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DYNAhtIC

ANALYSIS OF PERMANENT

MAGNET STEPPING MOTORS

by

David

J.

Robinson

L e w i s

Resedrch Center

CZeveZand, Ohio

N A T I O N A L A E R O N A U T I C S A N D S PA C E A D M I N I S T R A T I O N W A S H I N G TO N , D . C M A R C H 1 9 6 9



DYNAMIC ANALYSIS OF PERMAN ENT MAGNET STEP PING MOTORS

By David J . Robinson

L e w i s R e s e a r c h C e n t e r

Cleve land , Ohio

NATIONAL AERONAUT

ICs

AND SPACE ADMlN ISTRA TI ON

..

For sa le

by

the Clear inghous e for Federa l Sc ien t i f ic and Technic a l In format ion

Spr ingf ie ld , Vi rg in ia

22151

-

CFSTI

pr i ce

3.00



A B S T R A C T

A dynamic analysis of permanent magnet stepping motors for multistep operation is

described. Linearized tra nsf er functions

for

single-step respo nses ar e developed. Mul-

tist ep operation, when the motor i s driven by a current source,

is

analyzed by usin g

phase plane techniques.

Fai lu re of the motor to operate for fixed stepping rat es and

load torques is

discussed.

Dimensionless cu rves showing maximum stepping rat e

as

a

function of motor p ara met ers and load torque a r e developed and experimentally verified.

The curves allowed the maximum stepping rate of a motor with viscous, inertial, and

torque loads

to

be predicted from simple measurements of the motor parameter s.

ii



D Y N A M I C A N A L Y S I S OF PERMANENT MAGNET STEPPING MOTORS

by Dav id

J.

R o b i n s o n

L ew i s R e s e a r c h C e n t e r

S U M M A R Y

Permanen t magnet stepping motors a r e being applied

as

digital actuato rs. Analytical

models of stepping motors have been developed that analyze the dynamic response to

a

sing le step . They have not, however, been expanded to analyze stepping motor perform -

ance during multis tep operation.

An analytical study

is

presented herein that examines stepping motor performa nce

during both single-step and multi step operation. A line arized single-step model

is

developed which allows the stepping motor to be defined in te rm s of

a

natural frequency

and

a

damping ratio. A nonlinear analysis is

also

developed which, when a constant cur-

rent source

is

assumed, allows the stepping motor to be analyzed for fixed stepping ra te s

and applied load torques.

It

was found that the permanent magnet stepping motor cannot

respond to a st ep command when the applied load torque becomes g reat er than

0.707

of

the motor's stall torque.

Also

the motor cannot follow a sequential

set

of st ep commands

if,

during the sequence, the r otor lags the command position by m ore than two steps.

Based on these results, we developed and experimentally checked

a set

of dimension-

less

curves which express maximum normalized stepping

rate as

a function of normalized

damping and normal ized load torque. Thes e curves allow the maximum stepping rate of

a

motor with viscous, iner tial , and torque loads to be predicted from sim ple measurement s

of the motor parame ter s.

INTRODUCTION

In recent ye ar s, the us e of digital control has found incr eas ed application in the field

of automatic control sys tem s. One inherent problem in applying digital control is in se-

lected suitable

digital

actuators. One promising actuator

is

the permanent magnet (PM)

stepping motor.

The application

of

stepping moto rs

is

not new. Pr oc to r

(ref. 1)

rep ort s that stepping

motors were

first

used in servomechanisms

in

the early

1930's.

During the development

..



of closed-loop servomechan isms in the World War II years, proportional analog activa-

to rs largely replaced stepping motors in actuator applications. With the space

age

and

the development of au tomatic digital syst ems, prob lems ar os e in using analog closed-loop

servomechanisms because

of

the need fo r digital- to-analog (D/A) conversion techniques.

This problem was particu larly im portant in space applications because of the

extra

weight

requir ed for the D/A equipment.

does not requi re D/A conversion.

Nicklas

(ref.

2 )

utilized a stepping motor

as

an actua-

tor in

a

spacecraft instrumentation system. Giles and Marcus (ref. 3) reported on the use

of stepping motors

as

control drum actuators for the SNAP-8 program.

To

increase the application of the stepping motor, much analytical work ha s been

done. Bailey (ref.

4)

compares stepping motors with conventional closed-loop positioning

systems . The stepping motor was modeled,

for a

single-step response, as

a

second-

or de r linea r approximation. O'Donohue

(ref. 5)

and Kieburtz

(ref.

6) have developed

simi lar mathematical models. These particular second-order models cannot be extended

to adequately des cri be the dynamics of the stepping motor fo r multistep inputs. Bailey

analyzed multi step inputs on a limited bas is, by assuming that the time between applica-.

tion of the ste p command was long compared with the sett ling tim e of the respo nse to each

ste p input.

investigate multistep operation. The analysis

is

restricted to permanent magnet stepping

motors; however, the techniques can

be

extended to other forms

of

stepping moto rs,

electrical, pneumatic,

or

mechanical. Stepping motor operation

is

analyzed to determin e

an expression for developed torque .

A

linearized single-step analysis is presented to ex-

pre ss motor performance in term s of

a

natu ral frequency and damping ratio. Phase plane

techniques a r e used to extend

the

analysis to handle multistep commands supplied from

a

constant current source. From the analysis,

a

dimensionless curve

is

developed

that al-

lows

the

maximum stepping rate

of a

motor with viscous, iner tial , and torque loads to be

predicted from sim ple measurements

of

the motor parameters.

Simple experimental results from two stepping motors a re presented.

These

results

are

used to check the analytical analysis.

The stepping motor ha s found increa sed applications as

a

digital actuator be cause

it

The work des cribed in this rep ort was conducted to develop analytical techniques to

DESCRIPTION

OF

PERMANENT MAGNET STEPPfNG MOTOR O PERATION

The permanent magnet

( P M )

stepping motor is an incremental device that accepts

dis cre te input commands, and responds to these commands by rotating an output shaft in

equal angular incr ements, called ste ps; one ste p fo r each input command. The angular

position of the output shaft

is

controlled by the init ial location and the number of input

commands received . The angular velocity of the output shaft

is

controlled by the rate

at

which the input commands

are

carried

out.

2



S i m p l i f i e d P e r m a n e n t M a g n e t S t e pp in g M o t o r

A

simplified PM stepping motor is shown in f igu re 1.

The

P M

stepping motor con-

sists

of a sta to r containing two o r more ph ases wound on salien t poles and a permanent

magnet rotor of high permeabil ity .

When a stator winding is energized,

a

magnetic

f l u x

is

set

up which interacts with the permanent magnet rotor.

The rotor

will

move in such

a manner that the magnetic moment of the permanent magnet will aline with the field se t

up by the sta to r winding cur ren t.

In referring to figure 1, ass um e winding 1 is ener-

gized such that a magnetic flux

is

set up with a direction fr om the fac e of sal ient pole 1

and into the face of salient pole

3.

The rotor will aline

as

shown. If winding 2 is ener-

gized such that a magnetic field is

set

up with a direction fro m the face of salient pole 2

and into the face of sa lien t pole

4,

the rotor

will

turn such that the south magnetic pole

of

the rotor alines with sali ent pole face 2. If winding 1 is energized in the manner pre-

viously described, the rotor will move

so

that the south magnetic pole is again alined

with sali ent pole face 1. However, i f the current is rev ers ed in winding 1 such that the

magnetic field direction

is

from salient pole

3

and into

salient

pole

1;

the rotor

will

move

such that the south magnetic pole aline s with sali ent pole 3 ins tead of s al ien t pole 1.

Thus, the position of the rotor

of a

P M stepping moto r can be determined by a discrete

se t of stato r winding excitat ions and a discrete s et of current re ver sal s in the stator

windings.

Salient

pole

2

Permanent

magnet rotor

Salient

p1e3f l.

Salient -

pole

4

Figure 1. - Typical representation o f perman ent magnet stepping motor.

3



Synchronous Inductor Motor

There are seve ral different

PM

stepping motor configurations available comm er-

cially. The most common types are derived from two- or four-phase ac synchronous

moto rs. The synchronous inductor motor is one of t he se types and has received consid-

erable

attention

in

the lite ratu re fo r stepping motor applications

(refs.

1

and

6

to

9).

nous

inductor motor is examined. Figure 2(a) shows

a

simplified cross section of a two-

phase synchronous inductor motor with four salient poles and five rotor teeth. The rotor

tooth pitch is 72O, while the sal ient poles are located every 90 . One ste p corresponds

to one-fourth of the ro to r tooth pitch,

o r

a rotor movement of

18 .

One complete revo-

lution of th e rotor corresp onds to

2 0

steps. Figure

2(b)

shows an expanded layout of the

rot or and stat or. The expanded layout is used to simplify the graphical representation

of the developed torque as the rotor moves relative to the stat or salient poles.

Stepping the ro to r is accomplished by rev ersi ng the direction of the c urre nt in one

pha se while holding the other phase constant. F o r a two-phase, four-salient-pole motor,

the sa lient poles can for m fo ur possible magnetic pole combinations:

and PZB

are

south magnetic poles.

and PZB are south magnetic poles.

and P 2A are south magnetic poles.

and

P 2A

a r e south magnetic poles.

(1) .

The resultant curve

is

the summation of t he torque curves due to the individual

phases and, as will be discussed later, is ass um ed to be sinusodal. The equilibrium

points

of

the resultant curve lie midway between the equilibr ium points of the individual

phases. Fig ure 3 @) shows the developed torque as a function of r ot or position fo r com-

bination (2) . The equilibrium points of the resu ltant curve remain midway between the

equilibrium points of the individual phases. Comparing the resul tan t curves of fig ures

3(a) and

@)

show that the equilibrium points have shifted by 1/4 rotor tooth pitch. In like

mann er, fig ure s 3(c) and (d) show the developed torque fo r combinations (3) and

(4).

If

the motor is stepped by sequent ially repeat ing the magnetic pole combinations, the ro to r

equilibrium points

will

be shifted by 1/4 rotor tooth pitch for each combination applied.

As

an

aid

in

developing som e general r es ul ts , the torque developed by the synchro-

( 1 ) Salient poles P I A and P 2A are north magnetic poles, and salien t poles PIB

(2)

Salient poles

P 1 B

and

P 2A are

north magnetic poles, and salient poles

PIA

(3) Salient poles

P 1 B

and

P 2 B

are north magnetic poles, and salient poles PIA

(4)

alient poles

P I A

and

P 2 B

are north magnetic poles, and salien t poles PIB

Figure

3(a)

shows the developed torque

as

a function of ro to r position fo r combination

4



Phase 2

' 2 A 7 I I I

South magnetic

I I

section

of

mtor

- -______

Phase

1

-

4

Step= 18"

s ,

5 Salient

- Phase

South magnetic disk

-

Nort h magnetic disk

1

Axial maanet 1

' I

P 2 B J

Phase

21

(a) Cross section.

