intro to vector control of induction mc for electrical engg

Introduction to Vector Control of Induction

Machines

R.E. BetzDepartment of Electrical and Computer Engineering

University of Newcastle, Australiaemail: [email protected]

Last modified: February 24, 2000

Generated: February 24, 2000

Contents

Preface vi

1 Fundamentals 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Sinusoidal Assumption . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Winding Interaction with Spatial Flux Density Distribution 41.2.2 Winding Interaction with Temporal Flux Density Variation 7

1.3 Flux Linkage to Sinusoidally Distributed Windings . . . . . . . . 101.4 Other Important Assumptions . . . . . . . . . . . . . . . . . . . . 11

1.4.1 Properties of Three Phase Sinusoidal Windings . . . . . . 121.5 Torque Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.1 Torque of a Simple Reluctance Machine . . . . . . . . . . 141.5.2 Linear Torque Model . . . . . . . . . . . . . . . . . . . . . 241.5.3 The Ellipse Diagram for co-energy . . . . . . . . . . . . . 29

2 The Kron Primitive Machine 362.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.2 Model for the Doubly Fed Machine . . . . . . . . . . . . . . . . . 36

2.2.1 Zero saliency case . . . . . . . . . . . . . . . . . . . . . . 382.2.2 One degree of saliency case . . . . . . . . . . . . . . . . . 382.2.3 Torque expression . . . . . . . . . . . . . . . . . . . . . . 38

2.3 Commutator Machines . . . . . . . . . . . . . . . . . . . . . . . . 402.4 The Primitive Machine Concept . . . . . . . . . . . . . . . . . . . 42

2.4.1 Use of the Primitive Machine - the DC Machine . . . . . 462.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Frame transformations, DQ and Space Vector Models 503.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.2 dq Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2.1 Stationary Frame Transformations . . . . . . . . . . . . . 513.2.2 Rotating Frame Transformations . . . . . . . . . . . . . . 573.2.3 Example: SYNCREL Linear dq Model . . . . . . . . . . . 60

3.3 Space Vector Model . . . . . . . . . . . . . . . . . . . . . . . . . 653.3.1 Current Space Vectors . . . . . . . . . . . . . . . . . . . . 653.3.2 Flux Linkage Space Vector . . . . . . . . . . . . . . . . . 703.3.3 Voltage Space Vector . . . . . . . . . . . . . . . . . . . . . 723.3.4 Example: SYNCREL Space Vector Model . . . . . . . . . 733.3.5 Space Vector Power Expression . . . . . . . . . . . . . . . 74

CONTENTS ii

3.3.6 Example: Space Vector Expression for SYNCREL Torque 763.3.7 Relationship Between Space Vectors and dq Models . . . 78

3.4 Steady State Model . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4 Vector Control of Induction Machines 814.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.2 Vector Models for Induction Machines . . . . . . . . . . . . . . . 81

4.2.1 Flux Linkage Expression . . . . . . . . . . . . . . . . . . . 814.2.2 Magnetising Current . . . . . . . . . . . . . . . . . . . . . 844.2.3 Power and Torque Expressions . . . . . . . . . . . . . . . 844.2.4 The Space Vector Model of the Induction Machine . . . . 89

4.3 A Heuristic Explanation of Vector Control . . . . . . . . . . . . . 914.4 Special Reference Frames . . . . . . . . . . . . . . . . . . . . . . 94

4.4.1 The Magnetising Flux Linkage Reference Frame . . . . . 944.4.2 The Rotor Flux Linkage Reference Frame . . . . . . . . . 964.4.3 Stator Flux Linkage Reference Frame . . . . . . . . . . . 97

4.5 Derivation of Rotor Flux Oriented Vector Control . . . . . . . . . 994.6 Structure of a Rotor Oriented Vector Drive . . . . . . . . . . . . 103

4.6.1 Indirect Rotor flux Oriented Controller . . . . . . . . . . 1034.6.2 Direct Rotor Flux Oriented Controller . . . . . . . . . . . 104

4.7 Magnetising Flux Orientation . . . . . . . . . . . . . . . . . . . . 106

A Calculation of Inductances for Salient Pole Machines 107A.1 Calculation of Inductances . . . . . . . . . . . . . . . . . . . . . . 107

A.1.1 Self Inductances . . . . . . . . . . . . . . . . . . . . . . . 109A.1.2 Mutual Inductances . . . . . . . . . . . . . . . . . . . . . 117A.1.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

B Winding Functions 121B.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121B.2 Ideal Sinusoidal Winding . . . . . . . . . . . . . . . . . . . . . . 122

B.2.1 Conventional inductance calculation . . . . . . . . . . . . 122B.2.2 Alternative inductance calculation . . . . . . . . . . . . . 125

B.3 Non-sinusoidal winding . . . . . . . . . . . . . . . . . . . . . . . . 128B.3.1 Inductance Using Basic Principles . . . . . . . . . . . . . 128B.3.2 Inductance Using Winding Functions . . . . . . . . . . . . 131

B.4 Flux Linkage Expression . . . . . . . . . . . . . . . . . . . . . . . 132B.5 A Note on Winding Functions for Multi-pole Machines . . . . . . 133B.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

List of Figures

1.1 MMF calculation integration path. . . . . . . . . . . . . . . . . . 41.2 Dimensions of a single coil. . . . . . . . . . . . . . . . . . . . . . 81.3 Conceptual diagram of sinusoidally distributed three phase wind-

ings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Simple singly excited reluctance machine. . . . . . . . . . . . . . 151.5 Flux plots for static movement. . . . . . . . . . . . . . . . . . . . 171.6 Incremental energy change with small movement of the rotor. . 181.7 Area representing mechanical output energy. . . . . . . . . . . . 191.8 Energy with instantaneous movement. . . . . . . . . . . . . . . . 201.9 Flux versus current trajectory for typical real movement. . . . . 211.10 Flux versus current for linear magnetic material. . . . . . . . . . 241.11 Doubly excited reluctance machine. . . . . . . . . . . . . . . . . 261.12 Self and mutual inductance variation with rotor angle. . . . . . . 311.13 Flux linkage for the a-phase . . . . . . . . . . . . . . . . . . . . . 321.14 Total, self and mutual flux versus current loci. . . . . . . . . . . . 321.15 Segments used for ellipse area. . . . . . . . . . . . . . . . . . . . 341.16 Co-energy “ellipses” for a saturated SYNCREL . . . . . . . . . . 35

