Phase-Plane Methods

Home , Phase plane

NPS ARCHIVE 1969 HSIUNG, Y.

PHASE-PLANE METHODS

Yun-Lan Hsiung

DUDLEY KNOX LIBRARY NAVAL POSTGRADUATE SCHOOL MONTEREY, CA 93943-5101

United States Naval Postgraduate School

THESIS

PHASE-PLANE METHODS

Yun-Lan Hsiung

October 1969

Tlnu document, kcud been app\ove.d fan. pubtic *e- leo*e and 6aJU; itb duVubution

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DUDLEY KNOX LIBRARY SCHOOL N M/AL POSTGRADUATE MONTEREY, CA 93943-5101

Phase -Plane Methods

Yun-Lan Hsiung Lieutenant Commander, Chinese Navy B. S., Chinese Naval Academy, 1954

Submitted in partial fulfillment of the requirements for the degree of

MASTER OF SCIENCE IN ELECTRICAL ENGINEERING

from the

NAVAL POSTGRADUATE SCHOOL October 1969 c \ \

ABSTRACT

The phase-plane method is a graphical method for linear and non-linear second-order systems. There are many techniques for obtaining the phase portrait which consists of a number of phase trajectories on the phase- plane. From the summary and discussion the isocline method is a most general and useful method. The only difficulty is in the labor required. Solution by digital computer reduces this difficulty and makes the isocline method a much more useful method in non-linear systems analysis and design applications. LIBRARY IAYAL POSTGRADUATE SCHOOL MOHTJEBJSy, CALIF. 93940

TABLE OF CONTENTS

Page

I. INTRODUCTION 15

A. CONDITIONS FOR LINEARITY AND DEFINITION 15

OF NONLINEARITY

B. CLASSES OF NONLINEARITY 15

C. NONLINEAR ANALYSIS 19

D. THE GRAPHICAL METHOD - THE PHASE-PLANE 20

(PHASE-PLANE ANALYSIS OF NONLINEAR

SYSTEMS)

II. SUMMARY OF USEFUL GRAPHICAL METHODS FOR 22

CONSTRUCTING PHASE TRAJECTORIES ON THE PHASE-PLANE

A. ISOCLINE METHOD 22

B. LIENARD METHOD 28

C. DELTA METHOD 34

D. SZEGO METHOD 45

E. MURTHY'S NEW APPROACH TO PLOTTING OF 60 PHASE-PLANE TRAJECTORIES

F. THE ACCELERATION-PLANE METHOD (BY KU) 67

G. DEEKSHATULU ' S METHOD 71

H. PELL'S METHOD 78

I. SOMAYAJULU'S METHOD 83

J. THE E -FUNCTION METHOD 91

K. THE SLOPE-LINE METHOD 93 .

Page

L. COMPARISON

III. APPLICATION OF THE ISOCLINE METHOD 98

A. MANIPULATIONS ON THE PHASE-PLANE 98

B. COMMON PHYSICAL NONLINEARITIES 103

C. APPLICATIONS OF THE ISOCLINE METHOD 114

IV. GENERATING THE PROGRAMS TO DRAW ISOCLINES 136

AND TRAJECTORIES OF SECOND-ORDER DIFFERENTIAL

EQUATIONS

A. x + F(x)x + G(x) = TYPE EQUATION 136

2 B. x + x + G(x) = TYPE EQUATION 150

+ C. x x I x I +x=0, x+x1x|+x=0 154

= and | x | + x + x TYPE EQUATIONS

D. USE OF PROGRAMS 159

E. DISCUSSION 162

V. GENERATING PROGRAMS TO DRAW ISOCLINES AND 164

TRAJECTORIES WITH DIVIDING LINES OF THE

SECOND-ORDER DIFFERENTIAL EQUATIONS

A SATURATION 164

B. DEAD ZONE 17 2

C. IDEAL RELAY 174

D. RELAY WITH DEAD ZONE 187

VI. CONCLUSIONS 191

APPENDIX 193

COMPUTER PROGRAMS 193

BIBLIOGRAPHY 241 Page

INITIAL DISTRIBUTION LIST 242

FORM DD 147 3 245

LIST OF DRAWINGS

Figure Page

f ( z ) 1-1 A transfer function —*—*- which is 16 independent of time

1-2 The instantaneous nonlinear characteristic 17

is combined with inertia and damping

cannot be separated as in figure

2-1 Construction of a trajectory using 24

isoclines

2-2 Phase-plane plot of damped oscillation 27

obtained by isocline method for a

servomechanism

2-3 Lienard's construction of phase-plane 29

direction field

2-4 Lienard's construction of phase-plane 33

direction field

2-5 Direction field obtained by Lienard's 33 method

2-6 Exact relationship 36

2-7 Stepwise relationship for constant p 36

2-8a (a) Change with positive 6 38

(b) Change with negative 6 38

2-8b (a) Change with positive 6 40

(b) Change with negative 6 40

2-9 Example (I-B-2) 42 Figure Page

2-10 The operative displacement 6 for the 44

physical pendulum plotted in the phase-

plane. Center and correspond to

points P and Q„

2-11 The operative displacement 6 for a system 44

with restoration proportional to the cube

of the displacement. Centers and

correspond to points P and Q.

2-12 The operative displacement 6 for a system 46

with quadratic velocity damping. Centers

and correspond to points P and Q.

2-13 The operative displacement 6 for a system 46

with 6 for a system with positive linear

and negative quadratic restoration in

displacement. Center corresponds to

point P.

2-14 The two operative displacements 6 and £L 47

for a system with damping proportional to

the nth power of the velocity and with

restoration proportional to the first and

the mth power of the displacement. Center

corresponds to point Q. dv, 2-15 Curves v, = and -tt^ 54 1 dt dv 2 2-16 Curves v„ = and -rz~ 55 2 dt

8 Figure Page dv 2-17 Curves v = and -r-^- 55 3 dt dv 2-18 Curves v = and -—^ 56 u dt

2-19 Example (II-D-2) 56

2-20 Exact generation of trajectory from the 58

point P

2-21 Exact generation of trajectory from the 58

point P' by using v-function v = x

2-22 Example (II-E-2) 64

2-23 Example (II-E-2 66

2-24 Construction of trajectories by accelera- 68

tion plane method

2-25 Acceleration plane method 70

• 2 2-26 Phase trajectories for x + x + x = 76

2 .. . 2-27 Trajectories for tx+tx+x 3 =0 79

2-28 Pell's method for phase trajectory 84

construction

2-29 Example construction using Pell's method 84

for the hard spring system

2-30 Graphical determination of field 86

direction

2-31 Construction details 88

2 2-32 Example x + x + x = 90

2-3 3 The E-function method 94

2-34 Example (II-J-2) 94

2-35 Example (II-J-2) 96 Figure Page

2-36 The slope-line method 96 o 3-1 Differentiation procedure 102

3-2 Limiting of x on the phase plane 104

3 3 Piecewise shaping of trajectory 104

3-4 Common physical nonlinearities - Saturation 105

3-5 Common physical nonlinearities - Dead zone 108

3-6 Common physical nonlinearities - Backlash 110

3-7 Phase trajectory for an instrument servo 113

with Coulomb friction and stiction; step

response

3-8a Phase portrait of an instrument servo 115

with Coulomb friction in response to a

ramp input

3-8b Effect of stiction on the ramp response 115

of a servo with Coulomb friction

3-9 Example (III-C-1-a) 118

3-10 Example (III-C-1-b) 118

3-11 Isocline solution of x + x + sin x = 120

3-12 (a) Example x + xx + x = 122

(b) Example x + xx + x = 123

3-13 Example x + 0.2x + x = 0, x + 0.2y±0.3 = 123

3-14 Example 4: Oscillator by relay control 129

3-15 Example 5: x+ix + x = u 134

4-1 Example 1: (x + xx + x = 0) 137

2 4-2 Example 2 : (x - e(l-x )x + x = 0) 138

10 Figure Page

4-3 Example 3: (x +(cos x)x + tan x = 0) 139

2 4-4 Example 4: (x + 25(0.1x + l)x = 140

4-5 Example 5: (x + 0.5x + x = 0) 141

2 4-6 Example 6: (x(l-x )x + x = 0) 142

4-7 = 143 Example 7: (x + x + x | x | 0)

4-8 Example 8: (x + — x + x = 144 x 0)

4-9 Example 9: (x + x + sin x = 0) 145

= 4-10 (x 1 Example 10 +(1- x | )x + x 146

4-11 Example 11 (x + — sin x = 0, — = 2) 147

2 4-12 Example 12 (x +(l-x )x + x = 0) 148

4-13 Example 13 (x + 2xx + x = 0) 149 2 4-14 Example 14 (x + x + x = 0) 151

2 2 4-15 Example 15 (x + x + x = 0) 152

4-16 Example 16 (x + x + x 1 x | =0) 153

= 4-17 Example 17 (x + x | x | + x 0) 156

= 4-18 Example 18 (x + x 1 x | + x 0) 157

4 19 Example 19 (|x|x + x + x = 0) 158

,.. 4.88 . 4-20 Example 20 (X + r-r-r- X COS X + Sin X = 0) 161 1. 563 5-1 Example 1: x+0.2x+x=0, dividing 166

lines x = ± 0.3

5-2 Example 2: x + . 2x + x = , dividing 167

lines x = i 0.2

5-3 Example 3: x+0.2x+x=0 / dividing 169

lines x + 0.8y = ± 0.3

11 Figure Page

5-4 Example 4: x+0.2x+x=0, dividing 170

lines x - 0.8y = ± 0.3

5-5 Example 5t x+0.2x+x=0, dividing 171

lines x = J- 0.3

5-6 Example 6: x 4- 0.2x + x ± 0.3 = 17 3

5-7 Example 1% x + 0.2x + x + u = 176

u = - 0.3 Sgn(x)

5-8 Example 8: 177

(a) x + 0.02x + u = 177

u = - 0.3 Sgn(x)

(b) x + 0.02x + u = 178

u = - 0.3 Sgn(8x-y)

u = - 0.3 Sgn(8x+y)

(d) x + 0.02x + u = 180

u = - 0.3 Sgn(4x-y)

(e) x + 0.02x + u = 181

u = - 0.3 Sgn(4x+y)

(f) x + 0.02x + u = 182

u = - 0.3 Sgn(2x-y)

(g) x + 0.02x + u = 183

u = - 0.3 Sgn(2x+y)

(h) x + 0.02x + u = 184

u = - 0.3 Sgn(x-y)

(i) x + 0.02x + u = 185

u = - 0.3 Sgn(x+y)

12 Figure Page

5-8 (j) x + 0.02x + u = 186

u = - 0.3 Sgn(y), x = y

5-9 Example 9: x+0.2x+x+u=0 189

u = - 0.3 Sgn(x)

dividing lines: x = ± 0.2

5-10 Example 10 : x + x + u = 190

u = - 0.3 Sgn(x)

dividing lines: x = ± 0.2

13 ACKNOWLEDGEMENT

The author wishes to express his appreciation to Dr,

G. J. Thaler for the invaluable aid and counsel which he has offered during the preparation of this dissertation.

14 I. INTRODUCTION

A. CONDITIONS FOR LINEARITY AND DEFINITION

OF NON-LINEARITY

It is useful to define linearity and then any system, component, or phenomenon, which does not satisfy this definition will be considered non-linear. The simple mean- ing of linearity is that if the relationship between output y and input x is plotted the result is a straight line. Or that if the output is proportional to the input the process must be linear. So we can simply say the essential condition for linearity is that the law of superposition applies, and linearity is a specification which therefore limits the field of activity. Non-linearity is a "non-specification" and the field is unbounded, linearity being a particular part of it.

B. CLASSES OF NON-LINEARITY

Non-linearity in physical systems cannot be avoided, and in fact can sometimes be used to improve the system's performance. The non-linearities encountered in control systems may be classified as accidental non-linearities and intentional non-linearities. The accidental class consists of those non-linearities which are inherent in the components used in the system. Some of these are coulomb friction, stiction, backlash, binding, dead zone, saturation, etc. Their existence is accidental in the sense that

15 .

they are not deliberately inserted in the system. Inten- tional non-linearities are those which are introduced deliberately, either to obtain desired performance features or for weight, space, or economic advantages.

Clearly the simplest class of non-linearity is the instantaneous, a member of which only changes its input u(t) to some output x(t) without any delay, the ratio x(t)/u(t) being a function of u(t). Thus x(t) = f[u(t)] and it can be represented in schematic form (see Fig. (1-1)), having a transfer function f(z)/z which is independent of time.

Saturating amplifiers and relays are examples of instantaneous non-linearities. In practice, of course, such a non-linearity will almost always be associated with some linear (or non-linear) weighting function resulting from inertia and damping as in Fig. (1-2), so that

x(t) = f[u(t) ] (1-D

In general, the instantaneous non-linear characteristic is combined with inertia and damping in the same component, in which case it cannot be separated as in Fig. (1-2)

u(t) x(t)

(z) f ' Fig. 1-1. A transfer function jj which is independent of time.

16 u(t) y(t) ^--

Fig. 1-2. The instantaneous non-linear characteristic is combined with inertia and damping cannot be separated as in figure.