'1A '2B

i

b,.j 1

Rotor tooth Pitch

CD-9835-09

( c ) Expanded rotor-stator layout.

Figure 2 -Simplified synchronous inductor motor.



I 111111

111I I IWII II1 1 1 1 ~ IIIIIIIIIIIII.~11111111111 II II1111111

II

A Stable equil ibri um points

Unstable equil ibriu m points

Salient Salient Salient Salient

pole pole pole pole

rResul tant torque

(a) Stator currents I1 and

12:

salient poles P1A.and PZA are no rth

magn etic poles; P1B a nd PZB ar e sou th magne tlc poles.

r Resultant tnrni ie

(b) Stator currents -11 and IF salien t poles P14 an d PN are no rt h

m agne ti c poles; P I A and P 2 ~re south magnetlc poles.

Resultant toroue

(c) Stator currents -11 and -I2: s al ient po les P and P ZB a re no r t h

magnetic poles; PI A an d PZA are sou th magnetic poles.

Resultant

toroiie

114 Rotor tooth pitch

(90 )

(d) Stator currents 11 and

-I2:

salient poles P IA and PZB are nor th

magnetic poles; P1 B and PZA are sout h magnetic poles.

Figure 3. -Torq ue as function of rotor posit ion

-

both phases energized.

Arrows indicate direction of developed torque a b u t equ il ibri um points.

6



A n a l y t i c a l D e v e l o p m e n t of S t e p p in g M o t o r T o r q u e

Torque is produced on the rot or of th e P M stepping motor as the result of an in te r-

action between the flux creat ed by the st at or windings and the permanent magnet r otor.

In the P M stepping motor, the s tat or windings a r e wound in coils surrounding each salient

pole. Assum e that an energized st at or winding produces

a

paral le l magnetic field of den-

sity

B'.

Also assume that the rot or of the P M stepping motor is a thin bar magnet.

The

torque re lat ions of

a

thin ba r magnet in

a

parallel magnetic field

are

shown in f ig ure 4.

' D

la) General configuration.

b ) Maximum and min imu m torque developed.

Figure 4. -Torque relat ions of thin-bar magnet

in

parallel

magnetic field.

The bar magnet tends

to

turn in a direction such that its magnetic moment vector

line up with the para lle l magnetic field. The torque developed

??D

is

the cr oss product

of the magnetic moment vecto r

will

and the magnetic flux density

E:

(Al l symbols are defined in appendix

A .

) When figure 4(a)

is

used, the developed torque

becomes

7



Figure 4(b) shows that the maximum torque is developed when the permanent magnet

is

perpendicular to the paralle l field. When the permanent magnet is parallel with the mag-

netic field, no torque

is

developed.

and to the length 2 of the ba r magnet. Thus,

The magnetic moment of the thin ba r magnet

is

proportional to the pole strengt h m

lMl

=KmZm

( 3 )

If

the parallel magnetic field

is

produced by

an

ideal solenoid of n tu rns , the magnetic

density is proport ional to the number of tu rns and to the magnitude of the cu rre nt I(t )

flowing through the solenoid turn s.

=

KInI(t)

Thus, the torque developed on a thin bar magnet in a para llel magnetic field is given by

T ~ -K

Km

ZmnI(t)

cos a

(5)

or

TD = KTI(t) COS a

(61

where

KT = KmLmnKI

Since the maximum torque developed occu rs when

a

= 0,

Tmax

=

KTI(t)

Thus, the developed torque can be expressed by

cos a

T~ =

Tmax

(7)

Equation (8) may not be an exact model when applied to the P M stepping motor, sin ce

the stator windings are not ideal solenoids and the rotor

is

not a bar magnet. The mag-

netic

fluxset

up by the st at or windings has

a

distribution effect acro ss the s urfa ce of the

salient pole face. Similarly, the rotor has distributional effects due to

its

configuration.

8



A Stable equil ibr ium pi nt s

Unstable equil ibr ium pi nt s

c w torque

CCW mb r

-

displacement

Salient

p l e

Salient

p l e

Salient

p l e

Salient

p l e

p2A

'1A 2B

'1B

CW mtor

displacement

Rotor Ish ow i n

Reference

pint

~ ~

7

c w

torque

CCW m b r

-

displacement

CCW torque 1

tal Reference p si tio n: sd1ier.t

p l e s P

and

PZA

are no rth magnetic

ples;

PB

and

PPB

are south magnetic

pk .

Ibl

Step

1

Sdlinet pies

PB

ana

P2

are no rth nidynetic

p les ;

PIA and

PZ B

are

m t h

magnetic

pies.

CW robr

displacement

tcl Step

2

salient

p l e s

P 1 ~

nd

P Z B

are nor th magnetic

p l e s ;

PIA

and

P Z A

are south magnetic

poles.

CW rotor

displacement

Id1 Step 3: salient pies PZB and

PIA

are no rth magnetic

p l e s : PIB

and

PX

are south magnetic

p l e s .

CW lorque

CCWispldcementtor

c c h lorque

CW mtor

displacement

1 4

114 Rotor tooth pitch

L 18G 366

Rotor PIorition.

E.

d q

(el

Step

4:

salient

p l e s P I A

and

P ~ A

re nort h magnetic pies;

PB and

P p B

are south magnetic ples.

Figure

5

- Rerulldr.l torque as function

of

ro to r ps i l ion lor lour-step sequence.

Arrows

indicde direction

o f

developed torque a bu t equi l ibr ium p ints .

9



In

addition, the magnitude of the torque depends upon the number of phases energ ized,

since the magnetic str uc tu re of the moto r will be used m or e efficiently when

all

the

phases are energized.

an angular rotation q~ in mechanical degree s. Define an

angle

0 such that, when the

rotor moves 1/4 tooth pitch,

0

vari es by 9 The angle

0

is relate d to the mechanical

angle of rotation

q~

taken during each ste p by the number of teeth on the ro to r.

For a given stepping motor, each rot or movement of 1/4 tooth pitch corresponds to

The resultant developed stepping motor torque from each step command can be general-

ized by assuming that the developed torque varies sinusoidally when the rotor moves

1/4 ooth pitch. By superimposing the torque produced by each pole pai r, the developed

torque can be e xpresse d by

sin

0

T~

=

Tma.x

Figur e 5 shows the resul tant curves fo r developed torque

as a

function of position fo r

a four-step sequence in

a

synchronous inductor motor. In figure 5(a), equilibrium point

X

is

an arb itra ry re ference point. A s the rotor is moved, the resu ltant developed

torque is given by

T~

=

- T ~ %

in e

Figu re 5(b) shows the resu ltant torque curve for the

first

step command. The torque is

given by

TD = -Tma sin(8 - 9

cos 0

T~ =

Tmax

Figur e 5(c) shows the resultan t torque curve fo r the second step command. The torque

is

given by

TD = -TmZ sin(8

-

180')

T~

=

Tmax sin 0

(15)

Figur e 5(d) shows the re sultan t torque c urve for the third step command. The torque is

given by

10



Figur

TD = -Tm, sin(8 - 270')

TD

= COS

e

5(e) shows the resultant torque curve fo r the fourth st

is

given by

TD = -Tmm sin(8 - 360')

T

-

-Tmm sin

8

D -

command. The torque

This is the same resul t

as

fo r the refe ren ce point, equation (11). Thus, the torque de-

veloped in a P M stepping motor fo r any st ep command sequence can be given by a com-

bination of t he four equations:

TD = -Tmn sin

0

TD

=

Tmax

T~

=

Tmax sin 8

COS

e

TD = -Tmm

cos 8

Equation (7) stated that the maximum torque developed pe r p hase w a s related to the

stator current by

Tmax = KTI(t) (24)

If the time between st eps is long compared with the rise t ime for the stator current, the

equation for T can be written

Tmax = K 1 -- K 2 -- K I

(25)

The developed torque fo r one phase be comes

TD

=

KTI sin 8

11



TO

Resultant torque

t

/

W torque

CCW

rotor

displace- - /

\

ment

CCW torq ue

CW rotor

displace-

ment

-225 -180 -135 -90 -45 0 45 90

Rotor position, 8, deg

l l

135 180

225

Figure 6. - Torque as function of rotor position f or generalized permanent

magnet stepping motor - b t h phases energized.

With the aid of fi gure 6 , equation (26) can be extended to include the case when both

phases are energized and the magnitudes of the s ta to r cu rre nt s are assumed to be equal:

TD

= -KTIkin(8 -

4

+ sin(8 +

45Oj

(28)

TD = -

*KTI sin 8

The minus s ign in equation

(29) is

due to the reference point chosen.

It

indicates that

for a clockwise (CW)

8,

the developed torque will move the ro to r in a counterclockwise

(CCW) direction.

P M S t e p p in g M o t o r S i n g l e - S te p

A n a l y s i s

A

schematic representation of the

PM

stepping moto r

is

shown in fig ure 7. The

voltages supplied to the st at or windings a r e given

by

dIl(t)

l( t) =

FtIl(t)

+ L

t

12



dI2(t)

Ev

2(t)

=

m2(t)+ L

t

The induced voltage E, genera ted in the s ta tor is due to the magnetic ro tor moving

rela-

tive to the stator magnetic poles.

during th e ent ire step. However, due to the geometry

of

the stator, the magnitude

of

the

s ta tor flux density is distributed in the sta tor . When thi s distribution is assumed to be

sinusoidal and Maxwell’s equations are used, the induced voltage can be given by

It is assumed that the rotor velocity vector rem ains perpendicular to the st ato r

flux

Thus, the differential equation fo r the s ta to r voltage pe r winding becomes

In responding to

a

st ep command, the developed torque produced by the motor must

(1)

The inerti al torque Jb2 cp (t) /d tq , which includes the inert ia of the motor and

the ine rt ia of any load imposed on the stepping moto r shaf t

(2) The viscous damping torque D[dcp(t)/dg, which includes the motor damping and

any viscous damping on the motor shaft

(3)Torque due to Coulomb fricti on Tf {kq(t) /dtl/ Odcp (t)/dt n} (Coulomb friction is

defined as a friction independent of the magnitude of th e relat ive velocity of t he

su rfa ces in contact, but dependent on the direction of th e relati ve velocity.

