parameter plane techniques for feedback control systems · tableofcontents section title page 1....

Calhoun: The NPS Institutional Archive

Theses and Dissertations Thesis and Dissertation Collection

1965

Parameter plane techniques for feedback control systems

Nutting, Roger M.

Monterey, California: U.S. Naval Postgraduate School

http://hdl.handle.net/10945/12804

NPS ARCHIVE1965NUTTING, R.

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PARAMETER PLANE TECHNIQUES FORFEEDBACK CONTROL SYSTEMS

ROGER M. NUTTING

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ilil;

PARAMETER PLANE TECHNIQUES

FOR FEEDBACK CONTROL SYSTEMS

by

Roger M. Nutting

Lieutenant, United States Navy

Submitted in partial fulfillment of

the requirements for the degree of

MASTER OF SCIENCEIN

ELECTRICAL ENGINEERING

United States Naval Postgraduate SchoolMonterey, California

19 6 5

U. S. N aate '

'i I i<f

Monterey, Caliii,

PARAMETER PLANE TECHNIQUES

FOR FEEDBACK CONTROL SYSTEMS

by

Roger M. Nutting

This work is accepted as fulfilling

the thesis requirements for the degree of

MASTER OF SCIENCE

IN

ELECTRICAL ENGINEERING

from the

United States Naval Postgraduate School

ABSTRACT

Parameter plane techniques were first introduced in an IEEE paper

dated November 4, 1964. The paper dealt mainly with the theory of the

parameter plane whereby the roots of a polynomial could be determined

graphically in terms of two parameters which may appear linearly in

any of the coefficients.

In this text, parameter plane techniques are applied to the

compensation of linear feedback control systems by both graphical and

analytical means. Parameter plane equations are extended to include

parameter products and three parameters. An attempt is made to show

the complementary roles of the parameter plane and root locus.

The writer wishes to express his appreciation for the assistance

and guidance given him by Dr. G. J. Thaler of the U. S. Naval Post-

graduate School in this investigation.

ii

TABLE OF CONTENTS

Section Title Page

1. Introduction 1

2. Derivation of basic parameter plane equations 3

3. Application of the parameter plane to the compensa-tion of linear continuous systems 8

3-1 Algebraic solution 8

3-1-1 Feedback compensation 8

3-1-2 Cascade compensation 22

3-1-3 Combination cascade and feedback compensation 27

3-2 Dominancy of the specified roots 39

3-2-1 A method of employing a third parameter 40

3-2-2 Application of the dominancy technique 43

3-3 Some sketching techniques 56

3-3-1 Table of symbols 56

3-3-2 Basic derivations 58

3-4 Graphical solutions on the parameter plane 79

3-4-1 Advantages of the graphical solution 79

3-4-2 Some examples of the graphical solution 79

4 Miscellaneous aspects of the parameter plane 91

4-1 Some general comments 91

4-2 Normalized third order curves 93

4-2-1 Discussion of the normalized curves 93

4-2-2 Derivation of the normalized transformations 93

4-2-3 Application of the method 99

4-3 Normalized parameter plane curves of higher order 107

4-4 Three dimensional parameter plane space 108

1X1

Section Title Page

4-4-1 Discussion 108

4-4-2 Example problem 108

4-5 Characteristic equations involving product termsof alpha and beta 111

4-5-1 Basic derivations 111

4-6 Design of double section cascade compensators 115

4-6-1 Discussion 115

4-6-2 Design of a double section compensator on thebasis of given single section parameter values 115

4-6-3 Design of a double section compensator usinggeneral parameter plane methods 118

5 Root locus digital computer programs 124

5-1 A program to compute the coefficients of a 125

polynomial from the given factors

5-2 A program to plot root loci from given 128

characteristic equation in polynomial form

6 Parameter plane digital computer programs 133

6-1 A program to plot parameter plane curves with or 134

without a third parameter

6-2 A program to plot parameter plane curves fromcharacteristic equations involving product terms 146

of alpha and beta

7 The complementary roles of the parameterplane and root locus 157

8 Conclusions

Bibliography 176

Appendices

I Tables of Chevishev functions

. a. Tk

( j >

B. Uk ( J )

174

II The R.C. network as a lag-lead compensator 175

173

iv

List of Illustrations

Figure Title Page

3-1 A feedback compensated system 9

3-2 A third order system with tachometer andacceleration feedback 12

3-3 A third order system with tachometer feedback 16

3-4 A third order system with acceleration feedback 18

3-5 A third order system with tachometer and accelerationfeedback not enclosing an amplifier 20

3-6 A cascade compensated system 23

3-7 A third order system with single section cascadecompensation 28

3-8 A feedback and cascade compensated system 29

3-9 A third order cascade and tachometer feedbackcompensated system 35

3-10 Feedback compensation enclosing a cascade compensator 37

3-11 A second order system with single section cascadecompensation 45

3-12 A fourth order system with tachometer andacceleration feedback 47

3-13 A fourth order system with single section cascadecompensation 51

3-14 Third order root locus plot 54

3-15 Zeta equals and .5 curves for a second order system 68

3-16 Zeta equals and .5 curves for a third order system 69

3-17 Zeta equals and .5 curves for a fourth order system 70

3-18 Zeta equals and .5 curves for a fifth order system 71

3-19 A third order system with position feedback 73

3-20 A third order cascade compensated system 75

3-21 A fourth order system with position feedback 77

3-22 A fourth order system with tachometer andacceleration feedback 81

3-23 Parameter plane curves for the system of figure 3-22 83

Figure Title Page

3-24 A third order system with position feedback 85


3-26 A third order system with cascade compensation 88


4-1 Normalized B -B, curves 94o 1

4-2a Normalized B]-B. curves (large scale) 97

4-2b Normalized B. -B« curves (small scale) 98

4-3 A third order system with tachometer feedback 101

4-4 A third order system with tachometer andacceleration feedback 103

4-5 A second order system with cascade compensation 105

4-6 A third order system with tachometer feedbackenclosing a cascade compensator 110


4-8 A third order system with double section cascadecompensation 119

4-9 Constant zeta and omega curves for the system offigure 4-8 121

4-10 Constant sigma curves for the system of figure 4-8 122

7-1 A fifth order feed-forward compensated system (initial) 161


7-3 A root locus for the system of figure 7-1 163

7-4 A root locus for the system of figure 7-4 164

7-5 Root locus for the basic system of figure 7-1

with tachometer feedback added 165

7-6 A fifth order feed-forward compensated system (final) 166

7-7 Root locus for the system of figure 7-6 167

7-8 Root locus for the system of figure 7-6 168


vi

Figure Title Page

7-10 Transient response for the compensated systemof figure 7-6 170

II-a. Typical lead network 175

Il-b. Typical lag network 175

TABLES

Number Title Page

3-1 Error coefficients 30

3-1 List of symbols 57

viii

1 . Introduction.

The analysis and synthesis of feedback control systems, or the

compensation of same, can be effected by three general methods. The

first of these can be called the integral criteria. Here a cost function,

in which is inherent the system design specifications, is minimized with

respect to certain variable system parameters. This method is mainly

applied to the statistical properties of feedback control systems. The

second method is the Bode frequency response method whereby the system's

open loop transfer function is manipulated to obtain the desired system

response. This method has its inherent weaknesses, such as difficulty of

application to non-unity feedback control systems, difficulty in inter-

preting closed loop transient response in terms of open loop frequency

response, difficulty of varying more than one parameter, and the short-

comings inherent in the approximations which are required when applying

the frequency response method to compensation problems. Third are the

algebraic methods. Under this heading can be listed the root locus method,

The shortcomings of this widely known and widely used method are familiar

to its users. Its greatest disadvantages lie in plotting the actual locus

of roots , and in the fact that only one parameter can be varied conven-

iently. In references (1) and (2), Ross-Warren and Pollak respectively

developed algebraic methods of cascade compensation using root locus

techniques, but the inherent disadvantages of the root locus were still

present. In reference (3) Mitrovic developed an algebraic and graphical

A computer program is introduced in section (5) which makes thisdisadvantage less significant.

method of obtaining the roots of a polynomial in terms of two variable

parameters. Later in reference (4) Ohta developed some additional

sketching techniques which greatly facilitated the plotting of the

Mitrovic curves. In references (5) and (6) Choe and Hyon respectively

applied and extended the Mitrovic method to the compensation of linear

continuous feedback control systems. The inherent disadvantage of the

Mitrovic method is that the variable parameters may appear in no more than

two coefficients of the characteristic equation, which reduces the flex-

ibility of the method. In reference (7) Siljak introduced a method of

obtaining the roots of a polynomial in terms of two variable parameters

which can appear in any and all the coefficients of the polynomial. In

this text, Siljak* s method is applied and extended to the compensation

of linear continuous feedback control systems. General methods of com-

pensation will be developed and an attempt will be made to relate the

root locus, and the parameter plane as a set of complementary techniques

which when applied in conjunction with one another represent the most

adequate tool to date for solving the problem of compensation of linear

feedback control systems.

The relationship between being able to place the roots of a polynom-

ial at specified locations in the S-plane and the compensation of feed-

back control systems is as follows. The basic idea is that any feedback

control system, including any added compensators which may contain vari-

ables, can be reduced to or can be represented by a ratio of two poly-

nomials which is the closed loop transfer function. Well known methods

are available whereby a specified system response, in terms of over-

shoot, bandwidth, settling time, steady state accuracy, etc., can be

obtained by placing a pair of complex conjugate roots of the character-

istic equation at a specified location in the S-plane, while ensuring that

the real part of this complex root pair (called the dominant roots) is

smaller in magnitude than the real parts of the remaining roots of the

characteristic equation. The problem of compensation therefore reduces

itself to one of moving the roots of the characteristic equation to the

desirable locations. The usefulness of the parameter plane to achieve

this will soon become apparent.

2. Derivation of the basic parameter plane equations.

A feedback control system's characteristic equation can be re-

presented as a polynomial of the following form:

,kf(S) = £ a. S

K= (2-1)

K'-OK

Where the coefficients \(k = 0» l,....,m) are real, and S is the

complex variable S = -G+ jw - J w + jw y 1- J . (2-2)

w is the undamped natural frequency, and Tw is the relative damping

coefficient. It is noted in reference (7) that S may be represented

by the following:

.k k.„ , _, v , /, 7 27

S" » w (Tk(- J ) + j |/

1- J* Uk<- J )) (2-3)

where TR(- ^ ) = M)k

Tk < ^ ) and u*

k(-

J ) = (-l)k+1

Uk ( J

). \( ^ )

and U ( ?" ) are Chebishev functions of the first and second kind respect-

ively. Values of zeta will be considered such that 0-7-1 and

values of w such that s w s °a . The values of T. and U. are tabu-

lated in Appendix I for various values of zeta. But more useful to digi-

tal computer employment, they can be obtained from the following recursion

relations:

Tk + i<7> -

2 V?> + Tk-i<?>-° < 2 " 5 >

\ + 1< n -2 v f ) + \_i< ? > * °

Here TQ( f ) » 1, T, < J ) - ^ , U

q ( J ) = 0, U^ f) = 1

Substituting equation (2-3) into (2-1) and setting the real and imagin-

ary parts to zero independently one obtains:

k.

£ v\< -1 >

°

K, k

K:y\(- ? ) = (2-6)

Employing equations (2-5) one obtains from equations (2-6)

:

m k k| (-D v\-i ( ?> =

| o(-l)\„\(

? ) =0 (2-7)

Consider the coefficients a, of the characteristic equation (2-1) as

linear functions of the variable system parameters as follows:

V-V- + ekP+ -k

(2-8)

Employing the above relation for a, , equations (2-7) give the following

relations:

°^ B1+ ^ C

i+ D

is °

cLB + |3C, + D =0 (2-9)2 2

Where:

51 k k J^ k kB. = £ (-l)VX i

Bo " ^(-l)\w\

1 K=o k k-1 2 k=o K k

53 k 53 k kC - f. (-D c.U. , C = l(-l)c.w\ (2-10)

1 K=o k k-1 2 K^ k k

k. k„ _ J? . ,,k, k.Di

= g>« V\.i D2= £ <-1)dkwD

k

Since equations (2-9) are two linear equations in the two unknowns alpha

and beta, Cramer's rule can be applied to obtain:

<*-ClVC

2Dl (S =

B2DrB

lD2 (2-11)

B1C 2'B

2C1

B1C 2"B 2

C1

Equations (2-11) are now functions of zeta and w. Hence by fixing w and

varying zeta or by fixing zeta and varying w, the constant w or constant

4

zeta S piano contours respectively can be mapped into the real domain

of the alpha beta plane or parameter plane.

In reference (7) the following relationships are utilized:

Sk

= P + jw /l- J2

Q

Pk +1

+ 2W Pk+ Ac-1 - ° < 2 " 12 >

°k + l+2wQ

k+ W\.l =

°

PQ

. 1,?i

- -w? ,

o= 0, Q

x== 1

Pk

= -w^Qk-w\_

1

Here P and Q, are related to the Chebishev functions by

pk

= wklk ( " 7 >

=^ml^V 7 ) < 2 - 13 >

Qk- »

k"\<- } ) = (-Dk+1

wk_1

Uk

( J )

By using equations (2-12), and (2-13), one obtains proceeding as before:

1,-kir ^o\\'° (2- 14)

Employing equations (2-8), (2-14), along with Cramer's rule one again

obtains equations (2-11) where the following expressions now apply:

bi J. Vk-1 B

2 " | \\W, KM

ci

" | cA-i S £. \\ < 2 " 15 >

jfM KM

Equations (2-11) and (2-15) arc useful for mapping constant zeta-omega

curves from the S~planc into the parameter plane. As will be seen later

these curves play an important role in dominance considerations.

If the complex variable S is substituted in equation (2-1) by lett-

ing

S = - e£ (2-16)

where sigma corresponds to values of S along the real axis, then in

accordance with equations (2-8), the characteristic equation (2-1) be-

comes:

oL £ (-l)\^ k+ 8 f? (-D

kcw^

k+ I! (-D

kd,^

k= (2-17)

The above represents a straight line in the alpha-beta plane for a given

value of sigma. Hence a point on the real axis in the S -plane maps into

a straight line in the alpha-beta plane. Also for a given value of alpha,

beta, and sigma which satisfies equation (2-17), then the characteristic

equation (2-1) must have a real root at minus sigma. For the constant

zeta and omega curves as defined previously, if for certain values of

alpha and beta, say for a value obtained from equations (2-11) with a

certain value of zeta and omega, then the characteristic equation (2-1)

/ 21

has a pair of complex roots at S = - 7 w + jw J 1- T .

It is important to note that by applying equations (2-11) and (2-17)

one can, for a specified value of zeta, omega, and sigma, compute the

value of alpha and beta so that the characteristic equation will have a

I 21

pair of complex roots at say S = - T-.W. + jw J 1- J . , and a real

root at S = - ^ , The m-3 remaining roots of the characteristic equa-

tion can then be determined by dividing out the three known or specified

roots. This method where zeta, omega, and sigma, or just zeta and omega

are specified, and the computations for alpha and beta are done algebrai-

cally, will be referred to as the albegraic parameter plane solution.

6

To solve the problem in general, for all values of zeta, omega, and sigma,

it is necessary to plot a family of parameter plane curves for various

values of zeta, omega, sigma, and if desired, zeta-omega. On the result-

ing parameter plane plot one can, by picking an M point or operating

point, graphically read from the curves the values of the m roots corres-

ponding to the m order characteristic equation. This latter method will

be called the graphical parameter plane solution.

The algebraic solution has the advantage that the labor of plotting

2the curves can be avoided, but it has the disadvantage that without the

curves it is sometimes difficult to pick the most optimum value of zeta

and omega so as to ensure dominance and still meet the system specifica-

tions. The graphical solution has the advantage that one has a picture

of the way the roots of the characteristic equation move around in the

S -plane as alpha and beta are varied. This enables one to pick the values

of alpha and beta corresponding to the best values of zeta, omega, sigma,

and zeta-omega for all the roots of the characteristic equation. This

latter feature of the parameter plane points out a strong argument for

trying to obtain the parameter plane curves. If a digital computer is

not available then by using the relationships derived in section (3-3)

under sketching techniques, along with a desk calculator or slide rule,

the curves can be plotted with some labor. Under these circumstances it

is questionable whether the algebraic or the graphical solution would be

better. Which one is used is a matter of personal preference.

2The computer program presented in section (6-1) is most helpful in

reducing this labor.

3. Application of the parameter plane to the compensation of linear

continuous systems.

3-1. Algebraic solution.

In this section it will be assumed that the system performance specifi-

cations have been given in terms of placing a pair of complex roots at a

specific value of zeta and omega, say ? . and w , with the error coef-

ficient K being greater than or equal to a specified value. If the specify

ed location of the roots is such that after computation of the necessary

value of alpha and beta, the remaining roots of the characteristic equa-

tion are located so that the specified roots are not a dominant pair, then

either a different value of zeta and omega will have to be used (possibly

at the sacrifice of some measure of the system performance), or a differ-

ent method of compensation will have to be used.

In section (3-2) a method is presented whereby the dominancy specifi-

cation can be met by introducing a third parameter.

3-1-1. Feedback compensation.

Figure (3-1) represents a unity feedback control system. In order

to meet the system specifications a feedback compensator H will have to

be used. Let

G = K_ = K (3-1)

e(S> S

m +e ,Sm - l

+... +e^'m-1 L

Where K is the forward path gain which can be varied and e(S) is a poly-

nomial in S representing the poles of the open loop transfer function of

the uncompensated system. The letter L in equation (3-1) corresponds to

the system type. For a type system, L = 0, for type 1, L = 1, for type

2, L = 2, etc. By definition, the error coefficient is given as follows:

8

.

*

1

r

1

9-

i

.: i

/

•*

CD

K = lim SLG (3-2)

eS-»0

Here G is the open loop transfer function of the compensated system.

Sometimes, as for example in reference (8) the error coefficient is design-

ated as K for a type zero system, K for a type one system, and K for a

type two system.

Tachometer plus acceleration feedback .