By convolution integral

= y(t) j x(T)h(t-T)dT , and

changing the variable gives

y(t) = x(t-T )h(T )dr 1 1 1

Assume

x < for t < so that

x(t-T ) = for T ^ t 1 1

17 Then

= )dT.u, y(t) x(t-T 1 )h(Tw 1 J l'" .l' l o

f[u(t-T ) ]h(T )dT 1 1 1

r»t

KCuCt-T^^) , T '\dT (1-2) 1 1

K[u ( t-T, ) , T, ] is called the Kernel function and is seen to depend on both the dynamics of the system, f(z)/z and h(t), and the characteristics of the input process, u(t).

Zadeh has classified non-linear systems on the basis of multiway integrals and Kernel functions. Thus those that can be represented by a single-way integral belong to

the first class, r?, , which therefore includes the subclass of all linear systems and also the degenerate zero-order class of instantaneous non-linearities.

Then the first-order non-linear differential equation is

dx + F (x) = P [u(t)] H q 1 or

= If Fl [u( t) ] F (X)

Integrating gives

x(t) = K[u(t-T) , T]dT o

18 s .

Hence, the second-order non-linear differential equation is

f (x, x, x) = u(t)

Integrating twice gives

= x(t) K [uCt-T^, U(t-T ) ,T T ]dT dT 2 1# 2 2 o o 1

So, the second class, r\~, contains all those for which the input/output relationship can be expressed as a 2-way integral and this includes the degenerate cases belonging

to class r\ .

The nth class, T? , contains all those systems for which the input/output relationship is a n-way integral,, which means that they can be represented by nth order nonlinear differential equations.

Zadeh' classification is an important property in considering random processes in the presence of noise.

C. NON-LINEAR ANALYSIS

Many methods used in the analysis of non-linear prob-

lems are effective only for limited types of problems. A

few of them have been developed which have more general applicability to non-linear control systems:

19 1. Linear approximations.

2. Graphical method - The phase-plane.

3. Piecewise linear approximation.

4. Numerical methods.

5. Describing function.

6. The analog/digital computer.

D. PHASE-PLANE ANALYSIS OF NONLINEAR SYSTEMS

The phase-plane method is a graphical method. There are many techniques for obtaining the phase portrait which consists of a number of phase trajectories on the x vs x plane or in some plane related to this plane through a convenient transformation of variables. After the phase portrait is obtained, by means of it, analysis of the following aspects and items is possible:

1. Computation of the exact transient response for a given set of initial conditions.

2. Prediction of the type of transient response

(oscillatory or monotonic) with a certain range of initial conditions.

3. Estimates of the amount of overshoot in response to a step input of given magnitude.

4. Prediction of the system stability.

5. Prediction of the existence and/or amplitude of limit cycles.

6. Prediction of the existence of various modes of operation.

20 7. Study of the effect of a given type of non-

linearity on the performance of the system as compared with a linear system, with a system having a different

type of non-linearity, or with a system which is non-linear

in the same sense but in different degree.

8. Attempts to answer questions such as:

a. Does this non-linearity improve performance?

b. What changes can be made to suppress the limit

cycle?

c. How can we adjust this system to make it meet

specifications?

As stated above, we can understand a graphical method;

it is usually relatively simple to utilize, and may be

especially useful as an exploratory tool when a non-linear

equation is first being attacked. There are many ways to

obtain phase trajectories on the phase-plane. In the next

chapter, the most useful methods will be summarized and

compared. The advantages and disadvantages of the various

methods will be pointed out.

21 2

II. SUMMARY OF USEFUL GRAPHICAL METHODS

FOR CONSTRUCTING PHASE TRAJECTORIES

ON THE PHASE-PLANE

A. ISOCLINE METHOD

1. Method-description

The free motion of any second-order non-linear system can be described by an equation of the form:

x + $(x,x)x + >£(x,x)x = where

dx dt

d x x = — dt

Let y - f (2-1) then

~ = = X X < 2 2) * dt ~ *<*'¥>¥ " ^ 'Y)

Equation (2-2) divided by (2-1) yields

dy_ £(x,y)x dx y Let

-^- = M where M is a constant, dx

22 Then

My = $(x,y)y + >£(x,y)x (2-3)

Equation (2-3) is the equation of isoclines.

NOTES: a. If $(x,y) and ^(x,y) are constants, then isoclines are straight lines. Otherwise, isoclines will be curves.

b. For a first-order equation:

= f(x fc) dt '

Isoclines will be f(x,t) = M.

c. Equations more than second-order need some transformation technique, so this method is good only for second-order systems.

d. After obtaining the isoclines, phase trajectories are plotted on the phase-plane as follows:

M, , Mp, M_ represent the slope of each isocline. Start from same initial condition point, as in

Fig. 2-1 point A. From point A draw dotted lines parallel

to M.. , M_. These dotted lines intersect the M~ isocline at a and be. B is middle point of ab. Continuing in this way, obtain C, D ... points. Connect A, B ... to obtain a phase trajectory from the given initial condition A on the phase-plane.

2. Example

Consider a viscous-damped servo

2 d 9 d9 J —t~- + F -r-£ + K0 =0. dt dt o

23 -»-x

Fig. 2-1. Construction of a trajectory using isoclines.

24 Dividing through by J:

2 d e _ d0 —=S + £ —P- + =0. (2-4) ,,2 J dt |J dt o

Substituting

= "k J J and

F c = 2^/KJ into equation (2-4)

2 d d0 2 9 r - + 2co) + U> 6 = 0. ,,2 n -T7T-dt dt no

Let

° = dt z then

, , e -, ^ d0 x d dz d / o \ 1 o d0 d0 V dt J " z ,.2 o o dt so dz z = +T Z +T 05 = d0 ^"h"2CU> "h O

Let dz M = - dfl o

25 Isocline equations will be

2 ZM = -2cb) z - oj . (2-5) n n o \ -> i

If c = 0.25 (linear servo system) and 0) =1, n equation (2-5) will be ZM = -0.5z -0 ^ o If M is equal to 2,1,0, -0.5, -1, -2 we obtain straight radial lines (isoclines) as in Fig. 2-2.

Assume an initial condition represented by A in the figure, namely an output position = -2.7 and a position dB ° o rate of change -=77- = 0, we can plot the phase trajectory as in Fig. 2-2.

3 . Discussion

a. From Fig. 2-2 it is seen that by plotting some isoclines on the phase-plane (more or less depending on desired accuracy) then any phase trajectory is easily plotted from given initial conditions on the phase-plane.

b. By inspection of a few trajectories the sys- s tern's behavior is determined. For example: In Fig. 2-2 this system has damped oscillations and is a stable system.

Point B is an oscillation max. (0 = 1.2). o c. In this example, as applied to the case of a linear servo system (C is constant), isoclines are straight lines, so it is simple. If C is not constant more labor is required.

26 Fig. 2-2. Phase-plane plot of damped oscillation obtained by isocline method for a servomechanism.

27 B. LIENARD METHOD

1. Method Description

Lienard's method is a means for drawing a direction field in the phase-plane to guide the tracing of the trajectories of a differential equation.

Using the isocline method, (see Fig. 2-3) , draw the zero-slope isocline on the phase-plane. Let P be any point of the plane at which the slope of the direction field is desired. From P draw the line parallel to the z-axis, which intersects the zero-slope isocline at Q.

Let R be the projection of Q on the z-axis. Now, draw a short line segment through P, perpendicular to the straight line RP. This segment is tangent to the direction field of the integral curves of certain differential equations.

In practice, graphical construction of the entire direction field is quite easy since the short segments at all the points on the extension of the line PQ can be drawn by using a compass centered at the one point R. Note that this construction does, indeed, result in & = everywhere on the zero-slope isocline.

In Fig. 2-3 the two angles a. and (9 are equal because of the perpendicularity of their sides at P and Q. The tangent of each angle can be evaluated in terms of the coordinates of the points P, Q and R and the ratio of lengths of unit distance along the two axes. This length ratio is defined by

28 Fig. 2-3. Lienard's construction of phase-plane direction field.

29 _ unit distance for v ro_£\ unit distance for z *

tan oc = K ~ = KM (2-7) dz where M is the slope of the integral curves and/or of their

direction field

tan = = — (2_8) ^oR il ipt; ) ~k^

= the The expression Z_ f(VQ ) is the relation defining zero-slope isocline.

Since V = V , P is any point of the phase-plane. The

subscript can be dropped, giving

tan P = *a=* (2-9)

since

tan & = tan |9

then f " Z = = = tan ct KM tan (9 KV£$

Z M = f( ^" (2-10) K V

M = -=— dz

2 ,, dz dv dz d z fo -, - v dt dz dt -.,2

30 Combining equation (2-10) and (2-11)

- - k2 f + z - ° < 2 12 > ( ftac )" ^tdt^

Any servo which obeys a differential equation of this

form has a phase-plane representation with a direction

field which can be constructed by Lienard's method.

Note that the coefficient of z is unity and that K

specifies the ratio of vertical-to-horizontal unit distan- ces on the phase-plane. The zero-slope isocline is found d z directly from the differential equation by setting ~ dt Z equal to zero, since the isocline is the line for which 2 d z M = 0. So equation (2-11) gives o = 0. at 2. Example

Using the same example as in A, a linear servo-

mechanism having a constant damping coefficient C = 0.25,

and natural frequency oj = 10 radian/sec.

2 d fi dG « 9 x - + 2cui -rr^ + ti> 8 =0 ,.2 n dt n o dt

Changing to the required form (compared to Eq. (2-12))

2 . d e de ° + ^ TT^ + =0 2 , 2 ui dt o dtL n U)n 2 d 6 -, d9 100 2 20 dt o dfc

K = — = 0.1 n

31 , dft ~ d0 d0 /^ o N\ 2c_ o t f _ _ _1_ o _ V_ V dt J u) dt 20 dt 20

from which the equation for the zero-slope isocline is

obtained.

-¥- 6ft - - o " 20

(See Figs. (2-4) and (2-5)).

3 . Discussion

a. Any system which has a differential equation

like Eq. (2-12) can use Lienard's method to construct a phase-plane direction field. After the direction field is constructed, one can draw any phase trajectory from a given

initial condition on the phase-plane.

b. As compared with the isocline method in Sec-

tion A. If the differential equation can be manipulated

to (2-^12) form, a lot of labor may be saved, and one gets

the same result. It seems better than the isocline method

in this particular problem.

c. If C is not a constant, then the zero-slope

isocline will not be a straight line.

d. Lienard's method (even for drawing an indivi-

dual trajectory) is not applicable to all second-order

non-linear differential equations.

32 V ^0

4 *. Zero -slope isocline G *-V/20

Fig. 2-4. Lienard's construction of the phase-plane direction field.

\\\\\\\\V» N \\\\\\\\N *

Fig. 2-5. Direction field obtained by Lie'nard's method.

33 C. DELTA METHOD

1. Method Description

Given

x + G(x,x, t) = (2-13) where (x, x, t) may be given analytically or graphically and may be a very complicated function.

Rewrite Eq. (2-13) to the standard delta form. This is

2 done by adding and subtracting the term p x from Eq. (2-13)

2n 2 x + [G(x, x, t) - p x] + p x =

Let

2 2 p 6 = [G(x, x, t) , p x]

so that the linearity difference 6 becomes

2 6 = \ [G(x, x, t) - p x] (2-14) P

Eq. (2-14) will make (2-13) appear in the standard fi form

2 x + p (x + 6) =0 (2-15)

The geometric significance of (2-15) in a phase-plane representation is seen by introducing the phase-plane coordinates

x = x and — = v. P

Then

dv dv . 2 dv X = = X = V Pdt P dI P dx-

34 The standard second-order equation (2-15) then becomes a first-order differential equation in x, v, and fi, namely

v|+x+6=0 (2-16)

• ^Z = _ Ix±6Jl (2 _ 17) dx v

Fig. 2-6 shows in this case that the slope of the line x+6 QP will be ; consequently the equation of the normal

to QP will be identical with Eq. (2-17) . The instantaneous phase trajectory is therefore a circular arc passing through

P and with its center at x = -6 or at Q.

The next problem is therefore to relate the circular- arc segment beginning at P to the finite time interval At

Since x v = — P then

dx = pv dt

= — (2-18) dt pv

Fig. 2-7. Let the average radius of curvature of the step be given by QP = p.

35 Normol to OP P

Fig. 2-6. Exact relationship,

Fig. 2-7. Stepwise relationship

for constant p.

36 .

Then

d9 = %

-i 2 dx + dv 2

2 v +(x+6) 2

dx v

.*. dt = - d© (2-19)

Consequently, an integration will give

t - t = ^(9-e o p o )

pAt = A0 (2-20)

Eq. (2-20) shows, in regard to the angle subtended by the circular arc described by the phase point during the finite

time interval At .

A graphical integration of the differential equation in the standard delta form can then be effected by a successive construction of circular arc segments whose centers are located on the x-axis at various step value of x = -6 (see Fig. 2-8a)

37 r

(a) Change with positive 6. (b) Change with negative fi.

Fig. 2-8a.

38 . )

If x had been taken horizontal and positive to the right with v positive vertically, positive time would have been represented by a clockwise angular variation. (see

Fig. 2-8b)

The definition of 6 given by Eq. (2-14) transforms the original Eq. (2-13) into the standard delta form. Curves representing 6 in terms of the phase coordinates x and/or

— will be superimposed on the phase-plane so as to permit choice of a suitable average value of 6 for the construction of the circular-arc segments making up the phase trajectory.

