integer, is


which can be written as

since for , Eq.(14.37) becomes

Now, letting in Eq.(14.37) yields


( )F z

(14.35)

(14.36)

(14.37)

( )f t ( )f t

0{ ( ) ( )} ( ) k

kz f t nT u t nT f kT nT z

¥-

=

- - = -å

{ ( ) ( )} ( )nz f t nT u t nT z F z-- - =

( )

0{ ( ) ( )} ( )n k n

kz f t nT u t nT z f kT nT z

¥- - -

=

- - = -å( ) 0f t = 0t <

( ){ ( ) ( )} ( )n k nk n

z f t nT u t nT z f kT nT z¥

- - -

=

- - = -å

p k n= -

0{ ( ) ( )} ( )

( )

n p

p

n

z f t nT u t nT z f pT z

z F z

¥- -

=

-

- - =

=

å(14.38)




v Real Translation (shifting) Theorem-Left shift(Time Advance)§ The z-transform of a time function which is shifted to the left, or advanced in

time by nT, where n is a positive integer, is written


The sequence start at k=0

or


(14.39)

(14.40)

(14.41)

( )f t

0{ ( ) ( )} ( ) k

kz f t nT u t nT f kT nT z

¥-

=

+ + = +å

1

0{ ( ) ( )} ( ) ( )

nn k

kz f t nT u t nT z F z f kT z

--

=

é ù+ + = -ê úë ûå

( )

0{ ( ) ( )} ( )n k n

kz f t nT u t nT z f kT nT z

¥- +

=

- - = +å

1

0

{ ( ) ( )} ( )

( ) ( )

n k

k n

nn k

k

z f t nT u t nT z f kT z

z F z f kT z

¥-

=

--

=

é ù+ + = ê úë ûé ù= -ê úë û

å

å (14.42)




v Initial Value Theorem

§ If the function has the z-transform , then

if the limit exists.

§ Proof: From the defining equation of the z-transform, we write

Taking the limit on both sides of Eq.(14.44) as z approaches infinity, we get


( )f t ( )F z

0lim ( ) lim ( )k z

f kT F z® ®¥

=(14.43)

1 2

0( ) ( ) (0) ( ) (2 )k

kF z f kT z f f T z f T z

¥- - -

=

= = + + +å L (14.44)

0lim ( ) (0) lim ( )z kF z f f kT

®¥ ®= = (14.45)




v Final Value Theorem

§ If the function has the z-transform , and if the function does not have poles on or outside the unit circle in the z-plane, then

§ Proof: Let us consider two sequences,

and

where n is a positive integer.


( )f t ( )F z

(14.46)

(14.47)

(14.48)

1(1 ) ( )z F z--

1z =

1

1lim ( ) lim(1 ) ( )k z

f kT z F z-®¥ ®

= -

1 2

0( ) (0) ( ) (2 ) ( )k n

kf kT z f f T z f T z f nT z

¥- - - -

=

= + + + +å L

1 2

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

nk n

kf k T z f z f T z f n T z- - - -

=

- = + + + -å L




v Final Value Theorem

§ Comparing Eqs.(14.47) and (14.48), the latter can be written as

§ Taking the difference between Eqs.(14.47) and (14.49), and z->1, we get

§ Taking the limit as n approaches infinity on both sides of Eq.(14.50),

§ Interchanging the limits on the right-hand side of Eq.(14.51)

à


(14.49)

(14.50)

(14.51)

11

0 0[( 1) ] ( )

n nk k

k kf k T z z f kT z

-- - -

= =

- =å å

1 11

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

n n n nk k

z k k k kf kT z z f kT z f kT f kT f nT

- -- - -

®= = = =

é ù- = - =ê úë ûå å å å

11

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

n nk k

n n z k kf nT f kT z z f kT z

-- - -

®¥ ®¥ ®= =

é ù= -ê úë ûå å

1

0 0lim ( ) lim ( ) ( )

n nk k

n nk kf kT z f kT z F z

-- -

®¥ ®¥= =

= =å å (14.52)1

1lim ( ) lim(1 ) ( )n z

f nT z F z-®¥ ®

= - (14.53)




v Process in solution of Difference Equation§ First, the equation are transformed to the z-domain

§ Second, the variable of interest is solved for and z-transformed, by using the time delay or the time advance property.

§ Finally, the inverse z-transformation is performed to get solution of difference equations.

v Ex 14.3: z-Transform Solution of Difference Equation§ Solve the linear difference equation

with the initial conditions

z-Transform Solution of Difference Equation

( 2) (3 / 2) ( 1) (1/ 2) ( ) 1( )x k x k x k k+ - + + =

(0) 1, (1) 5 / 2x x= =

(14.54)




v Ex 14.3: z-Transform Solution of Difference Equation§ Sol)

z-Transform:We begin by z-transforming using Time advance theorem to obtain

§ Solve for We substitute the initial conditions and rearrange terms to obtain

then


[ ]2 2( ) (0) (1) (3 / 2) ( ) (0) (1/ 2) ( ) / ( 1)z X z z x zx zX z zx X z z zé ù- - - - + = -ë û