Here one lets

H = K S + K S2

(3-3)t a v '

The resulting compensated system's characteristic equation becomes:

e(S)+K+(K S+K S2) = (3-4)

or by expanding e(S), equation (3-4) becomes:

Sm

+ e .s"1 " 1

+...+(e_+KK )S2+(e.+KK )S+e +K = (3-5)

m-1 2 a v1 t o

where L is taken to be zero for a type zero system which one can consider

as the most general case. The following results also apply to a type

one system if e is set to zero, or to a type two system if both e and

e. are set to zero, etc. From equations (3-2) and (3-4) the error co-

efficient becomes:

K = lim S° K = K (3-6)6

S*o e(S)+K(K S+K S 2 ) ev vt a ' o

or if the uncompensated system is type one:

K = K (3-7)6

6l+KK

t

and if the uncompensated system is type two:

K = K (3-8)e ——-

—

KKt

Note: If the uncompensated system is type two, the compensated system

10

would be type one if tachometer feedback or tachometer plus acceleration

feedback is used.

In the compensated system's characteristic equation (3-5) let alpha

= KK and beta = KK . Equation (3-5) then becomes:a t '

Sm

-!- em _ 1

Sm " 1

+...+(e2+oOS 2

+ <e + )S + e + K - (3-9)

Recalling from equation (2-8) that in general the coefficients of the

characteristic equation are of the form:

\ = \o(. + ci. 6 + \i an<* correspondingly ih

equation (3-9), m = k, one finds that:

do

= eo+ k

>bo

= Co

=°» d

l= e

l'bl

= °» cl

= 1*

d2

= V b2

= l '

C2

=°> Vl = d

k-l« Vl =°' C

k-1=

°» etC '

Then from equations (2 10) one obtains:

B1

= (-l)2w2U1

= w2

B2

= w2U2

C. = -wU = C = -wU. = -w (3-10)I O Z L

Use was made of the fact that U ,= -1, U =0, and U, 1 (see appen-

-1 o 1

dix I-B). Using the expressions for alpha and beta as given in equations

(2-11) along with the above information the following relations evolve:

oc =ciVc

2Di = ; j; (-"Wi = - £ (-i)VV , (3-iD

B1C 2-B

2C1 -w

3 <= ° '

m0- €

5(-i)V <VU

2Uk-l>

At this point alpha and beta may be linear functions of K, the for-

ward path gain, and one can use the steady state error specification to

put K in terms of alpha and or beta. Since zeta and omega were assumed

to be specified, then equations (3-11) can be used to solve for alpha and

11

/'

/*

CMI

I*

/

(T

12

beta. Then K and K can readily be determined.t

J

Example 3-1

The system given in figure (3-2) is to be compensated using tacho-

meter plus acceleration feedback. The system specifications are as

follows:

1. Complex roots at f = .7, w = 10.

2. K — 6, not to be reduced.e *

Solution:

From equation (3-2):

K = lim SLG K ^>6

e ccS*o 2 + KK

t

or K — 12 + 6KK . The compensated system's characteristic equation is:

S3+ S

2(3 + KK ) + S(2 + KK ) + K = (3-12)

ct C

Lett5_ng alpha = KK and beta = KK equation (3-12) becomes:a t

S3+ S

2(3 + oC ) + S(2 + £ ) + K = (3-12a)

Employing equations (3-10) the following can be obtained:

B - 100 C =

D = -1100-K B = 140

C2

= -10 D£

= -1120

Using enuations(2-ll) one obtains:

oC = 10 (- 1100-K) 8 = 140 (- 1100-K) + 56000 (3-13)-1000 -1000

Note: The preceding expressions could have been arrived at directly by

employing equations (3-11). From the steady state accuracy specifica-

tions it is necessary that:

K - 12 +6^3 , hence let K = 12+6fg

(3-14)

13

Using equation (3-14) and (3-13) beta is found to be:

f= 140(-1100-12-6 ft ) + 56000 =625

-1000

Therefore K = 12 + 6(525) = 3762 and oC = .01(1100 + 3762) = 48.6

Since oC - KK then K = 48,6 - .0129a a

3762

Also from 3 = KK it is seen that K - .166.1 t t

The compensated system's characteristic equation becomes:

S3+ 51.62S

2+ 627S + 3762 (3-15)

2Now zeta = .7 and omega = 10 corresponds to S + 14S + 100. When this

quadratic is divided out of equation (3-14), the remainder is S + 37.62.

Hence zeta-omega of the desired roots 7 <<37.62, and the complex roots

are dominant so the problem is solved.

Example 3- 2

The problem presented in example (3-1) will now be solved by intro-

ducing the error specifications at the beginning of the solution instead

of at the end. Equation (3-12) is as follows:

S3+ S

2(3 + KK ) + S(2 + KK. ) + K =

a t

Again let alpha = KK and beta = KK . Then from the steady state error

requirement one obtains:

K - 12 + 6jS (3-16)

Substituting this expression for K into equation 3-12 results in:

S3+ S

2(3 +oL) + S(2 + £ ) + 12 + 6jB =

Therefore b =0, c =6,d = 12, and all other coefficients are aso o o

before. Hence from equations (3-10) it is seen that:

B = 100, Cx

- -6, DL

- -1112, B2

= 140, (?

2= -10, J>

2= -560

and oC = - 7760 =48.6 £ = 625 K = 3762-160 r

14

The above result agrees with example (3-1).

Note: Equations (3-11.) could not be used here since they are based on

applying the accuracy specifications at the end.

Tachometer feedback only .

Let H = K S (3-17)

The characteristic equation of the compensated system becomes:

S + %\-l+ °2 + ( e

1+o/-) s+e +(3 = (3-20)

o

From equations (2-10) one obtains:

B = B = -w

C = -1 C£

= (3-21)

°1=

Jo(-1)kV\-l °2 = | <" 1>V\

From equations (2-11) it is seen that:

* lo (- 1)kd

kwk"\ and P | <-«Wi <

3 - 22 >

If a given zeta and omega are specified, the alpha and beta can be

computed from equations (3-22). The error coefficient is then determined

directly from equations (3-6), (3-7), or (3-8). Hence the error coeffi-

cient is fixed once zeta and omega have been chosen, so if a certain

error specification is to be met, the specified values of zeta and omega

may have to be adjusted to meet it.

If the error specification was the overriding specification to be

met, then zeta could be fixed at some reasonable value. By means of the

given K , alpha could be computed from equations (3-6), (3-7), or (3-8).

Equations (3-22) could then be used to solve for first omega and then

beta. The calculations would be more tedious however.

Example 3-3 .

Figure (3-3) shows the same system as used in the previous two ex-

amples only now tachometer feedback alone will be tried. The same system

15

u

Gu '

I

|

or

16

specifications are to be met, i.e., K — 6, zeta = .7, and omega = 10.

The compensated system's characteristic equation then becomes:

S3+ 3S

2+ (2 + KK

fc

)S + K = (3-23)

Using equations (3-19) and (3-23) one obtains:

S3+ 3S

2+ (2 + cL )S +j9 = (3-24)

From equations (3-22) alpha is found to be:

cL = -2 + 30(1.4)-100(.96) - -56

Since alpha is negative it is seen that positive tachometer feedback is

required. Since the coefficient of the first power of S in the character-

istic equation would then be negative, the system would be unstable.

Hence the desired system specifications can not be met with tachometer

feedback alone.

Acceleration feedback only .

Let H = K S2

(3-25)a

The characteristic equation of the compensated system then becomes:

Sm + e .S

m ~ l+...+ (e. + KK )S

2 + e.S + e + K = (3-26)m-1 v 2 a' 1 o

Let alpha = KK and beta = K. (3-27)

Then from equations (3-26) one obtains:

Sm + e

1Sm_1

+. ..+ (e. +o£ )S2+ e.S + e +A=0 (3-28)

m-1 I i o i

Using equations (2-10) and (3-28) results in:

2 2 2Bx

= w Ux

= w B2

= w U2

Cx

= U_j = -1 C2

= (3-29)

"' V k W k kdi

= %y i> v\-i D2= $y i) v^

Solving for alpha and beta results in:

17

'•

•

I .

'.

1

4-1

iO

)C4-

s.

1

1

<8>

is-

> }

I

|«*!

QC

18

dL . _% = . _J_ £ C-l> V \ (3-30)

W U2

U2

VA

K-° U <ro2

Calculations for alpha, beta, and the error coefficient are performed

in the same manner as with the preceding tachometer feedback calcula-

tions.

Example 3-4

The system of examples (1) and (2) will now be compensated using

acceleration feedback as indicated in figure (3-4). As before K -^6,

zeta = .7, omega = 10. Therefore K = K/2 and the error coefficient is

unaffected by the acceleration feedback. Hence one can choose K = 12 to

meet the specifications. The compensated system's characteristic equa-

tion then becomes:

S3+ (3 + KK )S

2+ 2S + K = (3-31)

If K in equation (3-31) is set to its prescribed value of 12, only

one parameter remains and the parameter plane equations produce an indeter-

minate solution. Therefore K will be left as the variable beta. Since

beta is fixed by the chosen values of zeta and omega, then so is the error

coefficient, and it will most likely not agree with the error specifica-

tion. In view of this, the solution proceeds as follows. Making the

usual change of variables in equation (3-31) results in:

S3+ (3 +oC )s

2+ 2S + p = (3-32)

By employing equations (3-30) one can solve for alpha and beta.

°L = zJL t -2/10 + 3(1.4) -10(.96)] = 4

1.4

ft= 3(10)

2-(10)

3(1.4) -1/14 [-2(10) + 3(10)

2(1.4)-(10)

3(.96)]

or (3 = -700

19

u

4-

T

I

•H

•

i

20

From the negative value of beta it is concluded that the desired

values of zeta and omega cannot be obtained using acceleration feedback,

and of course neither can the desired error specification be obtained.

One would therefore choose another method of compensation.

If one chooses to use feedback compensation then perhaps tachometer

plus acceleration feedback should be tried first using equations (3-11)

and the appropriate steady state error specifications. If the specifica-

tions cannot be met in this manner then it is obvious that they cannot be

met by either tachometer or acceleration feedback separately. In this

case either the system's specifications must be modified or another

scheme of compensation must be employed. If it is found that the specifi-

cations err be met by combined tachometer and acceleration feedback, then

if desired, equations (3-22) or (3-30) can be used to see if tachometer or

acceleration feedback alone can be used.

Cases where feedback is not around the forward path amplifier .

Figure (3-5) illustrates a compensation situation that sometimes occurs

in practice. This is the situation where it is not possible or practical

to get at the input terminals of the error detector and the feedback has

to be inserted at the output terminals of the amplifier. The solution to

this problem is solved by means of an example.

Example 3-5

Figure (3-5) shows the system that was used in example (3-1) only

now the feedback is inserted at the output terminals of the amplifier

represented by the gain K. The same system specifications are to be met

i.e., K 5: 6, zeta .7, and omega = 10. The characteristic equation now

becomes:

S3+ (3 + K )S

2+ (2 + K. )S + K = (3-33)

21

Letting alpha = K and beta = K equation (3-33) becomes:a t

S3+ (3 +cL )s

2+ (2 + £ )S -1- K = (3-35)

Comparing equation (3-35) to equation (3-12a) it is seen that they are

identical, so the solution obtained for alpha and beta in example (3-1)

applies. This points out an important advantage of the parameter plane

method. This is that the solutions depend only on the characteristic

equation and not on the,, system that the characteristic equation was

formed from.

From example (3-1) it was found that alpha = 48.6, beta 625, and K

= 3762. In this example there are no additional computations necessary

to find K and K , since these are now the system's parameters alpha and

beta. So K = 625 and K =48.6.t a

This same general principal can be applied to control problems in-

volving tachometer feedback alone or acceleration feedback alone.

3-1-2 Cascade compensation .

Figure (3-6) represents a unity feedback control system which in

order to meet the system's specifications a cascade compensator G is re-

quired. Let G have the form of equation (3-1).

i.e., G = K = K (3-1)

e(S)Sm+ e

1Sm-X

+...+e SL

m-1 LK is the forward path gain which can be varied and e(S) is a polynomial

is S representing the poles of the open loop transfer function of the un-

compensated system. The letter L again corresponds to the system type.

Let:

G = P(S + Z ) (3-36)C

Z(S + P)

This compensator has a D.C. gain of unity so its placement in the forward

path will not affect the steady state accuracy. It is assumed that the

22

A'.•••

h'.. <.: :i\

V .'• : '. .••;

-v •'..

,-r •*

fr-ii'-

.

.,, / ••; -.•••• v

> •" i•

j ••;.'•."'

">.

..,-.1 >:•,.

,-...: .;!,..

v;^•^>:

-:

•:•'^•>'•;-:?'' :

l

•

'

1 ' ':'<

•

.•' ,:

.V'|S»t' *

;

. t.'.

•

1 ':

.'

' -:

•-'.

J

:••''

.

{''ft «%/•£$'

S '•' &1 ">- .v

"'• •••''•.• •;: >*' '•<

CD

.•<

f,

<8>

a ' ."«..

v .

('>K>\%'>

:.(.

1'

.':t

;• >. .•.- -:r

llil

SOI

:?

tfV

\l

ill:

A' j

v

I

i

*" ..:.-• Li

"..• ..'.'

> i,'. ., .. i. . «. . • ,. i, • '.• • • •.i

or

<r»...... .'••'

... ••

•

'

,\.:

- .•.«..

.. I

; ,

.,' ; '; :

23'

uncompensated system's forward path gain had previously been adjusted to

give the correct steady state accuracy. Using the form of G as indicated,

the values of 7. and P are computed to give the desired system response.

If P is less than Z a lag network is needed and the factor P/Z of the

compensator is inherently present due to the physical nature of the compen-

sator, which is assumed to be an R-C network. See appendix II concerning

the nature of lag-lead R-C networks. In this case all forward path ampli-

fier gains can remain unchanged to meet the stipulated accuracy specifica-

tions. If however, the computed values of Z and P are such that P is

greater than Z, a lead network is required and the compensated system's

forward path gain will have to be raised by a factor of P/Z. The physical

nature of the lead network is such that the factor P/Z is not inherently

present, so to maintain steady state accuracy this factor will have to

be provided either by adding an amplifier in cascade with the lead net-

work or by raising the gain K of the existing amplifier by this factor.

Continuing then, the procedure is as follows. The compensated system's

forward path transfer function is:

G = G G = K PtS + Z = TSfe + PA) , K (3-37)

CC ° e(S)*Z '

S + P S + P ' e(S)

Applying the definition of the error coefficient to the compensated system

one obtains:

K = lim S

S->0 L e(S) (S + P) J eL]

K_ *(S + P/-y ) ] = K (3-38)

Again assuming a type zero system where L = 0, the compensated system's

characteristic equation becomes:

e(S) (S + P) + K Y (S + P/Y ) (3-39)

or in general after expanding equation (3-39):

24

Sm+1

+ (P + e .)Sm+(Pe ,+e )S

mml+...+ (Pe + e.)S

2+ (3-40)

m-i m-i m-Z Z I

(KX + Pe. + e )S + P(e + K) -10 o

Letting alpha P and beta = If , equation (3-40) becomes:

S"*1+ <dC+ e )S

m U .oL+ em JS™"1 +...+(e_oC+ e.)S

2(3-41)m-i m-i m-Z Z 1

(K Q> + e. cC. + e )S + oC(e + K) =1 i o o

Comparing equation (3-41) to the general form of the characteristic equa-

tion as specified in equations (2-1) and (2-8), it is apparent that K

m + 1 (the order of the equation), b e + K, c d = 0, b, = e.

,

°1 = K»

dl

= V b2

= V %=

°- d2

= el»

etC *

It is important to note that the parameter plane variable beta re- .

presents the pole to zero ratio of the cascade compensator. In references

(1) and (2), Ross-Warren and Pollak respectively, utilized the concept of

root relocation zones to divide the S-plane into regions where lag compen-

sation or lead compensation is required. By assigning variables in the

above manner, the parameter plane is effectively divided into corresponding

regions above and below the straight line beta = 1. For values of beta

less than one, a lag network is required and for values of beta greater than

one, a lead network is required. In addition if beta is greater than 10

or less than .1, a multiple le^dor multiple lftg respectively is required.

A multiple section compensator will also be required if the computed value

of either P or Z turns out to be negative. A method is given in section

(4-6) for the design of double section compensators. If more than two

sections are required, the compensation has to be done in steps.

In this case complex roots are placed at some intermediate value of

zeta and omega and a new characteristic equation is then computed. An-

other compensator can then be designed on the basis of this new character-

istic equation to place the roots at the desired value. Thus by compensa-

25

tion in steps in conjunction with the double section design of section

(4-6), a four section compensator could theoretically be designed. The

use of more than four sections is questionable and in this event it would

be better to employ combined cascade and feedback compensation, or per-

haps feedback compensation alone.

On the basis of equations (3-41) and (2-10) it is found that:

B1

= -<eo+ K) + w

2e2+...+(-l)

k~ 2wk " 2

Uk _ 3

+ (-l^'V'^j

ct

=

Dl

=A +-"+ <- 1 >

k "2em-2

wk "2uk-3

+ <- 1 >

k 'lem-l

wk "luk-2

+

(-DVu^ (3-42)

B2

= -wex+ w

2e2U2

+. . .+(-l)k_2

Uk _ 2

+ (-D^V"' 1^C2

= -wK

D2

= -weo+ w

2eiU2 +...+(-D

k ' 2em . 2

wk" 2

Uk _ 2

+(-l)k ' 1

e ,wk"V , + (-D

kw\m-1 k-1 k

and from equations (2-11) it is found that:

<*- m "^1_ A -B2DrB

lD2 (3-43)

Bx

-ttKB

In equations (3-42), one sets e =0 for a type one system, e = e. =

for a type two system, etc. On the basis of equations (3-42) and (3-43)

a cascade compensator can be designed.