2. Example (II-B-2) , Mass on Non-linear Spring

The motion of a constant mass attached to a nonlinear spring which becomes relatively stiffer with increas- ed deflection is described by an equation of the form

2 ^-~ + 25(1 + 0.1 x )x =

Find a phase-plane solution curve for this equation by

6 method, with the initial condition that at t = 0, x = 3 and -rr = 0. dt Comparing to the standard delta-form (Eq. (2-15)

2 2 p (x+6) = 25(1+0.1 x )x

2 p =25OC

3 6 = O.lx

39 - 4 T P

• X

(a) Change with positive 6. (b) Change with negative 6,

Fig. 2 -8b.

40 .

First draw the § curve. Phase trajectory construction

proceeds as shown in the figure. The initial point is

x(0) = 3 and v(0) = 0, as the first step. It is assumed

that x decreases to the value 2.8 or Ax = -0.2 as shown.

The average value of x during this interval is x =2.9,

for which the average value of 6 can be read directly as

6 =2.4 approximately. Thus the center of the first

circular arc is located at point x = -6 = -24 and v= 0; * av • the radius is R = 5.4 and with this radius an arc is drawn

to the point where x = 2.8, indicated by the small dot marked x(0) + Ax in the figure. By continuing in this manner, the entire solution curve can be built up as a se-

quence of circular arcs. (Since the original equation involves no damping, the solution is a closed curve repre-

senting a periodic oscillation. (See Fig. 2-9)

3. Discussion

a. The delta method is limited to construction of

phase trajectories for second-order differential equations.

b. In starting a solution by the isocline method,

the entire plane must be fitted with line segments fixing

the slope of a solution curve. If only a single solution

curve is needed, only a few of these line segments are

actually put to use. Considerable simplification would

result if only information related directly to the desired

solution curve were used.

The delta (6) method leads more directly to the

desired solution.

41 Fig. 2-9. Phase-plane diagram constructed by fi method,

The curve for non-linear function fi(x) and the final solution curve are shown. Dots indicate junction of successive arcs used in building up solution.

42 c. The construction of phase trajectories will be carried out for various magnitudes of steps. So the accuracy may depend on different amounts of constructional work.

4. Notes

a. 6 may be either positive or negative. The following 6 curves for five systems may help to understand how plots of curves may be superposed on the phase-plane.

(1) The physical pendulum (See Fig. 2-10)

2 x + p sin x =

I o 6 = —j [G(x)-p x] = sin x - x P

(2) A system with restoration proportional to

the cube of the displacement (See Fig. 2-11)

x + x 3 =

o = —p X - X p

(3) A linear restoration system with quadratic velocity damping (See Fig. 2-12)

x + q|x|x+px =

6 = —5- [G(x,x) - p x]=-^r |x|x=q|^| P ^P P P

43 >. *"(sin;r-jr) •

/I/

Fig. 2-10. The operative displacement physical pendulum plotted in the phase- plane. Centers and correspond to points P and Q.

.— -~ y

Fig. 2-11. The operative displacement b for a system with restoration proportional to the cube of the displacement.

Centers and C" correspond to points P and Q.

44 x

(4) A system with positive linear and with negative quadratic restoration in x (See Fig. 2-13)

x + px-a|x|x =

= - - - —~ 6 —j(p x a I x I x p x)= I x I P P

(5) A system acted upon by nth -power velocity damping and restored by first and mth-power displacement restoration (See Fig. 2-14)

•• • I I «1 -n-1 I 2 a m-1 _ ff x + |x |x + px+ pri^x. | x =

2-n v p p 2 ' P ^ * P

6 + 5 = 1 2

(b) The b curve of a system indicates the system's type and degree of non-linearity. Thus, if a system is linear, 5 will be zero unless an artificial parameter has been introduced deliberately.

D. SZEGO METHOD

1. Method Description

Consider the second-order system:

(2-21)

45 Fig. 2-12. The operative displacement 6 for a system with quadratic velocity damping. Centers and correspond to points P and Q.

Fig. 2-13 The operative displacement 6 for a system with positive linear and negative quadra toe restoration in displacement. Center corresponds to point P.

46 Fig. 2-14. The two operative displacements

6, and fi~ for a system with damping proportional to the nth power of the velocity and with restoration proportional to the first and the mth power of the

displacement. Center o corresponds to point Q.

47 .

It is desired to identify the slope of the trajectory of this system at every point of the phase space. We consider the topographical system; that is a system of non- intersecting differential curves, which are symmetric with respect to the origin.

v = 0(x,y) (2-22)

In connection with the function (2-22), construct the curves defined by the equation

2 - + - ° (2-23)' &dt IBx &dt ¥by f£dt x and Eq. (2-21)

These curves we called contact curves. Substituting Eq.

(2-21) into (2-23)

r>v dy _|x_ _^t_ f(x = _ _ f y) dv dx y (v ) By dt

At every point of the curve (2-24) , the corresponding trajectory of the system (2-21) has the same slope as the curve of the family (2-22) at the same point; furthermore, the curves, representing in the phase-plane the solutions with odd-order of multiplicity of Eq. (2-23) divide the phase- plane into regions in which -rr has fixed sign. On the con- trary, -rr has the same sign on both sides of the curves, representing the solution with even order of multiplicity

48 . .

of Eq. (2-23) . Of course, in the latter case the slope of

the trajectories at every point on these curves is the same as the slope of the family (2-22)

In regions where -rr is negative, the representative point of the system (2-21) describes a trajectory in the phase-plane which crosses the curves of the family curves

(2-22) in the direction of decreasing v.

In the regions in which -tt is positive the trajec-

tories cross the curves of the family (2-22) in the direc-

tion of increasing v.

Choosing different families of curves (2-22) and plot-

ting the corresponding curves -rr =0 in the phase-plane, provides enough information to sketch the trajectories.

If the scalar function, v = 0(x,y) changes sign in the phase-plane, i.e., if the equation v = 0(x,y) has some real non-zero solution with odd-order of multiplicity, then the curves representing these solutions must also be considered

as boundaries between regions with different sign of -rr •

This must be done in order to take into account the varia-

tion in sign of -jr , due to variation in the sign of

the curves of the topographical system (2-22) . Notice that

the trajectories of the system (2-21) on the curves v = have not the same slope as curves of the family (2-22)

Consider the particular topographical systems:

v = x (2-25)

49 for which

dv = dt Y (2-26) and

= (2-27) y1 y for which

dv, f(x dt " ' y) (2-28)

The subsequent families of curves are now considered as topographical systems.

2 2 v = x + (2-29) 2 y

2 2 v = x - (2-30) 3 y

v = xy (2-31) 4

Using v, = y when v, = for y = (2-32) and from

dv Sv l **l dx l dy dy dt dx dt Sy dt dt then

f(x,y) = (2-33)

50 2 2 When considering v„ = x + y ; it is seen that v_ = for x = y = and

dv = dt from

dv _ 2 dx , dy n - " = 2x + dt —dt 2yJ dt

= 2xy + 2yf(x,y) =0 then

x + f(x,y) = and

y = (2-34)

For

2 2 v = x - 3 y

v = for x = y

and x = -y (2-35)

Also

dv_ = dt ,

But

dv„ Bv , av_ , 3 .. 3 dx _ 3 dy dt Bx dt By dt

„ dx « dy dt J dt = 2xy - 2yf(x,y) =

51 Hence

X = f (x,-y) and

Y = (2-36)

Finally, considering v. = xy, gives

v = 4 for x = and y = (2-37) and

dv = dt ,

But

dv. dv Sv. , 4 dx 4 dy_ 4 . dt ~ dx dt By dt

dx , dy J dt dt

= y • y + x f (x,y)

Hence

2 y = -xf(x,y) (2-38)

2, Example (II-D-2)

Consider Van der Pol's equation:

x = y 2 y = f(l-x )y - x (2-39)

52 dv. X (a) — = for y = 2 1 - x dv (b) ^ =0 for y = x = ± 1 dv 2 There is no variation in the sign of ,. between the two

2 sides of the curve y =0.

dv., (c) -r— =0 for y = dt 2x

1 - x

2 2 2 (d) ^=0 for y = | [(x -l)iy(x -l) + 4]

The next step is to plot the curves (a) , (b) , (c) and

(d), (see Figs. 2-15, 2-16, 2-17 and 2-18).

In order to use the method, it is necessary to super- impose the curves drawn in Figs. 2-15, 2-16, 2-17, and 2-18 as is shown in Fig. 2-19. dv. The best procedure now is to check the sign of ,.

(v = 1,2,3,4) numerically at one point of the phase-plane dv. and utilize the result for identifying the sign of ,.

(i = 1,2,3,4) in the different regions in which curves dv. -rr— = and v = 0(i = 1,2,3,4) divide the phase-plane.

Since

< v < + •

_oo < v < + °° 1

- 00 < V, < + °°

-« °° < v4 < + dv, dv..

It is necessary, before checking the sign of -,, , -., , dv4 ac ac and , at one point of the phase-plane, to identify the sign of v, , v_ and v. at that point.

53 v = for y = H dv 1 X — n for£ dt = 1-x'

M fro v,«o

generate regions M,N,T,Q,R and S

Fig. 2-15. Curves v, = 0, dv. and dF

54 dv, dT = for

4 y =

x = ± 1

• I- t II—— H 1 1 1 1 h- -I + 1

P generate regions

a, fi and v

A, 4 dt dt dv. Fig. 2-16 Curves v»= and -rH = 2 dt

I \ r

i V = for x = y \ i 3 X / / dv. I / 2x ^ \ = for y = dt~ : r H 1/ l-x %. $«> SI B generate regions v A,B,C,D,E,F,G,H, /\ c I,j, and K

I *• *I V^O I :\ t "3-0 A I. 4-o

at3«o dv. Fig. 2-17. Curves v = and -ttt = 0. J dt

55 v, =

for x , y = dv dt

2 for , 2 y = f |_(x -l) iv (K -l^ + 4

generate regions a,b>cd,e,f,g, and h

= 0.

VV$3a0 *-X

-«•

Fig. 2-19. Example (II-D-2)

56 Some information about the slope of the trajectories at every point of the phase-plane is thus provided. For example, consider the point P(-3; 1.5) in Fig. 2-20),

dv. from < , requires that the trajectory move into dt the region A'F'D'B*

dv. from dF > , the trajectory moves from P into the region E'C'A'F'

dv i from < , requires that the trajectory move into dTF the region C'A'F'D'.

These three conditions, taken simultaneously, define

the angle A'PF* in which the trajectory which originates at P must lie.

Since the slope of the trajectories is known exactly dv. at the points on the curves ;. = , it is possible to draw the trajectories starting from P with good accuracy.

(See Fig. 2-19)

Starting for example from point #(-3,1.5) contained dv.

in the region of fixed sign of , , namely T, a, I and G,

(See Fig. 2-19)

dv. < dt

dv. dF <

57 O • ii Ch

1 3fe JJ

in •H

c e •H JJ M fO M-l H i c M 0) o en JJ o ^fa JJ o •n fO fO X 5-1 o w JJ II \

o 1 II '

' slfc * I -a hi ^J& fa

58 dv. dF >

dv dF <

NOTE

dv. dF < in T, Q, M and S

dv. dF > in N and R

dv. dF < in oi and y

dv. dF > in j9

dv. dF < in B, D, F, H, J and L

dv. dF > in A, C, E, G, I and K

dv, > in b, d, f and dt k

dv, < in a, c, e and dt g

3. Discussion

a. This method is a simple and useful method for

drawing trajectories. Also it is essentially a generaliza-

tion of the isocline method.

59 b. Accuracy can be improved by considering further topographical systems.

c. It is worthwhile to notice that the isocline method is a particular case of this method. And it corresponds to the particular topographical system:

v = ax + a 2 y

(-•;i». J!+»;isl l 2)

NOTE:

dv _ jVv dx _>v dy_ dt ; ^x dt ^y dt

dx dy_ _ a + a " l dt 2 dt

dy_

dy dt . :: _ fl dx dx a„ dt

E. MURTHY'S NEW APPROACH TO THE PLOTTING OF PHASE-PLANE

TRAJECTORIES

1. Method-Description

This method is a new and simple approach for the isocline plotting of second-order time-implicit non-linear differential equations given by x = F(x,x').

) (x) = a. Isoclines of the equation x"+f (x ' +g 0:

x" + f(x') + g(x) = (2-40a)

60 ' .

Substituting Mx ' = x" into (2-24a)

dx -= where M = ; gives

Mx' + f(x') + g(x) =0 (2-4Qb)

(1) Plot from the given differential equation

(2-40a) the curves of x" vs x ' on the x"x' plane for constant (different) values of x. This is a simple matter since it amounts to shifting the curves of x" = f(x') vertically for different values of x.

(2) Draw a number of straight lines x" = Mx' through the origin (corresponding to the different values pf M) on the x"x' plane; this can be done quickly.

(3) With any particular straight line in

view, say for M = M, , determine all possible points of intersections of this straight line with the curves obtained from (a). Thus a number of values of x and x' can be

obtained corresponding to M = M, , which when plotted on the x'x plane and joined together represent the isocline for

M = M, . The procedure is repeated for M = M„, M_. ... until a fair number of isocline curves are obtained on the phase plane. The respective slope lines are marked on these isoclines. Joining these slope lines gives the different trajectories or the entire phase portrait for Eq. (2-40a)

(b) Isoclines for the equation

x" + f(x')h(x) + g(x) = (2-41)

61 :

The above procedure can still be applied except that the

curves of x" = f(x') are now multiplied by a scale factor

[due to the h(x) term] corresponding to different values

of x.