)

overcome the mechanical load placed on the rotor. The mechanical load includes

(4) xternal load torque applied to the motor shaft TL(t)

In refe rri ng to figure 7, the equation of motion of the PM stepping motor can be ex-

pressed by

TD = KT12(t) co s e(t)

-

KTIl(t) sin e(t)

de t)

(35)

- J d26(t) D d6(t)

N~~ dt2 NRT dt

-- -+-

13



Figure 7.

-

Schematic representation of permanent magnet stepping

motor

for

a

single step.

L i n e a r i z e d M o d e l

Equations ( 33 ) to (35)

are

nonlinear differential equations because of the torque-

These equations can be linearized about an

roducing te rm in each of the equations.

operat ing point by expanding them in a Taylor s er ie s (ref.

10)

of the fo rm

Ix2,o

IX2,O

Linearizing equations

( 33 )

to

(36)

resu lts in

KT12(t) cos O t ) - KTIl(t) sin O t ) = KT12,0 cos Oo

-

KT12,0 sin Oo AO(t)

+ KT cos

do A I 2(t)-

KTIl7 sin

Oo

14



J

d2e(t)

D

de(t) dt

----

NRT

dt2

NRT

dt

+ -?-

{

)o

+

P I

N~~

E2, + AE2(t)

=

R E 2

+

A12(tj + L

{

+

A [ y ] )

3

)o cos eo

NRT

90

E190 + AEl(t)

=

RIIl,o +

AIl(t)l

+

L

{(2)o

A +

(z)o

in eo

+- s i n 80

A - :

+-- 2TC:)o cos

' 0

Transfer Function About an Operating Point at Beginning

of

a

Step

The linearized equations are first considered about an operating point at the begin-

ning of a step. The constraining equation neglecting friction becomes

1 5



,

..

..

.

Now assum e that the vol tage applied to winding 1

is

held constant and the motor is stepped

by energizing winding 2. Assume the initial conditions

Also, since the voltage applied to winding 1 is being held constant, hEl(t)

=

0. From the

differen tial equation at the operating point

-KTI1,O sin eo = T L , 0

(42)

Equation (42) stablishes

eo

at the beginning of the s te p

T

(43)

o

=

sin

1

L,O

-KT1l,

0

F o r an excursion about the operating point, equations

(37)

nd

(38)

an be combined

as follows:

KT cos

eo

A12(t) - KTI1,

X AL + ATL(t)

(44)

Making us e of the ide nti tie s

TL, 0

sin

eo = -

KT1l, 0

2

COS

e,

=

+ X , O

-

TL,O

V

KT1l, 0

(45)

(46)

eqdation (44)can be written

16



+-

TL9

AIl(t)

-

ATL(t)

(47)

I1,O

Also, equations (39) and (40), fo r a n excursion about the operating point,

can

be written as

AE2(t)

=

RA12(t) + L A

A[y]

A tra nsfe r function relating r oto r position, st at or voltage, and load torque can be de-

rived by taking the Laplace tr an sf or ms of equations (47) to (49). The derivations of th e

Laplace transformations are found in appendix

B.

The resultant l inearized t rans fer func-

tion about an operating point at the beginning of

a

s tep

is as

follows:

AO(S)

=

JLS3 + (DL

2

KT

-

:io

[m2(S)

+%]

(R

+

LS)

KT’l. 0

+

J R ) S ~

t R +

TKV

N~~

+LKT1l, Of

=TI

1

,ofT

T’l,

0

is 0,

equation

(50)

can be reduced to

L ,

0

If the initial load torque

T

AO(S)

=

JLT3 +

‘21

-

(R + LS) ATL@)

KT k 2 ( s ) + S

=T1l,

0

(51)

17



l1111111 I I

Equations (50) and (51) show that the resulta nt lin eari zed t ran sfe r function about an op-

erat ing point

at

the beginning of a step results in

a

third-order system.

The resultant

lineariz ed tra nsf er function can

be

used to examine th e effect of

an

init ial value of load

torque on the ste p response.

It

can also be used to examine the effects of inductance,

generated back electromotive for ce (emf), ine rtia , and damping on the s tep response.

However, the choice of an opera ting point at the beginning of a step makes experimental

verification of th e resu lt ing model difficult.

Transfer Functions About an Operating Point at End of a Step

Consider the tra ns fe r function of the

P M

stepping motor about an operating point when

the motor

is

in equilibrium

at

the end

of a

ste p with both phases energized. Thi s opera-

ting point

is

chosen because the resulting

transfer

function can be easily verified experi-

mentally. A t the end of the step , the initia l conditions are Bdu)/(dt)lo = ed20)/(dt2)lo = 0

and 11, = 12,

=

I.

For an excursion about the operating point the constraining equation

at the operating point

is

KT12, cos

eo -

KTI1, sin

eo

=

TL,

(52)

or

L,

0

cos

eo

-

sin

eo

=

-

T1

(53)

Equation (53) establ ishes

eo. If

TL,

= 0,

then

eo =

45'. The linear ized equations

about the operating point equating friction are

-KTIIIsin

eo +

cos

eo] A e ( t ) +

K~ cos

eo

AI^(^)

- K~

sin

eo

Ml(t)

AE2(t)

=

R M2(t)

+

L

cos eo A

(55)

18



AEl(t) = R

AI l ( t )

+ L A

sin eo A

The Laplace transformations

of

equations (54) to (56)

are

KT cos Oo AI2@) - KT s in

Go

AI1@)

- ATL@)

A Q i S )

=

I

JS2

DS

+ K ~ Iin eo

+

cos eo]

N~~ N~~

Kv

os

OoS Ae( S)

AE2(S)

+

~ N~~ -

A q S )

=

R

+

LS

AE1(S)

A-

in BoS A @ )

s N~~

q s )

R + LS

(57)

Combining equations (57) to (59) gives the result ant linea rized equation fo r both phases

energized about an operating point at

the end of

a

step

(59)

S

KT

cos

€lo

~~

_ _ -

+ LS S R + LS

Ae(S)

=

JS2

+-

DS

KTKv

,os

2 eo

-

s in2 Bo) S

+

KTI(sin eo + cos

NRT NRT NRT(R+LS)

F o r no initial load torque

(TL,

= 0), the constraining equation (eq. (53)) yield s

eo = 45'. Thus,

0 * 7 0 m T [+ LS AE2(S)

L

]

- 0 ' 7 0 x T ~ 1 ( S )+ LS

+?d-

TL(s)

c

Ae(S)

=

J s L +-

DS +&KTI

NRT N~~

19



For load torque disturbances about the operating point at th e end of th e st ep and the initial

load torque equal to

0,

the tr an sfe r function is

a

second-order system

+ - + -

JS J

This result is important fo r two reasons. First, it allows the stepping motor to be de-

scribed in te rm s of a natural frequency and damping ratio. Second, these par ame ter s

can easily be determined experimentally.

Equation (62) ha s been developed in t e rm s of 8 , such that one ste p equals 9

Equation (62) can

be

written in te rm s of the actual mechanical degrees fo r

a

particular

stepping motor. Since

.

J J

it

follows that

JS J

The natural frequency obtained from equation (64) is given by

The natural frequency of the PM stepping motor is increased by

(1)

Decreasing the inertia

(2) Increa sing the number of ro to r teeth

(3) Increasing the magnitude of th e sta to r cu rr en t

The upper bound on the natural frequency due to iner ti a

is

often limi ted by the load iner-

tia. However, minimizing the rot or i ner tia in cre ase s the natural frequency when the load

20



inertia is not the limi ting factor. The natural frequency of the stepping motor can also be

expressed in te rm s of the stall torque TS with both phas es energized

TS = v 2 KTI

Thus,

wN

d,gTs

The damping rati o of equation (64) is given by

D

J

2 C W N

= -

The damping ratio is incr ease d by increasin g the viscous damping, o r by decrea sing the

inertia. The damping ratio

is

dec reased by incr easi ng the number of rot or teeth o r by

increasi ng the maximum torque by incre asing the sta to r current.

The natural frequency and the damping ratio of the lineari zed model of t he

P M

step-

ping motor can also be evaluated by considering

a

sma ll change in A 0 from the equilib-

riu m point. Conside r the linearized equations about the point of s table equil ibrium with

the st at or cu rre nt held constant and no applied load torque.

duce to

The linearized equations re-

J

-K I(sin eo + cos Go) AO(t)

=

T

N~~

The Laplace tr ansform of equation (70) is

21



J

For a small initial displacement AO(0) and initial conditions Oo =

45'

and

2 2

A[dO(O)/dt] = (dO/dt)O

=

(d O/dt )o

=

0, equation

(71)

can be rewritten

as

5

- N ~ ~K T I AO(S) = (JS2+ DS) AO(S)

-

(JS + D) O(0)

where

(S

+

AO(0)

AO(S) = .

(73)

- .

JS J

Equation (73)can be used to evaluate the nat ural frequency and the damping r atio of the

lin eari zed model about the st able equilibrium point.

PHASE PLANE ANALYSIS

The lineari zed mathem atical model of the PM stepping motor w a s developed to ana-

lyze the single-step response of the PM stepping motor.

For

multistep operation, the

resp onse of the PM stepping motor fo r

a

given load torque

is

governed by

(1)

The ra te of the input pu lses

(2)The curre nt tran sien ts in the sta tor windings

(3) The mechanical pa ra met er s of the roto r

The lineari zed model cannot be used to analyze multistep resp ons es because the model

fails to account fo r motor f ail ur es caused by exce ssive input stepping command.

22



The transien t response of the st ato r curre nt has a significant effect on the stepping

motor response

as

the stepping rate is increased. Even i f the back emf induced in the

sta tor is not significant, the curr ent turnon transient affects the maximum torque de-

veloped by the motor. Consider the elect ric al (L/R) time constant of the st at or windings.

If the time between the application of step commands approaches this time, the c urren t

w i l l not reach

its

expected value. The maximum torque developed by the motor is

re-

duced, and thus, the natura l frequency of t he stepping motor

is

reduced . The amount of

load torque which the motor can ste p against

is

also reduced.

rent transients. The driv e circuit tends to act as a constant curren t sour ce by control-

ling the stat or cu rrent independently of t he inductance

or

generated emf. A typical con-

stant current drive source used to compensate a stepping motor

is

presente d by Ze ller

(ref. 11).

a

current source for stepping motors. The stator current s

are

assume d to be constant

and to hzve magnitudes equal to

I

o r

-I

during each step.