Example 3-6

Problem:

Design a cascade compensator for the system shown in figure (3-7)

to place a pair of characteristic roots at zeta = .5 and omega = 1. The

error coefficient K should be 50.e

26

Solution:

From figure (3-7) it is apparent that K = K/10 = 50 or K = 500. Thee

characteristic equation is:

S4+ S

3(8 + P) + S

2(17 + 8P) + S(10 + 17P + KY ) + P(10 + K) = (3-44)

Letting alpha = P and beta = IS", equation (3-44) becomes:

S4+ S

3(8 +oC) + s

2(17 + 8o^) +S(10 + 17oC+ 500£ ) + (10 + 500 )t£ = 0(3-45)

Applying equations (3-42) and (3-43) one obtains:

Bl

-503

cl

=

Dl

9

B2

-9

C2

= -500

D2

= 6

o^ = .0179 = P (S = .0117 *t

But Y - P/Z so Z - 1.529.

G then becomes: .0117 (S - 1- 1.529 )

C(S + .0179)

This is a lag network where the factor .0117 is inherent in the R-C filter

design. Due to the small value for gamma, the size of the filter compon-

ents may be unreasonable, and a double lag network could be designed us-

ing the method of section (4-6-2). Instead the problem will be solved in

section (3-1-3) using combination cascade plus tachometer feedback compen-

sation which will permit the use of a single section lag network.

3-1-3 Combination cascade and feedback compensation.

Derivative signal not enclosing a cascade compensator .

In figure (3-8), G = K/e(S) and G TS S + P (3-46)c

S + P

First let H(S) K S. Then in view of equations (3-46) the compensated

system's forward path transfer function becomes:

G * S + P . K/e(S) = g S + P . K (3-47)cc

S + P 1 + KH(S)/e(S) S + P e(S) + KH(S)

27

''v.

.V >•

'.••.'.>..".•

•'

•.v ;• ,

.',: ,- '

.": ^ :: -r

-": •„• '-' '

'

v-'. •''•"• • >'V ;i

l •-... •

..>. : : .

'-.

•:" .''

•, '•, ;;,-. •',

/'• ''

•'•''

:.•'

(•\ .•':'

' :•?"''' •'.

:

'

•*•..' -'

'

' ' '' "- V

/.',;.

"; ''t ' '

'

•' " : - '"'"'

CD

+-HI

CO+

.^0

Q.+

;>

I '» Tl~|

"'

.'

1 ':

1 '.:'' ' '-.

-

Hi I 8>

1.

1

,..

1 ...

1 kV

*-'r

"".. 1 "5 .

,' C\ 2

:

;i » i

.1

1

''1i.

1 '

.1 .

' !,

": ' l&l 3, ;i

Si!

. I

'•

'. .

•.

>

. . \,

\*

- '

.'•, J. '

.

'•»1

.-

' i% ;•'

•

'' _. ;</ : ' :- >

;' •

. •j

*/

or

/

28

0^

i

fc

Ct

29

md K = lira S Ge

S*0K

e.(S) + lim KH(S)/S UL

S-»0

(3-48)

Note: The system type will change if the lowest order of the deriva-

tive signal fed back is less than the system type number. A few error co-

efficients are given in table (3-1) for different types of uncompensated

systems. In the table let L be the type number of the uncompensated

system.

H(SH

KtS

K Sa

Table of Ke

1 2

K/eo

K/(e1+KK

1) 1/K

X

K/eo

K/e- K/(e2+KK

2)

Table (3-1)

The compensated system's characteristic equation is:

Se(S) + Pe(S) +SKH(S) + PKH(S) + K X S + PK = (3-49)

Using equation (3-1) for e(S) and letting alpha = P, beta"ft , and

k = m+1, equation (3-49) becomes:

Sk +(o^+e JS

1"" 1 + (e ,o(.+ e „)Sk " 2

+. . ,+(e. + KK + e oC )S2

m-l m-i m-z i c c

+ (e oC + KK oC+ K(3 + eQ)S +q£ (K + e

Q) =

The parameter plane variables are then:

Bi = .eo-K + „

2e2+...+(-l)

k - 2em. 1

wk - 2

Uk .3+ (-l)*'

1^^cx-o

D, - (e. + KK )w e w2U +...+(-l)

k " 2e „w

k_2U,

(3-50)

2 "2 "m-2 k-3

+ <- 1 >k "

lem-l

wk"\-2 + <" 1 >kw\-l (3-51)

30

B2

- -(ej + Kt). ,- e

2,,2U2+...+(-l)

k - 2ein. 1

wk-\.

2

C -Ku

D2

= -v + (a, + KKt)w

2u2+...+(-»

k" 2am.2

»k"\.

2

In terns of equations (3-51) one can solve for alpha and beta

B2D

1- B,D

-wKB

oC= -d./Bj £ =B2D

1" B

1D2 (3-52)

1

One can now compare the above expressions for alpha and beta with

equations (3-42) and (3-43) to see the effect of tachometer feedback.

Let B.' , C' D*

, B', C', and D' represent the quantities given by equations

(3-42) where only cascade compensation was used. Then in terms of the

primed quantities, equations (3-51) become:

»! = BJ B2

= B' - KKtw

Cj = CJ = C2

= C2

(3-53)

Dx

= Dj + KKtw2

D2

= D2+ KK

tw2U2

Let the expressions for alpha and beta as given by equations (3-43)

be designated oL ' and A '. Then in terms of these quantities, equa-

tions (3-52) can be expressed as:

o^ = (^D| - KKtw2)/BJ =ot ' - KK

fc

w2/B| (3-54)

(3 - (-B2wK

t+ K

tDj + K

2Kw

2)/B» + K^ + £

'

Now it can be seen from equations (3-54) how tachometer feedback modifies

the values of cA ' and A * as computed for cascade compensation alone.

For instance if the pole to zero ratio 6 ' is too small, then tachometer

31

feedback can be used to increase this ratio, if for the specified values

of zeta and omega, Bi is negative, etc.

2Now letting H(S) = K S in figure (3-8), the characteristic equa-

a

tion becomes:

Se(S) + Pe(S) + KK S3+ Pkk S

2+ K ^ S + PK = (3-52)

a a

By making the same substitutions as in equation (3-49) one obtains:

Sk +(oC+e

1)S

k " 1+(e - o£ +e )S

k ' 2+. . .+(e. + e Q oC + kk )S

3

m-i m-i m-Z i. 5 a

(e + e o£ +kk oC)S2+ (e oC + K B + e )S + o£ (K + e ) = (3-53)

It then follows readily that:

B. = -(K r e ) + (e, + KK )w2 -e.wV +. . .+(-l)

k ' 2e ,w

k ' 2U,

"m-1wk-3

+ (-l)k'Kk"\_

2

C -0

D]L= e

xw2

- (e2-fKK

a)w

3+. . ,+(-l)

k' 2eB. 2

»k"\.

3 (3-53a)

*«wvi^V2 + (- 1)kw\-i

B2

= -e^-h (e2

-r KKa)w

2U2

- e3w3U3+ . . ^(-D^Vl^Va

/ 1N k-l k-1(-D w uk _ L

C2

= -Kw

2" '3 k " 2

.wk-VD

2= -e

ow + e

xv U

2- (e

2+ KK

a)w U3 +...+<-l) e^w ^ *

<- 1 >k "

lem-l

wk'\-l + ^Ac

After writing equations (3-53a) in terms of the primed quantities, which

correspond to the situation of cascade compensation only, it can be seen

that the following expressions result:

B- - Bi + KK w2

B = B' + KK w U.11a 2 2 a 2

32

C1

= C| = C2

= C2

(3 "54)

D. = D' • KK w3 D„= Dl - KK w

3U11a 2 2 a 3

Solving for alpha and beta one obtains:

o£.= -(Dj - KKw3)/(B{ + KK

aW3) (3-55)

8 = (Bl + KK w2U.)(D' -KKw3

) - (B' + KK w2)(Dl - KK wV)

z a z la la & a i

-wK(Bi + KKaW2

)

There is no straight forward way of observing from equations (3-55)

how cL ' andf3

• are modified by acceleration feedback. However, assum-

ing that one would logically try cascade compensation alone before attempt-

ing combination, one could easily compute the alpha and beta with accelera-

tion feedback added since the quantities B', B«, D' , and D' in equations

(3-55) would already have been computed.

Example 3-7

Problem:

Design a cascade compensator with tachometer feedback as shown in

figure (3-9), so as to place a pair of characteristic roots at zeta *.5

and omega =1. K should be 50.

Solution:

Referring to example (3-6) it is seen that this is the same problem

except that tachometer feedback has been added. As was found in example

(3-6) the value of beta using a single section compensator was too small.

The problem now is to see if tachometer feedback will increase this value.

As was determined previously, K = 500, Bj = -503, C' = 0, Dl = 9,

B* -9, C' -500, D' = 6, oL* - .0179, and fJ ' = .0117, where the primes

have been added to indicate cascade compensation only. Using equations

(3-55) one finds:

33

uCD

^

IT)

+

+

+in

+in

0.

m

I

©

i•H

ct:

Jls

P = (9Kt

-!- 9K + 5OOK^)/(-503) + Kt+ .0117 (3-55a)

It follows from the above equation that for beta to increase the following

inequality must hold:

K > (18K + K2)/503

tv

t t

or (3-56)

K* < (485Kt)/500

Since one excludes values of K less than or equal to zero, equation (3-56)

reduces to:

^ Kfc

< .97

K can arbitrarily be taken as .4. Beta can then be obtained from equa-

tion (3-55a) and is found to be .238 which is in the acceptable region of

.1 < £ -Y ^ 10.

Calculating alpha from equations (3-54) one finds that:

r

2/B| =

The cascade compensator becomes:

o£ = oC ' - KKw/Bj = .414 = P

.238 (S + 1.744) , which can be synthesized as an

(S + .414)

R-C lag network. The compensated system's characteristic equation is as

follows:

S4+ 8.415S

3+ 220. 3S

2+ 219. 06S + 211.69 = (3-57)

The roots of equation (3-57) are:

r, = -.499 - j.8661

r = -.499 + j.8662

r = -3.708 - J14.077

r. = -3.708 + J14.0774

The roots r. and r_ are the desired ones, and since ,499 <-£ 3.708, they

are also dominant.

35

Derivative signal enclosing a cascade compensator

.

In figure (3-10), G - K/e(S) and G = ^ S + P , where theseC

S + P

quantities have been defined previously. Let H(S) = K S, The forward

path transfer function is then seen to be:

G =GcG

= K >> (S + PA ) (3-58)1 + H(S)G

cG (S + P)e(S) + K *» (S + P/tf )H(S)

Since the D. C. gain of G is unity, the error coefficients for figure

(3-10) are the same as those given in table (3-1).

By expanding equation (3-58) into the characteristic equation one

gets after letting alpha P and beta = Y :

2Se(S) + Pe(S) + Ktf K S + KPK S + K ft S + KP = 0, or after expanding:

Sk+ (dL + e ,)S

k-1+ (e ,ot+ e )S

k " 2+...+(e. + KK ft + Pe.o^ )S

2

m-1 m-i m-z 1 tr I

+(e. oL . +KK> + K(S+e )S + °L (K + e ) =0 (3-59)llt'ooThen as before one obtains:

B1

= -(eQ+ K) + w

2e2

+. . 'H-^k'\.

x^'\

m3 + M^'V^U^

C = KK w2

l^-V +...+<-!) em _ 2

w Uk , 3

+(-l) mUmlw U

k _ 2

+(.l)kw\

-1

B2

= -(ei + KKfc

) w + e2w2U2+...+(-l)

k ' 2emlw

k " 2Uk2 + <-l)

k " 1wk " 1

Uk _ 1

C2

= -Kw + KKtw2U2

D2= -e

ow + ,f\ + (-l)

k - 2em .2

»k-\.

2+ M^V./V

+ (-l)kw\

With primed quantities corresponding to the case of cascade compensation

only it follows that:

Bi

= Bi

36

oHI

or

37

C, = C' + KK w2

- KK wlit t

Dl

= Di

B2

" B2 " «V (3-60)

C2

= C^ + KKtw2U2

D = D'2 2

and

o£ = KKfc

w2D^ - (C^ + KK

tw2li2)DJ

B|(C^ + KKfc

w2U2) - (B

2- KK

fc

w)KKtw2

(3-61)

B - (B£ - KKtw)D| - BJD^

B|(C^ + KKtw2U2) - (B^ - KK

tw)KK

tw2

2In figure (3-10) letting H = K S , the characteristic equation becomes:

s

Sk+ (oL + e ,)S

k ~ l+ (e .oC + e )S

k " 2+. ..+(e. + e,o^ + KK £)S

3

m-i m-i m-Z Z j a p

+ (e1+ e

2oC + KK

a<* )S + (e^ + K £ + e

Q)S + oC (K + e

Q) =0 (3-62)

Then in terms of the primed quantities it is found that:

B, = B' + KK w211a

Cl

* Ci

" **a"\ = -KK

aw3u

2

Dl

= Di (3-63)

B2

= B' 4- KKaw3U2

C2

= C' - KKaw3U3

D, - D-

and

o£ - -KK wJU Dl - (C' - KK wU,)D'

a 2 2 2 a 3 1 (3-64)

(B* + KKW2)(C

2- KK

aw3U3) + (B' + KK

aW2U2)KKw3

U2

38

B - (B' + KKw2U2)D| - (Bj + KKw2

)D'

(B' + KK w2)(Cl - KK w

3U ) + (Bi + KK w

2U-)KK w

3U„

1 a 2 a 3 2 a 2a 2

Comparing the expressions for alpha and beta for the case of the deriva-

tive signal not enclosing a cascade compensator to the case of the de-

rivative signal enclosing a cascade compensator one finds that the latter

are considerably more complex. So if one had a choice between the two

methods, the former could be tried first since it is easier to analyze.

If the values for alpha and beta thus obtained were still not acceptable

then the latter method could be tried.

3-2 Dominancy of the specified roots.

In the preceding examples nothing was done in the calculations to

make the specified roots a dominant pair. As was mentioned in section (1),

being able to predict a system's response on the basis of the location of a

pair of complex roots was based on the assumption that the magnitude of

the real part of the primary or specified roots was much less than the

magnitude of the real parts of all the other roots of the characteristic

equation. In most cases, if the real part of the primary roots is one

half to one fifth or less of the real parts of all the secondary roots,

the system is said to be dominant in the primary roots. In many cases

the system will still meet the specifications even if two pairs of complex

roots have the same real part, providing the zetas for both pairs of roots

meet the specifications, and the undamped natural frequencies are such

that the component time responses are not highly additive. The presence

of closed loop zeros will also greatly affect the dominancy factor needed.

For instance even if there exists a characteristic root whose real part is

closer to the origin than the real part of the primary root, the presence

39

of a closed loop zero could make the residue of the close in root negli-

gible as compared to the residue of the parimary roots. However, if possi'

ble, one tries to make the real parts of all secondary roots as large in

magnitude as possible.

In the preceding examples it should be noted that in many cases

there were actually three and sometimes four variable parameters. For

instance, the forward path gain was usually set at a fixed value in the

computations so as to meet the minimum steady state accuracy requirements.

There is, however, usually no reason why the gain cannot be raised above

the minimum value, thus permitting a third degree of freedom. When cas-

cade and feedback compensation are employed simultaneously, the forward

path gain and tachometer gain become the third and fourth parameters.

3-2-1 A method of employing a third parameter.

Recall that the system characteristic equation has the following form

•33 kf(S) = 2: \s = °» wnere sl = b ^ + c Q + d (equations (2-1) and

(2-8)). In order to meet the system specifications, one places a com-

plex root pair at S = - J .w. + jw / 1-?. , which implies that

S2+ 2 ^^S + w

2= 0. (3-65)

Since the coefficients of equation (3-65) are known, this quadratic can

be divided out of the characteristic equation, leaving a polynomial which

contains all the secondary roots of the characteristic equation. Since

only two of the degrees of freedom or variable parameters were used in

fixing the roots of equation (3-65), the remaining variable parameters

will appear in the coefficients of the quotient polynomial, and it is

these coefficients that can be varied to achieve dominance. Instead of

division to find the quotient polynomial, coefficients of like powers will

be equated to achieve a set of equations. Let the quotient polynomial be

40

given by:

fl<

S >= £- ^ = ° < 3 " 66 >

where n = m-2, i.e., equation (3-66) is order two less than the character-

istic equation. Using equations (2-1), (3-65), and (3-66) it is seen

that:

(S2+ 2 7 w.S + w

2)( £ f S

k) = £ a S

k(3-67)

Equating coefficients of like powers and taking a, 1, results in:

\ - fn ' l

Vl = fn-1

+2 ?1W1

ak-2 " f

n-2+ f

n-l2

?1W1

2(3-68)

a2=f

o+ 2

?lWlfl+ f

2Wl

2a, - 2 Z,w, f + f,e

, 2a f w,o o 1

The formulas (3-67) can be solved for the coefficients f in terms of the

coefficients a. The solution will be made for the following cases:

Case of k 3, n = 1

Equation (3-67) becomes:

(S2 + 2 z ^S + w

2)(f

LS + f

Q) = S

3+ a

2S2+ ajS + &

q

Equating coefficients of like powers one obtains:

a3

= 1 - i}

a2

" fo+ 2

J lelf1

al

= flWl+ 2

J iVo

Solving for the coefficients f results in:

41

fl

= 1

fo

= ao/W2 = (a

x- w*)/<2 ^w

x) - a

2- 2 ^ jWj

(3_M)

Case of k = 4, n = 2

Proceeding as before the a coefficients become:

V l - f2

a„ = 2 ? ,w, + f

.

flW

1

a2=W

1+ 2 ?lVl" + f

o (3-70)

al

= flWl+ 2 ?lVo

, 2a = f w,o o 1

When solved for the coefficients f, equations (3-70) yield:

f2=1

fl

= a3 "

2f 1

W1

" a2/2 ?1*1 " V 2 /l " V 2 /lWl

fx

= 1/w2

(ax

- 2 ^t^) (3-71)

fo

= ao/w

l= a

l/2

} 1W1

" a3Wl/2

? 1+ W

l= a

2 " 2? l

Wla3

-

w2 + 4 ^w2

= a2

- 2 ^w^+w^

Case of k = 5, n = 3

Proceeding as before the coefficients a are:

a5

= 1 = f3

a4

= 2 ?lWl+f

2

a3

- Wl+ 2

} lWlf2+ f

l

a2

= fo+ 2

} lWlfl+ w

lf2

al

' 2 } l6lfo+ f

lWx

c 2a f w,o o 1

42

Solving for the coefficients f one obtains:

f3

= 1

£2

= a4 " 2

f 1*1= a

2/w

l" "o<

47 l

/wl

- 1/wt> " 2

J 1S1/W

1

f2 " a

3/2

} 1*1 " al/2

?A + ao/wt " w

l/2

} 1

fj a,/w2

- 2? x

ao/w^ = a

3- 2

? ft«4 + UjW2

(3-73)

fo

= ao/w

?= a

l'2} l"l " a

3Wl/2

} 1+ Vl + s

} lwl

fo " a

2" a

32

} 1W1

In formulas (3-69), (3-71), and (3-73), the coefficients a are of the

form a, = b d- + c Q + d . In section (3-1), formulas for alpha and

beta were developed for various forms of compensation techniques. Since

the formulas for alpha and beta will in general contain other variable

parameters, then the coefficients f as derived above will be functions of

these parameters. In most cases the coefficients f will be functions of

only one parameter (or they can be made to be) . Hence the roots of f . (S)

can readily be placed algebraically for the cases of n = 1, and n = 2.