(1) This method is more convenient in deter-

mining the solution curves for the following differential equations

x"+ x'h(x) + g(x) =

x" + x' + g(x) = (2-42)

x" + xx' + g(x) = when compared with existing methods.

(2) This method can also be used for the following nonlinear differential equations:

h(x)x" + K(x) f(x') + g(x) =

h(x')x"+ K(x) f(x') + g(x) = (2-43)

x" = F(x, x')

(3) The Lienard method is a step-by-step con-

struction for drawing any particular trajectory correspond-

ing to a given set of initial conditions. The present method determines the entire phase portrait indicating the behavior for all initial conditions.

(4) Variations of the nonlinear restoring-

force term g(x) can easily be studied.

62 (5) The method is independent of the nature of the functions f,g,h,k, i.e., whether logarithmic, algebraic, or transcendental.

(6) For nonlinear differential equations of the type x" = f(x), the x"-x' curves become horizontal straight lines for constant values of x. The intersections

of these straight lines with x" = Mx ' lines give the points on the phase plane for any chosen isocline slope.

(7) The isoclines corresponding to M = and

M = °° can generally be plotted quickly.

(8) Sometimes an x"-x' curve (for a constant x) and an x" = Mx' line may run so close to one another that the isocline curve on the phase plane is nearly vertical in that range of x'.

2. Example

Consider a system with nonlinear damping and re-

2 2 storing forces described by x" + x 1 + x =0.

(a) Draw x"-x' curves (see Fig. 2-22).

(b) Draw Mx ' = x' corresponding to various values of M (a number of straight lines through the origin.

(See Fig. 2-22)

2 2 x" + x' + x = 0.

63 = lines for Fiq. 2-22. x"A' curves and x" Mx ' 2 2 the equation x" + x* + x =0.

64 Considering the line for, say M = + 5, the points of intersections are determined as

(x = , x' = 0)

' (x = 2 , X = -1)

(x = JT , x' = -7)

(x = Js t X' = -1.4)

' (x = Js , X = -2.0)

(x = J6 , x' = -2.5)

These points are marked on the phase-plane, joined and the

slope lines are marked on it (Fig. 2-23) . Similarly, different isoclines are obtained for different values of M.

The trajectories are then drawn as shown in Fig. 2-23.

3. Discussion

(a) This is a simple, all graphical, useful and general method to obtain the phase-plane trajectories corresponding to all pertinent initial conditions of second- order autonomous systems (which are not amenable to the usual isocline method) by making use of the isocline method. It reduces the labor of calculating isoclines for many non-linear differential equations and thus makes the isocline method a much more useful tool.

65 V +*+++ >+kH^co

Fig. 2-23. Isoclines and phase trajec-

2 2 tones for x" + x' + x = 0,

66 F. THE ACCELERATION-PLANE METHOD

1. Me thod-De sc r ip tion

It is based on the fundamental equation

= ~ (2-44) ^dx v , where w = w(t, v, x) or the acceleration is a function of t, v and x. For the sake of simplicity, assume that a constant driving force is applied.

Thus dv w(v,x) = (2-45) dx v

The method calls for preliminary plots of w vs x for constant values of v. This auxiliary w-x plot for constant values of v is not a trajectory. It characterizes the system just as the original differential equation characterizes the physical condition of the system.

This method is applicable to both linear and non-linear problems. In the w-x plot, though constant driving functions can be included, the time-varying forcing or driving functions must be added for different values of time.

2. Example x" + . 5x ' + x = 0.

For v = x' =0 , then x" = -x , and the x" - x plot or w-x plot is a straight line with a slope -1 .

For v = x* =1, then x" = -x -0.5, so this straight line is displaced from the first line by -0.5.

In Fig. 2-24 (a) six points of construction are shown for the v-x trajectory, starting from P where t = 0, x =2.5 and v = . o o

67 T3 O rC 4J CD 6

CD C rQ ro

i c o -H -U rC HCD Q) U U «J

to CD •H O -U O

JJ m o c o •H JJ U

JJ CO 05 c O U

I CM

01 •H P4

68 Taking x, as the center, and starting from w. , an arc can be drawn in the clockwise direction so that the arc strikes R, on the horizontal axis. From R, draw a line to

P, where x = x and v = v. ; the perpendicular line to

P-.R-, gives the correct slope for the v-x trajectory at P.. .

If the perpendicular line gives a slope that agrees fairly well with the assumed slope, point P, can be considered as

the desired point which falls on the v-x trajectory. W Slope of perpendicular line = l V l If the assumed slope does not agree with the slope of

the perpendicular line obtained from the above construction,

a new slope should be assumed and the above procedure repeated until the assumed slope agrees with the slope of the

perpendicular line (see Fig. 2-25). On Fig. 2-24 (a), P.

has the value x = x, and v = v, and v, is chosen as -0.5.

Thus if one prolongs the vertical line from x.. downward

through P,, this line should intersect with the v = -(h)

line in w-x plot (shown in dot lines) at a point w.. . By

connecting the points w , w. , w_ ... etc., the w-x trajec-

tory is constructed at the same time (see Fig. 2-24(b)).

3. Discussion

This method is applicable to both linear and non-

linear problems. One draws the w-x trajectory first, then

gets the v-x phase-plane trajectory. Each time only one

trajectory is obtained, so this is a useful method.

69 Fig. 2-25. Acceleration-plane method,

70 G. DEEKSHATULU ' S METHOD

This method suggests the use of simple transformations

for second-order non-linear differential equations to effect

rapid plotting of the phase-plane trajectories.

1. Method-Description

a. Transformations for second-order autonomous

systems

A general second-order non-linear autonomous

system is governed by a differential equation of the form

x = f(x,x) (2-46)

Let dx y = dt

then

2 d x _ dy_ _ dy_ dx ,,2 dt dx dt dt dx = M dt

Change Eq. (2-46) to

Mx = f(x,x) (2-47)

To draw the isocline for an assumed value of M, equal

to M, , say, x has to be found as a function of x and M, and

this in many cases is not possible; thus the only attempt

made is to find the points of intersections of such an

isocline with a given set of straight lines or with simple

71 .

standard curves. Consequently, on the phase-plane, if one considers the intersections of the isocline for slope M = M, and the set of straight lines x = Kx (K = ± 1, K = ± 2, ...),

Eq. (2-47) becomes

JV^Kx = f(x,Kx) or

M = F(x,K) (2-48) 2

For a specified K, the solution of Eq. (2-48) (algebraic equation with constant coefficients) for real roots of x gives the points of intersection. But it is easier to find the values of the slope M (for various values of x) on the lines x = Kx than it is to try to obtain the isocline curve for a given value of the slope. Even under this transformation x = Kx, every quadrant of the phase-plane is uniquely defined. It is advisable to find any symmetry present in the phase portrait for a given differential equation since, if it exists, it reduces labor and is tested as follows:

(1) Symmetry about the x-axis

If for K = positive (negative) K, and

x = positive (negative) x. , the slope M = M, , then for

K = negative (positive) K.. , and x = positive (negative) x^, the slope M = -M,

(2) Symmetry about the x-axis

If for K = positive (negative) K, and x =

positive (negative) x, , the slope M = M , then for K = nega- 2 tive (positive)K, , and x = negative (positive) x_, the slope = M -M2 . 72 The procedure for determining M, on the phase-plane

for any chosen value K. and x.. can be viewed in the phase

space (with x, x, and x as coordinate axes) . Determine the curve of intersection of the surface defined by Eq. (2-46) with the plane x = K, x, then determine the simple point of

intersection P of this curve with the plane x = x, . The

slope of the plane passing through the x-axis and the point

P with x-x plane determines the value of M, . Similarly, other values of M are obtained by taking different values of x and the procedure is repeated for each value of K.

Other transformations such as x = Kx , x = Kx , x or x = K, etc., can also be considered.

This procedure is general, systematic, and comparative-

ly simple when compared with the method of determining slope

lines arbitrarily at points for the phase portrait construc-

tion.

b. New Planes for Certain Second-order Time-

Varying Systems.

Time-varying elements occurring in physical

systems; for example, the variable mass of a guided missile

can be though of as function of time since its weight decreases as the fuel is consumed.

On the phase-plane, solution of this kind of problem makes use of step-by-step procedures by considering simul-

taneously the x-x and x-t (solution) planes. Professor

Deekshatulu suggests some new planes whose coordinate axes

73 are explicit functions of time and which effect an easy and rapid solution of at least certain classes of second- order non-linear time-varying systems not easily amenable otherwise by the phase-plane methods.

c. Suggestions for Solving Certain Higher-order

Systems

Assuming non-linear transformations of the form

£ = e + af(e) or t? = e + bf(e)+ ae (2-49) third and fourth-order time-varying systems described by the following equations:

2 t e + \ + f (4) = o

2~€i + \ + f U) = o (2-30)

t t? + tf?+ f (n) = o

\ 2trj + f? + f(r?) = o can first be solved by the method described and then be solved as a first-order or second-order nonautonomous system. Note that the equation of the form e + af(e) = h(t) can be solved on the e-t (solution) plane.

2. Examples

(Example 1) Consider a non-linear system with dis-

sipation proportional to x , given by x + hx + x =

74 Let

dx = dt Y

then

2 d x _ dy_ _ dy dx _ dx : ,,2 dt dx dt dt dt

Thus .2 x + hx, + x

= Mx + hx + x = or hx + Mx + x =

Both roots of this equation are valid. On the other hand,

if one substitutes x = Kx this equation becomes

2 h(Kx) + MKx + x =

-1-K2 hx M = K

which is simpler to plot for any given values of h and K.

Since the phase trajectories are found to be symmetrical

about the x-axis, only the upper half of the portrait is

shown in Fig. 2-26 for the case h = 1.

( «-^ for h = 1 <

( x = Kx

75 X + •X +

u mo w H

-U

(D w rd

tn -H fa

76 .

etc.

(Example 2) Let it be required to find the behavior of the following time-varying system:

2 3 t x + tx + x = for all pertinent initial conditions (with t 4 0)

The task would be enormous if it were to be attacked by the phase-plane method. Using the tx-x plane the differential equation reduces to

3 = x tx y = - or

Note:

Let

tx = y then

d = = Mx = x + tx dt£ ^^dx dt

2 3 Mtx = tx + t x = -x or 3 tx = y = - —

77 Taking different values for M the isoclines (enlarged

or reduced scale plot of the cubic curve) are drawn and the

respective slopes marked on them with their direction (see

Fig. 2-27) .

3. Discussion

2 a. Simple substitutions like x = Kx, Kx ,

o o o f* x + x — y in the isocline equations 6f ; second-order sys-

tems simplify plotting of the phase trajectories to a great

extent. The method is a useful supplement of the isocline

. method. , .,. ,

b. Since there are not many methods to evaluate

quickly the bahavior of second-order non-linear time-vary-

ing systems for various pertinent initial conditions and/or

for different-.- non+OLinea.'r characteristics, Deekshatulu

- suggests; in- general time-varying coordinate planes, the

. 2

use -of tx-x and tx ^x planes, .f.p.35 ;j obtaining.easily the solu-

tion of certain classes of, nonlinear* time-varying systems.

r C H. PELL'S METHOD ~ -f-

1. Method-Description

Quite generally, second-order nonlinear autonomous

equations of the form

d x , dx N = n n — 2 + 9(*< dt> dt

. may be written as f

' + w<* (#)v + f (x) = 0. (2-31) ,.2 dt dt io

x ... m , 78 79 .

Let

= dx y dt then

2 d x _ dy_ dy_ dx

: ,.2 dt dx dt dt

Thus

2 ^-f + (~) + f (x) dt c

./dx, , _ , . _ dy dx , = * + + f(x) = dxj T7Tdt 0("jr)dt or

-<*($:) - f(x) dy _ dt v ' dx dx dt

f til ~ M (2-32) y

Now, as shown in Fig. 2-28, plot -6(y) versus y and

-f(x) versus x. Given the initial conditions x(0) and y(0), the construction of a segment of the trajectory through the initial condition proceeds as follows:

a. Move vertically from IC and find f[x(0)] ; lay this off on the x-axis from x(0) by use of a compass or a 45° triangle. This establishes point A.

b. Draw line AB, where OB = y(0)

c. Find -#[x(0)] by moving horizontally and draw

CD parallel to AB.

80 d. Now the hypotenuse (D) (IC) of a right triangle whose base is the distance DE = #[y(0)] + f[x(0)] and whose altitude is the distance (E) (IC) = y(0) is established.

The slope of the hypotenuse is

(E) (IC) y(o) _ f2 _ 33>JJ/ DE " 0[y(O)3 + f[x(0)] U

e. The negative reciprocal of Eq. (2-33) is the

slope required in Eq. (2-22) ; thus a line perpendicular to

the hypotenuse, passing through the initial condition forms a segment of the required trajectory. The usual precautions must be observed as to the length of the extrapolated segment.

Note: The major advantages of this method are:

(1) Its simplicity. Only two triangles are needed to carry out the solution once the two functions are plotted.

(2) It eliminates the trial and error of the delta method.

2. Example

Consider the equation (the hydraulic speed control

system)

2 2 2 2 + Cl + U (l+a e )e + S| = (A) ^-f2 1 |fdt C dt ° 2 where

c, = viscous damping coefficient

81 c„ = inertial coefficient related to volume of nozzle chambers

K = gain of amplifier

Ui = natural frequency of vibration of flapper for

small amplitudes.