In general, stepping motor driv e circui ts are designed to compensate for sta to r cur-

For

multistep analysis,

it is

assumed that a drive circuit

is

used which approximates

Analytical

Model

Assuming Constant Current Source

The assumption of a constant curr ent sou rce simpli fies the analytical model of the

P M stepping motor. It also per mi ts the respons e of th e motor to be studied fo r a ser ies

of input pulses. Because the cu rre nt is independent of dI(t)/dt and dO(t)/dt, the st at or

winding equations become

E l

=

d R

(74)

E a =

*IR

(75)

The sign in each equation is determined by the input pulse sequence.

differential equation. For he rotor at the equilibrium position, the developed motor

torque h as been shown to be

The differentia l equation of rot or motion can be reduced to a second-order nonlinear

A

step command requ ires t hat the rotor angle

8 advance

9

The developed torque

generated by the motor to accomplish thi s ste p is given by

23



T~ = K ~ Iin

{

90' - [e(t) -

4 5 g }

+ KTI sin

{ g o o

- [O(t) + 45O]}

or

-

0(tU

+

s i n k 5 O

-

0(t]}

Equation

(78)

reduces to

TD = KTI cos 0(t)

Equating equation (79) to the mechanical load of the motor results in

Tf

J d20(t) D dO(t)

h K T I c o s 0(t) =

__

+--

dt

NRT dt2 N~~

-

Rotor command

\

position, step 1

dt +

TL(t)

Stator /

winding 2 7

\

I /

dt'

Rotor ini t ial posit ion

-

Rotor command

position, step 2

+

-sf

I \ I

Rotor command

- -

~ ~~

-

/ I

\45" posit ion, st ep4

/ \

/

Rotor command

position, step 3

winding

2

V

5

~

Stator

/

\

windingtator 1

\

(77)

(79)

Figure

8.

- Schematic representation

of

rotor command posit ions. Mult i step operation.

24



M u l t i s te p A n a ly s i s A s s u m i n g C o n s ta n t C u r r e n t S o u r c e

Equation (80) can be extended to include multis tep operation. Each st ep command

ca lls fo r the rot or equilibrium position to shift by 90'. Examine figure 8 and assume the

rotor is initially at 0 The equations for the developed torque for a se ri es of ste ps are

then

Step 1:

T~

=

K ~ Iin { 90' - [e(t) -

45q}

+

K ~ I

in { 90'

-

[e(t) +

45'1)

(81)

Step 2: T D = KTI sin { 180' - [e(t) -

45.1)

+

KTI sin { 180' - [O t ) +

45.1)

(82)

Step 3:

T D =

KT

I sin

{

270'

- [O t ) - 4501) +

KTI sin

{ 270' -

[e(t)

+ 45j}

(83)

Step 4: (84)

Step 5: T D = KTI sin

{450°

- [O t ) -

45.1)

+ KTI sin {

450'

-

[O t )

+

45q)

(85)

and so forth.

These equations reduce to

Step 1:

Step 2-

Step 3:

Step 4:

Step

5:

T~ = f i ~ ~ ~in e(t)

T~

= - I ~ K ~ Iin e(t)

and so forth.

The basic equation for multistep operation becomes

25



d W )

(91)

d20(t) D dO(t)

G K ~ IinEc - e(t,l

--

NRT dt2 NRT dt

N o r m a l i z e d M o d e l for M u l t i s t e p O p e r a ti o n

The second-orde r nonlinear di fferential equation obtained in equation (91) lends itself

to phase plane analysis. This equation can be considered

as

the tran sient solution

to

a

s e t of ini tial conditions. By specifying the initial conditions

O 0)

and dO(O)/dt, the solu-

tion for all positive time is completely determined.

An advantage of t he phase plane

analysis is that trajectories representing the transient solution of the equation can be de-

termined without solving the equation fo r the dependence of position on time .

Equation (91) can be normalized to reduce the number of param ete rs. By using the

linea rized analysis, the normalization can be perfor med to allow the use of experimental

data. Dividing equation (91) by the stall torque TS = 2 K I results in

( d- T )

(92)

in[Oc

-

e t j

=

J d20(t)

D

d W )

Tf

dt

TL

dt

d-

K

I

G K T I

I

dt

1

i K T I d t2 NRT G K T I

N~~

Recall that the natural frequency of t he linearized second -order model is given

by

Equation (92) can be simplified by defining a new time parameter in ter ms

of

the natural

frequency:

26

T = W t

N

(94)

(95)



Equation (97) can also be defined in te rm s of the damping rat io

J

or

-

D =

2{

Also, normalizing the friction and load torques gives

Tf

-

Tf

Tf = _ -

-

KTI TS

and

TL

=

+KTI TS

Thus, equation (92) can be writt en with 8 and BC specified in radi ans

(99)

(102)

Time can be eliminated from equation (102)

by

defining

27



Rewriting equation (103) n te rm s of position 8 and velocity V result s in

v -

-

dV

sin(OC-

6 ) = V

DV

+

Tf

de

The phase plane solution to equation (104) s a plot of velocity V as a function of po-

sition 8. The initial conditions V(0) and

e 0 ) locat e an initi al point in the phas e plane.

The trajec tor y through this point desc rib es the res pon se of equation

(104)

or

all

positive

time. The slope of the trajectory is given by

F o r

a

given motor with a specified value of fric tion, the slop e of th e tr aj ec to ri es is

a

function of three

variables,

Thus, the unique respon se to

a

s e t of initial conditions must be specified in ter m s of the

load torque, assuming that the motor is loaded by a speed independent torque of magni-

tude TL.

In

the normalized equations,

8

and 8, mu st be speci fied in te rm s of radians.

P h a s e P l a n e A n a l y s is by D e l ta M e t h o d

There

are

sever al graphical methods to determine the phase plane trajec tories . Fo r

single-step analysis, the most convenient method

is

the delta method (ref. 10). The del ta

method assu mes that the trajectorie s can be approximated by arcs

of

cir cle s with origins

on the

axis.

In fig ure 9(a),

6

gives the cen ter of a circ ula r a r c through point P1.

Comparing simil ar triangles,

28



dV

dB

-6 = e + v -

To determine a trajectory using equation (108), first select

a

point

P1

and solve fo r

del ta. The value of de lta will determ ine t he cente r of an a r c rc that w i l l pa ss through

point

PI.

Notice that the cen ter

of

the a rc always

lies

on the &axi s. After drawing a

segment through point P1, select a point P 2 on the segment. Compute a new de lta and

new a r c center, etc.

Figure 9(b) ll us tr at es th e method. Using equation (104) with

V

I

r-

(a) Determination of center

of

c i rcu lar arc

t h r ough p i n t P i .

1.

O

-.51

I 1

0

. 5

1.0 1.5

2.0

Rotor position, rad

I

d

0

45 90

Rotor position, deg

(b) Use of

delta method.

V dVld0 = coS(0

-

V).

F igure 9. - Graphical solution of a phase plane trajectory by delta method.

29



-

V(0) =

e ( 0 )

= Tf

=

TL = 0, Bc

=

a /2 ,

and D = 1 . 0 results in

- -

dv

- cos

e - v

de

dV

de

- 6

= e + v- = e + COS e - v

The pro ces s fo r computing the traj ect ory continues until th e equilibrium point

is

reached.

Tim e can be computed along the trajec tory (ref. 12).

The relation between time, po-

sition, and velocity

is

given by

T = /L

de

V

For a

segment along the traje ctor y,

or

The tota

e

TN+l - TN =

'av

N+1 - %J

- (vN+l + vN)

N+1 - N =

time is the summation of the inc rements of ti me along the en ti re trajectory.

Single-step response.

- For a

single-step response t he phase plane

is

useful in ana-

(1)

The transient response to a se t of ini tial conditions

(2 ) The effect of load torque TL on the tra nsi ent response

(3)

Other nonlinear effects, in par ti cul ar Coulomb fric tion , provided they ar e built

lyzing

into the model

The equation for

a

single-s tep response was given in equation (104) as

v -dV -

sin(Qc

- e = V -

+ DV + Tf

d e

- -

The st ep response

will

first be analyzed fo r Tf

=

TL

=

0. Equation (114) reduces to

30



-

dV

d e

sin(OC -

0 )

= V DV

Assume that the rot or

is

initially

at

rest and that the step command will move the rotor

from 0 to 90 as shown in figure

8.

The equation for the step becomes

sin(:

- e) =

v V

+-

V

d e

or

dV

-

d e

COS

e =

v- +DV

-

For

a

given value of the normal ized damping rat io D, equation (117) de sc ri be s

a

trajec-

tory in the phase plane

for

a given set of initial conditions.

Normal zed

time,

Normalized T

-1.01

0

.5 1.0

1.5

2.0 2.5

Rotor position, 8, r ad

I I I

Rotor position,

8,

deg

0 45 90

Figure 10. - Effect of norm alize d damping on single- step response. Nor -

malized frict ional torque Tf, 0; rotor command position BC, 1.57 radians.

31



- -

Figure 10 shows th e effect of D on the st ep respons e. Increasi ng D makes the

re-

sponse le ss oscillatory and increa ses the

rise

time for each step response. In figure 10,

the normalized time

is

given for each ste p respon se when 8 reaches 1.50 radians o r

about 95 perce nt of t he command

value of 1.57 rad ians (90'). The normal ized ti me var-

ies

from 1.97 to 5.74 as

D va rie s from 0.25 to 2.0.

Effect of load torque.

-

The addition of load to rque cau ses an offset in the equilib-

rium position of the roto r. Load torque also slows down the st ep response.

With load

torque applied, the initial position of the ro to r

is

dete rmined by the equilib rium point of

the previous step,

Figure 11 illust rates this effect. Figure l l ( a ) shows the developed

motor torque when TL = 0. Assum e the equilibrium position with no load torque before

the s te p command is given is 0'. The previous ste p command required to re ach th is po-

sition is given by

-

dV

d e

sin(O0

- e = v DV

o r

-

dV

dB

-sin

0 = V - + DV

The equilibrium position is given by

1

sin e = 0

or

J

=

Oo(O

radians)

These fo rm the initial conditions fo r the next step.

Notice that at th is position no motor

torque is developed. If the rotor is moved about thi s point, the motor develops a torque

which tends to d rive it back to the point of equi librium. When the next ste p is com-

manded, maximum moto r torque is applied to the rot or fo r acceleration.

A s

the rotor

moves toward 90°, the amount of available torque

is

reduced unt il the 90' posit ion is

reached.

A t

this

point, the developed torque

is

again

0.

F o r the previous s tep , the equilibrium position ha s shifted 23.6'.

fact that som e motor torque

is

required to hold the load torque.

position of -23.6' w a s determined from

Figure l l ( b ) shows the developed motor torque fo r a normalized load torque of 0.4.