For the case of n = 3, a root locus sketch will usually suffice. Although

the coefficients f have been derived for only up to a fifth order case,

they could easily be obtained for higher order cases if necessary.

3-2-2 Applications of the dominancy technique.

Example 3-8 (Third order characteristic equation)

Problem:

Compensate the system of figure (3-11) with a cascade compensator to

obtain:

1. Characteristic roots at zeta .5 and omega = 40.

2. K * 250.e

43

3. The specified roots are to be made dominant.

Solution:

The characteristic equation of figure (3-11) is:

S3+ (4 + P)S

2+ (4P '+ K t )S + KP =

3 2or S + (4 + oL )S + (4o£ + K £3 )S + Ko<- =

where alpha = P and beta = tf

Now G = K/e(S) = K/(S2+ 4S), so e • 0, e. - 4, e„ = 1, U 1, U. = 0.

o 1 i. 2 3

Equations (3-42) are now applied to obtain:

B = -K + w2 = -K + 1600

ci-°

Dl = 4w2

- w3U2

= -5.7xl04

62

- -4w + w2U2

= 1440

C2

= -wK = -40K

D2

= <tw2U2

- w3U3

= 6400

From equation (3-43) are obtained:

°^= 5.76xl04

-K+1600

6 = (1440)(-5.76xl04) - (-K + 1600)(6400 ) (3-74)

~40K(-K + 1600)

Now for any value of K, equations (3-74) will provide a value of alpha

and beta to provide characteristic roots at zeta .5 and omega = 40.

The value of K will now be chosen on the basis of dominance and steady

state error considerations. To satisfy the error specifications it is

necessary that K £ 1000. Since f - (S) is of order one, the appropriate

equations are (3-69), hence:

44

u

v: +to

10 HHI

CV

©

aft,

CU

+I/)

+LO

or

45

£1- l

£o

= ao/w^ - (aj - v*)/2 jf ^ - a

2- 2

J ^From the characteristic equation it is seen that

a2

= 4 +oC

a = UoL + K(3

a = KoLo

The real part of the specified roots is ?iw i= 20. Arbitrarily choos-

ing a dominance factor of five, the dominance criteria becomes:

f > 5 "? _w = 100. To satisfy this requirement, the simplest form of

2f will be chosen, namely f = a /w, . Therefore it is seen that f =o o o 1 o

Ko£ /1600 = (5.76xl04) = 36K > 100, where equation (3-74)

1600( K + 1600) (1600 - K)

was employed.

The above inequality implies that K > 1180. Since K >1180 also

satisfies the error specification, a value of K = 1200 is chosen arbi-

trarily. Using this value of k, the following quantities are computed:

alpha = 144, beta =4.2, and f = 108. As a check, the expression f = a«

-2J jW can be employed. Hence:

f = (4 + 144) - 2(.5)(40) = 108.

Example 3-9 (Fourth order characteristic equation)

Problem:

Compensate the system shown in figure (3-12) employing tachometer

plus acceleration feedback to obtain:

1. Characteristic roots at zeta = .5 and omega = 2.

2. K = 12.e

3. The specified roots should be dominant.

46

C\JHI

o

ft,

or

47

Solution:

The characteristic equation from figure (3-12) is:

S4+ 16. 5S + (73 +o£ )S

2+ (82.5 + (3 )S+25 + K =

Here alpha KK and beta = KK .

a t

G = K/e(S) = K/(S4+ 16. 5S

3+ 73S

2+ 82. 5S + 25)

By inspection it is seen that:

eQ

= 25, eL

= 82.5, e2

= 73, e3

- 16.5, e4

= 1, Ug

= 1, U3

= 0,

and U, = -1. By employing equations (3-11) one can find:

alpha = (185 + K)/4

and

beta = (K - 23) /2

From the characteristic equation it is seen that:

25 + K.a, = 1, a = 16.5, a * 73 + «/ , a = 82.5 + jg , a

2Since the quotient equation f . (S) is a quadratic, i.e., S + f-S +

f = 0, it would be desirable from the dominancy standpoint If f. > 5 ? ,w

= 5. However, looking at the dominancy equations for this case (equations

(3-71)), it is seen that one of the several expressions for f.. is f = a. -

2 J .w . Since all the expressions for f1have to be simultaneously

satisfied, it is seen in this problem that f- is a fixed constant since

a., and ? w are fixed. This is true even though the remaining express-

ions for f1

involve one or more variable coefficients. Therefore f- is

found to be 14.5. Observing the pertinant expressions for f it is seen

that none of them contain only constants, so the simpler expression f =

2a /w is chosen. Now since f. = 14.5 ^5, a dominant situation alreadyo * 1

exists, but the system's performance can be further improved by choosing

a reasonable value of zeta and omega for the secondary roots. From the

error specification it is necessary that K/25 - 12, or K ^300. Now

48

f (S) = S2+ 14. 5S + a

Q/w

2or f

1(S) = S

2+ 14. 5S + 6.25 + .25K. For

2K = 300, f (S) becomes: S + 14. 5S + 81.25. Therefore, 2 ^ w =

1f l 2

214.5, w

9= 81.25 orw. = 9. Then zeta = .806. These are reasonable

values for ~j and w_ since the secondary roots taken by themselves would

produce much less overshoot and a much smaller settling time than the

primary roots. Using this smaller value of K one can compute alpha and beta.

alpha = 121.2 beta = 138.5

Since alpha - KK + 300K and beta - KK + 300K then K = .405 andrt t a a t

K = .462.a

As an added bonus of the method, all roots of the characteristic equa-

tion are now known and the time response could be computed if desired.

Example 3-10 (Fifth order characteristic equation)

Problem:

Compensate the system shown in figure (3-13) to obtain:

1. Characteristic roots at zeta * .5, and omega = 4.

2. K i8.e

3. The specified roots should be made dominant.

Solution:

The characteristic equation is after letting alpha = P and beta = tf

S5+ ( o^ + 17)S

4+ (17c* + 84)S

3+ (148 -f KK

fc

+ 84(*)S2+ (148c/ +

KK est + K (2 + 80)S + cJ- (K + 80) = 0.

G m K/e(S) = K/(S4+ 17S

3 + 84S2+ 148S + 80)

From the above it follows that: e = 80. e, = 148, e = 84, e = 17,o 1 2 3

e = 1, U = 1, U = 0, U= -1, and U\ = -1. Employing equations (3-42)

one obtains:

49

B, - K + 1196 B2

- 492

C, - C2

= -4K

D = -1986 D2

= -1276

Therefore:

alpha = 1986/(K + 1196) beta = (1276K - 555000)/[4K(K + 1196)]

For beta to be positive it is necessary that K be greater than 435. The

accuracy specification implies that K be greater than 640. For K = 640,

beta .0555 which is out of the desired range of . 1 ^ (S - 10.

Of course, increasing K makes beta even smaller. At this point one could

design a multiple section compensator in accordance with section (4-6-2)

or feedback compensation could be added. The latter course of action

is chosen and tachometer feedback not enclosing the cascade compensator

is chosen. The applicable equations are (3-54). From them are obtained

the following, where the above alpha and beta now become of and ^ '.

. . 1 (1986 - 16KK )alpha=K + 1196

fc

beta = -492(4)K - 1986K + 16KK? ,_ Q ,vt t t + 4K + pK + 1196

To simplify the analysis let K = 640 which meets the accuracy specifica-

tion. Then alpha and beta become

. , 1986 - 10210K ,_ __.alpha = t_ (3-75)

1836

Here alpha is positive for K less than .193.

beta = 5.64K^ + 1.83K + .0555 (3-76)

The appropriate dominancy equations are (3-73) where it is seen that

neither f , f., or f. are restricted to constant values. From the charact-

eristic equation it is seen that:

50

o

1

©

I

3T

a5

= 1

\ = o£ + 17

a3

= YJdL + 84

a - 148 + KK + 84oC

a - 148<^ + KKtoC+ K (9 + 80

a = oL (K + 80)o

On the basis of these equations the following expressions are chosen

from equations (3-73) for the f . (S) coefficients:

2 = a4 " 2

} 1W

1f„ = a, - 2 3 ,w, = oL -|

21"1"4 " "3"f = a„ - 2 ? ,w.a, + U w 13c*. + 16

i 3 r

f = a /w? = 180o^o o 1

Then f - (S) becomes:

S3+ (d + 13)S

2+ (13oi+ 16)S + 180oC = (3-77)

Equation (3-77) contains only one variable so it can be put in root locus

form as follows:

S3+ 13S

2+ 16S +^(S 2

+ 13S + 180) =

3 2After dividing by S + 13S + 16S the above equation becomes:

(S2+ 13S + 180) = (S + 6.5 + ill. 7) = -1 (3-78)

S(S2+ 13S + 16) S(S + 1.37)(S + 11.6)

By plotting the root locus poles and zeros of equation (3-78) or

by inspection if one wishes, it is apparent that there will be a real

root of f (S) . Also the magnitude of the real part of the complex roots

can never be greater than 6.5, and it approaches this value as alpha tends

to plus infinity. For dominancy therefore, a large value of alpha is de-

sired. From equation (3-75), the maximum value of alpha is for K = 0,

52

but then beta is too small since K = implies the case of cascade com-

pensation only. It is obvious that K should be as small as possible,

which from equation (3-76) implies that beta should be as small as possi-

ble. Applying the lower limit of beta = .1 to equation (3-76) one can

show after solving the quadratic in K that K = .023 is the necessary

value.

From equations (3-75) and (3-76) one can solve for alpha and beta

obtaining:

alpha = .955 = P beta = .1 = tf

The cascade compensator is then:

.1(S + 9.55)(S + .955)

This can be realized by an R-C lag network. Equation (3-77) for f. (S)

becomes

S3+ 13.955S

2+ 28. 4S + 172 = (3-79)

A root locus of equation (3-77) for < oC <o0 is shown in figure (3-

14). The root locus was done by the computer program presented in sec-

tion (5-2). The roots of equation (3-79) were obtained from the printed

out data from the program and are as follows: for alpha .955; S- -

.585 + J3.64 and -12.8. Since .585 is less than ^-|w i which is two, the

primary or specified roots are dominant. Now the maximum value of alpha

is obtainable when K = K and even for this extreme case, the speci-

fied roots will not be dominant. To solve the problem, therefore, a

compromise on the specifications must be made. One approach would be to

specify omega as some value less than four and repeat all the preceding

calculations, but one could also do the following:

From equation (3-75) observe that alpha can be decreased by in-

creasing K . From the root locus data it is observed that if alpha =

53

•

^_— -fn

•

'

•1

1

<

•

QOC

1

<

'

»

OQ

'

O

V V.i >,

i

,

,1 r •,

3f|5 oio 008 D06 00*4- ~^002*" ' 1 r/

oQ

boo

1

X-SCflLE = 2.G0E+0O UNlTS/.lrtCH.

Y-5CALE = 2.00E+0O UHITS'INCH.FIG. 3-14

RM NUTTING5 * »3-KA+ 1 3)S *2+( 1 3fi F 1 6)S+ 1 80A=Q

54

.117, then ??

= .5, and w = 1.34. If this reduced value of omega

c; n be tolerated then the problem is solved since the secondary roots

located at ?9

and w_ are now dominant and only the settling time has been

modified (due to the decreased value of omega). On the other hand since

the roots ?1

and w are still located at zeta = .5 and omega = 4, the

overall system response will probably be acceptable to the designer.

The final results are therefore as follows:

K = .173 obtained from equation (3-75)

alpha - P « .117

beta = gamma = .5405 obtained from equation (3-76)

Characteristic roots are located at zeta = .5, omega = 4; zeta .5, omega

= 1.34; and a real root at S = -11.8. The cascade compensator is of the

form

:

.5405(S + .216 )

(S + .117)

This can be synthesized as an R-C lag network.

55

3-3 Some sketching techniques. \

The graphical solution will be discussed in following sections and

of course this involves plotting curves. Due to the complexity of the

parameter plane equations, two methods of curve plotting are recommended.

The easier and faster method is by means of a digital computer. A com-

puter program which will do this is presented in section (6). The second

method is by means of a desk calculator or slide rule. To facilitate the

use of the latter method, the parameter plane equations are manipulated to

put them into forms more suitable to non-computer plotting.

Due to the time limitations usually imposed on digital computer

operation and due to the time consumed in plotting the curves by hand, it

would be desirable to just sketch, say the zeta equals zero and the zeta

equals one-half curve for various values of omega to see if the type of

compensation proposed will do the job. This type of rapid sketching is

also helpful when choosing a scale for the digital computer graph plot.

For this reason, in the following sections, special characteristics of the

zeta equals zero and the zeta equals one-half curve have been derived for

characteristic equations of order two through five.

3-3-1 Table of symbols.

In table (3-2) are listed various symbols which apply to the deriva-

tions made in section (3-3-2). The change of variables as indicated in

table (3-2) is useful because many of these products and sums appear re-

peatedly. When doing hand computations of the parameter plane curves it

is only necessary to compute these quantities once, so substantial labor

is saved.

56

J = -b^Cj + bLc2

K Vl blCo

L = Cld2- c^ - c

o

M c d - c do 1 1 o

N = b^ - bodi

P = b2dl - bld2+ b

o

R = doC2 " d

2Co

T = b2Co " b

oC2

U = bod2

- b2do

A = b2C3

" b3C2

B m b3Co " b

oC3

C = C2d3 " C

3d2 * C

l

D = c„d - c d_3 o o 3

E = b3d2

- b2d3+ bl

F = b d, - b.do 3 3 o

G = B + J

H - D + L + co

I = P - b + Fo

X = c d.. - c.d_ + c3 1 13 o

TABLE (3-:I)

Y » b3cl

- ble3

Z bld3 "Vl

9< = C4d2

- c2d4

C**- SB C2d4

" d2C4

9n = C2d3

• c3d2

*n« Cld4 • C

4dl

jP = doC4

- c d.o 4

A- Vl " cld3

T- Cod3

" c3do

0- b4C3

" b3C4

e « b2C4 2 4

*- b3C2

" b2C3

e m Cob4-be.

o 4

S M b2d4

- b4d2

M. = bod4

- d b.o 4

P = b3d4

- d3b4

f- d3C4 -=3d

4

T- Vl " bld4

57

3 -3 -2 Bnsic derivations.

Case I (Second order characteristic equation)

The characteristic equation is of the form:

S2+ (b* + c.8+ d,)S + b oL+ c ft+ d =0

v1 lr 1 o or o

Employing equations (2-10) one can obtain:

Bl

" "bo

B2 * "V

Cl

= " Co

C2

= " C1W

D = -d + w2

D = -d. + w2U

1 o 2 12Solving for alpha and beta using equations (2-11) results in:

alpha = (c w2

- cqU w + M)/K (3-80)

beta = (-b,w2+ b U.w + N)/K

1 o 2

To find the maximum and minimum points, the first derivatives of alpha

and beta with respect to omega are taken and set equal to zero.

d o<- /dw = (2c.w - c U„)/K = or w = c U /2c. (3-81)1 O Z Oil

d (S /dw - (-2b.w + b U )/K - or w = b U,/2b-r x1 o 2 o 2 1

In equations (3-80) after letting w = one obtains:

oL = m/K P = N/K (3-82)

Letting w tend to plus infinity one obtains:

alpha —> +o0 beta—-> ± ^ (3-82)

Equations (3-80) through (3-82) are valid for all values of zeta

between zero and one. Since U = when zeta = 0, then equations (3-80)

and (3-81) become:

alpha = (c w2+ M)/K beta = (-b w

2+ N)/K (3-80a)

d^/dw = 2c w/K w - (3-81a)

d p /dw - -2b w/K w -

3The relative magnitudes of the coefficients determine whether alpha

and beta approach plus or minus infinity.

58

Equations (3-82) remain unchanged. Also since U- = 1 when zeta = .5

one can obtain:

2 2alpha = (c w - c w + M)/K beta - (-b w + b w + N)/K (3-80b)

d d /dw = (20^ - cq)/K w = c

q/2c

1(3-81b)

d B /dw = (-2b,w + b )/K w = b /2b,v 1 o o 1

Equations (3-82) remain unchanged.

Case II . (Third order characteristic equation)


S3+ (boDi+ c. ft+ d )S + (b.oL+ c. 8+ d,)S + b oL + c P> + d =0

Z ZrZ i 1~1 o o v o

proceeding as before the following expressions are obtained:

alpha = "Vc2U3+ C

2U2> ' Cl^ + (L + C

o+ C

oU3)w2 + U

2Rw + M

Jw2+ U

2Tw + K

beta . w4(-b

2U2

-f b2U3) + blU 2

w3 + (P - bQ

- bQU3)w

2+ U

2Uw + N

Jw2 + UTw + K

For w = 0,

alpha = M/K beta = N/K (Same as equation (3-82))

For w -?• + o*>

alpha = +06 beta = + <*> (Same as equation (3-82))

Due to the increased complexity of the expressions for alpha and beta,

the derivatives are only computed for the zeta equals zero and the zeta

equals one-half curves.