2.1 Magnetic circuit conceptual diagram of a double fed machine . . 372.2 Torque plot for the double fed machine with DC rotor and stator

currents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.3 A two pole commutator machine . . . . . . . . . . . . . . . . . . 402.4 Equivalent circuit of a DC machine in motoring mode . . . . . . 422.5 Primitive dq machine . . . . . . . . . . . . . . . . . . . . . . . . . 432.6 Separately excited DC machine with a compensating winding . . 47

3.1 Three phase to two phase transformation . . . . . . . . . . . . . 513.2 Two phase stationary to two phase rotating frame transformation 593.3 Conceptual diagram of a three phase SYNCREL . . . . . . . . . 613.4 Model for the ideal dq equations . . . . . . . . . . . . . . . . . . 633.5 Resolving the current space vector onto the abc axes . . . . . . . 673.6 Relationship between the dq axes and the current space vector. . 693.7 Space vector rotating frame transformations . . . . . . . . . . . . 693.8 Phasor diagram for a steady state SYNCREL . . . . . . . . . . . 80

4.1 Conceptual diagram of an induction machine. . . . . . . . . . . . 824.2 Relationship between stationary and rotating frames and the gen-

eral reference frame for the induction machine . . . . . . . . . . . 89

LIST OF FIGURES iv

4.3 Conceptual diagram of an induction machine with quadrature-phase stator windings. . . . . . . . . . . . . . . . . . . . . . . . . 92

4.4 Space vectors in quadrature induction machine at time t+o . . . . . 934.5 Space vectors in quadrature induction machine at t > to. . . . . . 934.6 Position of the space vectors after the stator has been rotated. . 944.7 Relationship between the dq frame and the special xy frame. . . 954.8 Relationship between various space phasors in the stator and ro-

tor flux linkage reference frames . . . . . . . . . . . . . . . . . . . 964.9 Relationship between the stationary reference frame and the spe-

cial reference frame fixed to the stator flux linkage space phasor. 984.10 Block diagram of a indirect rotor flux vector oriented control scheme1044.11 Block diagram of a direct rotor flux field oriented vector controller.1054.12 Flux model in a rotor flux reference frame . . . . . . . . . . . . . 105

A.1 Two pole three phase SYNCREL – conceptual diagram . . . . . 108A.2 Developed diagram of a SYNCREL. . . . . . . . . . . . . . . . . 111A.3 d axis developed diagram for SYNCREL . . . . . . . . . . . . . . 112A.4 ‘a’ phase inductance plot. . . . . . . . . . . . . . . . . . . . . . . 117

B.1 Two pole sinusoidal winding layout . . . . . . . . . . . . . . . . . 122B.2 Calculation of the MMF for a sinusoidal winding. . . . . . . . . . 124B.3 Different methods of calculating the flux linking a coil. . . . . . . 126B.4 Cumulative number of turns for a sinusoidally distributed winding.127B.5 Turns function and mmf distribution for two fractional pitch

windings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129B.6 Physical layout of the non-sinusoidal winding. . . . . . . . . . . . 129

List of Tables

3.1 Summary of Stationary Frame Transformations . . . . . . . . . . 573.2 Summary of Rotating Frame Transformations . . . . . . . . . . . 60

Preface

These notes contain material for the first half of Elec414 - Energy Flow Control.Their objective is to give students a basic understanding of vector controlledinduction machines. The approach taken is to start off from first principles anddevelop the basic expressions for a machine. Next the primitive dq machineis developed and this is applied to the DC machine and the synchronous re-luctance machine. The next stage is to develop stationary and rotating frametransformations for AC machines, and then show that these machines can alsobe developed as dq machines.

The last part of the notes introduces space vector representations for ma-chines, and shows how they are related to dq representations. These models arethen used to develop various forms of vector control for the induction machine.

Robert E. Betz – Newcastle, Australia, 1997

Chapter 1

Fundamentals

1.1 Introduction

This subject provides an introduction to Generalised Machine Theory and thedynamics of electrical machines. This area has become increasingly importantdue to the prevalence of variable speed drive systems in industry. As industrycontinues to modernise the penetration of AC variable speed drives into ap-plications which were previously considered to be the exclusive domain of DCmachines is continuing. Consequently an understanding of how these drive sys-tems work and what their capabilities are is becoming increasingly important.

Variable speed drive systems first began to appear in the late 1960’s early1970’s. During the early years many of these drive systems were based on chop-per fed DC machines, this being especially true for applications that requiredgood transient performance. AC machine drive systems were also availablearound this time, but generally these were limited to low performance applica-tions where the transient performance was of little importance. The other issuethat tended to limit the application of AC variable speed drives during this erawas the reliability of the systems. The dreaded “shoot through” problem in thepower electronics was a frequent occurrence.

Whilst reasonably high performance drives were developed based on DCmachine technology they still had the problem that the machine was a DC ma-chine. These machines are inherently costly, they require a lot of maintenance,and they are less reliable than other machine types (mainly due to the presenceof the commutator and brushes). Therefore there was a motivation to developdrive systems based on AC machines, and in particular the induction machine.This machine is extremely reliable and low cost. If has often been said that theinduction machine will continue to work even if an axe has been sunk into therotor.

Two problems initially held back the development of AC induction machinevariable speed drive systems:

• the previously mentioned lack of reliability in the power electronics.

• the lack of control capable of giving good performance from the inductionmachine.

1.1 Introduction 2

Now let us examine these in more detail. The early development of variablespeed drive systems was dogged by the poor reliability of the power electronics.This was largely to do with the fragility and difficulty of use of the early powerdevices used. Consequently variable speed drives got a bad reputation amongstmany industrial people, which took more than 10 years to overcome. Virtuallyall of these initial problems have been overcome with the latest power electronicdevices. Also one can not underestimate the influence that the microprocessorhas had on the development of drives systems. The availability of low costcomputation has allowed cost effective implementation of sophisticated controlalgorithms for drive systems.

The other factor that limited the development of AC drives was the control.The control for high performance DC drives was easy since this machine struc-ture was very easy to control. In fact with a separately excited DC machine thesystem dynamics were a first order linear differential equation. However, the ACmachine was a very complex fifth order non-linear system. For this reason earlyAC drives were designed to operate at steady state for most of the time, thevariable speed capability only being used to move to a different set point. Thedevelopment of the concept of vector control and the early to 1970’s overcamethese limitations, and similar performance to the DC machine can be obtainedfrom the induction machine. Research into various aspects of vector control hasbeen active ever since this time.