The equivalent flapper spring is assumed to be linear

Eq. (A) may be rewritten as

2 d e . de ,, 2 ,,2 2, _ ,_. c , M e )e = (B) —2 l dt 1 ^ dt

Let T = o>, t to change time scale; thus Eq. (B) becomes

2 c _ , ,, 2. d e de , ,,2 ,_. + +(1+b e )e = (C) —2 1Z d? ° dT 1

Choose the following more or less arbitrary set of coefficients designed to emphasize the construction tech-

nique :

c, = 164 U) = 377 1 o

2 c = 0.01 a = 0.858 2

K = 1000 on a consistent set of units. Then

2 2 K + C % 2 2 b 2 = o 0.58 2 *i

2 a = 0.5

fi 0.33

82 Let

x = e

Eq. (C) becomes

2 dy .. -0.33y -(1+ 0.58x )x dx v

0(y) and f (x) may be identified and plotted and the construction completed as shown in Fig. 2-29.

3. Discussion

This method is satisfactory:

a. If only a single solution curve is needed.

b. Only a few of these line segments are actually put to use.

I. SOMAYAJULU'S METHOD

In this method a graphical procedure is suggested for

drawing phase-plane trajectories making use of tangent

slopes instead of normals (i.e., the negative reciprocal of

tangent slope values) as is done now in the Lienard con-

struction.

1. Method-Description

A straightforward construction is presented for

more general cases to which the isocline method is not

applicable and construction of individual trajectories is

required.

83 \/C[x (0), 2.(0)]

Fig. 2-28. Pell's Method of phase .-trajectory construction.

CB=AD

Fig. 2-29. Example construction using Pell's method for the hard-spring system.

84 a. Let the slope of a second-order system be given by dx x M = -z— = — dx k

f (x) + f (x) 1 2 (2 " 34) (x) + (x) g x g 2

Assume f , f„, g, , and g„ are all continuous, single-valued 1 and of the odd characteristic type; these curves are first drawn on the phase-plane.

Referring to Fig. 2-30, the initial point P(x ,x ) is

marked. Make PQ = f .. (x ), QR = f (x ). Similarly the ?

length RST is made equal to g, (x ) + g„ (x ). The angle PRT

is a right angle. Note that the horizontal line RST is

drawn to the left of PR since the slope is known to be po-

sitive from the nature of the functions f -. , f~, g. , and g~

and the initial conditions x and x . A small length of the o o J straight line joining P and T gives the required slope at P.

The field direction is marked on the basis that with x posi-

tive x should increase. It is seen in the procedure that

it is necessary to first determine the direction of the

normal to the phase trajectory.

b. Consider equation

x + f(x) h(x) + g(x) = (2-35)

This equation is considered to be of practical impor-

tance because of the presence of nonlinear damping and

nonlinear restoring forces.

85 86 Let

dx , y = he and rewrite the equation (2-35) in form (2-36)

f(x) h(x) + g(x) M = _ (2-36) x

The plots of f(x), h(x) and g(x) are shown in Fig. 2-31.

• 2 2 Also a parabola x or x (only one half will suffice) is drawn in addition to the above curves.

Let the initial point be at P (x , x ) ; r o o

then:

OS = x , OQ = x o o

QR = h(x , SB = f (x ** o ) o ) and SA = (x g3 o ) Make SM = QR.

Draw a circle with BM as a diameter and let it intersect

the x-axis at the points C, C.

Then

SC = Jf (x ) h(x ) o o

With the help of the parabola the squared length of SC

is obtained and is marked as PN.

Make NL = g(x 3 v o )

and LK = x o

87 w •H •H tO -U Q) T3 C O •H JJ U M JJ to a o u

•H

88 .

The segment line through P joining P and K represents the

required tangent slope of the phase trajectory at P. The

direction of motion is decided to be downwards. The new values x, and x, at the end of this segment are determined

and the procedure repeated.

Note: (a) In both the above mentioned methods it is

necessary to take equal scales on the x and x axes.

(b) The accuracy in plotting is enhanced by

considering small segments.

2. Example

x + x + x = with initial condition x = -4 and x = o o 2 Rewrite x + x +x=0 to the form

- M = -x x = x+x X -x

and draw

f = x f = x 1 2

= -x q 1

on the phase-plane (see Fig. 2-32) (using Somayajulu's

method as in I-l-a)

3 . Discussion

This method is based on geometrical relationships,

so a precise step-by-step procedure must be followed.

89 90 J. THE E-FUNCTION METHOD

1. Method-Description

^-| + f (|* , x) + g(x) = h(t) (2-45) dt

Let the Kinetic energy and potential energy be defined as

2 [%)x and Q(x) where

«x Q(x) = g(x)dx. (2-46) o

Define the total energy as

2 E = -|(x) + Q(x). (2-47)

Differentiating E with respect to time

DE dx d x dQ(x) _ o dt dt" ,,2 dt dt (2-48)

E = x • x + Q(x)

Differentiating E with respect to x

dE _. d dQ(x) (X) + dx dx L 2 _ dx

dx [ d dx, I dQ(x) f (2-49) dt L dx^dt ; J dx

= x + g(x)

91 From Eq. (2-49)

2 d x dE . . — = " g(x) dt 2 dx"

Substituting this into Eq. (2-45) yields

- x) = || g(x) + f(|f , + g(x) h(t). or

|| = h(t) - f (x, x) (2-50)

From Eq. (2-47) , an E-x plot can be made for different values of x. These curves may be called equi-velocity curves, as x = v, and they represent different kinetic energy levels. For x = 0, the E-x curve is simply E = Q(x)

where Q(x) represents the potential energy. At x = x, , the difference between any two E-x curves, say, between the curves for v = v, and v = v is equal to the difference 1 o ^ 2 2 (v, - v )/2. o The value of v = x is, from Eq. (2-47)

x = v = ±

Substituting Eq. (2-51) into (2-50),

|| = h(t) -f["x, +V2[E-Q(x) ]] for v > (2-52)

|| = h(t) -f[x, -72[E-Q(x)]] for v <

92 . .

The E-function method is thus a method for construct-

ing the E-x trajectory according to Eq. (2-50) . See Fig.

2-33, the slope of AB is equal to h(t)-f(x,v ), where

V + V l 2 v = x av 2

X + X 2 . . l ~ and x should be corrected to , even though it is initially taken as equal to x,

2. Example: Van der Pol Equation x-(l-x)x+x=0o

E-x plot (see Fig. 2-34)

From E-x plot get the v(v = x)

Note: As compared with the isoclines and trajectories of the Van der Pol equation obtained from the digital computer, errors exist in this method. (See Fig. 2-35)

K. THE SLOPE-LINE METHOD (See Fig. 2-36)

Given an equation

f = '« or dy = f(x)dx, the incremental relations may be written

Ay = f Ax (2-53) * av f + f l 2 where f = represents the average value of f (x) av 2 ^ during the interval.

Thus - A Ax - , Ax Ay = f + f — l T 2 (2-54) = A + Ay Yl 2

93 .

AO»l CO«h(t)-| (X .-V5-)

Fig. 2-33. The E-function method,

•** -t -t$ •! S a -s i »s « a

Fig. 2-34. Example (II-J-2)

94 Ax Let an angle 6 be chosen such that tan 9 = -~- , for a chosen value of Ax, then

Ay, tan = —-±- r l

= (2-55) -^r 2

Starting from an ordinate f. , make an angle and draw a dotted line be to intersect the horizontal line at c.

Then ac = Ay. From c, draw another dotted line, making the same angle with the vertical line erected from c, until another ordinate f_ is intersected. As shown de = fp can be projected from the f (x) curve on the left of the figure.

Then ce = Ay„, and ae = Ay, corresponding to the increment Ax shown on the left of the figure.

This method can be applied to the solution of simulta- neous first-order differential equations, linear or nonlinear. For a first-order nonlinear equation

dy + g(x) = h(x) (2-56) dx the incremental relations give

Ay = [h(x) - ]Ax (2-57) g(y) d V

95 isoclines

and trajectories

-I ; Mill * -1

Fig. 2-35. (Example II-J-2)

Fig. 2-36. The slope-line method,

96 For example, in a problem of surge tank transients, the equations are of the following form:

- f|= y + f (u) (a)

|* = v(T, y) - u (B)

When the equations are in normalized form both the horizontal and vertical systems of slope lines can be drawn with the same slope. Eq„ (A) represents acceleration.

Eq. (B) is obtained from the condition of continuity.

L. COMPARISON

It is readily seen that some methods are easier than others but may be of limited usefulness, applying only to some particular problems. Some of them are useful but complicated, so that it is easy to make a mistake. It seems

the most useful and general method is the isocline method,

when one solves problems by hand labor. Deekshatulu ' s and

Murthy's methods are useful supplements. The difficulty of the isocline method lies in the labor required to get

the isoclines. So if we can use the digital computer to generate a program, call upon it to draw the isoclines and phase trajectories on the same phase-plane, then the isocline method is more useful and general than other methods,

For time-varying systems, the Delta method and Pell's method are useful. Deeskshatulu" s transformation method

(by using a new plane) is a useful supplement.

97 III. THE APPLICATION OF PHASE-PLANE METHODS

Phase-plane analysis of nonlinear system problems not only gives a convenient display for interpreting results, but also permits use of different techniques for shaping the phase trajectory to make it satisfy design requirements.

So, it can be said that the phase plane has become a logic- al tool in the solution of problems.

Generally, phase-plane analysis is limited to second- order systems. Since higher-order derivatives cannot be plotted on the phase plane, higher-order systems would not be completely defined. Of course, the method might be extended from phase plane to phase spaces of n dimensions to handle nth-order systems, but then it loses some of its simplicity and convenience.

In this chapter, the manipulations of the phase plane are discussed, then the uses of different graphical methods and techniques for phase-plane analysis of nonlinear system problems. All the problems are of second order or can be approximated by second-order differential equations.

A. MANIPULATIONS ON THE PHASE PLANE

1. Elementary Algebraic Manipulations

a. Addition

Addition of phase-plane trajectories is accom- plished in a fairly straightforward manner. Suppose we have

98 two quantities x. and x„, and it is desired to find the phase-plane trajectory of x, + x~.

Let x and x represent phase-plane coordinates:

therefore

X X + X t " l 2 (4-1) X X + X t ~ l 2

On the graph, simply add the x's to obtain the new x at a particular time and the x*s to obtain the new x. The time for each plot must be known at each point so that addition takes place at points of isochronism.

b. Multiplication:

If the trajectories are x, and x„ then the products are:

X = X * X (4-3) t l 2

x. = X-.X,, + XpX- (4-4)

Again the addition and multiplication must take place at points of isochronisnu

c. Raising to a power:

For nth power

n x = x (4-5) fc

x. = nx x (4-6) The plot is taken point by point and the x coordinate

is raised to the nth power to obtain the new x ; the x co-

ordinate is raised to the x power and multiplied by nx

in order to obtain the new x, . t d. Operation on the trajectory with a function:

If the trajectory for x is known, then the

trajectory for f(x) can be founded as follows:

x = f(x) (4-7) fc

_ df (x) dx X : t dx dt (4-8) df (x) dx

The new x is the value of f(x) and the new x is the

slope of the f (x)-vs-x curve times x.

2. Differentiation

The trajectory x is known and the trajectory of

the derivative of x is desired. This has a practical application. Suppose a device or system output is to be display- ed as a phase-plane trajectory. This can be done by differentiating the output x, thus obtaining -rr or x. Then x

is plotted against x. Unfortunately, it is very difficult

to differentiate properly; and alternate is to take the in-

tegral of the trajectory, then differentiate the trajectory graphically to obtain the true x-x plot.

100 For illustration purposes, let

u = fxdt ; (4-9) where du x " dt

Then

* dx x _ ~ dt

But

dx . . dx du . dx dt du dt du

So in order to find the new x for each point x and u: dx a. Find the slope -r— .

b. Multiply this, value by the value of x at that point.

c. Plot the new x on the -u axis. (See Fig. 3-1)

3. Transformation

It is possible to transform the phase-plane tra-

jectory by applying some transformation to x and x quanti-

ties for some purpose.

For examples

Xp r vX-j g x. j

X = X X 2 g ^ l' l^ An example of this might be

x~ = x. sin x,

X~ = x, cos x,

101 Sg-M

-V

Fig. 3-1. Differentiation procedure

102 )

4. Methods of Shaping Phase-Plane Response

There are several ways in which a phase-plane re-

sponse can be shaped to conform to design requirements.

These are:

(1) Choosing specific nonlinearities.

(2) Limiting.

(3) Piecewise shaping.

a. Choosing Specific Nonlinearities:

For the following second-order differential

equation:

x + f(x, x)x + g(x, x) = various combinations f(x,x) and g(x,x), may be chosen to

obtain a required response.

b. Limiting:

The values of x and x may be restricted to certain values as shown in Fig. 3-2. The limiting may be

extended to x also or even to some function of x and x.

c. Piecewise Shaping:

See Fig. 3-3.

B. COMMON PHYSICAL NONLINEARITIES

1. Saturation: (the most common of all nonlinear

phenomena )

Consider the case of a saturating electronic ampli-

fier driving a motor in an instrument servo. (See Fig.

3-4 (a) and (b)

103 X

-J ~ J - -*c/ •*

Ul'd

Fig. 3-2. Limiting of x on the phase-plane,

*-/ f(x)+ X -o /+ £6t'/ + x=o 3

(*) 4- X'O x + (M+x-o X+fy*)+K-0 * +f5 f6

Fig. 3-3. Piecewise shaping of trajectory,

104 .