Letting zeta 0, equations (3-83) become:

4 2 4 2, . c w + Lw + M , -b.w + Pw + N /-. oo \alpha: 2 beta = _2 (3-83a)

Jw2+ K Jw

2+ K

For w = and w = o°, equations (3-82) apply.

Taking derivatives of equations (3-83a) one obtains:

59

«2*/*.2

- "4

C2J + "2 2c

2K + Kt - ^

(3-84)2 2

(Jw* + K)

The derivative goes to zero for:

w2 = -2c

2K + ^ (2c

2K)

2- 4c

2J(KL - JM)

__ (3-84)

d2 g /dw

2 = w4 (-b2J) - w2 (2b

2K) + KP - NJ

(Jw2+ K)

2<3 -84 >

The derivative goes to zero for:

w2 _ 2b

2K + J(2b

2K)

2+ 4b

2J(KP - NJ)

2V(3 "84 >

Letting zeta = .5, equations (3-83) become:

4 3 2alpha c_w - c.w + (I + c )w + Rw + M—

\2 (3-83b)

Jw + Tw + K

beta - -b2w4+ b^3

+ (P - b )w2+ Uw + N (3-83b)

Jw + Tw + K

For w = and w ot> , equations (3-82) apply. Taking derivatives of

equations (3-83b) one obtains:

d cL /dw = [2c2Jw

5+ (-3c J + 4c T - c

2T + 2CjJ)w

4+ (4cJC -

2c T)w3+ (JR + LT +c

qT - 3c

xK + 2JR)w

2 + (-2JM +

2LK + 2c K) + KR - MT]/(Jw2+ Tw + K) (3-85)

d £ /dw = [-2b2Jw

5 + (-2b2T + b^w4

+ (-4b2K + 2b T)w

3+

(3bxK + PT - b

QT - UJ)w

2+ (2PK - 2b

QK + 2JN)w

+ (UK - NT)]/(Jw2 + Tw + K)

2

In the general case, equations (3-85) involve fifth order poly-

nomials in omega. In specific problems, however, some quantities in

these equations will be zero, enabling one to more readily solve for the

critical values of omega.

Case III . (Fourth order characteristic equation)


S4+ (b

3oC + c

3(S + d

3)S

3+ (b

2o<-+ c

2 (2+ d

2)S

2+

(bl0

t + c,_ (3 + dps + bQo6+ c

o £ + dQ

=

Proceeding as before the following expressions are obtained:

alpha = [ w6c3(U3 - U^) + w5 c

2(U4

- U^) + w4(-C - c^

+ cxU3+ U

2(C + c

x)) + w

3(U

2(X - c

o) - c

QU4) +

w2(-U,D + L + c ) + wU,R + M]/A

beta = [w6b3(U

2U4

- U3) + w5b

2(U

2U3

- U4) + w

4(U

2(E -

bx) - b

tU3

- U3(E - b^) + w

3U2(Z + b

Q) + b

QU4)

+ w2(P - b - UJ?) + wU.U + N]/„

O 3 e-

A = w4A(U

2- U

3) + w

3YU

2+ w

2(-BU

3+ J) +wTU

2+ K

For w and w - °°, equations (3-82) apply. Letting zeta = 0,

equations (3-86) become:

6 __ 4 , __ 2 , u, , c„w + Cw + Hw + M

alpha = 3,

4 2Aw + Gw + K (3-86a)

6 4 2-b w + Ew + Iw + N

beta = _34 2

Aw + Gw + K

For w = and w = **>, equations (3-82) apply. Taking derivatives of

equations (3-86a) one obtains:

d2oC /dw

2 = [w8Ac + w

63c

3G + w

4(AH + CG + 3Kc

3) + w (2CK +

GH - 2A(H + M)) + KH - G(H +M)]/(Aw + Gw + K)

(3-87)

61

d2£ /dw

2= [-w

8Ab

3- w

62bG + w (EG - AI + 3Kb3> + w

2(GI +

2EK - 2AN) + KI - NG]/(Aw4+ Gw

2+ K)

2

In the general case, equations (3-87) involve eighth order poly-

nomials in omega. In specific problems, however, some quantities in

these equations will be zero, enabling one to more readily solve for the

critical values of omega.

Letting zeta = .5, equations (3-86) then become:

„i~u~ c ow " c ow + (C + c.)w + Xw + (L + c )w + Rw + Malpha = 3 2 1 o

4 3 2Aw + Yw + Jw + Tw + K

(3-86b)Ac / o o-b.w + b.w + (E - b, )w + Zw + (P - b)w + Uw + N

beta = _3 2 1_ * '

Aw + Yw + Jw + Tw + K

For w and w =<*>, equations (3-82) apply. Due to the increased

complexity of equations (3-86b), it does not appear practical to compute

the derivatives for the general case.

Case IV . (Fifth order characteristic equation).


S5+ (b

4d- + c

4 p + d4)S4 + (b

3cL + c

3 jg+ d

3)S

3+ (b

2<* + c

2 p

+ d.)S2+ (b.<* + c, /3 + d,)S + b ^ + c 8 + d =0

l 1 1«1 O O < O

Proceeding as before the following expressions are obtained:

alpha = [w8c4(U

5+ U

2) - w

?c3(U

2U5+ U^) + w

6( ^ (U^

-U2) + c

2(U

5+ U

2U4)) + w5 (U

2U3X+ U

4^- c

xU4)

+ w4(U

2>l- U

3^+ U^ + c

q) + w

3(U4<?+ V^) +

w2(U

3T+ L + c

q) + wU

2R + M]/A (3-88)

62

beta = [-w\(uj - U3U5) + w

7b3(U

3U4+ U^) + w

6

f (U^

"U3 " b

2(U

2U4+ U

5}) + w5((U

2U3 " V § + blV +

w4(-U

3(E - b^ + U

3T + U

2(E - b

x) + b

QU5) + w

(U4/^+ U

2(Z + b

Q)) + w

2(-U

3F + P - b

Q) -1- wU

2U + N2/A

/ 9

A = w60(u

2u4

- u2) + w

5(u

4- u

2u3) + w -% (u

3- u

2) +

w3(U

4€ + U

2Y) + w2 (J - U

3B) + wU

2T + K

For w = and w = ^ , equations (3-82) apply. Letting zeta = 0, equa-

tions (3-88) become:

"

alpha = w8c4

+ w6(c

2 £) + w4(<ty J + cj + W

2(L + c

p- T) + M

Al (3-88a)

beta = w\ - v6p (b

2+ 1) + w

4(E -b^ - P + b

Q) w

2(F + P-b

p + N— si

A = -w6

- w4X + w

2(J + B) + K

Letting zeta = .5, equations (3-88) become:

alpha = [w7c3

- w6(^ + 2c

2) + w5 ( C]L - & ) + w ft + w (R -

^ ) + w2(L + c

q) + wR + M]/ A 2 (3-88b)

beta = [-w\ - w7b3+ w

6f> (2b

2- 1) + w

5( S -*£> -H*

4

(E - b - bo) + w3 (Z + b

Q- M ) + w

2(P - b

Q) + wU + N]/ A 2

A 6rt

- w59 - w

4K + w

3(Y - 6 ) + w

2J + wT + K

A2

= -w

Example 3-11 (Case I example)

Problem:

Sketch the zeta = and the zeta .5 curves for the following

63

characteristic equations. Let the abscissa variable be alpha and the

ordinate variable be beta.

S2+ ( oC + 2)S + (3 + 1 =

Solution:

From equations (3-80) it is found that:

2alpha = U_w - 2 beta = w - 1

From equations (3-81)

:

d oL /dw = U2

d jS /dw - 2w

From equations (3-82):

alpha = -2 and beta = -1 for omega = 0.

Therefore for zeta the following relations apply:

2alpha =-2 beta = w - 1 and d ^3 /dw = 2w,

which implies a minimum for beta at omega 0. The zeta = curve is

therefore a vertical line in the alpha-beta plane, going from alpha = -2,

beta -1 to plus infinity as omega increases from zero to infinity.

Setting zeta to .5, one finds that:

2alpha = w - 2 beta w - 1

or

w cL + 2 beta = (ot+ 2) - 1.

The alpha-beta curve is therefore a parabola with vertex at alpha -2

and beta 0; symetric about the line alpha = -2, and opening in the plus

beta direction. The curve can readily be plotted after calculating a

few critical points.

The above curves are plotted in figure (3-15).

Example 3-12 (Case II example)

Problem:

Repeat example (3-11) for the following equation:

S3+ ( oC+ 1)S

2+ ( 8 + 2)S + 3 =

64

Solution:

From equations (3-83a) for zeta = 0, it is found that:

2 2alpha = -1 + 3/w beta = w - 2

For omega = 0, alpha plus infinity and beta = -2.

From equations (3-84):

d2oL /dw

2= -3/w

4d2£ /dw

2= 1

But d ^ /dw = 2w, so alpha has a minimum at omega equals infinity. Beta

has a minimum at omega = and it is -2. The curve is asymptotic to the

lines alpha -1 and beta = -2.

The following points are readily obtained:

omega alpha beta

1.414 .5

1.732 1

From equations (3-83b) for zeta = .5, it is found that:

2 2alpha = w - 1 + 3/w beta = w -2 4- 3/w

For omega 0, alpha = + cP , and beta = +oO

Also,

d oc/dw „ i . 6 /w

3d (3 /dw = 2w - 3/w

2

It is therefore obvious that:

3 3d ^/doC = w(2w - 3)/(w - 6), from which the critical points are

1/3found to be: w = and w = (3/2) = 1.145. This point is a minimum

since the second derivative is negative.

The following values can be obtained:

omega alpha beta

1.145 2.435 1.93

.

1

300 2813 2

3 2.33 8

The curves are sketched in figure (3-16).

Example 3-13 (Case III example)

Problem:


54+ (c^+ 1)S

3 +2S 2 +o^S+jS =

Solution:

From equations (3-86a) for zeta=0:

alpha = -w2/(w

2- 1) beta = (-w

6+ 3w

4- 2s

2)/(w

2- 1)

As omega tends to plus infinity, both alpha and beta tend to minus

infinity. As omega tends to zero, both alpha and beta tend to zero.

Also:

d2

<±- /dw2

= [-(w2

- 1) + w2]/(w

2- l)

2

d2

(S /dw2

= [(w2

- l)(-3w4+ 6w

2-2) - (-w

6+ 3w

4- 2w

2/|/(w

2- l)

2

It can be seen from the above derivatives that d P /dC* = 0, implies six

critical points. It appears unpractical to compute them.

From equations (3-86b) for zeta = .5, it can be shown that:

3 2alpha = -w + 2w = w(2 - w )

6 4 3 2 2 4 2beta = w -2w -w +2w =w(w -2w - w + 2)

As omega tends to plus infinity, alpha tends to minus infinity and beta

tends to plus infinity. As omega tends to zero, alpha and beta both tend

to zero.

Using a slide rule or desk calculator along with the above informa-

tion the curves can be plotted as shown in figure (3-17).

Example 3-14 (Case IV example)

Problem:


55+ 3S

4+ S

3+ 2S

2+ c/ S +(9 =

Solution: •

From equations (3-88a) for zeta=0:

2 4 2 4alpha = w - w beta = 2w - 3w

66

exists at omega .578. The following points can readily be calculated:

omega alpha beta

.573 .222 .333

.816 .222

1 -1

From equations (3-88b) it is seen that:

3 5 3 2alpha = -3w + 2w beta = w - w + 2w

Therefore:

d <A- /dw - -9w2+ 2 d (9 /dw = 5w

4- 3w

2+ 4w

Hence d @ l&d- - 0, implies that w(5w - 3w + 4) = 0. It is not

worthwhile to factor the above cubic in omega to obtain the three corres-

ponding critical points. The critical point at omega equal zero indicates

that the curve starts at the origin. A slide rule or desk calculator can

now be employed along with the above information to plot the curves as

shown in figure (3-18).

_l_ _r [j

f-

—1

/\ -/» |.| / 7\ LJ

t \ /

Y JX 1

V ,m /_T 3 ' /Y 7Jy 7\ /\

~JT 7\

1\ Jf I

\ Ji Jif

i

7"

i Jr\ /

Y jY 7

A • _1 /\ /

/

f

\ _ _ _ . :s_ .

I ,

<-r 7Y /-LjV ,"l

... ^ . ^

r^r c>j

L \ L- .• _ " 7 ±1 VI

-5 " vv. L ~A-^n ' 4 Irx^' _7 _A _ ZLT

I '-"1 r

L jYj

i ; " .

Yj

L i J

Y y i

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\~ii

V>

YjY / r

j C J i- >n L. -±sJ 1_4-^^ v 4 - V^r —u y IL 3LL <5 1 ^X..^y i«» |= _ -.^..^^_ — -taa

*̂k\^s zs ?

s >''' I""" 1

it _. 2 Z

- Fl <niT»» o_T iffigure ,?""J.p

ZVLj^.- _ — _ _.

-* ~i

68

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f

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r~*

I.

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Ii_ o «.^_^r ^3 >•\, ?i

î >ît ~ - '-£+- **-i d *- ^r<-X X t^zZ*-4 <^

=Su I ^n*£. - -I -1*2-- - - - -

LAl-t - I U, *

it a1

if

VIv\

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4J. ^_ Q_ \i

12 3 -4 ,, ,5 <*£,

\J\X[\ *

V . /TT " _x Figure 3-loI ,

\ j.__. _.

v^v

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-,—o~ s rSJZ _ ±S1 _

Nîs,

r "^^^^

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j&a.

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J^P^y

ss

**

J?~s

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r-

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y~?

^ 7~7

y.' ^

y% ' jrif / ^— -w -K — — —^i,/ r "

J3c> •y-^ _

r/

7~X ' -2.-ZZi:y /

/,r

7~7

j.' .

7 __ . _ _«5 7 — j Figure 3-17

1

4 C "" / ~ '" "f

/yr*

/

y(T

VL" j «— *"""*

ii V «-it"™I

2 't""' ' **"'3"

> ? ^""s""

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rrt

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f\— *"" ;S, ^*-

/ £•-•" -i-Z---. (?4;

Hn_ .£ ^j^ *^ !*: ~ .£ :

-?e-

.: "- , ; : ::-+=. :| ! ! / i :

ji

j

JX £

—

-"——-— jL/_)

'

A'

>iVj

N• .

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-

~- —- ^i--: ——« -- A

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f

.

4- 1 C. <.0

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a, i - -

/j

i^ ,/£-

'Py-,&Is'/TV -

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< 1- . X 1 ,

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- .£5 .£; - a2Il

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8

5 x zz\ «-«,„« •4 iO .S Figure 3-lo r\ - _ _. yj

O ..... _ -X = = — .-- .....,, ... . .

~ x x x~Ql

Mo ~ .

7]i

Example 3-15

Problem:

For the system shown in figure (3-19) let alpha = KJK., and beta

= K-,K9 , and satisfy the following requirements:

1. Assuming that a computer will plot the curves for zeta =

0, .5, and 1, determine an appropriate graph scale.

2. Only first quadrant values of alpha and beta are of

interest and the graph is eight inches wide and fourteen

inches high. Alpha is the abscissa and beta is the

ordinate.

3. Due to bandwidth consideration's, omega should be less than

1500.

Solution:

From figure (3-19) the characteristic equation is determined to be:

S3+ 1100S

2+ (10

5+ c£ )S + =0

From this it is seen from table (3-2) that:

L = 1 P =-1100 J =

M = 10 N = K = -1

R = -1100 T m U =

Using equations (3-83a):

alpha (Lw2+ M)/(Jw

2+ K) = w

2- 10

5

beta = (Pw2+ N)/(Jw

2 + K) = llOOw2

For omega = 1500:

alpha - 21. 5x 105

beta 24. 7x 108

From equations (3-84) or by taking derivatives directly:

d2 cL /dw

21 d

2p /w

2= 1100

Hence:

d £ /doL = 1100

Letting 2eta ,5 and employing equations (3-83b) :

72

'1

s

ft

to

[:1

ri

i

pi

i

'i

HI

o

CC

u 73

alpha = (Rw + 105)/(Tw - 1)

beta - (w3

- llOOw2+ Uw)/(Tw - 1)

Therefore:

alpha = llOOw - 105

beta = -w2(w - 1100)

It can be seen that the zeta - .5 curve is not in the first quadrant for

omega greater than 550, so this curve does not influence the scaling pro-

blem.

The origin point of the curves for omega = 0, is obtained from equa-

tions (3-82) resulting in:

alpha = -10 beta =

From the above data it is concluded that it would be best to scale

using the values of alpha and beta for the zeta = curve corresponding

to w 1500.

Therefore:

alpha-scale = 3x10 units per inch

Q

beta-scale = 2x10 units per inch

Example 3-16

Problem:

For the system shown in figure (3-20) it is desired to have roots

with zeta greater than .5 and a maximum error coefficient. Plot para-

meter plane curves to find the best values for K and Z.

Solution:

The characteristic equation is:

S4+ 15S

3+ 150S

2+ (100 + K)S + KZ =

Let alpha K and beta KZ to obtain:

S4+ 15S

3+ 150S

2+ (100 + o^ ) S + (3 =0

From equations (3-86a) corresponding to zeta = 0:

4 2 4 2alpha (Cw + Hw + M)/(Aw + Gw + K)

4 2 4 2beta = (Ew + Iw + N)/(Aw + Gw + K)

74

CD<J

+

2

oO

+

CM

+

oI

©

Or

75

Using table (3-2) one can obtain:

C = K - -1 R =-150

H = -15 E = 1 Z = 15

M = 1000 I = -150 P = -150

A = N = U =

G = X = 1

Then:

alpha = 15w2

- 1000 beta = -w2(w

2- 150)

Hence:

2Beta is greater than zero for w less than 150 and alpha is greater

2than zero for w greater than 66.5, so both alpha and beta are greater

2than zero for 66.5 < w < 150.