This set of notes will gradually work its way towards the development ofvector control. The material to presented will follow the outline below:

1. Examination of the fundamental assumptions underlying the developmentof the dynamic models.

2. Development of the rotating field concept.

3. Introduction to the concept of the Generalised Machine

4. Detailed development of the Generalised Machine.

5. Development of the conversions between the three phase machine and thegeneralised machine.

6. Detailed development of the concept of space vectors.

7. Development of DQ and space vector models for several different types ofmachines.

8. Concepts of field orientation.

9. Vector control of the synchronous and induction machine.

10. Introduction of load dynamics.

11. Hardware and software structure for variable speed drive systems.

Modelling of machines is very complex due to their highly non-linear natureand the difficulty in obtaining analytical models that reflect the underlyingphysics. Consequently all the modelling approaches used involve assumptions.It is important to understand what these assumptions are, and how they affectthe validity of the models produced. We will not con-

sider all the as-sumptions here, butwill concentrate onthe main ones.

1.2 The Sinusoidal Assumption 3

The next section will examine one of the fundamental assumptions of ma-chine modelling, the sinusoidal winding distribution assumption, and considerits implications.

1.2 The Sinusoidal Assumption

One of the main assumptions that is used in the modelling of many types of ACmachines is the sinusoidal assumption. Essentially the assumption is that thewindings in the machine are arranged so that the resultant mmf has a spatiallysinusoidal distribution. A normal AC machine usually has three windings spacedat 120◦ electrical, each producing a spatially sinusoidal mmf when fed with acurrent. An amazing property of this arrangement is that if it is fed with threetemporal sinusoidal currents, separated temporally by 120◦, then the resultantmmf is a spatially moving sine wave around the machine (this will be shownmathematically in later sections in this Chapter).

Why is this assumption so important? From a modelling point of viewthe sinusoidal functions have a rich set of mathematical properties which makethe modelling of machines analytically tractable. One of the key properties ofsinusoidal functions is their connection with vectors, and the consequent abilityto take orthogonal components of them.

In reality the mmf produced by real windings are not pure sinusoids. Mostwindings for real machines are confined to slots in the stator. This leads to anmmf that has step changes in it, and consequent higher order spatial harmonics.However, the winding configuration is designed to minimize these harmonics. Aswe shall see below, the significance of these winding harmonics on the perfor-mance of the machine also depends on the harmonics in the flux waveforms thatinteract with the windings.

The sinusoidal assumption is not only applied to the mmf produced by thewindings, but it is also applied to the resultant fluxes produced by the action ofthe mmf on the iron circuit of the machine. In the case of the SYNCREL theiron circuit reluctance varies in a complex fashion due to the rotor saliency. Thismeans that the flux density produced by the mmf is in general not spatially si-nusoidal. However, the harmonics in these waveforms are usually neglected, andonly the fundamental component is considered from an analysis point of view.This may seem to be a gross approximation, but models developed using thisapproach have been shown to give reasonable representations of the behaviourof real machines.

In the remainder of this section we shall look at some of the propertiesof a sinusoidally distributed winding. Specifically, the characteristics of a non-sinusoidal flux density interacting with a sinusoidal winding shall be considered.This is of particular relevance to the SYNCREL and other salient pole machinesas their flux density distributions in general are not sinusoidal.

The remainder of this section discusses the foundations of the sinusoidalassumption, and why it can be used successfully to simplify the modelling ofmachines, with special emphasis on the SYNCREL. Specific issues addressedare:

• Consideration of some of the general properties of sinusoidally distributedwindings (e.g. only link with fields of the same pole number).


Figure 1.1: MMF calculation integration path.

• Detailed analysis of the variation of inductance with rotor position for atwo pole axially laminated rotor would be beneficial.

1.2.1 Winding Interaction with Spatial Flux Density Dis-tribution

In this sub-section we shall consider the interaction of a spatially non-sinusoidalflux density distribution with an ideal sinusoidal winding. Such an ideal wind-ing will produce a temporal sinusoidally varying current density around themachine. The following equation can be written for the conductor density as afunction of the angle θp around the periphery of the machine:

n(θp) = na sin θp (1.1)

This waveform has an amplitude of na conductors, and goes positive and neg-ative. How can one have positive and negative numbers of conductors? Thesign convention is based on the direction of the current in the conductor. Thepositive part of this conductor distribution carry currents in one direction, andthe negative part carry the return currents [4].

Given this winding distribution, the mmf spatial distribution readily follows.If the a-phase is carrying ia amps, then the mmf can be calculated by imple-menting Ampere’s Law. This is achieved by carrying out a closed path integral


over the full coil span (see Figure 1.1 for the path of integration):

FaT (θp) =∫ θp+π

θp

naia sin θp dθp

= 2naia cos θp

= 2Fa cos θp

∴ Fa(θp) = Fa cos θp where Fa = naia (1.2)

The ‘2’ factor in the front of the right hand side of the above expression isthere because the total mmf is expended across two air gaps, and the Fa(θp)expression represents the mmf expended per air gap.

The total number of coils in the winding is simply the sum of the numberof coils at each θp position. Due to the continuous nature of the proposeddistribution this sum becomes an integral:

Na =∫ π

0

na sin θp dθp

= 2na (1.3)

Therefore the peak mmf for the winding may be written as:

Fa =Naia2

= Naia = naia

Now let us consider some general flux density waveform that varies in thefollowing way spatially with respect to θp around the machine, and also has atime varying spatial phase angle δ(t):

B(θp) = Bn sinn(θp − δ(t)) (1.4)

This flux waveform is a non-sinusoidal waveform as it contains a number ofharmonics denoted by the integer value of n.

Furthermore assume that the winding is on a machine with the followingphysical dimensions:

l � the length of the machine.

vB � the linear velocity of the B field.

r � the radius of the stator of the machine.

Therefore the B(θp) field phase is changing in the following fashion:

δ(t) =vBt

r(1.5)

This expression implies that the B(θp) field is spatially moving with respect totime.