The initial equilibrium

This shift is due to the

-

-sin e = V ?

+DV +

0.4

(121)

d e

32



Developed motor tor que

against position before

step command

Developed motor torque

for step command

Usable torque fo r

e

Initiall, Fin al

-90 0 90

Usaole torq ue for TD

acceleration

1

\

\

- e

+

\

-90 0 90

(b) TL = 0.4.

No

torque available

for acceleration.

Motor cannot s tep7 lT D

In i t ia l

-90 90

Rotor psit ion, deg

(c)

TL

=

0.707.

Figure

11.

- Effect of.normalized load torque

TL

on developed

motor tor que for a step command. Arr ows indicate direction

in wh ic h developed motor tor que acts.

/

33



When the rotor reac hes equilibrium,

-s in 0 = 0.4

8 =

-

23.6'(0.645 radians)

The shifted equilibrium position reduces t he amount of motor torque that is available to

acce lera te the rotor to the next step. The torque that

is

available for acceleration is

given by

-

dV

de

COS

8 - T = V - +DV

The amount of motor torque available fo r acceleration de cre ase s with increasi ng load

torque until

TL

0.707. From figure ll ( c ) for

a

load torque of 0.707, the rot or position

offset

is

45'. When the next s tep

is

commanded,

cos(:

-

0.707)

= V

+

G V

d8

dV -

0

=

V DV

de

there

is

no motor torque available to accel erat e the ro to r, and

a

step command cannot

be accomplished. The motor can sta tica lly hold

a

load torque equal in value to the

stall

torque of the moto r. But the motor cannot step

a

load torque that

is

equal to

o r

greater

than 0.707 of the motor's stall torque TS. Thus, a normalized load torque

of

0.707 will

cause the

P M

stepping motor to fail to respond to a st ep command.

Figure

1 2

shows the st ep respon se in the phase plane for various values of load

torque. Normalized damping

is

given as

1.0.

Notice that although the initial and final

roto r positions are offset

by

the load torque, the d istance between these positions

is

90'

(1.57 radians). Thus, load torque does not al ter the s iz e of the st ep, just the initial and

final positions of the rotor .

Figure

1 2

il lu str ates the effect of load torque on the speed of response . The maxi-

mum velocity reached is reduced with incr easing load torque. Thi s is a direct result of

the available motor torque to acceler ate the rotor. Notice that with increasing load

torque the amount of posit ion overshoot

is

reduced. Also shown in figure

1 2

is the nor-

malized time for the r ot or to move

50

percent of

its

step. The time is incre ase d with in-

creasing load torque.

34



1.0-

-45

T

B (1.33) 7

\

A

(1.581

0 45

Rotor position, deg

5

2

90

Figure 12.

-

Effect of n ormal ized load torqu e o n single-step response. Normalized damp-

ing,

1.0;

fr ictional torque, 0.

Normalized

Rotor position, 8, rad

I I I I

0 45 90 135


wit hou t load torque. Norm alized damping, 0.5; nor mal ize d load torque ,

0.

f i g u re

13. -

Effect of n ormal ized Coulomb fr ic tion on single-step response

35



Effect of

-

oulomb friction. - Coulomb fric tio n changes

a

given ste p size. Figur e

13

illus trates the ste p response for various values of Tf with TI;

=

0 and D

=

0.5.

N e a r

the command position, the developed motor torque becomes sm all . If Tf

>

cos 0 when

V

=

0, the friction will hold the roto r, causing an offset from the command position.

Thus, the equilibrium position lie s in

a

band about th e command position. The

size

of

the band

is

determined

by

the

amount of fric tio n pre sen t. The speed of response

is

also

reduced by friction. The reduction is small, however, unless a la rg e amount of frict ion

is present.

- - -

1 . o r I

>

-

Normal zed

-.

-.

5 0 . 5

1.0 1.5

2.0

Rotor position,

0,

rad

I I I

0

45

90

Figure 14. -Effect of normalized Coulomb fric t ion o n single-step response

wi th load torque. Normalized damping,

0.2;

norm alize d load torque,

0.5.


Combined effects.

-

Figure 14 il lu st rat es the effect of fri cti on combined with load

torque.

The effects are si mil ar to those experienced with load torque alone. The final

rotor position for a given tep is determ ined by the values of load torque and fric tion

present. The effect of Tf on stepping motor operation will not be considered further in

thi s repo rt since the frictional torque magnitude is generally sm all and

its

effects ar e

somewhat si mi la r to the load torque effects.

Multistep analysis. - The real advantage of using th e phase plane i n analyzing the

stepping motor is that the analysis can be extended to include multistep operation. In

multistep operation, each input ste p to the motor commands the rot or to move to a new

position 90 away fro m the existing command position. The dynamic equation tha t re-

lates the motor movement is equation (104). If th e ti me between the application of input

s teps is long, the ro to r may r each the new point of equilibrium befo re the next step com-

mand is applied. A s the rate of step commands is increased, the rotor may not reach

the new equilibrium point. At the instant the next ste p command is applied, the position

36



and velocity of the ro to r fo rm init ial conditions fo r the new equation of ro to r motion. The

step commands

are

applicable to unidirectional o r bidirectional r otor movements.

The phase plane fo r multistep analysis can be considered as

a

ser ies

of

trajectories

fo r each possible dynamic equation given by the ste p commands. Tim e is computed along

the traje ctor ies and

is

used to indicate the time

for

the application

of

the next st ep com-

mand.

A s

an illustration, consider th e following se ri es

of

input commands with Tf

=

0:

-

Step 1:

Step

2:

Step

3:

Step 4:

Step

5:

-

dV

d e

C O S ~ = V - + D V + T

-

sin(n- - e = sin

e

= v d

+ DV

+ T~

dB

-

dV

de

L

in(27r

- e) =

-sin

e = v DV

+

T

V -

sin(: -

> =

cos e = v- B +Dv + TL

-

igure

15

shows

a

phase plane port rait fo r this se ri es of steps.

The trajectory is for

D = 0 . 2 5 and TL = 0.

Consider

a

normalized stepping period of AT = 1 . 3 1 .

The s tep commands a r e applied

at

the following normalized times:

Step 1:

T = , 6 = 90

C

Step 2:

T =

1 . 3 1 ,

BC =

18Oo(n radians)

Step 3: T

= 2 . 6 2 , 8 = 270

C

Step

4:

=

3 . 9 3 ,

Bc =

360°(277 radians)

Step 5:

T =

. 2 4 , 8

C = 4

radians)

37



5

-2

e

Command position: Fir st step

(90 )

Second step (180") Th ir d step (270") Fo ur th step

(360 )

Fifth step

(450")

~~ - I

2 -

-1

Step 5

Command

First step

Second step Th ir d step

Fou rth step Fifth step

position:

(90")

(

180")

(270") (360") (450")

I I

5

I

4

I

3

I

2

Rotor position,

8,

rad

I

6

I

7

8

Figure

16. -

Phase plane portrait for mult istep inp ut with normalized load torque equal

to

0.2.

38



The initial conditions for step

1

are

V(0) = 0(0)

=

0. The trajectory is governed by

equation (126) nd is shown in figure 15 between 0

=

0,

V

= 0, and point

A.

A t

point

A ,

the total response tim e

T

is

1.31,

and step

2 is

applied. The values of

V

and

0

at

point

A

become the initial conditions for the r espo nse of ste p

2.

The equation governing

step

2 is

equation

(127).

The trajectory moves from point A to point

B. A t

point

€3,

the

total time T is

2.62.

The third step command is applied and is governed by equa-

tion

(128).

The values

of

V and

0

at point B become the initial conditions for equa-

tion (129).

changed by simply changing the ti me between steps AT

ceeding commands are not applied. Curve

OA

is for

a

single-step command, OB' is for

a two-step command, and

so

forth. The response is oscillatory in each case,

but

each

trajectory reaches

its

cor rec t command position. The oscillations

of

the trajectories

as they approach the equilibrium position are a function of the maximum velocity reached

during the stepping sequence.

torque TL of 0.2. The addition of load torque cause s the ro tor velocity to build up at a

slower ra te than in the ca se of no load torque. Without load torque, the rotor velocity

peaks during the third st ep. With the load torque, the velocity does not peak until the

f i f t h

step. The dashed traj ecto ries in figure

16

show the phase plane t raje ctor ies i f the

succeeding commands

are

not applied.

A

comparison of the se tra jec to rie s clearly shows

the applied load torque slows down the speed of response .

This process

is

continued fo r each st ep applied. The ra te of stepping can be

The dashed traj ecto ries in figure 15 show the phase plane traj ecto ries if the suc-

Figure

16

is

a

phase plane por tra it of the s am e stepping sequence but with

a

load

P h a se P l a n e A n a l y s i s b y M e t h o d

of

I s o c l i n e s

The phase por tra it f or the differential equation of the

P M

stepping motor can be used

to determine whether the stepping motor will fail for a given stepping rate with a given

value of load torque. The de lta method prov ides a convenient way of constructing the

phase trajectorie s for a given

set

of initial conditions and lends itself to computer solu-

tion. However, the enti re family of phase traject ori es can be graphically examined by

developing the phase por tr ai t using the method of isoc lines (ref .

12).

This method de-

velops the phase plane by plotting curves, called isoclines, of constant traj ectory slope.

The slope of the stepping motor differential equation is given by

d0 V

-

-

The slope of the differential equation mu st be specified in te rm s

of

a given

D

and TL.

39



I11111111111111111

11111111

1 I IIIII

-2

The value of

5

is determined by the specific system in which it

is

applied. Fo r the fol-

lowing analysi s,

D is chosen as

0.25.

The load torque will be neglected i n the following

analysis.

For a st ep command of

90°,

the slope

is

given by

I

Figure 17 shows a plot of the isoclines f or equation

(132). A

phase trajectory, for a

given set

of

initia l conditions,

is

plotted by connecting sh or t line segments with thep rop er

slope

at

interva ls along th e isoclines.

A

trajectory with initial conditions

O(0)

=

0

and

V(0)

= 0 is

plotted i n fig ure 17.

Figure 17 shows the existence

of

singula r points in the phase plane.

A singula r point

is

a representative point (eo, Vo) in th e phase plane fo r which (dO/dt)

=

(dV/dt)

=

0

(ref. 12).

For

a

singular point, the direction

of

its tangent

is

indeterminate, and

its tra-

jectory degenerates into the singular point itself.

From figure

17,

the singular points

I

1

I

2

I

3

I

4

Rotor position, 8, rad

I .I

-90 0 90 180 270


Figure

17. -

Isoclines for dV / = cos 8

0.25.