Letting zeta = .5 and employing equations (3-86b) it is found that:

alpha = -w(w2

- 15) - 1000 beta - -15w2(w - 10)

From these it is seen that for omega greater than ten, beta is less than

zero and alpha is less than zero. For omega greater than zero and less

than ten, beta is greater than zero and alpha is less than zero. Hence

the zeta = .5 curve is not in the first quadrant. Since the order of

the characteristic equation is only four, it is safe to assume from the

above data that all constant zeta curves for zeta greater than or equal

to .5 are not in the first quadrant. Since the specifications of the

problem cannot therefore be met, it is unnecessary to plot the curves.

Example 3-17

Problem:

Referring to figure (3-21),

1. Determine a scale for alpha and beta and plot the curves.

2. Place a pair of roots with zeta - .5.

3. Make the roots dominant.

4. Due to bandwidth considerations, omega must not be greater

than 4000.

76

C2><o

4-—

—

00)£

+LO

t

OOC3 .

OJ

+no

«

CO

/<

V!>

I

•rl

cr

77

Solution:

From figure (3-21) the characteristic equation is found to be:

S4+ 1005S

3+ (5100 + oC )S

2+ (10

5+ 5 0^ )S + 100 cL + /3 =

Here alpha = KJC- and beta K.K^.

Since the purpose of plotting the curves is to determine alpha and

beta to meet specifications 2,3, and 4, the sketching techniques are

first employed to determine if curve plotting is necessary.

From table (3-2) one can find:

c =o

L = -1 R = -5100

X - 1 M = 105

E = -1000

Z - 4925 U = 5.1xl05

N = -107

A = Y = J -

T - 1 K = -5 P - 74600

From the above values along with equations (3-86b) one finds:

alpha - [w2

(w - 5100) + 105]/(w - 5)

It can be seen that for omega greater than 5100, alpha is positive,

and for omega less than 5100, alpha is negative.

beta [w4

(w - 1005) + 4925w3+ 74500w

2+ 10

?(.051w - 1)]/

(w-5)

Hence for omega greater than or equal to 5100, beta is positive.

To avoid positive feedback, it is concluded that zeta = .5 can be

obtained only for values of omega greater than 5100, which violates

specification 4. It is therefore not necessary to plot the curves.

78

3-4 Graphical solutions on the parameter plane.

3-4-1 Advantages of the graphical solution.

The previously discussed algebraic solutions have the disadvantage

that a fixed value of zeta and omega must first be chosen. In some cases,

the remainder polynomial can then be modified to ensure that the specified

roots are dominant. However, it is not always possible to guarantee that

roots placed at a specified location can be made dominant, and a trial and

error procedure may have to be employed to achieve the best values for the

various parameters. Trial and error may also have to be used in the de-

sign of cascade compensators where a specified root location may require

parameter values that are not physically realizable.

In these instances, the calculations have to be redone in terms of

slightly modified specifications or a different means of compensation may

have to be used. In general, system specifications are not rigidly fixed,

but can be met by a given range of values or by some upper or lower limit.

To avoid this trial and error analytical procedure one can employ the

graphical solution. If a net of curves is plotted by a computer or other-

wise, one can, by picking an M point or operating point in the parameter

plane, read from the curves the n roots of the n order characteristic

equation. The trial and error procedure can then be done visually to pick

an operating point which best meets the given specifications.

3-4-2 Some examples of the graphical solution.

In this section only first quadrant values of alpha and beta are

assumed to be of interest. The graphs were plotted with the aid of the

computer program presented in section (6-1). At the end of each curve on

the graph is a letter and a number. The letters are abbreviations for the

following quantities:

Z(zeta), S(sigma or the real part of the complex variable S) , ZW

(zeta-omega or the real part of the complex roots) , w(omega or the undamped

79

natural frequency of the complex roots).

Example 3-18

Problem:

For the system shown in figure (3-22) set K at the stability limit.

Place a dominant root pair within the following region of the S -plane:

.4 — f ~ »7» an<* 2 < w £ 6. Both tachometer and acceleration feed-

back can be used. If possible use only one or the other.

Solution:

From figure (3-22) the uncompensated system's loop transfer function

is:

GH - -1 - K/[S(S + 10) (S2+ 5S + 100)], which when expanded be-

comes:

S4+ 15S

3+ 150S

2+ 1000S + K =

To determine the value of K at the stability limit the Routh array is

formed

:

1 150 K

15 1000

1250 15K

1.25xl06

- 225K

15K

From the Routh array the stability limit is seen to be:

K = 5555.5

If both tachometer and acceleration feedback are used the compensated

system's characteristic equation becomes:

S4+ 15S

3+ (150 + 5555.5 c^ )S

2+ (1000 + 5555.5 (S )S + 5555.5 =

where alpha = K , beta K and K has been set at the stability limit.

Parameter plane curves for this characteristic equation are shown in figure

(3-23).

80

o

(t

CMC\J

1

CO

©

fe

81

From these curves, the following analysis can be made. The origin

of the parameter plane corresponds to the roots of the uncompensated

system. Since the zeta = curve passes through the origin, two roots

are located on the jw axis of the S-plane as was to be expected from the

Routh array. The other two roots are also complex and are located at

zeta = .8, and omega 5.

It is important to note that when an operating point involves two

different pairs of complex roots, then the curves for two different values

of omega and two different values of zeta will have to pass through the

point. To determine which value of omega goes with which value of zeta,

it is necessary to refer to the computer printout data. Due to lack of

space, this data is not provided herewith.

With K set to zero, the effect of tachometer feedback alone corres-a '

ponds to movement of the M point along the Y-axis. In figure (3-23) the

unstable region is determined by an inspection of the way the constant

zeta curves tend as zeta increases. Since the Y-axis is always in the

unstable region, it is concluded that tachometer feedback alone cannot

stabilize the system.

The effect of acceleration feedback alone can be observed by moving

along the X-axis of figure (3-23). If K is varied between .01 and .06,

the system will have two pairs of complex roots with the following ranges

of values for zeta:

.3 c zeta < .5 and .25 < zeta < .32

If both tachometer and acceleration feedback are used, it is concluded

that tachometer feedback will in general cause the zeta of one pair of roots

to increase while the zeta of the other pair decreases. It appears that

acceleration feedback alone would be better.

From the graph it is determined that with K =0 and K = .012, thatt *

complex roots are located at zeta = .45, omega = 4, and zeta .32, omega

82

FIG. 3-;:

83

= 13. Since it is true that (.45) (4) = 1.8 << (.32) (13) = 4.15, it

is apparent that the roots at zeta = .45 and omega = 4 are dominant.

The specifications have therefore been met and the problem is solved.

Example 3-19

Problem:

With the K-K?product set at the stability limit of the uncompensated

system, use position feedback as shown in figure (3-24) to obtain domin-

ant characteristic roots with maximum possible zeta and omega, and with a

dominancy factor of about two to one (i.e., the ratio of the real part

of any secondary root to the real part of the primary roots is about two

or greater)

.

Solution:

From figure (3-24) the characteristic equation becomes:

S3+ 2100S

2+ ( oL+ 1.2xl0

6)S + 100 c^ + (2+ 10

8=

where alpha KJC- and beta K-K . Setting K. and using the Routh

criteria, the value of K.K- at the stability limit of the uncompensated

9system is found to be: 2.42x10 .

In figure (3-25) are plotted parameter plane curves for the above sys-

tem. Constant zeta-omega curves have also been included to assist in the

dominancy considerations.

The analysis proceeds as follows. The unstable region is to the

left of the zeta = curve. Root values for the uncompensated system

are obtained for M point values along the Y-axis, since this corresponds

to K~ = 0. The stability limit of the uncompensated system corresponds

to the intersection of the zeta curve and the Y-axis. The value of

beta K K at this point is observed by the Routh check. The M points of

interest therefore lie along a horizontal line drawn through this point as

shown in figure (3-25).

The maximum zeta obtainable for any value of alpha along this line is

about .33. As alpha increases then zeta-omega increases and the magnitude

©

Cvi

oo<\i o

*: +UO

1

nk:

I

<D

fa

or

.£5-

of the real root decreases and the root becomes more dominant. Omega

also increases with alpha. Therefore the overriding criteria is the

dominancy factor. On this basis an M point located at alpha KJC

6 9= 2.5x10 and beta = K..K- 2.42 x 10 is chosen. Characteristic roots

are read from the graph and are located at: zeta = .32, omega = 1680,

zeta-omega 550, and a real root at -1000. The dominancy factor is

computed to be 550/1000 or 1.82. Since this is about two and the maxi-

mum zeta and omega have been obtained, the solution is complete.

It can be remarked that since KLK« was set at a specific value for

analysis purposes, the problem was then reduced to only one variable and

root locus techniques could have been employed. This example illustrates

the flexibility of the parameter plane and it also points out the interest-

ing fact that root values determined along either a horizontal or a verti-

cal line in the parameter plane constitute a root locus in terms of the

variable parameter. Root values along a sloped straight line in the para-

meter plane constitute a sort of hybrid root locus where the two root locus

parameters are linear ly related.

Example 3-20

Problem:

1. Set K such that the system of figure (3-26) will follow a

ramp input, 9 = l.Ot.R

2. Steady state error should be less than two degrees.

3. Step response overshoot should be less than 30%.

4. Find values of Y> and P to meet the above specifications.

Solution:

It is known that steady state error 0_/K , where in this problemR e

K = K/10, and 0_ = 1.0. Error - 2° .0349 rad. Therefore, K =28.6e r e

and K 286. If the above system can be made second order dominant, the

second order curves can be used. For an overshoot of 307o , a zeta of .35

CD

k:

+10

10

4-a.+

&

OCT

I

I-

88

is required.

The characteristic equation of the compensated system then becomes:

S4+ (11 +c^)S 3

+ (10 + 11 06 )S2+ (10 o£. + 286 @ )S + 286oC =

In the above equation, alpha = p and beta V . Fourth order parameter

plane curves are plotted in figure (3-27). It is seen that the zeta =

.35 curve has a minimum value of beta = 25,5. Since beta is greater than

ten, a multiple lead compensator or perhaps a lag-lead compensator is

indicated. To enhance the chances of having to use no more than two

sections, the minimum value of beta 25.5 is chosen for the single section.

From figure (3-27) the following root locations can be read off correspond-

ing to beta =25.5 and alpha 87: zeta = .35, omega = 8.2, and real roots

at -90 and -4.5. Zeta-omega for the complex roots is three. Since 4.5 =

1.5x3, the compensated system is second order dominant with a dominance

factor of 1.5.

At this point either a multiple section compensator can be designed

from the above values of alpha and beta, or another scheme of compensation

can be used. Both alternatives are considered in section (4).

89

X-3CALE • X98E+8J UMJTV1NCH.

1-iCflLt - 3.e8E-t-88 UNITS'INCH.

RM NUTTIMG FIG. 3 "2,

X -SCALE =P=flLPHA , Y-SCflLE=GflMMfl=BETfl

4 Miscellaneous aspects of the parameter plane.

4-1 Some general comments.

In section (3-2-2) it was shown that constant zeta parameter plane

curves of order two through five originate at a point where alpha = M/K

and beta N/K, where M, N, and K are determined by the zero and first

power coefficients only. Inductive reasoning can be used to conclude that

constant zeta curves of any order originate at this common point which is

determined only by the zero and first power coefficients. An exception

is when K = 0. In this case the origin point depends on higher order co-

efficients and its location will be obvious given a specific equation. If

K is not zero, the origin point is independent of the order of the charact-

eristic equation.

Inspection of the expressions for alpha and beta in section (3-2-2)

indicates that the shape of the constant zeta curves as omega becomes

larger is primarily determined by the coefficients of higher power, and

in general the curves become more complex and less well behaved as the

order of the characteristic equation increases. For a given character-

istic equation, an increase in complexity can be observed as alpha and

beta appear in more coefficients. As indicated in section (3-2-2) all

constant zeta curves tend to plus or minus infinity. The relative magni-

tudes of the coefficients determine whether the limit is plus or minus

infinity. It is therefore necessary to choose a frequencyrange of inter-

est before plotting the curves, thus limiting the analysis to one "window"

of the infinite plane.

Mitrovic curves, since they involve equations which have only one

parameter appearing in each of two coefficients, are in general simpler

and more well behaved than the parameter plane curves. It is also interest-

ing to note that when parameter plane curves are plotted for characteris-

tic equations of the Mitrovic type, the resulting curves are identical to

91

those plotted from the Mitrovic equations. This can be seen as follows.

From reference (8) page 349, B and B. are given by the following

relations where notation has been changed to conform to this text.

Bo

- -

|edkM

k

k. 1BX

= ^ i^'\

wherek

= -(2 J kl + k2 ) for k > 2

and = 0, = 1o l

Comparison of the tabulated functions in table (10-1) of reference (8)

with appendix (IB) of this text leads to the following relation:

The B -B, Mitrovic characteristic equation is of the form:o 1

Sn+...+d S

2 +8,8+6 =02 1 o

If B = alpha and B = beta, alpha and beta can be computed from the

parameter plane equations (2-10) and (2-11) and are as follows:

alpha = £_ (-l)kd, w\ beta = £ (-l)

kd, w

k-1U

If the relationship between the 0, and U, functions is employed in the

Mitrovic expressions for B -B. , one obtains:

11 k k 1 k-1 k^V-dVi bi

" &z VM<-i>*«

It is seen that B = alpha and B- = beta. The above procedure can be

repeated for variables appearing in any two of the coefficients of the

characteristic equation. This duality property is employed in section (4-2)

to compensate parameter plane type characteristic equations employing normal-

ized Mitrovic B -B, and B,-B third order curves which have been plottedo 1 1 2

using the parameter plane computer program presented in section (6-1)

.

Since in the parameter plane, a negative value for alpha or beta

does not necessarily imply a negative coefficient of the characteristic

92

equation, one is not restricted to first quadrant values of alpha and

beta. In most applications however a negative value for alpha or beta

corresponds to a right half plane pole or zero in a cascade compensator

or to positive feedback. For all of these cases only the first quadrant

of the parameter plane is of interest.

Since no stability criteria, either relative or absolute, has been

established for the parameter plane, it is necessary to base the stability

analysis on observing which way the curves tend as omega and zeta are

varied. For this reason it is desirable to plot curves for as many values

of zeta, omega, sigma, and if desired, zeta-omega, as is necessary to fix

the pattern.

4-2 Normalized third order curves.

4-2-1. Discussion of the normalized curves.

Third order curves are rvailable for finding the roots of polynomials

where the coefficient of zero power is one variable and the coefficient of

the first power is the other variable. Such a set of curves is presented

in reference (8), and the curves are called normalized Mitrovic B -B. curves'

'

o 1

In this section a method is presented whereby parameter plane type

characteristic equations of third order can be compensated on two different

types of normalized curves. One type is the B -B. curves of reference (3),

and the other type is the B=-B_ curves which are presented in this section.

The method here presented is divided into three cases.

4-2-2 Derivation of the normalized transformations.

Case I

Characteristic equations of the type

S3+ d

2S2+ (b

l(yL + c

t |3+ d

x)S + b

QdL + cfl + d =0 (4-1)

are considered, where alpha and beta are the variable parameters.

In equation (4-1) letting S d.s one obtains:

93

X "SCALE - J.80E-9J UflJTS^-JMCh.

Y-3CALE - 5.00E-0O LKIT3/-]NCh.

RM HUTTING , NORMALIZED BO-B: CURVE'S

d3s3+ d.s

2+ (b. oL +c, (2 + d

n)d s + b ©L + c fi + d =0

2 2 1 II 12 o o I O

3The above equation is divided by d resulting in:

s3+ s

3+ s(b

l0C + c

xjS+ dj.)/^ + (b

Q0^ + c

q(2+ d

Q)/<i2 = <*-2)

Let:

B1

= (b^ + Cl (S + d^/d2

(4-3)

Bo=^ B.P + V /dl

Rearranging equations (4-3), two linear equations in alpha and beta are

obtained:

bl°

L + Cl£ " d2Bl

" dl

(4 "3a)

"2Lob d~ - c jg

= d^B - d

Equations (4-3a) can be solved by Cramer's rule or by the inversion of

the matrix of coefficients to obtain:

alpha = [(d^ - dl)Co - (d\lo

- d )c1]/(b

1co

- boex )

beta = [(d^ - djbj - (d^ - VV&A ' Vl>

Substituting equations (4-3) into equations (4-2) one obtains:

s3+ s

2+ B.s + B =0 (4-5)

1 o

Equation (4-5) is a normalized B -B type equation.

In figure (4-1) parameter plane curves have been plotted for equation

(4-5) where B. is the alpha or X-axis variable and B is the beta or Y-

axis variable.

Systems whose characteristic equation is of the case I type can there-

fore be compensated on the normalized curves of figure (4-1), and the

necessary values of alpha and beta can be found from equations (4-4).

Case II .

Characteristic equations are considered of the type:

S3+ (b

2<* + c

2 (3+ d

2)S

2+ (b

1<JC +C

t (3+ d

x)S + d

Q= (4-6)

95

3/1

Letting S =\J d s in equation (4-6) and dividing through by d

one obtains:

s + (b2

d- + c2 |9 + d

2)s /d^ + (b

1</- + c p + d )s/d 3 +1 = (4-7)

Let:

B2

- (b2cL + c

2|S + d

2)/d^

Va(4-8)

Bl " <

bl°

i + clf-

+dl>

/do

Equations (4-8) when rearranged, reduce to two linear equations in alpha

and beta as follows:

(4-9)

(4-10)

b2cL + c

2(s=d;B

2- d

2

V + ciP m 4Bi

- di

Solving for alpha and beta as in case I results in:

alpha = [(d^B2

- d2)c

x- (d^B

1- V^^Vl " b

ic2)

2 X

beta = [(d^B1

- d1)b

2- (d

QB2

- ^V^i ' b^)

As a result of equations (4-8), equation (4-6) becomes:

s3+ B

2s2+ B

xs + 1 = (4-11)

Equation (4-11) is a normalized B.-B„ type equation. In figure (4-2),

parameter plane curves have been plotted for equation (4-11) where B_

is the alpha or X-axis variable and B. is the beta or Y-axis variable.

Systems whose characteristic equation is of the case II type can

therefore be compensated on the normalized curves of figure (4-2), and

the necessary values of alpha and beta can be found from equations (4-10).