From basic physics we can say the following – the voltage induced in a lengthof conductor l, moving with a velocity of vB perpendicular to a magnetic fluxdensity of B is:

e = BlvB (1.6)


In the case of a sinusoidally distributed coil the length of conductor for oneside of the coil at some position θp is:

lT = nal sin θp (1.7)

therefore the induced voltage in the conductors at this angular position is:

e(θp) = BvBnal sin θp (1.8)

The flux density at this position at a particular instant of time can be determinedfrom (1.4), and consequently (1.8) becomes:

e(θp) = vBnalBn sin θp sinn(θp − δ(t)) (1.9)

To simplify the following manipulations let Kn � vBnalBn. In order to cal-culate the total voltage produced by these conductors we have to add up thecontributions of all the conductors in the coil. This involves integrating thevoltage at each position θp for the circumference of the machine. Thereforeassuming a single pole pair machine we have:

eT =∫ 2π

0

Kn sin θp sinn(θp − vBt

r)dθp (1.10)

Using the trigonometric relation sinx sin y = 1/2[sin(x+y)+sin(x−y)] one canwrite:

eT =∫ 2π

0

Kn

2

[sin

((n+ 1)θp − nvBt

r

)+ sin

((n− 1)θp − nvBt

r

)]dθp

(1.11)

For the specific case of n = 1 (i.e. only the fundamental harmonic present)then (1.11) can be integrated and becomes:

eT = K1π sin(vBt

r

)= K1π sin(ωBt) (1.12)

i.e. the voltage induced by the winding is a temporal sinusoidal voltage (asexpected).

Now consider what happens to the higher order harmonics in the flux densitywaveform. If we carry out the integration of (1.11) for the case of n > 1 wehave:

eT =Kn

2

[− cosn+ 1

(2π(n+ 1)− nvBt

r

)− cosn− 1

(2π(n− 1)− nvBt

r

)

+(

1n+ 1

+1

n− 1

)cos

(−nvBtr

)](1.13)

If we consider the various terms in (1.13) using the trigonometric relation:

cos(x− y) = cosx cos y + sinx sin y (1.14)


we get the following:

− cosn+ 1

(2π(n+ 1)− nvBt

r

)= − 1

n+ 1

[cos 2π(n+ 1) cos

nvBt

r

+ sin 2π(n+ 1) sinnvBt

r

]= − 1

n+ 1cos

nvBt

r(1.15)

and similarly:

− cosn− 1

(2π(n− 1)− nvBt

r

)= − 1

n− 1cos

nvBt

r(1.16)

Therefore (1.13) can be written as:

eT =Kn

2

[−

(1

n+ 1+

1n− 1

)cos

nvBt

r+

(1

n+ 1+

1n− 1

)cos

−nvBtr

]= 0; ∀ n > 1 (1.17)

Remark 1 The implications of the above expression are that the higher orderharmonics in the flux density spatial waveform do not link to the sinusoidallydistributed winding. In other words the pole number of the flux density waveformhas to be the same as that of the winding. This is a very important propertyof sinusoidal windings. One can then consider the flux density harmonics to becontributing to the leakage flux.

Remark 2 Real machine windings are not exactly sinusoidally distributed asin the ideal case above. Therefore there are spatial harmonics in the windingdistribution itself. Consequently it is possible for higher order harmonics in theflux density waveform to link with same pole number harmonic in the windingdistribution, resulting in a harmonic voltage. For example, most winding con-figurations contain a significant third harmonic spatial component, therefore thethird harmonic in the flux density waveform (introduced by saturation effects)can link with the individual windings. Consequently third harmonic voltages canbe seen in the phase voltages.

1.2.2 Winding Interaction with Temporal Flux DensityVariation

In this section we consider a non-sinusoidal, spatially stationary flux densitydistribution which has a sinusoidal temporal variation, interacting with a sinu-soidal winding distribution. For the sake of the following argument consider theflux density to have the following form:

B(θp) = Bn cosnθp (1.18)

Let us firstly consider the n = 1 case. Consider a single coil which has thedimensions shown in Figure 1.2. One can calculate the flux linking a coil at anyposition using the general expression:

φ =∮

B.dS (1.19)


Figure 1.2: Dimensions of a single coil.

Consider the situation where there is only the fundamental flux density dis-tribution. The above surface integral can be written as follows (using Figure 1.2)for the flux at angle coil position θp1:

φ(θp) = B1r

∫ θp+π

θp

∫ l

0

cos θ dl dθ

= B1r

∫ θp+π

θp

l cos θ dθ

= −2B1rl sin θp (1.20)

The dot product is eliminated in this situation as the flux density is perpendic-ular to the integration surface.

In order to get the voltage induced in the coils at a particular position aroundthe machine the following calculation has to be carried out:

e(θp) = n(θp)dφ(θp)dt

(1.21)

= na sin θpd

dt

(−2B1rl sin θp

)= −2narlB1 sin2 θp

ωcosωt (1.22)

Remark 3 Equation (1.22) is obtained by realising that we are dealing witha sinusoidal temporal variation of a sinusoidal spatial distribution. Thereforethe amplitude of the flux density is varying with respect to time in a sinusoidalmanner. Therefore:

B1 = B1 sinωt (1.23)1Note that the θp in the following expression is the angle of the most clockwise side of the

coil.


where ω is the frequency of the temporal variation. It is important to realisethat the θp angle in (1.22) is constant with respect to time in this case.

To find the total voltage for the whole winding the individual contributionsfor the number of turns at each position θp have to be added:

eT =∫ π

0

e(θp) dθp

=−2narlB1 cosωt

ω

∫ π

0

sin2 θp dθp

=−narlB1π

ωcosωt (1.24)

i.e. a temporal sinusoidal voltage is produced from the winding as one wouldexpect.

The more interesting case is when the flux density spatial distribution isnon-sinusoidal as in (1.18). In this case the flux for a single coil is:

φ(θp)n = Bnr

∫ θp+π

θp

∫ l

0

cosnθ dl dθ

=Bnrl

n[sinn(θp + π)− sinnθp] (1.25)

Clearly φ(θp)n = 0 for n even. Therefore the even harmonics do not link to asingle coil.

For the n odd case it can be seen that the expression for the flux becomes:

φ(θp)n =−2Bnrl

nsinnθp (1.26)

To calculate the voltage in a single coil at some position θp we again apply(1.21). Carrying out the differentiation on (1.26) we get:

e(θp)n =−2naBnrl

nωsin θp sinnθp cosωt :n is odd (1.27)

To get the total voltage due to the winding the individual contributions areintegrated as in the previous case:

eT =−2naBnrl cosωt

nω

∫ π

0

sin θp sinnθp dθp (1.28)

It can be shown that∫ π

0sin θp sinnθp dθp = 0, therefore the total voltage due to

the odd harmonics is zero. Therefore, as with the spatially moving flux densitycase, only the component of the flux density that has the same pole number asthe winding links with the winding, even if the harmonics are space stationaryand have a time varying amplitude..