Amplifier ]';';r and gears

0u<(A. . K,E r x \ kjtf *. K. y -m£0- y ${*ru + l) — | *.*i Linear gain = X C

(a) Block diagram of a servo with saturation.

Output Linear ^ approximation ^

Input

(b) Input vs output characteristics.

Fig. 3-4. Common physical nonlineari ties - Saturation,

105 The equation in the linear region is

i K K K E + -£- E + -^—- E = Tm NTm

-E < E < + E

For saturation operation:

1 • E+^^E=+c E<-E n m

E+-^E = -c E^+E n Tm 1 where

KK K - s m _ C " NT *! ' m

Using the isocline method, draw isoclines on the phase- plane E vs E. In the linear region the isoclines are radial

straight lines (see Fig.3-4c), emanating from a focus at

the origin.

While in the saturation region the isoclines are horizontal, the transition from linear to nonlinear operation is indicated by a vertical line at E = E.. Dividing lines on

the phase plane may be obtained from many nonlinear phenomena.

106 -os W—M'tf.-1.5' ^M._7

Fig. 3-4.

107 «

putput

Error transducer Am plifiar. motor Output \ * 1/ ?. l, kwul Input 9 ion* / 0«»d H Untsr gain-JC, and geart

(a) (b)

M--

(c)

Fig. 3-5. Dead zone - (a) Input vs output characteristics (b) Block diagram of a servo with dead zone (c) Isoclines, dividing lines and trajectory.

108 .

2 . Dead Zone

This arises in mechanisms which are spring-loaded

to minimize backlash and in many other devices which are

insensitive to small signals. Consider a servo with a dead zone in the error transducer.

E+ — E + E = Tm

- E.. < E < + E

. K KK E + — E + -|—-(E - E..) = Tm Nt m 1

E | > E,

In the dead zone, there is no control forcing- function,

so in this case isoclines are straight lines with slope

• 1 -E -E f- M = 53-

(See Fig. 3-5)

3. Backlash

Most mechanical linkages exhibit a certain amount

of backlash. (See Fig. 3-6)

Consider a simple positioning servo with backlash.

When the backlash is taken up, the operation is linear. The

equation of the system is

2 E + 2tiX> E + Ui E - n n

109 i . Output position

(a) (b)

=0 - 5 M-0,25 M

=-0.2

(c)

Fig. 3-6. Common physical nonlineari ties Backlash.

110 . .

When zero velocity is reached and the motor reverses, the output shaft remains stationary until the backlash is taken up. Thus the error remains fixed at some value E.

The equation of this motion is

B + 2Tu = E m n m 1n

4. Instrument Servos with Coulomb Friction and

Stiction

a. A second-order servo with coulomb friction,

(see Fig. 3-7 (a) )

The error equation for a step-displacement input is

j£ + fE + KE + C Sgn(E) = from which the isocline equation becomes

• KE C Sgn(E) MJ+f MJ+f

Note: Let E = y

*L *£. dE E15 = = = MEu dt dE dt

The second term introduces the effect of the nonlinearity. This effect is seen to be translation of the focal C C point to E = + — when E is negative and to E = - — when

. • E is positive. E = (or E axis) is a dividing line. (See

111 .

Fig. 3-7). The trajectory terminates at D because the drive torque produced by the error at D is less than the coulomb torque C.

b. If stiction is present the isocline equation becomes

• _ KE S Sgn(E) MJ+f MJ+f

s This. moves the focal point to - — . (See Fig. 3-7 (b) ) is. A small initial disturbance (such as might be represented at point A) would not produce sufficient torque to pull the system free from the stiction. For a large disturbance

(such as A") the system does pull free and begins to acquire a velocity.

c. When a ramp input is applied and coulomb friction is present but stiction may be neglected, the isocline equation becomes

fOJ. + C Sgn(0 - E) E = - r^-r + MJ+f MJ+f

where U). = 8 is the input ramp velocity and the coulomb

-E) = ( friction is designated by C Sgn(0R C Sgn ) because the coulomb friction effect is controlled by the relative velocity between the friction surfaces at the output.

The focus is plit by the C Sgn(0 -E) term, and is

translated by both C and f U). . The foci thus appear at

E = (fa*. ^ C)/K; they are displaced i C/K from the point fO). l K *

112 - f Motor Ampfifier and gears \i 7zr/z -ot

(a)

Trajectory

(b)

Fig. 3-7. Phase trajectory for an instrument servo with Coulomb friction and stiction: Step response.

113 . .

The dividing line is a horizontal straight line at

E = oj.

Note j

Q ~ E = 8^ Rr> C

e = o , ft = e = « c R

(See Fig. 3-8 (a)). For a ramp input suddenly applied at

t = 0, the initial conditions are E(0)=0,E(0) = 03., which

correspond to point P on the figure.

d. When stiction is present but discontinuous,

the isocline system of Fig. 3-8 (a) is not altered. But the phase trajectory cannot break free from the dividing line C S at E = — . It must remain on the dividing line until E = —

(See Fig. 3-8 (b) )

C. APPLICATIONS OF THE ISOCLINE METHOD

Since there are many methods of solution of differen-

tial equations which are comparatively simple on the phase-

plane but difficult or impossible otherwise, it is rather

obvious that the phase-plane provides a very convenient

means of analysis of extremely nonlinear first and second-

order differential equations. It seems that the isocline method is a general and useful method with which to approach

problems by using the phase-plane method. Application of

the isocline method to specific problems illustrates the usefulness of this method.

114 Fig. 3-8a. Phase portrait of an instrument servo with Coulomb friction in response to a ramp input.

Fig. 3 -8b. Effect of stiction on the ramp response of a servo with Coulomb friction.

115 1. First-Order System

The first-order equation has the form

ff = f(x,t) (3-1)

Assume that the function f(x,t) is continuous and single- valued with the possible exception of certain singular points. The isocline method requires a graphical construction performed on axes of x and t and yields a solution as a curve. For any given point on the x-t plane, the numeric- dx al value of f(x,t), and thus of -rr , can be calculated.

Example 1: Find a solution for the equation

dx — = - x + t dt

Consider the initial conditions x =0 at t =0, and o o also x = 0.2 at t =0. For this linear equation, the o o •* » isoclines are straight lines given by the relation m = x + t.

A family of isoclines, each carrying its line segments of proper slope is shown in Fig. 3-9. Solution curves are sketched in the figure, starting at each of the specified initial conditions.

Example 2: RL circuit with nonlinear L / R S /yV\A.

— E L ~ + iR = Eu(t)

116 Since the inductor is nonlinear, and the coefficient L therefore variable, it is more convenient to express the relationship as

iR = Eu(t) M^+dt from which

The equation of the isoclines is obtained by letting

-r-r = M, some constant, and noting that Eu(t) = E for t > 0, then

E - MM., i = ^

Knowing the magnetization curve of the inductor, one can obtain the transient current variation of the nonlinear

RL circuit, i vs t. (See Fig. 3-10)

2. Second-order System

(Example 1) The equation for a damped pendulum is

. . _ d x dx , +, KT -tt + sin x = dt dt

Let

dx v = dt

2 d x dv dv dx _ dv m v ,,2 dt dx dt dx dt

117 . .

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Fig. 3-9. (Example III-C-1-A)

0.0175 ,01-00172 V4.-0.-o i-0,-00170 M 000}° »Oh-00167 E/.-Oo°o« 01-OO1M "•-0006r„S-0 »i-ooi5« M,-ooo« 0.015- «>,- 00149 0.010

(6,-0.0137 M,-o.012

0,-0.0115 M,-0.014

0010

0,-0 0087 N|-0.01»

0.007$

0,-0.0048 M, -0.018

0,-0 M,-0.02

Fig. 3-10. Example (III-C-1-B)

118 Thus

d x dx . , T/ 7T + K -rr + sin x ,,2 dt dt

-=— + Kv + sin x = dx or

dv (Kv + sin x) dx v

Let

K = 1

dv _ (v + sin x) dx v

j— is the slope at any point x, v of a possible phase- plane trajectory.

Now assume a number of values for -r— dx

-t— Equation of v vs x

v = - sin x

1 v = - y sin x

-1 = - sin x

2 v = - -r sin x

-2 v = sin x 1 3 v = - sin x t4 1 -3 v = 77 sin x 2

(See Fig. 3-11)

119 Fig. 3-11. Isocline solution.

120 )

Example 2% Nonlinear system x + xx + x =

Let x = Mx ;

then

Mx + xx + x =

-x x = M + x

Draw a number of straight lines x = Mx through the origin (corresponding to different values of M) on the x-x plane (as in Fig. 3-12 (a)).

Plot x vs x from equation x + xx + x = for con-

stant (different) values of x (see Fig. 3-12 (a)).

With any particular straight line in view, say WL — M_,

determine all possible points of intersections of above plots. The procedure is repeated for M = M ?# M_, etc., until a fair number of isocline curves are obtained on the phase plane. The respective slope lines are marked on

these isoclines. Joining these slope lines gives the dif-

ferent trajectories or the entire phase portrait (as in

Fig. 3-12 (b)).

Example 3: Consider the case of a saturating elec-

tronic amplifier driving a motor in an instrument servo.

(See Fig. 3-13 (a)

The equation in the linear region is

, K KK E + ~ E + -~ E = -E.< E < +E.. Tm NTm 11

121 t^'-to

*=-.*

Fig. 3-12 (a) Example x + xx + x =

122 l^^rtpi M=c^. IMf . fl

1 t

Ht+J..

liHtr'^titlrnfHrrrHffr^TH'rT'rfi^rtlntilri-t- -7 -6 -5 -« -3 3 4- 5 © X - 'Hi&tjL&i&jif'tiMiritiMtii.uiH'i*'*

-tt V ; ^//"//^//^/^ SS\-Vv\VN>*V^ v\S\N \-V^*

Ms-3

Fig. 3-12 (b) . Example x + xx + x =

Amplifier Motor and gears Output 3-J$y\£J7X- K~ KmA/ **. i v Input ^ t »{»r +l) u

i k Oc Linear gain -if

Fig. 3-13 (a)

123 .

where E is the value of | E | which causes saturation and N

is the gear ratio.

For saturation operation, the equations are;

E + it- E = + C E < -E. Tm

E + ^- E = - C E £ +E. Tm

where K K K s m C = NTm

On the E vs E phase plane the isoclines are all

straight lines; in the linear region the isoclines are

radial straight lines emanating from a focus at the origin,

While in the saturation region the isoclines are horizon-

tal, the phase trajectory is constructed in the usual

manner (as in Fig. 3-13 (b) )

Example 4s Oscillator with relay control.

mx + Kx = F / / / //

—K —F x + m x = m

Let

£- i m

A —F = m u

then x + x = u

124 ))

Let

x = y then

Mx dt dx dt where

dx thus

My + x = u

u - x M = —

a. If u = K the phase trajectory is as in

Fig. 3-14 (a).

b. If u = -K Sgn(x)

x f K Sgn(x) M = -

x = or y-axis is a dividing line.

(See Fig. 3-14 (b)

x "I" K If x is positive M = - y x - K if x is negative M = -

c. If u = -K Sgn(y)

M = _ x + K Sgn(y) y y = or x-axis is a dividing line,

(See Fig. 3-14(c)

125 Lr -9

-15

5 4

Fig. 3-13 (b)

X-Scale 2.00E-01 Units Inch. Y-Scale 2.00E-01 Units Inch.

126 : )

X + K If y is positive M = - Y x - K If y is negative M = - Y

c. If u = -K Sgn(x+y)

- M = x K Sgn(x+y) Y

x + y = is a dividing line,

(See Fig. 3-14 (d)

e. If u = -K Sgn(x-y)

- = x K Sgn(x-y) Y

x - y = is a dividing line.

(See Fig. 3-14(e)

Due to the changing dividing line the system became unstable.

Example 5

u x

(s+2) (s+ |)

Given:

x < 0.5 u =' ^ -Sgn(x) , I x I 0.5

127 .

Since

x (sj u(s) 2 (s+2) (s+ -|) s + |- s + 1

5 • x+-jx+x = u

Let

x = y then

dy dy dx X = -rr = " -rr = MX dt -T-dx dt thus

- - y x + u M = 2

At u = 0, the equation of isoclines is

- - My = t2 yY x

At u = - Sgn(x)

If x is positive, the equation of isoclines is

My = - - x - 1 -J y

If x is negative, the equation of isoclines is

My=--~y-x + l

(See Fig. 3-15)

128 X-Scale = l.OOE+00 Units Inch. Fig. 3-14(a) Y-Scale = l.OOE+00 Units Inch.

129 * J-Im ; , t

X-Scale = 2.00E-01 Units Inch. Fig. 3-14 (b) Y-Scale = 2.00E-01 Units Inch.

130 . .

Y /

Isocline is the sam as 3-14 (b)

»*•

/ dividing line \

-••

*•>*

Fig. 3-14 (c)

131 .

Isocline is the same as 3-14(b)

Fig. 3-14 (d)

132 .