Case III .

Characteristic equations are considered of the type:

S3+ (b

2d + c

2 jg+ d

2)S

2+ (b

1o^ + Cj (3 + d

x)S + b

Q&- + c

q f^+ d =

(4-12)

Letting S (b.o^- + c (3 + d )s and dividing through by (b c/L + c_ £ + d )"

S i~ *, s» t •'l f

X JCAIE » l.eeE+80 IMTS/'JNCH. -«

Y-ac/iLE = |.eoE-t«e uiiTS^irot e

H- «£Tfl

S - W-mA*/- OMf&d

Rh NUTTING , NORMALIZED MITROUIC B1~B2 CURUESS«»3+AS*"2+BS+1=0

<*7

030 035 O40

FIG 4-2b.

x-scale • SjOoe-oi unitvinch. - ay-scau - 6.oee-oi uhits^inch. - s

? - terh,

S -SIG-MflW - OHt li

RM NUTTING , NORMALIZED M1TR0U1C B1-B2 CURUESS""3+AS--2+BS+l=0 98

equation (4-12) becomes:

s3+ s

2+ (b oi + c

l(9 + d^s/Cb^ + c^ + d

2)

2(4-13)

+ <bo°^+ C

o f

2 + do)/(b2^ + C

2 I

3 + d2}

= °

Let:

Bl

" <bl^+ c

lP+ d

l>/(b

2°^ + C

2 ^+ d

2)2 (4 " 14)

Bo

= (bo^ + Co P

+ do)/(b2^ + C

2 £+ d

2)3

In equations (4-14) after rearranging, expanding, and collecting terms,

one obtains:

tflcL1

* *Ap2 + (2BiV2- b

l)oC+2BlV2^ +

< 2Blc2d2 " ^(g +»

14" -l"°

and

Bob2^

3+ B

o°2 ?3

+ 3Bob2C2^ ^ + 6B

ob2C2d2^^ +

3Bob2c^ 2

+ 3Bob2d2o^

2+ 3B

oC2d2<g

2+ (4-15)

< 3Bo»2d2

" bo>^

+ (3BoC2d2

" <0$ + Bod2

"do " °

When the substitution of equation (4-14) Is made In equation (4-12), a

normalized B -B, type equation results. Systems whose characteristico 1

equation is of the case III type can be compensated on the curves given

in figure (4-1), to obtain a value of Bj_ and Bqwhich can be substituted

into equations (4-15). Equations (4-15) contains the solutions for alpha

and beta. In the most general case where none of the coefficients in

equation (4-15) are zero, a graphical solution of the second and third

order polynomials can be made. That is the polynomials of equation (4-15)

can be plotted on an alpha-beta plane and the necessary values of alpha

and beta can be read from the curve intersections. When some of the co-

efficients of equations (4-15) are zero, an analytical solution may be

easiest, or perhaps a combination of the two can be used.

4-2-3 Application of the method.

Example 4-1 (Case I type)

Problem:

In figure (4-3) find the values of K and K to obtain:

1. Characteristic roots at zeta .5, and omega 10.

2. Error coefficient should be greater than or equal to 6.

Solution:

The appropriate characteristic equation is:

S + US 2+ (30 +o* )S +fi

= 0, where alpha = KK beta = K.

Specification 2. can be satisfied if:

K = K/(30 + KK ) = (2 /(30 +^ ) ± .5e t i

The characteristic equation is of the case I type, and the appropriate

frequency transformation is therefore: S = d_s = lis

Letting w represent the undamped natural frequency on the normalized plane,

one can obtain via the frequency transformation: w = .91

The value of zeta is unaffected by the transformation. From the

B -B, curves of figure (4-1), a value of w„ = .91 and zeta = .5 is seeno 1 N

to correspond to B = .075 and B- = .91. Employing equations (4-4) along

with the appropriate coefficient values one can solve for alpha and beta.

The result is:

alpha = 121(.91) - 30 = 80 beta = (11)3(.075) = 100

Since alpha = KK = 100K _ ort .. . .. ,. , „ . nnKt t

= 80 it is seen that: K = .8 and K = 100 =

beta.

The error coefficient is therefore:

K = /2 /(30 + cL ) = .91, which is greater than .5 so the specifi-

cations are satisfied.

Note: If the error coefficient was required to be greater than .91, the

specified value of zeta and omega would have to be modified so as to in-

crease beta, decrease alpha, or both.

Example 4-2 (Case II type)

100

CDu

I

©

g!

•H

101

Problem:

In the system shown in figure (4-4) find K, K , and K to obtainC cl

the following:

1. Characteristic roots at zeta = .7, omega = 10.

2. The error coefficient should be greater than or equal to 6.

Solution:

The characteristic equation is:

S + (3 +cA )S2+ (2 + fS )S + K = 0, where alpha = KK and

beta = KK . The equation is of the case II type and the appropriate

j,

frequency transformation is: S = d3s, where d is as yet unknown.

Therefore:_i

wM = 10d"3

(4-16)NoEmploying equations (4-10) with the appropriate coefficients one obtains:

alpha d B. - 3 beta - d3B. -2 (4-17)

o Z o 1

The error coefficient restriction can be satisfied if:

Jd = 12 + 6/3 or (d - 12)/6 = beta = (TB, - 2 (4-18)o ' o o 1

From equation (4-16) it is seen that:

d3 = 10/w„, d5 = 100/w„ , and d = 1000/w„ (4-19)O No N o N

Employing equations (4-19) and (4-18) one obtains:

1000/6wT̂

- 2 = lOOOB./w2

- 2 or B, = 1.67/wXT (4-20)N IN 1 N

The B.-B- curves of figure (4-2) can now be employed.

In figure (4-2) one can pick of various values of w along the line

zeta = .7. The desired value of w is one that satisfies equation (4-20).

Referring to figure (4-2), the following values can be obtained when zeta

.7:

w = .8 and B. =2.2 and using equation (4-20) it is found

that B = 2.1.

Also from figure (4-2) when zeta = .7 it can be seen that:

102

<~D.-J

OJ

©cr

103

WN = .76 and B = 2.2 and from equation (4-20) : B = 2.2.

The latter set of values are the desired ones and the corresponding value

of B« is 2.5. From equation (4-19):

d = 1000/(.76)3

= 2280 = Ko

From equations (4-17)

:

alpha =29.9 beta = 379

Therefore: beta = 379 = KK„ = 2280K or K = .166 and alpha = 29.9 =t t t

r

KK = 2280K or K = .0131.a a a

The error coefficient specification should automatically be satis-

fied due to the way the analysis was performed, but as a check:

K = K/(2 + KK ) = K/(2 + 8 ) ^6.e t i

Now:

K/(2 + @ ) = 2280/381 - 6 which confirms the result.

Example 4-3 (Case III type)

Problem:

In figure (4-5) find values of gamma, P, and K to obtain:

1. Characteristic roots at zeta = .7, omega = 15.

2. The error coefficient should be equal to 20.

Solution:

The characteristic equation becomes:

S3+ (oL + 5)S

2+ (5</ + 100 f3 )s + 100^ = 0, where beta =

gamma, alpha P, and K 100 to satisfy specification 2. This is recog-

nized as a case III type and the appropriate frequency transformation is:

S = (o£ + 5)s. The appropriate curves are the B -B. curves of figure (4-1).

Employing equations (4-15) after dropping the terms with zero co-

efficients one obtains:

2BXU + (10B

1- lW ' 100 £ + 25B

1= (4-21a)

B oC3+ 15B oC

2+ (75B - 100) o/ + 125B =

o o o o

The latter equation when divided by B becomes:

104

CDu

1

©u

rffc;

105__

c^3+ 15 04

2+ (75 - 100/B )oC + 125 = (4-21b)

From the frequency transformation it is seen that:

wN

= 15/(oC + 5) (4-22)

The procedure is to find values of B and w , employing the zeta = .7

curve of figure (4-1) such that equations (4-21b) and (4-22) are simultan-

eously satisfied. This can best be done by trial and error.

Trial 1 .

From the curves, B = .065 and w = .58. From equation (4-22) one

finds that alpha 20.9. These values are used in equation (4-21b) to

see if the left side of the equation becomes zero. Hence:

(20. 9)3+ 15(20. 9)

2- 1465(20.9) + 125 = -14895?*

Trial 2 .

Try B = .07, then from the curves, w = .55. From equation (4-22)

one finds that alpha 26. Using equation (4-21b)

:

(26)3+ 15(26)

2- 1355(26) + 125 = -17775 t

Trial 3 .

Try B = .075, then from the curves: w = .5. From equation (4-22)

one finds: alpha = 25. From equation (4-21b)

:

3 2(25) + 15(25) - 1255(25) + 5(25) - -250, which can be consider-

ed to be close enough to zero for the size numbers involved. Therefore

from the curves, B. = .45.

Equation (4-21a) can now be employed to find beta:

beta = [(Bjot2+ (10B

1- l)c< + 25B 1/100

or

beta = [(.45)(25)2+ (10x.45 - 1)25 + 25(. 45)1/100 = 3.8

The final results are:

gamma =3.8 P 25 K = 100

106

4-3 Normalized parameter plane curves of higher order.

In the preceding jection, essentially two types of transformations

were made to normalize the third order characteristic equations. The

first being the magnitude scaling, was tacitly assumed since the b co-

efficient of the characteristic equation was taken to be one. Dividing

through by b_ constituted the magnitude scaling. The second transformation,

which was frequency scaling,was used to make one of the remaining coeffi-

cients unity. This left only two variable parameters, and normalized

curves could readily be plotted. If the characteristic equation is higher

than third order, more than two parameters are involved since only two

coefficients can be made unity, so general normalized curves of higher

order than three are not feasible except in the special form as follows.

In reference (5), Choe introduced normalized families of fourth order

curves where the family parameter was taken as the B« coefficient. Para-

meter plane transformations similar to those given in section (4-2) could

be derived for the fourth order case, but they would be too complex to

be of practical use. Obviously normalized curves for higher than fourth

order are impractical.

107

4-4 Three dimensional parameter plane space.

4-4-1 Discussion.

Many compensation problems involve finding values for more than just

two parameters. Such is true in multiple section cascade compensation and

combination feedback and cascade compensation. This type of problem can

be solved by conventional parameter plane methods if all but two para-

meters are set at some arbitrary value. A more illuminating approach,

however, involves the use of three dimensional parameter plane space.

(When the problem involves or can be reduced to three parameters)

.

Since the parameter plane equations are obtained by equating the

real and imaginary parts of the characteristic equation to zero, unique

solutions exist for only two parameters. However, a third parameter can

be intorduced by plotting families of alpha-beta curves in two dimensional

space. Theoretically a three dimensional parameter space surface could

be plotted;but interpreting the results would be difficult.

4-4-2 Example problem.

In section (6-1) a computer program is presented which will plot

families of parameter plane curves in terms of a third parameter as a

variable. The third parameter may appear linearly or non-linearly in any

of the coefficients. The following example shows one application of the

method.

Example 4-4 .

Experience has shown that if the parameter values necessary to com-

pensate a system using a single section cascade compensator are not reali-

zable, then the application of tachometer feedback will often permit the

use of values within the acceptable limit. The system shown in figure

(4-6) is the problem of example (3-20) but with tachometer feedback employed,

Problem:

Using the same criteria as in example (3-20) use tachometer feedback

108

to find more reasonable values for the cascade compensator parameters.

Solution:

From figure (4-6) it can be seen that:

Kg

= KP/(10P + KKtP) = K/(10 + KK

t)

From example (3-20), K =28.6.

Hence: K = 286 + 28.6KK .

The characteristic equation is found to be:

S4+ (11 +o<- )S

3+ (10 + 11</ + KK

tjS)S

2+ (100^ + KK^ +K

j

6)S + Kr^=

where alpha = P and beta = gamma. The third parameter is seen to be KK .

The above characteristic equation is seen to be identical to that of ex-

ample (3-20) when KK is set to zero. The error coefficient restriction

can be incorporated into the above equation by letting K = 286 + 28.6KK .

This results in:

S4+ (11 +<* )S

3+ (10 + 11^ + + KK

tp)S

2+ (10c/+ KK^ + 286/S+ 28.6KKS

)S + 286o^ + 28.6KKt<* m

Third parameter values of 10, 50, 100, 200, 300, 400, 500, and 1000

are investigated, employing the computer program of section (6-1). The

resulting constant zeta and constant omega curves are shown in figure

(4-7). For simplicity the only constant zeta curves that are plotted

are for the required value of zeta of .35. If the M-point shown in

figure (4-7) is chosen, the following values are applicable:

KK = 300 gamma =10 P = 113

This results in the compensator zero being located at S = -11.3, which is

also a closed loop system zero. For the above parameter values the closed

loop system poles were found using the digital computer and are as follows:

-11.056 -90.835 -11.054 + J29.56

The residue of the real root at S = -11.056 is approximately zero due to

the closed loop zero at -11.3, so the complex roots which give the zeta

109

CD

v

or

of .35 are dominant and the problem is solved. It is seen that tacho-

meter feedback increases the undamped natural frequency of the complex

roots by about a factor of three so the settling time is decreased by a

factor of one- third. It is noted however, that settling time is not of

interest in this example.

At this point one should investigate to see if tachometer feedback

alone could produce the desired results. The Routh check is first employed.

The applicable characteristic equation is:

S3+ US2

+ (10 + KK )S + K =

and

Kg

- 28.6 = K/(10 + KKfc), hence K - 286 - 28.6KK .

Using this value for K, the characteristic equation becomes:

S3+ US 2

+ (10 + KKt)S + 286 + 28.6KK

fc

-

The Routh array is:

1 10 KK

11 286 + 28.6KKt

-176 - 17.6KKt

286 + 28.6KKt

Since a negative value of KK is required to even stabilize the system ,

it is concluded that tachometer feedback alone will not work.

4-5 Characteristic equations involving product terms of alpha and beta.

4-5-1 Basic derivations.

Assume that the coefficients of the characteristic equation are of the

form:

• •k-Vt+ck P+ Vt

f3 +d

k(4" 23)

Employing equations (2-7) one can obtain:

d- Bx+ |S c

1+ oL{9 A

x+ D

x= (4-24a)

<* B2+ P C

2+ °^ A

2+ D

2=

° (4-24b)

111

000 002 coT

X-SCflLE '- 2i>0E-tOI CHITS' INCH.

~f SCALE = 3.0OE+OO UNITS' INCH.

006 006 oio 012 cm cW

FIG. 4-7

RM NUTTING , X-AXIS = P , Y-AXISTHIRD PARAMETER = KKT

GAMMA

Where:

k. k„ J5 . ,,k, k.Al

=go M V U

k-1A2 "

O<"» V \ <4 " 25 >

U, , i U. , E. , C. • D. , B„, C_, and D are as defined in section (2).k-1 k 1 1 12 2 2

Equations (4-24) contain in general, two unique solutions for alpha

and beta. Note that if A1

= A_ = 0, equations (4-24) reduce to equations

(2-9) and determinants can be used to obtain the one unique solution.

Solving equations (4-24a) and (4-24b) for alpha and equating the

results one obtains:

Dl+ P C

l ="D

l" P C

l (4-26)

Bx+ (S A

xB1+ jS A

x

Equation (4-26) can be solved for beta resulting in:

beta- < A AD+ A

BC> ± J < A AD+A

BC>2

'4 A

ACA

BD (4-27)

The deltas in equation (4-27) are shorthand notation for terms of the form:

ABC " B

1C2 " »2C1«

AAD

= A1D2 " A

2D1'

eCC -

Proceeding in a similar manner, alpha is found to be:

alpha --< A

DA+ A BC>±/ j

ADA

+ ABC>

2- 4 A

BAA

DC'

(4-28)

2 A* BA

Equations (4-27) and (4-28) contain four solution pairs for alpha and beta,

only two of which satisfy equations (4-24). To find the two correct solu-

tions, equation (4-27) can be used to find two values of beta which can

then be substituted into equation (4-26) to find the corresponding values

of alpha. The two solutions can also be found from equations (4-27) and

(4-28) where the two correct solution pairs are the ones that make equa-

tions (4-24a) or (4-24b) go to zero.

Constant zeta and constant omega curves can be plotted from equations

(4-27) and (4-28). Proceeding as in section (2), conscant zeta-omega

curves can also be plotted from equations (4-27) and (4-28) where B- , C.

,

113

D. , B , C_, and D_ arc defined by equations (2-15) and where:l 7 z z.

w m

Constant sigma curves however are no longer straight lines when an alpha-

beta product is involved. In this case equation (2-17) becomes:

£<;0

at £ C-l>\* + £ ^ (-Dkck ^ + ^ f

2 ^ <-»\* <4"30 >

+ ^ (-i)\^ K= o.

fc-0 k

If one assumes the following notation:

DDD = Jl (-l)kd. G k

CCC = £. (-l)kc. ^ k

K=0 kJc = *

BBB = ^ (-DV ^ kBC = £ (-l)

kh. £ k

K^O k /c=o k

Then equation (4-30) becomes:

^ BBB + ^ CCC + o£ (S BC + DDD = (4-31)

Equation (4-31) is a special form of a conic section and can be plotted

by solving for alpha and incrementing beta over a range of values of interest,

The above equations are programmed for the digital computer in section (6-2).

Example problems involving the above concepts are presented in the

following section.

4 6 Design of double section cascade compensators.

4-6-1 Discussion.

Double section compensators can be designed by use of the parameter

plane in two ways. The double section compensator can be made equivilant

to a single section at a specific complex frequency of interest. Here

the system is first designed using a single section compensator but the

single section parameter values turn out to be physically unrealirable.

This method has the disadvantage that control is maintained over only one

pair of complex roots and dominance is difficult to ensure.

The second method involves writing the characteristic equation with

a double section compensator inserted and drawing the parameter plane

curves. Control over all the roots is obtainable and the dominance pro-

blem is much simplified.

Both the above methods involve characteristic equations with alpha-

beta product terms appearing in the coefficients, and new parameter plane

equations have been derived to handle this situation in section (4-5-1).

4-6-2 Design of a double section compensator on the basis of given single

section parameter values.