Remark 4 The main implications of the above analysis is that the flux densitycomponent with the same pole number as the winding links with the winding.Therefore, for a pure sinusoidally distributed winding, the harmonics in the fluxdensity only contribute to the leakage flux, and do not have a role in determiningthe performance of the machine. However, in reality a pure sinusoidal winding

1.3 Flux Linkage to Sinusoidally Distributed Windings 10

cannot be produced, and there are spatial harmonics in the winding distribution.Therefore, harmonics in the flux density waveform can link with similar polenumber harmonics in the winding distribution resulting in higher order voltageharmonics being produced in the winding. These harmonics will also have aninfluence on machine performance.

Remark 5 The fact that even non-sinusoidally distributed windings react pri-marily to fluxes of the same pole number as the spatial fundamental of the wind-ing means, from a machine modelling perspective that the models can produceperformance results that are relevant to real machines.

1.3 Flux Linkage to Sinusoidally Distributed Wind-

ings

The flux linkage properties of sinusoidal windings are very important in themodelling of machines with sinusoidally distributed windings. In these sectionwe will evaluate the flux linkage expression for the winding specified by (1.1)except that it is shifted by the angle α. Therefore the winding equation is:

n = na sin(θp − α) (1.29)

This winding is subjected to a sinusoidally distributed flux density distribu-tion of the form:

B(θp) = Bm cos θp (1.30)

We wish to work out what the total flux linkage to this winding is. Applyingthe fundamental definition of flux (1.19) to a single turn of the winding at someangle θp, similarly to (1.20) we get:

φ(θ) =∫ θp+π

θp

B(θ)lr dθ

= 2Bmlr sin θp (1.31)

Therefore for the turns distribution given by (1.29) the total linkage at a par-ticular position θp is:

ψ(θp) = nφ(θp)= 2naBmlr sin θp sin(θp − α) (1.32)

To find the total flux linking the coil integrate over the entire positive half cycleof the flux density waveform:

ψT =∫ π

2

−π2

ψ(θp) dθp

= 2naBmlr

∫ π2

−π2

sin θp sin(θp − α) dθp

= naBmlrπ cosα (1.33)

1.4 Other Important Assumptions 11

Remark 6 The above expression for ψT says that the flux linking a sinusoidalwinding varies co-sinusoidally with the angle between the axis of the winding andthe vector of the sinusoidally distributed flux density distribution. Another wayof interpreting this is to say that a flux linkage vector of amplitude naBmlrπlies along the flux density vector, and the component of the flux linkage vectorthat lines coincident with the axis of the winding is the flux linking the winding.Therefore, we are creating the view that there is a flux linkage that is sinusoidallydistributed in space. This concept is important in space vector modelling ofmachines.

1.4 Other Important Assumptions

Whilst the sinusoidal assumption is very important, other assumptions are alsomade in order to make the modelling of the machine tractable. These assump-tions are:

1. The stator windings are assumed to be sinusoidally distributed. Whenexcited with current a sinusoidal spatial distribution of mmf is produced.

2. The machine does not exhibit any stator or rotor slotting effects.

3. The machine iron is a linear material, i.e. it is not subject to magnetic sat-uration effects. The permeability of the material is very large in compar-ison to air. Therefore the permeance of the magnetic paths is dominatedby the air gaps.

4. The airgap flux density waveforms can be adequately represented by thefundamental component.

Let us examine some of these assumptions. The first assumption says thatwe have ideal sinusoidally distributed windings. As mentioned previously thisnever occurs in practice because the machine windings have to be made upfrom finite diameter wires, and generally speaking these wires are placed formechanical and magnetic reasons in slots in the stator or rotor of a machine.However, because of the properties of sinusoidally distributed windings (evenapproximately sinusoidally distributed windings) the effects of slotting only pro-duces secondary effects in terms of induced voltages. The second assumption isclearly related to the first assumption.

The third assumption would seem to be rather restrictive at first glance.Almost all practical machines are constructed using iron which exhibits a non-linear flux density versus magnetising force characteristic – . the iron saturates.Most real machines exhibit saturation in their normal operation regimes. Thisassumption under the circumstance of high saturation makes a model generateerroneous results. However, in many situations the saturation is such that thelinear approximation still gives reasonable results. In any case, the essentialcharacteristics of the dynamic performance of the machine is preserved underthe linear material assumption. This is fortunate, as the models can be analysedusing standard linear analysis because of this assumption.

The final assumption essentially says that the modelling will assumed thatthe fluxes in the machine are spatially sinusoidal. This assumption does notcause too many problems for reasons stated previously, and is important becauseit allows powerful analysis techniques to be applied to machines.

1.4 Other Important Assumptions 12

120�

�

ia

ib

ic

va

vb

vc

Figure 1.3: Conceptual diagram of sinusoidally distributed three phase windings

1.4.1 Properties of Three Phase Sinusoidal Windings

Most practical AC machines use three phase windings on their stator, and somealso have these types of winding on the rotor (e.g. wound rotor induction ma-chine). If one is seeks to understand AC machines it is important to understandthe basic properties of these windings.

Figure 1.3 shows a conceptual layout of a set of three phase windings. Notethat in this diagram the concentrated coils drawn represent sinusoidally dis-tributed windings. The axes of each of the coils are 120◦ apart in space. If eachof these coils was fed with a DC current then each winding would give a mmfdistribution that is sinusoidal in space. The three mmf waveforms can be addedto give a single sinusoidal resultant mmf waveform. By controlling the currentsin these three winding the resultant mmf waveform can be made to move inspace. This phenomena becomes interesting if the windings are fed with threephase currents.

Given that the space distribution of the mmfs for windings a, b and c canbe modelled similarly to (1.2) then the following expressions can be written for

1.5 Torque Expressions 13

the mmfs:

Fa(θp) = Fa cos θp (1.34)

Fb(θp) = Fb cos(θp − 2π

3

)(1.35)

Fc(θp) = Fc cos(θp +

2π3

)(1.36)

where θp is as defined in Figure 1.3 and Fa, Fb, and Fc are the peak mmfs.The resultant mmf distribution at any point θp around the three phase machineperiphery is:

FT = Fa cos θp + Fb cos(θp − 2π

3

)+ Fc cos

(θp +

2π3

)(1.37)

Assuming that the three phase windings have identical turns, and they are beingdriven by three phase currents of the form:

ia = Ipk cosωt (1.38)

ib = Ipk cos(ωt− 2π

3

)(1.39)

ic = Ipk cos(ωt+

2π3

)(1.40)

then (1.37) can be written as:

FT = NIpk[cosωt cos θ + cos

(ωt− 2π

3

)cos

(θ − 2π

3

)

+ cos(ωt+

2π3

)cos

(θ +

2π3

)](1.41)

=32NIpk cos(ωt− θ) (1.42)

where N = N/2. As mentioned previously the division by 2 is to make themmf expression the mmf per airgap (i.e. the total mmf is expended across twoairgaps).