I

5

/”

6

40



,,-Trajectory

&

(a) Focal p in t. Indicates

s in g ul ar p i n t

of

stable

b)Saddle pint. Indicates

s in gu la r p i n t

of un-

stable eguil ibriu m.

Rotor oosition. 0. rad

.

I I I I I I I I I I I

-270 -180

-90

0

90 180

270 360 450 540 630

Rotor ps i t ion . 0, deg

(cl Singula r p i n t locations for stepping motor equation.

Figure

18.

-Types and locations of singular p i n t s for normalized stepping motor

equation.

are found, for example, at e =

*goo, *270°,

and *450°. The singular points for these posi-

tions a re either focal points or saddle points (ref. 13). Focal points ar e singularities in

which trajectories approach in a decreasing s D i r a l manner, as shown in figure 18(a).

Foca l points form points of stable equilibrium for the stepping motor. Saddle points

are

singularities which are approached by tr aje cto rie s forming distinct sectioned curves, as

shown in figure 18(b). Saddle points fo rm points of uns table equi librium.

shows the alternat ing locati ons of focal points and saddle points. The focal points are

360'

apart and 180' fr om the sad dle points.

Multistep analysis. - Multistep operation can be analyzed for a given stepping se-

quence by plotting the isoclines of the resulting equations for the d esi red st ep commands.

Fo r

a

given stepping sequence,

set

of initial conditions, and stepping

rate,

the phase

tra-

jectory

is

developed by selecting the set of isocli nes corresponding to the given ste p com-

mand. The phase trajectory

is

then drawn fro m the initial conditions. Tim e

is

computed

incrementally along the trajec tory. When the computed time equals the stepping period

for

the given command step, the next st ep command in the sequence is applied. The point

on the trajecto ry at which the computed tim e equals the stepping period becomes the ini-

Figu re 18(c)

4 1



I I I I - 1 I I I I

( a1 Ph a se p l a ne p r t r a i t f o r r e s p n s e

to

s tep wmmands

1

and 5.

Wid6

=

w s

9 IV -

0.25.

-

1

J

11

1 I

I

I 1

450

90 180

270

360

R o to r p s i t i o n .

6.

d q

[ b l P h as e pl an e p r t r a i t f o r r e s p n s e

to

step command

2

dVldB

= sin BIV -

0.25

F i g u r e

19.

-Mu l t i s tep sequence

lor change

in

normal ized t ime h Tof

1.31.

540

42



4

3

i

1

I

>

c

m

>

0

n

m

m

-

-

.-

E

z

I I I

I

I

I I

I

I

( c ) P h a s e p l d n e p o r t r d i t f o r r e s p o n s e

to

s t e p c o m m a n d

3.

d V l d 8 =

-cos

8 I V - 0.25.

I ~ I

I I I

5

b

7

8 9 10 11

I I

I

1 1 3 4

R o t o r position, 8. r a d

0 . 9 0 180 270 360 450

R ot or p s i t i o n , 8. deg

( d l P h a s e p l a n e p o r t r a i t f o r r e s p o n s e t o s te p c omm a n d 4 . dVl f f i = - s i n 0 I V - 0.25

Figure

19.

-

C o n c l u d e d .

43



tial condition fo r the next st ep command. The tra jecto ry fo r

this

st ep command

is

plotted

by using its corresponding set of isoclines. Time is computed along the tra jec to ry and

compared with the stepping period to dete rmine the application of the succeeding

step

command. This method is continued fo r the given stepping sequence.

Effect of stepping rat e.

- -

Figures 19(a) o (d) show the effect of stepping rat e. on the

ability of the

P M

stepping motor to follow

a set

of sequential st ep commands. These fig-

u r e s show phase plane plots developed

by

the method of is oc li ne s and consider the

set

of

sequential st ep commands tha t were developed in equations (126) to (130) . The equations

fo r the slope of the tra jec to rie s with

=

0.25 and

TL = 0 are

Step 1: 8 = 90 (i radians) - :---os e 0.25

V

0

dV sin

8 o.25

8 = 180

T

radians); --

d8 V

Step 2:

Step

3:

Step

4:

Step 5:

8 =

270 adians

cos e

0.25

C

o( V

0

dV sin 8

- o.25

8 =

360

(277

radians);

--

de V

radians

*

, :----os

e

0.25

V

For

the sequential ste p commands, again consider

a

normalized stepping period

of

A T = 1 . 3 1 . Recall that the ste p commands ar e applied

at

the following normalized times:

Step

1: T

=

0; 8 =

90

C

Step

2:

= 1 . 3 1 ; 8, = 180°(r radians)

Step 3:

T =

2.62; 8

= 270

C

0

Step

4:

T =

3 . 7 3 ;

OC =

360 (277 radians)

Step 5:

44

T =

5.24;

8 =

450

C



Step

1

calls for a position command of 90'.

The trajectory fo r the response to this

command is shown in figure 19(a). The roto r is assumed to be initially in equilibrium at

8 ( 0 )

= 0 and

V(0)

= 0. The trajectory is drawn from 8 ( 0 ) and

V ( 0 )

to point A. At

point A, the normalized time

T

is 1.31, and the command fo r ste p 2 is applied. Step 2

calls

for a

position command of 180'. The tr ajec to ry fo r th e response to th is command

is

shown in figure 19(b) from point A to point B. At point B the computed time

is

2.62,

and the command for ste p 3 is applied. Step 3 cal ls fo r

a

position command of 270'.

The trajectory fo r the response to t his command is shown in figu re 19(c) fro m point B t o

point

C.

At point

C

the computed time

is

3.93, and the command fo r ste p 4 is applied.

Step 4 calls for

a

position command of 360'. The tr ajec to ry fo r the response to th is com-

mand is shown in figure 19(d) from point

C

to point D. At point D the computed time is

5.24, and the command fo r ste p 5 is applied. Step

5

calls for a position command of

450'. The slope of the tr ajec to ry in response to th is command is given by equation (137).

This equation gives the sa me r esu lt

as for

the slope of th e traje ctor y in response to the

90' st ep command given by equation (133). Thus, the response fo r the 450' position com-

mand can be shown in figure 19(a) from point

D

to point

E .

This procedure can

be

con-

tinued fo r plotting the tra jec tor y in response to any additional st ep commands desi red.

Fai lur e to execute commands. - Increasing the stepping r at e decre ases the amount

of time in which the motor can respond to the step command.

A s a result,

the rotor

tends to lag behind the command position until sufficient velocity

is

built up to allow the

rot or to reach the command position. The rotor can dynamically

lag

behind the command

position by

as

much

as

180'

(two

ste ps) and

still

follow the command. But once the rot or

falls

behind by mo re than 180°, the motor cannot reach the cor re ct command position.

As an illustr ation, consider the previous st ep sequence with the normalized stepping

period

AT

reduced from 1.31 to 0.92. The ste p commands a r e applied at the following

times:

Step

1 :

= 0; 8

=

90

C

Step 2 :

=

0.92;

OC =

180°(a radians)

Step 3':

T =

1.84; 8,

= 270

Step 4':

=

2.76;

OC =

36Oo(2a radians)

Step 5':

T

= 3.68; 8, = 450

45



-2 I I 1 I I I

la1 Phase plane portrait for resprnse to step commands 1'and 5'. dVld8 = cos 0IV - 0. 25.

>

-

-

J

'

I

I I - 1 - -

0

-

80 270 360 4

M0

Rotor ps i t ion. 0. deg

l b l Phase plane portrait for response to step command 2'. dVld0 = sin 8IV

-

0.25.

Figure 20. -Mul t is tep sequence f o r change i n normal ized t ime bh~of 0 .92 .

46



S t e p

3

I I I I I 1

I

I

_ I -:

-2 I

I I

>

IC)

P h a se p l a n e p r t r a i t f o r r e s p n s e io s t ep c o m m an d

3 .

d V l d 0 = -ms IV - 0.25.

0

n

Step 4 '

I

I I I

1 -

I

1

I

6

7 8

9 10 11

1

0

1 2 3 4

5

R o to r p s i t i o n . 6. r a d

-2

I

- 1

I

I

I

0 90 180 27L 36C 45c 54c

R o t o r p s i t i o n . 6. deg

dl P h a s e p l an e p r t r a i l f o r r e s m n s e to s t ep c o m m a n d 4 ' . dVld8 - - s i n

B I V - 0.25.

F i g u r e 20. - C o n c l u d e d .

47



The response to ste p

1

is shown in figure 20(a) fro m V(0)

=

0

and O(0) to point A'.

At point A', = 0.92, and the 180' st ep command

is

applied. Notice that the value of

velocity and position reached is considerably lower than in the previous case f or = 1.31

Figur e 20(b) shows the traj ect ory f or the re sponse to the 180' st ep command between

points A' and B'.

Note in figure 20(b) that th e rotor position at point B' is lagging the

command posit ion by more than 90. At point

B',

T = 1.84, and the ste p command fo r

270' is applied. Fig ure 20(c) shows the tra jec tory fo r the respon se to the 270' ste p com-

mand, between points B' and

C .

The velocity of the ro tor actually dec reas es during this

step, and the rotor lags fu rt he r behind the command position. At point C ,

=

2.76, and

the 360' st ep command is applied. Note by examining th e isoclines , that had th e 360'

st ep command not been applied, the ro to r could still have reached the correct command

posi tion of 270'.

Figure 20(d) shows the trajectory fo r the response to the 360' st ep com-

mand between points C' and D'. Again during the step th e ro to r velocity decrea ses. At

point D' the rotor is approx imately 180' behind the command position. However, the

rotor can

still

reach the pr op er command posit ion of 360'. At point

D ,

=

3.68, and th e

450' s tep command is applied.

Fig ure 20(a) shows the re sponse to the 45OOstep com-

mand between points D' and E'.

The rotor

is

mor e than 180' behind the command posi-

tion during the ent ire step. Clearly, the motor cannot reach th e desired command posi-

tion of 450'.

If no fu rt he r st ep s

are

applied, the

rotor will

come to res t at 8 =

90°, o r

four s teps behind the desi red command position.

In general, the rotor

wi l l

fail to execute the

step

command i f the stepping motor tra-

jectory cr os se s the position axis before the desired command position is reached. This

can only occur i f the position of the rotor la gs the ste p command by mor e than 180'. The

fai lure analysis must be car rie d out by using enough step s to show that either t he motor

fails to follow the desi red command

o r

that the motor reaches a steady-state condition

fo r the given command ra te.

commands reg ard less of the number of s te ps applied.