This method was proposed by Hyon in reference (6) but Hyon used the

Mitrovic equations to obtain a solution. Parameter plane techniques will

now be employed.

Let the open loop transfer function of a double section cascade

compensated system be:

* i* 2 <s + yr^g + yy 2

>, gmml (4 .32)

(s + ppcs + p2)

The uncompensated system's forward path transfer function is G,

where unity feedback is assumed. The open loop transfer function of a

single section compensated system is:

tt (S + P/ ft ) . G = -1 (4-33)(S + P)

115

Assume complex roots are required at:

si - - f i»i ± jw

i [wT <4 -34 >

Equations (4-32) and (4-33) are equated and G is divided out:

*Lll(S + V fri><S + Vjf 2> = < S + P'X > (4-35)

(s + pi)(s + p

2)

~(S + P)

After rearranging, collecting terms of like power, and dividing by S,

equation (4-35) becomes:

S2( * ' ^]1 2

) S[(PX+ P

2) X + P(l - V ! K 2

> " PlY

2"

(4-36)

P2 * il + P^Y + P(P

1+ P

2- P

xY 2

- P2 K !> " *

x*2

" °

The unknowns in equation (4-36) are P.. , P_, tf , ^ «, and S.

By substituting a specific frequency for S and equating the real and imagin-

ary parts of equation (4-36) to zero separately, two equations in two un-

knowns can be obtained. The parameters % .. andft « can be pre- set to a

fixed value, thus determining whether the compensator will be double lead,

double lag, or lag lead. Values for the remaining unknowns P. and P can

then be obtained from the resulting two equations.

It can be noted however, that when ft and ft are given fixed values,

the parameter plane techniques of section (4-5) can be applied directly

if one lets P =c/ , P = (S , and P P oi |g . Equations (4-27) and

(4-28) give the desired values of P and P^ when the specified value of

zeta and omega for the complex roots is substituted. Only one or at most

two points in the parameter plane are of interest thus obviating the need

for plotting curves. Hence straight analytical techniques can be used.

Example 4-5 .

Problem:

Apply the above techniques to design a double section filter equivalent

116

to the single section filter of example (3-20) at the specified frequency

of zeta = .35 and omega =8.2. Of the possible solutions, choose the one

that makes the specified complex roots most dominant.

Solution:

From example (3-20) a value of gamma = 25.5 was needed, indicating a

double section lead filter might work. The value of P was 87. Let ft=

Y) « = 5, which is a reasonable value for a lead filter. When the above

values are substituted, equation (4 36) becomes:

S2+ (21 </- + 21 (2 - 2118)S + 25^ (2 - 348( oL + |2 ) =0 (4-37)

when zeta = .35: U . - -1; U - 0; D- - 1; U„ - .7-1 o 1 2

From equation (4-37) the coefficients can be used along with equations

(2-10) and (4-25) to obtain the following quantities:

Ax

= -25 A2

=

B - 348 B2

= -172

Cx

= 348 C2

= -172

T>x

= 67.2 D2

= 17397

The deltas are then found to be •

AAD " "4 - 34S I X 10

5 A = -6.066 x 10

ABC ' ° A_ - 4.349 x 10

5

DA

AA(;

= 4300 ABA

= -4300

ABD

= 6.066 X io6

Using equations (4-27) and (4-28) the solutions are:

o£ ,= 16.7 c/

2- 84.4 (8 - 84.4 fS 2

16.7

From equations (4-24) it is found that cJ. paired with £ and &L

paired with & are the consistant solutions.

Since in equation (4-37), the coefficients of alpha and beta are

identical, both solutions produce the same result. cA . and Q are arbi-

trarily chosen. The characteristic equation of the compensated system

117

then becomes:

S5+ 112. 14S

4+ 2533. 2S

3+ 2.3678 x 10

4S2+ 1.587 x 10

5S + 4.0344

x 105

=

Using the digital computer the roots are found to be:

-4.205 -16.79 -85.51 -2.82 + J7.674

The complex roots are at the specified value of zeta and omega.

Comparing the above roots with those obtained in example (4-4) where

the single section plus tachometer feedback was employed, one concludes that

the double section compensator produces dominant complex roots whereas the

tachometer feedback and single section scheme does not. In example (4-4)

however, the complex roots were found to be effectively dominant since the

residue of the nearby real root was about zero. In this example, closed

loop zeros are located at -3.34 and -16.9. Therefore, the magnitude of

the residues of the complex poles and the real pole at -4.205 are almost

the same so it is fortunate that the complex roots are dominant.

4-6-3 Design of a double section compensator using general parameter

plane methods.

This method involves incorporating the double section cascade compen-

sator equations into the uncompensated system's characteristic equation.

This technique has the advantage that parameter plane curves can then be

drawn and control xl maintained over all the characteristic roots rather

than only the two specified complex roots as is the case with the method

of section (4-6-2).

Here the technique can best be explained by an example.

Example 4-6

Problem:

Figure (4-8) shows the system of example (3-20) but with a double

section compensator employed. Solve example (3-20) using a double section

compensator as indicated in figure (4-8)

.

©

CO

iI*

v^

q:

119

Solution:

For comparison with the results of example (4-5) let:

tl-V

i= 5.

The characteristic equation then becomes:

s5+ (oc+ (3+ ids

4+ ( o^-p + n^. + ii /a + io)s +

(llcLp + 10<*- + 10 jS + 7150)S2+ (10ot p + 1430od +

1430 (2. )S + 286 oC (3, =0 (4-38)

Here: o* = P^ (3 = ?r and ot p « P^.

Parameter plane curves for equation (4-38) have been plotted in

figures (4-9) and (4-10) utilizing the computer program of section (6-2).

Since the curves are rather complex, the constant zeta and constant omega

curves are plotted in figure (4-9) and the constant sigma curves are plot-

ted in figure (4-10). In figure (4-9) it is noted that there are two

sections of constant zeta curves for each value of zeta and two section of

constant omega curves for each value of omega. This agrees with the re-

sults proven earlier that two unique solutions exist for each value of

zeta and omega.

In figure (4-10), the discontinuities of the constant sigma curves

are indicated by the horizontal straight line. The straight line is not

part of the curve and should be ignored. A study of the curves indicates

that for a given curve, if one chooses a point, say ( oZ ^, |S j) »tnen

there exists a corresponding point ( j2 ^, cL ) . This is due to the

fact that the coefficients of alpha and beta are identical in the

characteristic equation.

If the M-point indicated in figures (4-9) and (4-10) is chosen then

the following characteristic roots are indicated:

-9.21 -26.14 -65.34 -2.1 + J5.61

The accuracy of the root values was obtained from the printed out

computer data, given that alpha =31.2 and beta =62.5. The complex roots

are located at zeta = .35 and omega = 6.

120

88Q 884 88€ 818 812 814 816

\SCPLl - 2.88E+81 UNITS'JNCH

^f-SCflLE - L88E-f81 UNITS'INCH,

RM MUTTING FIG. 4-9XSCflLE = PI = ALPHA p YSCflLE = P2 - 3E^fi

/2.i

x-scrli - .o.eeE+ej umjts/jnch.

Y-3CALE - i.eeE+ei UWT3'INCH.

RM NUTTING FIG, 4 -10KSCflLE - PI = ALPHA , YSCPLE - P2 = BETA

It*

The roots obtained in example (4-5) are:

-4.205 -16.79 -85.51 -2.82 + J7.674

Comparison of the two sets of roots indicates that in example

(4-5) the ratio of the nearest real root to the real part of the complex

roots is 1.44 whereas in this example it is 4.4. Hence the dominancy

factor has improved by a factor of three in this example.

The effectiveness of the method of this section is thus apparent.

5 Root locus digital computer programs.

The programs of this section were written using Fortran 60 along with

subroutines available at the computer facility of the U. S. Naval Post-

graduate School. It is necessary to have the coefficients of the poly-

nomial or characteristic equation to use the root locus program of section

(5-2), so a program is presented in section (5-1) to compute the coeffi-

cients in case the equation is in factored form.

The use of the programs is explained in the comment cards at the

beginning of the programs.

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122

6 Parameter plane digital computer programs.

The programs of this section were written using Fortran 60 along

with subroutines available at the computer facility of the U. S. Naval

Postgraduate School.

The use of the programs is explained in the comment cards at the

beginning of the programs.

133

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7- The complementary roles of the parameter plane and root locus.

As was mentioned previously, parameter plane curves are infinite in

extent. The curves also exhibit discontinuities and become less well

behaved as the order increases. In addition, choosing a good graph scale

involves a good deal of trial and error or pre-plotting calculations.

The root locus can be very useful in overcoming some of the above diffi-

culties.

The general procedure is to use the root locus to limit the range

of parameter values of interest, and to gain additional insight into the

problem. This is particularly true if a computer is used to plot the root

locus. Since a correct root locus graph scale can be fixed in one or two

computer runs, several root loci can be plotted in fairly short time.

The parameter plane can then be used to complete the problem.

The technique can best be illustrated by the following example.

Example 7-1

Problem:

For the system shown in figure (7-1) find values for K.. , K„, K , K ,

and K to give a good transient response. Low settling time is the

primary consideration. Suggest modifications to improve the system.

The following parameters have fixed values:

P = 6.28 x 106

P3

= 2.5 x 105

w = 6.28 x 106

zeta = 0.5

Solution:

It is known that the block diagram elements of figure (7-1) that

involve K, and K c were added on the basis of physical reasoning to help4 5

reduce the settling time. The basic uncompensated system consists of the

single loop unity feedback control system involving K. , K_, and K_. From

the block diagram reduction of figure (7-1) the characteristic equation

157

is computed and the given parameter values are substituted. Letting

alpha - K K and beta = K.KJC. one then obtains:14 1 z j

S5+ 1.281 x 10

7S4+ (8.194 x 10

13+ o^)S

3+ (2.678 x 10

2°

+ p+2.5x 105o<L )S

2+ (6.18 x 10

25+ 6.28 x 10

6|S +

K1K2K3K4K5)S + 3.94 x 10

13 # = (7-1)

To study the effect of the K, compensator, K_ is set equal to zero

in equation (7-1) and parameter plane curves are plotted in figure (7-2)

for the remaining system.

The unstable region for these curves is above and to the right of

the zeta equals zero curve. From the Routh check, the stability limit

19gain of the uncompensated system, i.e., LKL, is equal to 1.025 x 10

This value corresponds to point B in figure (7-2). The effect of the K,

compensator is to make the system less stable as KR, is increased. With

19K..K K set to 1.025 x 10 , a root locus is plotted in figure (7-3) with

K K, as the variable. This plot shows the system to be unstable for all

K.K, greater than zero. Gain changes have small effect on the right half

plane root locations.

o

Now K.K. is arbitrarily set to 9.1 x 10 and K-K-K- is left unchanged,14 I Z j

The effect of the K,. compensator is seen from figure (7-4) to have a

stabilizing effect. It is known that settling time is inversely propor-

tional to the undamped natural frequency of the complex roots. In figure

(7-4), point A represents a zeta and omega that would be satisfactory for

the complex roots, but the location of the real root shows the latter to

be dominant.

Two avenues of approach appear to be open. Try a different means

of compensation, or modify the suggested means. The former is investi-

gated first.

The effect of tachometer feedback around the entire forward path,

158

with K, = K = 0, is shown in figure (7-5). From the root locations as

tachometer gain increases from point A to B to C, it is seen that the

real root is highly dominant. Tachometer feedback would therefore be un-

satisfactory. Since the system's operating frequency is in the mega-

cycle range, higher forms of derivative feedback would not be practical.

The fact that the K, compensator, which is the feedback path around

KK. in figure (7-1), makes the system less stable, suggests that this

path could be eliminated. When this is done the resulting system is

shown in figure (7-6) and the characteristic equation is as follows:

S5+ 1.281 x 10

7S4+ 8.194 x 10

13S3+ (26.67 x 10

19+£ )S

2+

(6.18 x 1025

+ 6.28 x 106

(3 + oL )s 3.94 x 1013^ = (7-2)

In equation (7-2), alpha - K.KKKK and beta - KKR.

The advantage of this scheme over the original system is that the

K.K compensator can be realized by an R-L-C circuit, whereas due to the

physical nature of the problem, the original scheme would require electro-

mechanical implementation with the inherent disadvantages.

In figure (7-6), KKX is again set at the stability limit of the

uncompensated system and a root locus with variable K.K is plotted in

figure (7-7). A study of figure (7-7) shows that a slightly better domin-

ance factor can be obtained with the modified system. It is concluded

that the latter system not only performs as good or better than the former,

but it is simpler and cheaper to implement.

To see if system performance can be improved by increased gain, the

forward path gain K.KJK. is increased by a factor of ten. The resulting

root locus in figure (7-8) indicates that the system is unstable for all

K,K . This illustrates the difficulty in choosing the best values for

system parameters by root locus techniques when more than one parameter

is involved.

The parameter plane can now be advantageously employed. The resulting

159

curves are plotted in figure (7-9) with variables K-.K.K and K-KJC-K.K,..

A close study of the curves indicates that the M-point shown is perhaps

the best one. The five characteristic roots can be read directly from

the curves and are as follows:

zeta = .55 omega = 3 x 10

zeta = .71 omega = 6.5 x 10

S = -2.2 x 106

The damping factor of the first pair of roots is 1.65 x 10 which

when compared to 2.2 x 10 , indicates that these roots are dominant.

In the original uncompensated third order system, for a zeta of .55,

the maximum obtainable omega for the dominant roots is only about .3 x

10 . On the basis of the dominant roots only, one can conclude that the

compensation has reduced the settling time by a factor of ten. The trans-

ient response of the compensated system to a unit step input was computed

by digital computer and is given in figure (7-10). The settling time is

about four microseconds and the overshoot is 50%.

160

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170

8- Conclusions.

Parameter plane techniques have been applied to the compensation of

linear feedback control systems. In particular, general equations have

been derived for the cases of feedback, cascade, and combination feedback-

cascade compensation, to enable one to analytically place a pair of com-

plex roots at a specific location in the S-plane, while simultaneously

satisfying the steady state accuracy requirements.

A dominancy technique has been introduced whereby once the specified

pair of complex roots is fixed, the remaining roots of the characteristic

equation can be placed to ensure that the specified roots are dominant.

Sketching techniques are developed enabling one to quickly sketch the

zeta equals zero and zeta equals one-half curves. This can be useful in

determining the type of compensation to use and in choosing an appropriate

graph scale for the digital computer plots.

Graphical solutions on the parameter plane are discussed in terms

of engineering example problems.

Miscellaneous aspects of the parameter plane are discussed. In parti-

cular, transformations are derived to enable one to compensate parameter

plane type characteristic equations on normalized Mitrovic third order

B -B and B -B curves. Three dimensional parameter plane space is dis-

cussed and applied to an example problem.

A derivation of parameter plane equations involving product terms of

alpha and beta is made and the results are applied to the design of double

section cascade compensators. Double section compensators are designed on

the basis of unrealizable single section parameters and by incorporating the

double section compensator into the characteristic equation and plotting the

parameter plane curves.

Digital computer programs are introduced that one can use to plot root

loci and parameter plane curves. Parameter plane curves can be plotted for

171

characteristic equations involving three parameters and alpha-beta pro-

duct terms.

Finally an engineering example is presented that points out the

complementary nature of the root locus and parameter plane.

A basis for further investigation involves plotting constant band-

width curves on the parameter plane and determining the nature of para-

meter plane curves resulting from characteristic equations containing

squared terms of alpha and beta.

172

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174 W&&.

APPENDIX II

A. Synthesis of R-C lead network

&IH

n c

-AAA-

R» iz-

In figure (a), — -

Fig. a

5+ l/<?,e

S + ef-cf-*U+f?C

~1

S + r

where ^s l?i Ci and °^ =^z.

/

( /? -f /? > ) • Tne aD°ve transfer function

has a D. C. gain of c/~ so to make the D. C. gain unity an amplifier of ^jT

gain will have to be added. The pole to zero ratio, ^ is greater than

one, indicating that the circuit of figure (a) is a lead network.

If the pole to zero ratio and T become very large the filter be-

haves like a pure differentiator and noise problems arise. If the pole to zero

ratio s kept less than ten, the noise problem is reduced.

B. Synthesis of R-C lag network

-W\Aa-

iKl

Fig. b

In figure (b) the transfer function iS

:

c ^ 2 4- I~]

l

_ ^LfR.+ Rz) C i (R l +Rj)J c£S+ ^

where f= C*(fl,+ f?4 )

*«<* d = Ri / ( ft ,+ f? z )^

T^e D. C. gain of the above transfer function is unity and the pole to

zero ratio is alpha. Since alpha is less than one, the circuit of figure

(b) is a lag network.

175

BIBLIOGRAPHY

1. Ross, E. R. , Warren, T. C, and Thaler, G. J., Compensation Basedon the Root Locus Approach, AIEE Paper, Sept. I960.

2. Pollak, C. D. , and Thaler, G. J., S-Plane Design of Compensatorsfor Feedback Systems, IRE Transactions, May 1961.

3. Mitrovic, D. , Graphical Analysis and Synthesis of Feedback ControlSystems, IEEE, Theory and Analysis, II Synthesis. AIEE Trans-actions, Part II, Jan. 1959.

4. Ohta, T., Mitrovics Method - Some Fundamental Techniques, ResearchPaper No. 39, U. S. Naval Postgraduate School, Jan. 1964.

5. Choe, H. H. , Some Extensions of Mitrovic' s Method in Analysis andDesign of Feedback Control Systems, M. S. Thesis, U. S. Naval Post-graduate School, 1964.

6. Hyon, C. H., A Direct Method of Designing Linear Feedback ControlSystems by Applying Mitrovic' s Method, M. S. Thesis, U. S. NavalPostgraduate School, 1964.

7. Siljak, D. D. , Analysis and Synthesis of Feedback Control Systems inthe Parameter Plane, Part I, Linear Continuous Systems, IEEE Trans-actions, Nov. 4, 1964.

8. Thaler, G. J., and Brown, R. G. , Analysis and Design of FeedbackControl Systems, McGraw-Hill Book Co., Inc., 1960.

176

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