Equation (1.42) means that the resultant mmf has a spatial sinusoidal distri-bution which is rotating around the machine at ωt electrical radians per second. This rotating mmf

property is funda-mental to the oper-ation of all AC ma-chines

If the mmf of (1.42) is acting on a uniform air gap between two iron structures(such as occurs in an electrical machine with a round rotor) then the flux pro-duced by the mmf will also be spatially sinusoidal. If the mmf is acting on anon-uniform air gap between two iron structures (such as occurs in a salientpole machine) then the resultant flux will most probably not be sinusoidal.2

1.5 Torque Expressions

One of the most fundamental quantities of any machines performance is thetorque. After all one is usually using a machine to produce torque and power

2Note the the amplitude of the mmf waveform is 34NIpk because half the mmf is expended

across each of the airgaps in the machine.


from electrical energy. Consequently it is essential to have techniques in anymachine model to allow the torque that will be produced to be accurately esti-mated.

The approach taken in this section is to firstly consider a very general tech-nique for estimating torque. This technique, based on the concept of co-energyis capable of accounting for the saturation non-linearities in the machine. Itsgenerality extends further, since it can be successfully applied to a variety ofdifferent machines. The following development will place special emphasis onthe reluctance machine.

The classic way of calculating the power or torque produced by an electricalmachine is to write the equation for the energy balance in the machine. Thefollowing discussion is based on that in [2]. The expressions developed are forelectrical to mechanical energy conversion. Mechanical to electrical conversionis the dual of this, and will not be discussed separately. The energy balanceequation is based on the conservation of energy principle – i.e. the input energymust be balanced by the losses (both electrical and mechanical), any energytransiently stored in the system (both mechanical and electrical), and the outputenergy in the form of mechanical work. Therefore the energy balance can bewritten as:[

Electricalenergy input

]=

[Electricallosses

]+

[Stored energy

in fields

]+

[Mechanical

energy

](1.43)

In symbol form this may be written as:

Ee = Ele + Efe + Eme (1.44)

The mechanical energy component may not appear as mechanical work, butsome of it may be stored in forms such as kinetic energy and various forms ofpotential energy. Therefore the actual output mechanical energy is:

Emo = Eme − Elm − Esm (1.45)

where:

Elm � the mechanical losses

Esm � the stored mechanical energy

Therefore (1.44) may be written as:

Ee = Ele + Elm︸︷︷︸Losses

+Efe + Esm︸︷︷︸Stored

+Emo (1.46)

1.5.1 Torque of a Simple Reluctance Machine

The following discussion is with reference to Figure 1.4, which is a schematicdiagram of a simple reluctance machine. The following assumptions are madein the analysis:

1. The iron circuit exhibits saturation, i.e. it has a nonlinear flux vs currentrelationship.


Figure 1.4: Simple singly excited reluctance machine.

2. There is no leakage flux.

3. Hysteresis and eddy currents are ignored.

4. Mechanical energy storage and losses are ignored.

Using standard circuit analysis on circuit in Figure 1.4 gives:

v = Ri+dψ

dt(1.47)

Therefore the power being input into the circuit is:

vi = Ri2 + idψ

dt(1.48)

To get the total energy input into the circuit over a time t integrate (1.48)assuming that the initial energy storage is zero:∫ t

0

vi dt =∫ t

0

[Ri2 + i

dψ

dt

]dt (1.49)

Equation (1.49) can be broken up into two sections – the Ri2 term is clearlyrelated to the resistive losses in the coil. Therefore the remaining term must berelated to any stored field energy and mechanical energy. Considering this termin detail, it can be seen that it is in a form that enables it to be transformed by


a change of variable. Since i = F(ψ), where F denotes a non-linear function.Therefore the second term in (1.49) can be written as:

i(F(ψ))dψ

dtdt = i(ψ)dψ (1.50)

Therefore (1.49) can be written as:∫ t

0

vi dt =∫ t

0

Ri2 dt+∫ ψ

0

i(ψ) dψ (1.51)

The second term therefore can be written in terms of the energy equation as:∫ ψ

0

i(ψ) dψ = Efe + Emo (1.52)

Equation (1.52) contains the output energy term, and therefore can be usedto calculate the output power and torque if there is mechanical movement inthe system. However, before this can be achieved the field energy term must beseparated out. If there isn’t mechanical movement then this term would be thestored field energy in the coils.

Consider the rotor in Figure 1.4. If it is held stationary, and a voltage andcurrent are applied then there can be no mechanical energy output. Thereforeignoring the resistive losses all the input energy must be stored in the magneticfield of the stator coils. Consider two different rotor positions; one where therotor is aligned with the axis of the windings, and the other where it is at someangle to the windings. In the first situation there will be more flux for a givenvalue of current as compared to the second case. A sketch of the flux versuscurrent plots are shown in Figure 1.5. Notice that the unaligned flux plot ismore linear than the aligned flux plot due to the fact that the flux path isdominated by air in the former case, and consequently the flux density does notget high enough to cause saturation.

For the aligned position the integration of (1.52) is represented by the shadedarea in Figure 1.4. Therefore this area represents the field energy stored in thesystem, and is the useful electrical energy applied to the system. Similarlythe area represented by PA, PB , PE , and PA, is the field energy stored in theunaligned position. The alternative integration

∫ i20 ψ di, is called the co-energy

and, as we shall see, is also important in the determination of the mechanicalenergy output of the system when the rotor is allowed to move.

In order to calculate the mechanical energy the rotor is allowed to move. Inorder to find the mechanical energy a thought experiment involving two differenttypes of movement is carried out. The movements are:

1. Very slow movement of the rotor which does not produce any voltageacross the stator coil due to dψ/dt. This means that the current flowingin the coil is determined by its resistance, and is therefore constant.

2. Very fast movement of the rotor. Since the flux linkage cannot changeinstantaneously then the flux is considered to be constant throughout thisprocess.

Ii should be noted that the above two movements are idealised and it is im-possible to carry out these experiments with a high degree of accuracy. Howeverit is possible to approximate them.


Figure 1.5: Flux plots for static movement.