Also,

if it

cannot follow a s tep

command, the motor will continue to lo se steps. The st ep s that it lo se s can be applied

after the ste p commands have been completed. This can be accomplished by feedback but

introduces more complications (ref. 1 ) .

The method of isoclines is not limited to analyzing step commands in one direction o r

to

a

constant time between step s. Any step sequence o r stepping rat e can be considered

by selecting the proper set of isoclines corresponding to the ste p sequence. In genera l

any step sequence o r stepping rat e can be considered by selecting the proper s e t of iso-

clines corresponding to the st ep sequence. Thus, any input ste p sequence can be inves-

tigated by this method.

If a motor will follow step commands

at a

given stepp ing rate, it will follow the step

48



'

DIMENSIONLESS CURV ES FOR EXPRESSING P M STEPPING

MOT0R PERFORMANCE

Multistep operation in the phase plane has been shown to be a function of three di-

mensionless param eters:

(1)Motor damping parameter, = 2{ (138)

t = o

N

2)

Time parameter, T

=

m m

-

I L

(3)

Load torque parameter,

TL

=

--

G K T I TS

(139)

Using these pa ramet ers,

multistep analysis in the phase plane can be used to specify the

maximum stepping ra te f or any given

PM

stepping motor and desired value of load torque.

The maximum stepping ra te is defined as the

rate

in which the motor can step with a

given value of load torque and not mis s

a

single step. The

P M

stepping motor

is

assumed

to be initially

at

rest

at a

point of sta ble equilibrium for the given value of load torque.

By determ ining the maximum stepping ra te fo r various values of and

TL,

a set of di-

mensionless curves can be developed to predict PM stepping motor performance.

A computer program w a s developed which uses the delta method to map the maxi-

mum stepping rate as a function of and TL and

is

desc ribed in appendix

C.

In the

computer program , the maximum stepping rate is determined by an iteration proces s.

A normalized stepping

rate

T is

chosen and exp ressed in te rm s of the time between

application of s te ps AT.

The motor is then stepped

at

the given rate for a number of

steps. The phase portrai t is examined to se e i f the motor follows the command without

missing a step.

If the m otor follows the command, a higher stepping rat e is chosen, and

the procedure

is

repeated until a stepping rate

is

found in which the motor cannot follow.

mance derived from the computer program. The cur ves show that, fo r a given value of

load torque, the maximum stepping ra te of a given P M stepping motor decreases

as

the

damping ratio increas es.

This has

a

di rec t bearing on the application of the

P M

stepping

motor to system design. The motion of the

PM

stepping motor for

a

single step is highly

underdamped, yielding low values of damping ratio. The motor can be compensated to

obtain a higher damping ratio and a bet ter single-step response. But raising the value of

Figure 21 shows the resul ting d imensionless curves of P M stepping motor perfor-

49



Single-step damping ratio, 5

L

1

I I I

0

1

2

3 4

Normalized motor damping,

Figure 21. - Dimensionless parameter curve s for

permanent magnet stepping motor.

the damping ratio decreases the maximum stepping

rate

that can be obtained.

multistep operation, compensating for the single- st ep response may not be desirable.

st ep commands was, in pa rt , determined by the position of the ro to r corresponding to the

command position during each succeeding step.

If

the r ot or position falls fur the r behind

with each succeeding step, the motor will eventually miss a step. Thus, the ri se time

and, hence, the damping rat io of the rot or during each step affects the maximum stepping

rate

of the

PM

stepping motor.

Thus, for

The phase plane analysis demonstrated that the ability of a motor to follow a

set

of

The damping ra tio

is

given by

Thus, the damping ratio can be lowered by dec reasing the viscous damping o r increasing

the motor ine rtia

or

the natural frequency. However, increasing the ine rti a decr ease s

the natural frequency since

50



If the natural frequency of t he moto r

is

incre ased , the tim e between ste ps can be de-

creased, and the maximum stepping

rate

can be increased since

A T = w N A t

1 -

wN

A t

A T

Thus, for a given load torque, maximum stepping

rate is

obtained by minimizing the

damping ratio and maximizing the natural frequency.

cr ea se s the maximum stepping ra te that can be achieved. Since the dimensionless curv es

assume a constant curre nt source , the maximum stepping ra te

will

be decreased if the

motor

is

driven from a voltage source. The stepping ra te in this case

is

decreased

as

a

re su lt of the sta to r winding

L/R

tim e constant and induced

emf,

both of which cause the

rotor position to further

lag

behind the intended command position.

Figure 21 also shows that, fo r a given damping ratio , increasing the load torqu e de-

pulse

logic

for arma-

Ammeter

28

v

d c

servo-

motor

Gear ratio, 6 (load

torque)

Emitter

follower

ratio,

Gear L

ro

oscilloscope

Figure 22.

-

Experimental setup

for

veri ficat ion of analytical model.

51



A

Experimental results-

Computed cur ve for

TL

=

0

I I .. I -1

1 2

3

4

Normalized motor damping,

0

=

2 5

Fi gu re 23. - Experimental verification

of

dimensionless

parameter cur ves for normalized applied load torque

TL of 0.

A Experimental results

- omputed curve

I I

1 . _

u

0

. 2

. 4

.6 .8

Normalized load torque, TL

Figure

24.

- Experimental verification of normalized step-

ping rate for vg rious values of load torque with norma l-

ized damping

D

o f

0 9

52



EXPERIMENTAL VERIFICATION OF DIMENSIONLESS C URVES

Some experimental mea surements wer e taken to check the validity of the dimension-

le ss curves developed fo r the P M stepping motor. Figu re 22 is a schematic diag ram of

the apparatus used in the experimental work. The meas ureme nts

were

taken on syn-

chronous inductor stepping motors. The stepping motor dri ve circu it use d approximated

a cu rre nt source. Damping torque was obtained by adding weights to the motor's output

shaft. Additional damping torque

w a s

obtained by coupling to the stepping mo tor output

shaft a dc servom oto r whose field windings were open-circuited. Load torque w a s ob-

tained by using the servomotor in the conventional manner.

derived curve for TL = 0. The experimental data show reasonable agreement with the

analytical curve.

Figure 24 shows the comparison of experimental data with the co rresponding ana-

lytical data from the dimensionles s curves fo r values of

TL

of

0,

0.2, 0.4, and 0.6.

The normal ized damping

w a s

held at

0.9.

The results show good agreement.

Figure 23 shows a comparison of the experimental values obtained to the analytically

CONCLUS IONS

An ana lys is of the permanent magnet

( P M )

stepping motor can be used to study the

effects of inert ia , damping,

current transients,

etc ., on the stepping motor re sponse to

a single-step command.

A

nonlinear an alysis which use s phase plane techniques w a s de-

veloped. This analysis allows the

P M

stepping motor to be analyzed fo r failu re to re-

spond to given stepping rates with various values of load and friction.

analysis

w a s

simplified by assuming that the st ator windings are suppl ied by a constant

current drive c ircuit that could compensate for the el ectric al (L/R) time constant.

Thus, the res ul ts re pres ent an upper bound on stepping motor performance.

Based on the nonlinear analysis, it w a s found that the P M stepping motor cannot re-

spond to a step command when the applied load torque becomes g re at er than 0.707 of the

motor's

stall

torque. The P M stepping motor cannot follow a sequential

set

of step com-

mands i f , during the sequence, the rotor lags the command position by more than

two steps.

maximum normalized stepping

rate

as a function of normalized damping and normalized

applied load torque. The se curve s allow the maximum stepping rate of a permanent mag-

net stepping motor to be determined fr om simple measu remen ts of

its

parameters or

from manufacturer's data.

The nonlinear

This result w a s extended to develop a se t of dimen sionless curv es which expr ess

5 3



Because the dimensionless curves assume a constant cu rre nt sourc e, the maximum

stepping rate w i l l be decreased i f the motor is driven from a voltage source. The step-

ping rate in this cas e

w i l l

be decreased as a result of the stator winding electrical (L/R)

time constant and induced electromotive forc e, both of which cause the ro tor position to

further lag behind the intended command position.

However, the maximum stepping rat e curves show that improving the single -step re-

sponse by adding damping will lower the maximum stepping rate that can be achieved.

Stepping motor response to a single-step command is usually highly underdamped.

Lewis Research Center,

National Aeronautics and Space Administration,

Cleveland, Ohio, November 19, 1968,

P22-29- 3-01- 2.

54



APPENDIX A

SYMBOLS

B

D

-

D

E

I

KI

Km

KT

KV

L

9[1

I

M

m

magnetic flux density, T

viscous damping of motor and

2

load, (g)(cm )/s ec

normalized damping

constant value of st ator input

voltage, V

Laplace tra nsf orm of p(t) , V

stator input voltage, V

stat or induced electrom otive

force (emf), V

constant value of st at or input

current, A

Laplace trans for m of

Z(t)

stator input current, A

in er ti a of motor and load,

(g)(cm2)

2

solenoid constant, (Q)(sec )/m

bar magnet constant, m/(Q)(sec)

motor torque constant, (N-m)/A

constant relating induced voltage

and stator flux density,

(V)sec)/deg

stat or inductance per phase, H

Laplace transformation

length of bar magnet, cm

2

magnetic moment, (A)(m

ba r magnet pole strength,

W b

N~~

n

R

S

Tmax

TS

6

constant relating rotor position

in degrees

8

to mechanical

degre es of rotation

number of solenoid tu rns

stat or resistance per phase, 5

Laplace operator, l/ se c

developed motor torque, N-m

frictional torque, N-m

normalized frictional torque

applied load torque, N-m

normalized applied load torque

maximum motor torque pe r

phase, N-m

motor stall torque, both phases

energized, N- m

time, sec

norma lized velocity

variable

angular measurement variable,

deg

line segment, cm

change in a variable about an

operating point

center of cir cul ar ar c, deg

damping ratio

Laplace transfo rm of e(t) , spe-

cified in degrees except where

noted in radians

5 5



rot or position defined

so

that

1/4

rotor

tooth pitch equals

90°,

specified in de gree s ex-

cept where noted in radia ns

rot or command position, spe-

cified in degrees except

where noted in radians

7 normalized tim e

AT

change in norm alized tim e

l / A r

normalized stepping rate

q(t)

roto r position in mechanical

degrees

natural frequency, rad/ sec

in it ial condition of

a

variable

wN

(0)

Subscripts

:

0

value

at

operating point

1

stat or phase 1

2 stat or phase 2

56



A P P E N D I X

B

L A P L A C E T R A N S F O R M D E R I V A T I O N S O F L I N E A R I Z E D F U N C T IO N S

The Laplace transform ation can be applied by considering the following development.