I, Y

Isocline is the same as 3-14(b)

Fig. 3-14 (e)

133 Fig. 3-15 X-Scale = 5.00E-01 Units Inch. Y-Scale = 5.00E-01 Units Inch.

134 3. High -Order Systems

The phase-plane, as normally used, is restricted

to problems which can be described by first and second- order differential equations. This is a fairly serious re-

striction, since many practical systems require higher-order differential equations for an adequate description. In the discussion of numerical methods for solving differential equations, first-order equations are studied in the beginning. Then equations of higher order are solved by replac-

ing each equation of nth order by n equations of first-order and solving these simultaneously. In like manner, the discussion of graphical methods has begun by considering the

isocline method of solving a first-order equation first,

then second-order equations. But further consideration may

not be possible since there is not enough information avail-

able initially to construct isocline curves for the several

first-order equations that would arise from a single high-

order equation. So, if we can factor high-order equations

into several second-order equations, these can be solved by

phase-plane method. The difficulty is in converting a

high-order equation into several second-order equations.

No convenient long hand method is available, but use

of the digital computer permits study of such phase space

problems.

135 :

IV. GENERATING THE PROGRAMS TO DRAW ISOCLINES AND

TRAJECTORIES OF SECOND-ORDER DIFFERENTIAL

EQUATIONS

A. x + F(x) x + G(x) = type equations (See Appendix,

Program 1) (Where F(x) f G(x) can be any function of x)

x + F(x) x + G(x) =0

Let

y = x

dy_ dy . dx dt dx dt

Let

= M ¥dx then

x = My thus

x + F(x) x + G(x)

= My + F(x) y + G(x) =0

v = - — r equation of isoclines 1 M + ^BF(x)— ^

Trajectories

ZD0T(1) = Z(2)

ZD0T(2) = -F(Z(1))* Z(2) - G(Z(1))

(By using subroutine program RKLDEQ.)

136 Example 1: x + xx + x =

li*uti

Fig. 4-1. X-Scale = l.OOE+00 Units Inch. Y-Scale = l.OOE+00 Units Inch.

137 *2 Example 2: x - e(l-x )x + x =

Fig. 4-2. X-Scale = 8.00E-01 Units Inch Y-Scale = 8.00E-01 Units Inch

138 Example 3: x" +(cos x)x + tan x rH

Fig. 4-3. X-Scale = l.OOE+00 Units Inch, Y-Scale = l.OOE+00 Units Inch,

139 Example 4: x+25(l+0.1x)=0

Fig. 4-4. X-Scale = l.OOE+00 Units Inch, Y-Scale = l.OOE+00 Units Inch,

140 Example 5: x + 0.5x + x =

-3

2b Fig. 4-5. X-Scale = l.OOE+01 Units Inch. Y-Scale = l.OOE+01 Units Inch.

141 Example 6: x -(l-x 2 )x + x =

Fig. 4-6. X-Scale = l.OOE+00 Units Inch. Y-Scale = l.OOE+00 Units Inch.

142 =0 Example 7 : x + x + x | x |

Fig. 4-7. X-Scale = 2.00E+00 Units Inch. Y-Scale = 2.00E+00 Units Inch.

143 Example 8: x+ — x + x =

Fig. 4-8. X-Scale = l.OOE+00 Units Inch. Y-Scale = l.OOE+00 Units Inch,

144 Example 9: x + x + sin x =

Fig. 4-9 X-Scale = 9.00E-01 Units Inch, Y-Scale = 9.00E-01 Units Inch,

145 ) = Example 10: x + (1- ] x | x + x

: n

ifl

Fig. 4-10. X-Scale = 1.00E+00 Units Inch, Y-Scale = 1.00E+00 Units Inch,

146 Example 11 x + sin x =

Fig. 4-11 X-Scale = 2.00E+00 Units Inch. Y-Scale = 2.00E+00 Units Inch.

147 2 Example 12: x +(l-x )x + x =

Fig. 4-12 X-Scale = 5.99E-01 Units Inch. Y-Scale = 5.00E-01 Units Inch,

148 Example 13: x + 2 xx + x =

Fig. 4-13. X-Scale = l.OOE+00 Units Inch. Y-Scale = l.OOE+00 Units Inch.

149 :

• 2 B. x + x +G(x)=0 TYPE EQUATIONS (SEE PROGRAM 2)

(WHERE G(x) IS ANY FUNCTION OF x)

2 x + x + G(x) =0

Let

x = y then

dy dx .. x = = My -& dt thus

x + x + G(x)

2 = My + y + G(x) =

2 y + My+ G(x) =

-M + j¥?- 4G(x) ^ ^ Y ^ 2 — ^ a j ^equations of -" isoclines -M -MZ- - 4G(x) Y = M5

Trajectories

ZD0T(1) = Z(2)

ZD<£T(2) =-Z(2) ** 2 - G(Z(1))

150 Example 14: x + x + x =

-3

Fig. 4-14 X-Scale = 5.00E-01 Units Inch, Y-Scale = 5.00E-01 Units Inch,

151 • 2 2 Example 15: x+x +x =0

Fig. 4-15 X-Scale = 5.00E-01 Units Inch, Y-Scale = 5.00E-01 Units Inch,

152 .2 Example 16: x + x +x|x! =0

Fig. 16. X-Scale = 5.00E-01 Units Inch, Y-Scale = 5.00E-01 Units Inch,

153 :

C. |x|x+x + x = TYPE OF EQUATIONS x + x|x|+x = (SEE PROGRAM 3) x = x|x1+x =

In Section A and Section B the equations of isoclines are solved by y in terms of x. In this section we will solve x in terms of y to prevent imaginary values.

= 1. | x | x + .x + x

Let

y = x , then

= * = 1^ It M^ thus

|x |x + x + x = |y|My + y + x =

- x = - | My | y y

= - of isocline. 1 M + 1) equation y( I y

Trajectories

ZD<£T(1) = Z(2)

" z 2 z(1 ( ) - > ZD0TU)zDrtTm =- DABS ( Z (2))

154 :

2. x +x|x|+x = x+x|x|+x=My+x|y|+x=0

My x = - equation of isoclines

1 + I y I

Trajectories

ZD0T(1) = Z(2)

ZD0T(2) =-Z(l)* DABS(Z(2)) - Z(l)

3. x + x | x | +x =

x + x | x | +x=My+y|y| +x=0

x = - - = - My 1 (M + ) y 1 y y 1 y | equation of isoclines.

Trajectories

ZD0T(1) = Z(2)

ZD0T(2) =-Z(2)* DABS(Z(2)) - Z(l).

155 Example 17: x + x|x|+x =

Fig. 4-17. X-Scale = l.OOE+00 Units Inch, Y-Scale = l.OOE+00 Units Inch,

156 Example 18: x + xlx|+x =

Fig. 4-18, X-Scale = l.OOE+00 Units Inch, Y-Scale = l.OOE+00 Units Inch,

157 Example 19: |x|x + x+x =

Fig. 4-19 X-Scale = l.OOE+00 Units Inch, Y-Scale = l.OOE+00 Units Inch,

158 : .

D. USE OF PROGRAMS

1. Table 1.

Type of Equation How to Set the Subroutine and Function Programs

x + F(x)x + G(x) =0 (1) Need to compare standard form (Refer to Section A) to give real function F(x) and G(x) in each function subprogram.

(2) Rewrite the given -equation x + F(x)x + G(x) = into a system of equations in subroutine program.

ZD0T(1) = Z(2) ZD0T(2) =-F(x)*Z(2)- G(x)

(F(x), G(x) should change to z form)

2 x + x + G(x) = (1) Need to give real function (Refer to Section B) G(x) in function subprogram.

(2) Same as above (2)

= 1 x 1 x + x (1) Isoclines

x + x|x|+x = • Let x = y

x + x|x|+x = solve x in terms of y and give (Refer to Section C) this equation in main program.

(2) Same as above (2)

Table .1.

159 Input deck, See Table 2.

Sequence of Cards Purpose Punch in Column

Scale One significant number 1st Card Column 1-10

( Number of Cards (column 11-20

2nd Card Title Column 1-48

3rd Card Title Column 1-48

4th Card Different ranges Column 1-10 Ini-

• tial

• value Column 11-20 N Increment cards No. of Column 21-30 Final depends on value number of different ranges of slope cards.

(N+l) th Card Time increment Column 1-10

Number of )

trajectories j Column 11-20

(N+2) th Card Initial time Column 1-10 • Final time Column 11-20

• Initial x Column 21-30 K Initial y Column 31-40 No. of cards depends on number of trajectories.

(K+l) th Card Blank Blank

! 1

Table 2.

160 NOTE: FOR SHIFTING THE CENTER OF COORDINATES. SEE

AND COMPARE PROGRAM 10.

4- RR = Example 20. x + J^63 * cos x + sin x °

n. ! 1 w

i ! \\

- i hH-i -

l. \\Y\\ Uito

^pfcg» „«

Fig. 4-20. X-Scale = 2.00E+00 Units Inch. Y-Scale = 2.00E+00 Units Inch.

161 E. DISCUSSION

In Chapter II it was shown that the isocline method is a very useful general method of phase-plane analysis, the only difficulty being the labor required to get the isoclines. This difficulty has been reduced by developing computer programs to draw both isoclines and trajectories.

In Section A, a program works for x + F(x)x + G(x)=

type equations. This is the general case, and also can be done by hand labor. In Section B, a program works for

• 2 x + x + G(x) = type equations. Working this kind of

equations by hand labor, some supplementary methods are

needed, such as Deekshatulu ' s or Murthy's, but use of the computer saves time and labor.

In Section C, a program works for some special prob-

lems. In this program the equation of isoclines is first

solved for x in terms of y (Note: In Section A and B, iso-

cline equations are solved for y in terms of x) . After

calculating the required points the plotting routine treats

y as a function of x„ The reason for this is to avoid the

problem of complex values in the computation of y. This

procedure is not always successful; there are at least two

problems for which complex values apparently cannot be

avoided. They are:

+ x x 1. x I =0 I +x

= 2. x + xlx + x |x |

162 For example: Equation x + x | x | + x =0

Let

• X = y then

X = My thus

= x + x 1 + x + + I x My y I y I x =0

Case a : Solving for y in terms of x

2 2 y + My + x =0

- M ± Jy? 4 x y = M when

M < 4x 2 y is complex.

Case b : Solving for x in terms of y

= - + x [y | y | My]

x = J-[y I y I + My]

when | + is [y | y My] positive x becomes complex,

163 V, GENERATING PROGRAMS TO DRAW ISOCLINES

AND TRAJECTORIES WITH DIVIDING LINES

OF THE SECOND-ORDER DIFFERENTIAL

EQUATIONS

SATURATION

1. Program for Vertical Dividing Lines only

(See Program 4)

Refer to Chapter III, Section B-l.

164 \ : :

Example 1. Tra jec tories : x + . 2x + x = Vfor Inner

regions ) Dividing lines

x = ± .3

Isoclines

/ „ „. c n Inner linear region^ ( See Fig. 5-1) x + 0.2x + x =

My + 0.2y + x =

= -x y M + 0.2

Saturation regions:

x + 0.2x ± 0.3 =

My + 0.2y i 0.3 =

= T 0.3 Y M + 0.2

Trajectories

Inner region

ZD0T(1) = Z(2)

ZD<2T(2) = -0.2* Z(2) -Z(l)

Saturation region

ZDtfT(l) = Z(2)

ZD$T(2) = -0.2* Z(2) ± 0.3

165 u

Fig. 5-1. X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

166 Example 2: x+0.2x+x=0 Dividing lines x = + 0.2

Fig. 5-2. X-Scale = 2.00E-01 Units Inch Y-Scale = 2.00E-01 Units Inch,

167 :

2. Program for Vertical Dividing Lines which can be

Rotated. (See Program 5)

Equation of dividing lines:

Ax + By = ± C

Isoclines in linear region:

x + F(x)x + G(x) =

assume F(x) = 1, G(x) = x

Let y = x , then x = My

: Y M + 1

Isoclines in saturation regions:

= ± C Y M + 1

Trajectories

Inner fzD0T(l) = Z(2)

region |^ZD0T(2) = -'1 * Z(2) - Z(l)

= l. ZD0T(1) Z(2) Saturationa i region I Z D0T(2) = -Z(2) ± C

168 Example 3: Dividing lines: x + 0.8y = ± 0.3

Trajectories : x + 0.2x + x =

Fig. 5-3 X-Scale = 2.00E-01 Units Inch, Y-Scale = 2.00E-01 Units Inch,

169 Example 4: Dividing lines: x - . 8v = i 0.3

Trajectories : x + 0.2>: + x =

Fig. 5-4. X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

.170 Example 5: Dividing lines: x = ± 0.3 (Y = 0)

Trajectories : x + 0.2x + x =

2':. ?>

I.' -33

X-Scale = 2.00E-01 Units Inch. Fig. 5-5 Y-Scale = 2.00E-01 Units Inch.

171 ) :: ::

B. DEAD ZONE. (SEE PROGRAM 6)

(Refer to Chapter III

Section B-2 .

Example 6: (See Fig. 5-6)

Inner region:

Isoclines x+0.2x+0.3=0

My + 0.2y + 0.3 =

-0.3 M = M + 0.2

Trajectories

(zD<£T(l) = Z(2)

(zD0T(2) =-0.2* Z(2)

External region:

Isoclines

Right and left x+0.2x+x^o.3=0

Trajectories

ZD#T(1) = Z(2)

ZD(JT(2) = - 0.2 * Z(2) - Z(l)± 0.3

172 I

\ \

Fig. 5-6. X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

173 : :

C. IDEAL RELAY. (SEE PROGRAM 7, 8, 8A)

For x + 0.2x + u =

u = - K Sgn(x)

Switching line is x =

or y axis.