Slow Rotor Movement

Assume that the rotor is moved from the unaligned position to the alignedposition very slowly. The point PB will move along the vertical line to theposition PC as the rotor moves. The initial energy in the system is that energystored in the unaligned position (represented by the area PA, PB, PE , and PA inFigure 1.5). The final stored energy in the system is represented by the shadedarea of Figure 1.5. The difference between these two areas is the change in thestored energy of the system. If the change in the stored energy is subtractedfrom the energy input to the system during this movement then the differencemust be the mechanical energy output during the movement. The energy inputcan be ascertained in the following manner. Assume an incremental movementof the rotor is made as shown in Figure 1.6.

If the current is constant then the shaded area is the incremental electricalenergy added to the system. If these areas are integrated for a movement fromPB to PC then the energy applied to the system is:∫ ψ2

ψ1

i2 dψ = i2∫ ψ2

ψ1

dψ = i2(ψ2 − ψ1) (1.53)

This is represented by an area that is a square region of height (ψ2 − ψ1) andwidth i2. Therefore the energy which is transferred into the mechanical move-ment is:

Emo = i2(ψ2 − ψ1)︸︷︷︸Elec input energy

−[∫ ψ2

0

i dψ −∫ ψ1

0

i dψ

]︸︷︷︸

Change in stored energy

(1.54)


- incremental energy change

Unaligned

Aligned

Figure 1.6: Incremental energy change with small movement of the rotor.

The sign convention of positive power for energy flowing out of the machineshall be taken. This convention is consistent with the normal convention usedfor motoring machines. Power flowing into of the system is negative and powerflowing out of the system is positive. In the particular case above electricalenergy is flowing from the supply into the rotor, and mechanical energy is flowingout of the rotor into an external load via the rotor shaft. The calculation in(1.54) can be seen in Figure 1.7. The output energy area is the differencebetween the co-energy at the aligned position and the co-energy at the unalignedposition. Therefore the mechanical output energy can be written as:

Emo = E′fe2 − E′

fe1 = δE′fe (1.55)

where:

E′fe1 =

∫ i2

0

ψunalign di � the co-energy in the unaligned position.

E′fe2 =

∫ i2

0

ψalign di � the co-energy in the aligned position.

If we adopt the convention that the energy changes are calculated by subtractingthe initial energy from the final energy then the co-energy calculation naturallygives the correct sign for the mechanical output energy using the energy signconvention that we have defined above.

If the energy for a mechanical movement is known then the torque can befound using the expression:

Tave =mechanical energyangular movement

=Emo

∆θpd(1.56)


Figure 1.7: Area representing mechanical output energy.

If one considers a very small movement of δθpd, and take the limit as δθpd → 0,then one can find the expression for the instantaneous torque under the slowmovement condition:

Te = limδθp→0

(δE′

fe

δθpd

)i constant

=∂E′

fe

∂θpd

∣∣∣∣i constant

(1.57)

Instantaneous Rotor MovementOne can think ofthe instantaneousmovement occur-ring because therotor has virtuallyno inertia. Therotor will tend toalign in the positionto maximise theflux.

The second thought experiment is to imagine that the rotor is moved virtuallyinstantaneously from the unaligned to the aligned position. During this move-ment the flux linkage cannot change. The reason for this is that Lenz’s Lawstates that the induced voltage during a rate of change of flux linkage is such asto oppose the rate of change of flux linkage. The constant flux linkage duringthese movements is known as the law of constant flux linkage. In this particularcase, the flux linkage would be attempting to increase as the rotor aligns withthe stator pole. Consequently the induced voltage is such so as to reduce thecurrent from i2 to i1 to keep the flux constant at ψ1. Once the rotor stops at thealigned position the induced voltage due to rotor movement disappears and thecurrent can increase back to the steady state value of i2 (which is determinedby the resistance of the stator winding). The path followed by the flux linkagesis shown in Figure 1.8.

During the movement from the unaligned to aligned position the flux link-ages remain constant (from points PB to PF in Figure 1.8). At position PF therotor has reached to aligned position and the movement stops. The flux thenfollows the PF to PC path. During the movement phase there is no electricalenergy flowing into the system (except to supply the resistive losses). This can


Figure 1.8: Energy with instantaneous movement.

be concluded because∫i dψ = 0. However the stored energy in the field has

changed considerably. Using conservation of energy arguments the conclusionis that the stored field energy has been converted to mechanical energy. Thisenergy is presented by the area PA, PB , and PF in Figure 1.8. Once the rotor isstationary, the current will increase from the i1 value at PF to i2, at a rate de-termined by the time constant of the stator winding. During this time electricalenergy is flowing into the system and being stored in the magnetic field. Thisstatement can be deduced from the fact that the rotor is not moving, thereforethere can be no mechanical energy.

The expression for the mechanical energy can be written as:

Emo = −[area (PAPFPEPA) - area (PAPBPEPA)]

= −[∫ ψ1

0

ialign dψ −∫ ψ1

0

iunalign dψ

](1.58)

The mechanical output energy is therefore the change in the field energy Efe,during this movement. The negative sign results so that the mechanical energyfrom this expression is consistent with that calculated using co-energy. Remem-ber that the convention for the sign of the energy is that energy flowing into thesystem is negative, and out is positive. In the case of both the slow and factrotor movements the mechanical energy flows out of the system (i.e. the energyis being applied to the shaft load on the rotor).

The torque expression can be found for this case as it was for the case usingslow movement. Assume a small increment, δθp, of the movement shown inFigure 1.8. The instantaneous torque over such a movement can be found by


Figure 1.9: Flux versus current trajectory for typical real movement.

letting the angular movement approach zero:

Te = limδθ→0

−(δEfe

δθpd

)ψ constant

= −(∂Efe

∂θpd

)ψ constant

(1.59)

Real Movement

Real movements usually don’t consist of either a very slow rotor movement oran instantaneous movement. They are usually somewhere in between. However,as will be seen in a moment, real movements can be analyzed by using the twoidealised movements in combination. The following discussion is with referenceto Figure 1.9, which shows the ψ versus i trajectory for a typical real movement.

Consider a small elemental movement of the rotor. The area PAbdPA repre-sents the actual amount of mechanical output energy. With a slow movementthe mechanical output energy is PAbcPA, and with instantaneous movementPAbaPA. Heuristically one can see from Figure 1.9 that as the movement an-gle becomes less, the slow movement and instantaneous movement areas willapproach each other, and in the limit they will be equal. In other words theareas bcdb and bdab tend to zero as δθpd → 0. This means that the areas forthe slow movement and instantaneous movement converge to the shaded area,which is the area representing the mechanical energy for the actual movement.