( )X z

2 2(3 / 2) (1/ 2) ( ) / ( 1) (5 / 2 3 / 2)z z X z z z z zé ù- + = - + + -ë û

3

2

[1 ( 1)( 1)]( )( 1)( 1)( 0.5) ( 1) ( 0.5)z z z zX zz z z z z

+ + -= =

- - - - -

(14.55)

(14.56)

(14.57)




v Ex 14.3: z-Transform Solution of Difference Equation§ Partial Fraction Expansion

The partial fraction of is

where

To obtain B, multiplying by the denominator and get equation as

If z=0, àB=0 (A=2,C=1)


( ) /X z z2

2 2

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

X z z A B Cz z z z z z

= = + +- - - - -

2 2

2 20.5 0.5

( ) (0.5)( 0.5) 1( 1) (0.5 1)z z

X z zC zz z= =

= - = = =- -

22

1 1

( ) 1( 1) 20.5 1 0.5z z

X z zA zz z= =

= - = = =- -

2 2( 0.5) ( 0.5)( 1) ( 1)z A z B z z C z= - + - - + -

0 0.5 0.5A B C= - + +

(14.58)

(14.59)

(14.60)




v Ex 14.3: z-Transform Solution of Difference EquationSo, we have Two terms only

§ Inverse z-TransformationFrom the Table 1. the Inverse z-transform of is


2

2( )( 1) 0.5z zX z

z z= +

- -

( )X z

( ) 2 (0.5)kx k k= +

(14.61)

(14.62)




v Stability

§ The glaring difference between analog feedback control systems and digital feedback control systems, which is the effect that the sampling rate has on the transient response.

v Digital System Stability via The z-Plane

§ In the s-plane the region of stability is the left half-plane.

§ If the transfer function, G(s), is transformed into a sampled data transfer function, G(z), the region of stability on the z-plane can be evaluated from the definition, z = .

§ Letting , we obtain

Stability

Tses ja w= +




§ Letting , we obtain

§ Since .

Stability

s ja w= +

( )

( )

cos sin

Ts T j T j T

T T

z e e e ee T j T e T

a w a w

a aw w w

+= = =

= + = Ð

(cos sin ) 1T j T Tw w w+ = Ð

(14.63)

Figure 14.7 Mapping regions of the s-plane onto the z-plane




v Stability criterion

§ Thus, a digital control system is

• Stable if all poles of the closed loop transfer function, T(z), are inside the unit circle on the z-plane.

• Unstable if any pole is outside the unit circle and/or there are poles of multiplicity greater than one on the unit circle.

• Marginally stable if poles of multiplicity one are on the unit circle and all other poles are inside the unit circle.

Stability




v Bilinear Transformation

§ Bilinear transformations give us the ability to apply our s-plane analysis and design techniques to digital systems. We will treat it in chapter 16 more specifically.

§ We can analyze and design on the s-plane as we have done in Chapters 8 and 9 and then, using these transformations, convert the results to a digital system that contains the same properties.

§ What we would like is a simple transformation that would yield linear arguments when transforming in both directions through direct substitution and without the complicated z-transform.

§ Bilinear transformations of the form

Stability

dcsbasz

++

= (14.64)




and its inverse

have been derived to yield linear variables in z.

§ Different values of a, b, c and d have been derived for particular applications and yield various degrees of accuracy when comparing properties of the continuous and sampled functions.

Stability

(14.65)aczbdzs

-+-

=




v Digital System Stability via the s-plane

§ In this subsection we look at a bilinear transformation that maps -axis points on the s-plane to unit-circle points on the z-plane.

§ Further, the transformation maps right half-plane points on the s-plane to points outside the unit circle on the z-plane.

§ Finally, the transformation maps left-half-plane points on the s-plane to points inside the unit circle on the z-plane.

§ Thus, we are able to transform the denomination of a continuous transfer function, D(s), and use the Routh-Hurwitz criterion to determine stability.

§ The bilinear transformation

§ And its inverse

Stability

jw

(14.66)11

zsz+

=-

11

szs+

=-

(14.67)




§ Perform the required transformation

§ We can show this fact as follows: Letting and substituting into Eq. (14.67)

§ From which

§ Thus

Stability

s ja w= +

( 1)( 1)

jzj

a wa w+ +

=- +

2 2

2 2

( 1)( 1)

za wa w+ +

=- +

1 0

1 0

1 0

z where

z where

z where

a

a

a

< <

> >

= =

(14.68)

(14.69)

(14.70a)

(14.70b)

(14.70c)




v Ex 14.4 Stability via Routh-Hurwitz

§ Given , where, use the Routh-Hurwitz criterion to find the number of z-plane poles of T(z) inside, outside, and on the unit circle. Is the system stable?

§ Sol) Substitute Eq.(14.67) into D(z)=0 and obtain

Stability

3 2( ) 0.2 0.1D z z z z= - - +( ) ( ) ( )/T z N z D z=

3 219 45 17 0s s s- - - = (14.71)




§ The Routh table for 14.3 shows one root in the right half-plane and two roots in the left half plane.

§ Hence, T(z) has one pole outside the unit circle, no poles on the unit circle, and two poles inside the unit circle.

§ The system is unstable because of the pole outside the unit circle.

Stability

TABLE 14.3 Routh table for Example 14.4




TTHHAANNKK UUYYOO

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Chapter 14 Intelligent Robot Lab z-Transform and Stability...

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