Isoclines

At +x x + 0.2x - K =

At -x x + 0.2x + K =

Trajectories

At +x x + 0.2x - K =

(zD0T(l) = Z(2)

( ZD#T(2) = - 0.2 * Z(2) + K

At -x x + 0.2x + K =

(zD0T(l) = Z(2)

(ZD0T(2) = - 0.2 * Z(2) - K

174 :

Example It x+0.2x+u=0, u=0.3 Sgn x

Switching line: x = or y-axis.

Isoclines -0 Right of y-axis: = ' 3 y M + Q<2

• 0.3 LeftT CU of4= y-axis: y = - + Q ^ 2

Trerjec tories :

Right ( ZD(4T(1) = Z(2)

ZD*T(2) = - 0.2 * Z(2) + 0.3

Left fzD(JT(l) = Z(2)

\zD<£T(2) = - 0.2 * Z(2) - 0.3

(See Fig. 5-7)

175 _ ;s

*0 1 3

Fig. 5-7. X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch,

176 .

Example 8(a) : x + 0.02x + u = u = - 0.3 Sgn(x)

Fig. 5-8(a) X-Scale = 2.00E-01 Units Inch, Y-Scale - 2.00E-01 Units Inch,

177 :

Example 8 (b) x + 0.02x + u = u * - 0.3 Sgn(8x-y)

(x = y)

+- -•

Fig. 5-8(1?) X-Scale - 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

178 .

'- = Example 8(c) x 4- 0.0 2x u d = - 0.3 Sgn(8x+y)

(x = y)

Units Inch, Fig. 5-8(c) X-Scale 2.00E-01 Y-Scale 2.00E-01 Units Inch,

179 Example 5-8 (d) : x + Q.02x + u = u = - 0.3 Sgn(4x-y), x = y

vj 1

Fig. 5-8(d). X-Scale = 2.00E-01 Units Inch Y-Scale = 2.00E-01 Units Inch

180 :

0.02x + u = Example 5-8 (e) x + (4x+y) x = y u = - 0.3 Sgn ,

2.00E-01 Units Inch. 5-8(e) X-Scale Inch, lig. Y-Scale 2*.00E-01 Units

181 :

Example 5-8 (f) x + 0.02x + u = u = - 0.3 Sgn(2x-y), x = y

inj-l

Fig. 5-8(f). X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

182 Rx ample 5-8 (g) x + 0.02x + u =

- - 0.3 Sgn(2

Pig. 5-3 (g) X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

183 '

Example 5-8 (h) : x + 0.02x + u = u - - 0.3 Sgn(x-y), x = y

r r,-rM

. s

1 .M

• : ' 1

-~ . - . -\25 ~ '"t'-7^-7~ ~- ~_—— - - — =d 1 __jF' _|^. f J j J -JJ :rj _y izt m nu 1 - - —', *~"— ""'' * " "r ' / _ ~ ~~ 7 : / / / / /

| /

r r^ '" ! s /

, -f J

• i • i Fig. 5-8(h) X-Scale 2.00E-01 Units Inch, Y-Scale 2.00E-01 Units Inch,

184 .

Example 5-8 (i) : x + 0.02x + u = u = - 0.3 Sgn(x+y), x = y

. .. . _ ti_

i —.2^ — . . r

i := : 1 3q= -Trz^-==r^.^_ zir-_= ?7 - ". i • ^ i 1 5 ' i 30'

S^ L __ _ . 1 ZZj

/ / 7 ,! M -•:•• I i ^v 7" A < /

^ %.

\ \^ /

I r

IT)

T n

Fig. 5-8(i). X-Scale = 2.00E-01 Units Inch, Y-Scale = 2.00E-01 Units Inch.

185 '

Example 5-8 (j) : x + 0.02x + u = - u - 0.3 Sgn(y) , x = y

1.0

'."

;

> . ,

i J JO -Quv ::u ]£. ZZ1 30>j •• - ,-—.--, -

/ 1 M • / 1 / / / /

.y' /• y' y' U •1 £ 1

Fig. 5-8(j). X-Scale = 2.00E-01 Units Inch, Y-Scale = 2.00E-01 Units Inch,

186 : :

D. RELAY WITH DEAD ZONE (SEE PROGRAM 9)

For x+x + x + u =

u = -K Sgn(x)

Isoclines

At dead zone region:

x + x + x =

My + y + x =

-x = Y M + 1

At right:

x+x+x-K =

My + y + x -K =

-x+K Y M + 1

At left:

x+x + x+K = My+y+x+K=0

-x-K y M + 1

Trajectories

At dead zone region: ZDtfT(l) = Z(2) x + x + x — 0. ZD0T(2) =-Z(2) - Z(l)

87 At right:

(zDdTCL) = Z(2) x+x+x-K=0, I lzD*T(2) =-Z(2) - Z(l)+K

At left:

(ZD0T(1) = Z(2) x+x+x+K=0, ) (zD0T(2) = -Z(2)- Z(l)- K

188 :

Example 9 x+0.2x+x+u=0 u = - 0.3 Sgn(x) dividing lines: x = ± 0.2

Fig. 9. X-Scale = 2.00E-01 Units Inch Y-Scale = 2.00E-01 Units Inch

189 : 'I}

Example 10 x + x + u = u = - 0.3 Sgn(x) dividing lines: x = ±0.2 -36

\ *J>

-75 v

\ \ \ / \ s \ \\ \ f -2 \ \ \

-A iisii i£g^k ii \\\\P55

/ / i ' / / 1 '?

/ I \ \

/ Cl ~ \ \ / \ 75 / kO / ( / a

/ m \ 25 f * Fig. 10 X-Scale = 2.00E-01 Units Inch. Y-Scale = 2.00E-01 Units Inch.

190 IV. CONCLUSIONS

Chapter III summarizes eleven useful methods. Some of them draw direction fields to get trajectories. Some of them are good supplements to the isocline method. Some of them, by using geometrical relationships, draw the trajectories step-by-step. In general, none of these methods is a perfect one. For specific problems one may be better than the others in terms of the labor required. But if we want to look at the overall picture of some nonlinear system characteristics such as stability limit cycles, etc., none of them can do this easily. All need lots of time and hard labor.

The isocline method has many advantages. Programs have been developed to compute and draw both isoclines and trajectories. All the programs in Chapter IV and V, first draw isoclines, then draw trajectories from given initial conditions. In analysis and design work, isoclines may not be needed for all problems. Conversely, sometimes only isoclines are needed and we do not need trajectories. Use of isoclines with superposed trajectories gives a global pitcure. That is, it indicates what can happen to trajectories for all initial conditions, and tells whether the system is stable or not, has limit cycle or not, ... etc.

191 The programs generated in Chapter V add dividing lines to the isoclines and trajectories. By changing or shifting dividing line position, the trajectories are shaped. This is helpful for analysis and design work.

In this thesis, all graphs are plotted by machine and all are two dimensional. The variable symbols y vs x or x vs x are used. There are no time variables in the plot.

If the system equation involves time as a variable, we may choose the ft method to work it or apply some transformation, changing the equation into the standard type forms.

In general, machine time required to compute the isoclines and trajectories of a second-order nonlinear differential equation was about 60 sec. This is an appreciable saving compared with hand labor.

I 9 2 APPENDIX

COMPUTER PROGRAMS

•> o m O »- o

X < ^ X h- >"

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~~242 .~~

~~BIBLIOGRAPHY~~

~~1. Robert Lien Cosgriff, Nonlinear Control Sys terns , McGraw-Hill Book Co., Inc., 1958.~~

~~2. Irmgard Fliigge-Lotz, Discontinuous Automatic Control , Princeton University Press, 1953.~~

~~3. John E, Gibson, Nonlinear Automatic Control , McGraw- Hill Book Co., Inc.~~

~~4. Lydik S. Jacobsen and Robert S. Ayre, Engineering~~

~~Vibrations , McGraw-Hill Book Co., Inc., 1958.~~

~~5 Y . H . Ku , Analysis and Control of Nonlinear Systems (Nonlinear Vibrations and Oscillations in Physical~~

~~Systems) , The Ronald Press Co., 1958.~~

~~6. LeRoy A. MacColl, Fundamental Theory of Servomechanisms, D. Van Nostrand Co., Inc., 1945.~~

~~7. George J. Thaler, Elements of Servomechanism Theory , McGraw-Hill Book Co., Inc., 1955.~~

~~8. George J. Thaler and Marvin P. Pastel, Analysis and~~

~~Design of Nonlinear Feedback Control System , McGraw- Hill Book Co., Inc., 1962.~~

~~9. George J. Thaler and Robert A. Brown, Analysis and De-~~

~~sign of Feedback Control Systems , McGraw-Hill Book Co., Inc., 1960.~~

~~10. John C. West, Analytical Techniques for Nonlinear~~

~~Control Systems , D. Van Nostrand Cp., Inc.~~

~~11. Anthony Ralston and Herbert S. Wilf, Mathematical~~

~~Methods for Digital Computers , John Wiley & Sons, Inc., 1960.~~

~~12. Deekshatulu, B. L. , Techniques for Analysis of Certain~~

~~Nonlinear Systems , IEEE Trans. Applic.Ind., July, 1964.~~

~~13. B. L. Deekshatulu and Murthy, New Approach to the~~

~~Plotting of Phase-Plane Trajectories , Proc. IEEE, Vol. Ill, No. 10, October 1964.~~

~~14. M. Makinoutz Tumura and N. Hakuo, E-Function Method , Trans. Japan Society of Mechanical Engineers. 17(1951): 69-80 (In Japanese).~~

~~243 INITIAL DISTRIBUTION LIST~~

~~No. Copies~~

~~1. Defense Documentation Center 20 Cameron Station Alexandria, Virginia 22314~~

~~2. Library, Code 0212 2 Naval Postgraduate School Monterey, California 93940~~

~~3. Chinese Navy 1 Training Section Headquarters Taipei, Taiwan REPUBLIC OF CHINA~~

~~4. Dr. George J. Thaler 10 Electrical Engineering Department Naval Postgraduate School Monterey, California 93940~~

~~5. Dr. B. L. Deekshatulu 1 Electrical Engineering Department Indian Institute of Science Bangalore 12, INDIA~~

~~6. Dr. D. D. Siljak 1 Electrical Engineering Department University of Santa Clara Santa Clara, California~~

~~7. Dr, A. G. Thompson 1 Mechanical Engineering Department University of Adelaide Adelaide, SOUTH AUSTRALIA~~

~~8. Dr. S. H. Liu 1 P.O. Box 1-3 Lung -Tan, Taiwan REPUBLIC OF CHINA~~

~~9. Dr. Sydney R. Parker 1 Electrical Engineering Department Naval Postgraduate School Monterey, California 93940~~

~~244 No. Copies~~

~~10. Professor Abraham Sheingold 1 Electrical Engineering Department Naval Postgraduate School Monterey, California 93940~~

~~11. LCDR Hsiung, Yun Lan 2 337, Section 1, Mou-San Road Mou-San, Taipei, Taiwan REPUBLIC OF CHINA~~

~~245 . 1~~

~~Security Clarification i£ DOCUMENT CONTROL DATA R&D~~

~~(Security classification of title, body of e*a tract and Indexing annotation must be entered when the overall report la classified)~~

~~I, originating AC Tl VI TY (Corporate author) 2a. REPORT IfCURITV CLASSIFICATION Naval Postgraduate School Unclassified Monterey, California 93940 26. GROUP~~

~~3 REPORT Tl TLE Phase-Plane Methods~~

~~4 descriptive NOTES (Type ol report and,lnclusive dates) Master's Thesis; October 1969 5- AUTHOR(S) (Flrat name, middle initial, last name) Hsiung, Yun-Lan~~

~~6 REPORT DATE 7a. TOTAL NO. OF PAGES 7b. NO. OF REFS October 1969 246 14 8a. CONTRACT OR GRANT NO. 9a. ORIGINATOR'S REPORT NUMBERO)~~

~~b. PROJEC T NO.~~

~~eb. OTHER REPORT NOISI (Any other numbera that may be assigned thlw report)~~

~~a^->cl ^&~jU;~~

~~j'pL. CUq-hu h ulu/h^ . L&, ULsdti ftMb(J Distribution of this do ( _~~

~~11. SUPPLEMENTARY NOTES 12. SPONSORING MILITARY ACTIVITY Naval Postgraduate School Monterey, California 93940~~

~~13. ABSTRACT~~

The phase-plane method is a graphical method for linear and nonlinear second-order systems. There are many techniques for obtaining the phase portrait which consists of a number of phase trajectories on the phase plane. From the summary and discussion the isocline method is a most general and useful method. The only difficulty is in the labor required. Solution by digital computer reduces this difficulty and makes the isocline method a much more useful method in non-linear systems analysis and design applications.

~~""" (PAGE 1) DD i nov •• 1473I *T I ej Unclassified S/N 0101 -807-681 Security Classification A-ai4oa 247 1 Unclassified - Security Claaaification~~

~~KEY W O * D~~

~~Phase-plane~~

~~Isoclines~~

~~Trajectories~~

~~Isocline Method~~

~~DD .'2?.. 14 73 «>*«> Unclassified S/N 0101 -807-6821 Security Classification A- 3 I 409 248~~

~~thesH8297 Phase-plane methods.~~

~~3 2768 001 01547 2 DUDLEY KNOX LIBRARY~~