F o r a var iab le X(t) about

an

operating point,

and so fort h. Taking the derivative of equation (Bl) and assuming that X, is a constant,

d

d

d

dt dt dt

-X(t)

=-

X,)

+

U t)

Combining equations

(B2)

and

(B4),

Taking the Laplac e transform of equation (B5),

In the sa me manner, taking the derivative of equation (B2),

037)

57



Combining equations (B3) nd (B7) esults in

Taking the Laplace transformation of equation (B8),

/d2X\

When the initial conditions are used, the Laplace transforms of the excursions about

the operating point at the beginning of the ste p a r e

= S Ae(S)

-

Ae(0)

-

-

S A @ )

S

58



0

= S A I 2 ( S ) - AIZ 0) - -

S S

= S

AI1(S)

-

AI1(O)

-

___ - s

AI1@)

S

The derivative (d12/dt)0 in equation (B15) can be determined by noting that s ince

(dB/dt)O =

0,

the induced voltage E

=

0. The voltage supplied to winding 2 becc A l e s

v, 0

AE2(t)

=

A12(t)R

+

A [ y ] L

A E 2(S )

= (R

+

LS)A12(S)

Assuming that the stator input voltage to winding

2 is

a step function res ul ts i n

12(S)

=

-

or

When the initial value theorem (ref.

14)

s applied,

t-O

dt

S-00

R + L S

L

\ d t

/

0

59



Thus, the Laplace tran sfo rm s of the equations for the exc ursion s about the operating

point at the beginning

of

the step become

=

T’90 S AO(S)

2 2

KT1l.0

R +

LS

R

+ LS

’

Combining equations (B22) to (B24) gives the resu ltant linea rized t ra ns fe r function about

an operating point

at

the beginning

of

a step:

JLS

N~~

- (R + LS)ATL(S)

S

60



APPENDIX C

DIMENSIONLESS PARAMETER CURVE COMPUTER PR OGRAM

A

digital computer progra m

w a s

written to develop the dimensionless curves using

The phase trajectory w a s

the multi step phase plane analysis. The maximum stepping

rate w a s

determined by step-

ping the motor a number of ste ps

at a

constant stepping

rate.

calculated and checked

to

see

i f

it followed the command without missi ng a step.

The

stepping

rate w a s

inc rea sed until the motor failed to follow a command.

The computer program developed

is

shown in table

I.

The program w a s developed by

using the delt a method fo r computing the phase tra jecto rie s. The delta method allowed

the traj ecto ry to be calculated by using smal l incremen tal sections. This method

w a s

also convenient for calculating time along the traject ory. The computer input variable s

were

(1)

Initial

velocity,

VIN

(2)

Initial position, THIN

( 3 ) Initial value of

6, DELTIN

(4)Total time fo r the number of ste ps, TTOT

(5) Total tim e between st eps , DTAUS

(6) Position incremen t used in computing tra jecto ry, DTH

(7)

Normalized motor damping par am ete r, D

(8) Normalized load torque, TL

Twenty steps were commanded for each stepping rat e. Thes e were enough

step s to

determine whether the motor could follow the stepping ra te with the given and

TL.

Motor fai lu re was dete rmined by checking fo r the r otor velocity to change sign (motor

rev ers es it s direction) before the d esired step command

w a s

reached. The stepping

rates

w e r e

vari ed in incremen ts of 0.01, and the values of position were compared along

the tra jecto ry. The normalized time was computed, and a new value for delta was com-

puted

f o r

each time the position of the rot or changed by

a

predetermined increment.

This increment affects the overall accuracy of the traj ecto ry and of the calculated time.

Thus,

it

was made an input variable (DTH), and its effect on the res ponse

w a s

checked.

A value of DTH = 0 . 0 5 was found to give resul ts with sufficient accu racy

for

the program.

The maximum stepping ra te

w a s

computed, using thi s prog ram , fo r values of

D

=

0 . 2 5 ,

0.

5,

1.

0,

1 . 5 , 2 .

0,

and

4 . 0

and fo r values of

TL

=

0,

0 . 2 ,

0.

4,

and 0.6.

61



TABLE I . - DIGITAL COMPUTER PROGRAM

100

101

1 0 2

FORMAT (8E

10.5)

FORM AT( 1H

1,5X,

3HVI N, 9X,4HTHI N, 7X, 6HD EL TI N,9X,4HTTOT18X, 5HDTAUS,

l O X 1 , 3 H D T H , I I X , I H D , l l X , 2 H T L / ( 8 E 1 3 . 5 ) )

FORMAT (3E 20. 6)

103

FORMAT( lH0,5X,8HVELOCI TY, 14X,SHTHETA, l 5 X , 3HTAU)

500 READ 5,100) V I N TH IN, DE L Tl N, TTOT, DTAUS, TH D, T L

WKI TE 6,101

) V I

N,TH I N, DE LT l N,TTOT, DTAUS, DTH, D,TL

W R I T E ( 6 , l O j )

TAU=O.

0

S I G N = l . O

V I = V I N

TH

1

= TH I N

WR

ITE (6 , 102 ) V

1,

TH

1,

TAU

D E LTA = D E LT I N

N = l

M=

1

TH2=THI+DTH

TEM

1=

DELTA+TH 2)* (DELTA+TH2)

I

F(TEM1 .LE .RADSQ)

GO

TO 20

TH2=BM2-DTH

DTH=- DTH

S I GN=-S IGN

i o

RADSQ=

( V I*V

I+(DELTA+TH

) *

DELTA+TH

I ) )

T EM 1= DE LTA+ TH

2 )

* DE LTA+TH 2)

IF(TEM1.GT.RADSQ) GO TO 21

DTAU=2.O; t (TH2-THl ) / (V2+Vl )

v 1

= v 2

T H l = T H 2

K=M-

(M/5)9:5

i

F

K)

0,60,70

20 V2=S IGN*SQRT(RADSQ-TEMI)

2

1

TAU=TAU+DTAU

60 W R I TE (6,102) V

1

,TH

1

,TAU

70 M=M+l

I F ( T A U . G T . T T Q T ) GO TO 500

3 0 T E k T A U - F L O A T (N-

1

)+:DTAAUS>t4.0

TEM2=TEM/DTAIJS

I F ( T E M 2 . G T . 4 . 0 )

GO

TO

4 0

GO TO

5 0

40

N=N+I

GO

TO 30

50

I F ( T E M 2 . G T . 3 . 0 ) GO TO

14

I F ( T E M 2 . G T . 2 . 0 ) GO TO 13

I F(TEMZ.GT.

1

.o)

G O TO 12

1 1 V D V = C O S (TH

1

) D : v 1 -TL

2

V D V = S

I

N

(TH

1

) - D - ” v I

- T L

13

VDV=-COS(THI ) -DWI-TL

1 4 V D

v=-

s

I

N (TH 1 ) -Dv

1 -

T L

G O TO

15

G O

TO

15

GO TO

15

15

DELTA=-VDV-TH 1

GO

TO

10

E N D

62

. -



REFERENCES

1. Pro ctor, John: Stepping Motors Move In. Product Eng., vol. 34, Feb. 4, 1963,

pp. 74-78.

2. Nicklas, J. C. : Analysis, Design and Testing of

a

Position Servo Utilizing

a

Step-

per Motor. Tech. Rep. 32-206,

Jet

Propulsion Lab., Califo rnia Inst. T ech. ,

Jan. 25, 1962.

3. Giles, S.

;

and Marcus, A.

A.

:

SNAP-8-Control-Drum Actuators. Rep. NAA-SR-

9645, Atomics Inte rnat ional, Dec. 15, 1964.

4. Bailey,

S.

J. : Incremental Servos. Part

I -

Stepping v s Steples s Control. Contro l

Eng., vol. 7, no. 11, Nov. 1960, pp. 123-127; Part 11

-

Operation and Analysis,

vol.

7, no.

12, Dec. 1960, pp. 97-102; Part

III

- How They've Been Used, vol. 8,

no.

1,

Jan. 1961, pp. 85-88;

Part IV -

Today's Hardware, vol. 8, no.

3,

Mar.

1961, pp. 133-135;

Part V

- Interlocking Steppe rs, vol.

8,

no, 5, May 1961,

pp. 116-119.

5. O'Donohue, John P.

:

Tran sfer Function for a Stepper Motor. Contro l Eng., vol. 8,

no. 11, Nov. 1961, pp. 103-104.

6. Kieburtz, R. Bruce:

The Step Motor - The Next Advance in Contro l Sys tems. IEEE

Tr ans. on Automatic Contro l, vol. AC-9, no.

1,

Jan. 1964, pp. 98-104.

7. Snowdon, Arthur E. ; and Madsen, El me r

W.

: Cha ract erist ics of a Synchronous In-

ductor Motor. AIEE Trans . on Applications and Industry ,

vol.

81,

Mar.

1962,

pp. 1-5.

8.

Morgan,

N.

L.

:

Versatile

Inductor Motor Used in Solving Industria l Con trol P rob-

lems.

Plant Eng., vol. 16, no. 6, June 1962, pp. 143-146.

9. Swonger, C.

W.

:

Polyphase Motors as Digital-Analog Stepping Devices. Data Sys-

tems Eng., vol. 18, no. 6, July-Aug. 1963, pp. 52-56.

10. Graham, Dunstan; and McFuer, Duane: Ana lysi s

of

Nonlinear Control Systems.

John Wiley

&

Sons, Inc., 1961.

Inst. Tech., 1966.

12.

Cunningham,

W.

J.

:

Introduction to Nonlinear Analysis. McGraw-Hill Book Co. ,

11.

Zeller, Donald

F.

:

Digita l Control of

a

Stepping Motor. MS Thesis, Massachusetts

Inc., 1958.

13. Minorsky, Nicholas:

Introduction to Non-Linear Mechanics. J.

W.

Edwards,

Ann

Arbor, 1947.

14. Savant,

C .

J. ,

Jr. :

Bas ic Feedback Control System Design. McGraw-Hill Book

Co., Inc., 1958, p. 64.

E-4479

ASA-Langley, 1969

63



ERONAUTICSND SPACE ADMINISTRATION

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Postal Manual) Do Not Re

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the Un ite d States shall be

conducted

so

as

t o

contribute

. . . t o

the expans ion

of

h a ~ z a n no wl-

edge

of

phenoiiiena

in

the atmosphere and space. Th e Administration

shall provide f o r the widest prcrcticable aud appropriate disse?Jzi?za?ion

of

in fo r imt ion

concerning its actkities and

the resvlts

thereof.

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