Therefore the following expression can be written:

Te =∂E′

fe

∂θpd


= − ∂Efe

∂θpd

∣∣∣∣ψ constant

(1.60)

In other words the same value of torque is obtained if the rate of change ofco-energy is calculated with a constant current, or if the rate of change of fieldenergy is calculated with constant flux linkage.

The same result may be obtained via a much more formal route. Both thefield energy and the field co-energy can be expressed as functions of the followingform:

Efe = −G(ψ, i) (1.61)E′

fe = H(ψ, i) (1.62)

where G(ψ, i) and H(ψ, i) represent non-linear functions of the variables ψ andi.

Similarly the current, flux and rotor angle can be expressed as the followingfunctional relationships:

i = f(ψ, θpd) (1.63)ψ = g(i, θpd) (1.64)θ = h(ψ, i) (1.65)

where f, g and h are non-linear functions of their respective variables.If the current and flux functional relationships are substituted into (1.61)

and (1.62) then it can be seen that Efe and E′fe can be expressed as functions

of (i, θpd) or (ψ, θpd), i.e.:

Efe = −I(ψ, θpd) (1.66)E′

fe = J (i, θpd) (1.67)

where I and J represent the new non-linear functions after the substitution.Let us consider the co-energy expression. Since i is constant for this expres-

sion we can take the partial derivative with respect to θpd:

Te =∂J (i, θpd)∂θpd

=∂J (i, θpd)∂θpd

∂θpd∂θpd

=∂E′

fe

∂θpd


(1.68)

In a similar fashion it can be shown that:

Te = − ∂Efe

∂θpd

∣∣∣∣ψ constant

(1.69)

Example 7 As an example of the use of the above expressions assume that thecurrent flux relationship is as follows:

ψ = (L1 + L2 cos 2θpd)i0.9 (1.70)


This expression is based on the self inductance expression for a single windingas obtained in (A.28), with the addition of a non-linearity to approximate thesaturation of the steel in the stator of the machine. Note that this is not anattempt to accurately model saturation, but is simply an artifice for this example.Using the expression for the co-energy we have:

E′fe =

∫ i1

0

ψ di

=∫ i1

0

(L1 + L2 cos 2θpd)i0.9 di

=11.9

(L1 + L2 cos 2θpd) i1.91 (1.71)

Using (1.57) on (1.71) one gets:

Te =−2L2i

1.91

1.9sin 2θpd (1.72)

If one now uses (1.52) and rearranges (1.70) with i as the subject of the expres-sion, then the field energy can be calculated as follows:

Efe =∫ ψ1

0

i dψ

=∫ ψ1

0

ψ1

0.9

(L1 + L2 cos 2θpd)1

0.9dψ

=0.91.9

[ψ

1.90.91

(L1 + L2 cos 2θpd)1

0.9

](1.73)

Using (1.59) and (1.73) one obtains:

Te = −(

11.9

2L2ψ1.90.91 sin 2θpd

(L1 + L2 cos 2θpd)1.90.9

)(1.74)

Using (1.70) in (1.74) the following expression can be derived:

Te = −(2L2i

1.91

1.9sin 2θpd

)(1.75)

which is the same as (1.72) as expected.As a final check on this result, the field energy and the co-energy should add upto be ψ1i1. Substituting (1.70) into (1.71) gives:

E′fe =

11.9ψ1i1 (1.76)

Similarly substituting (1.70) into (1.73) gives:

Efe =0.91.9ψ1i1 (1.77)

Adding (1.76) and (1.77) one gets ψ1i1 as expected.


Figure 1.10: Flux versus current for linear magnetic material.

1.5.2 Linear Torque Model

Singly Excited System

An important special case of the above co-energy based analysis is for linearmodels. Linear models are no particularly accurate, but they are useful becausethey can be analysed mathematically and therefore aid our qualitative under-standing of a machine. We shall firstly analyse the simplest possible reluctancemachine – the singly excited reluctance machine.

Consider a linear magnetic system. It can be characterized by the followingexpression:

ψ = Li (1.78)

where L � the inductance of the system. The flux versus current plot for sucha system is shown in Figure 1.10. If the field energy of the system is calculatedone has:

Efe =∫ ψ

0

i dψ

=∫ ψ

0

ψ

Ldψ

=12ψ2

L(1.79)

Given that L = ψ/i then (1.79) can be written as:

Efe =12ψi =

12Li2 (1.80)

Clearly from Figure 1.10 it can be seen that the co-energy and the fieldenergy are equal, i.e.:

Efe = E′fe (1.81)


If the linear system is of the type shown in Figure 1.4 then as the rotor of themachine aligns with the stator poles the inductance of the system will increase.This is indicated in Figure 1.10 by the steeper flux line. For such a movementwe can calculate the torque using a formal method based on (1.57):

Te =∂E

′fe(i, θpd)∂θpd

∣∣∣∣∣i constant

=∂

∂θpd

[12Li2

]∣∣∣∣i constant

∴ Te =12i2dL

dθpd(1.82)

A less formal and heuristic approach is to realise that the co-energy under theless aligned flux line is:

E′fe1 =

12ψ1i (1.83)

and under the more aligned line after a δθ rotor movement is:

E′fe2 =

12ψ2i (1.84)

Since:

ψ1 = L1i and (1.85)ψ2 = L2i (1.86)

we can write:

δE′fe = E

′fe2 − E′

fe1

=12(L2 − L1)i2

=12δLi2 (1.87)

Therefore:

δE′fe

δθpd=

12i2δL

δθpd(1.88)

As δθpd → 0 we end up with (1.82).

Doubly Excited System

Now let us consider a machine system with two windings, instead of one. Weshall consider the situation where we have a stator winding as in the previoussingly excited case, and in addition we also have a winding on the rotor itself.This is a similar situation to a synchronous machine. One might be tempted toask “Why are we considering a winding on the rotor, after all we are supposed tobe concentrating on reluctance machines without such a winding?”. The reasonfor considering this case is that the expressions for the field energy obtained are


Figure 1.11: Doubly excited reluctance machine.

also applicable to the two phase reluctance machine – i.e. a reluctance machinewith two orthogonal stator windings.

Consider a machine of the form shown in Figure 1.11. The expressions forthe flux linking the windings is:

ψ1 = L1i1 +Mi2 (1.89)ψ2 = L2i2 +Mi1 (1.90)

where:

L1 � the self inductance of the stator winding

L2 � the self inductance of the rotor winding

M � the mutual inductance between the statorand the rotor.

The instantaneous voltage equations for the coils are:

v1 = R1i1 +dψ1

dt(1.91)

v2 = R2i2 +dψ2

dt(1.92)

Substituting (1.89) and (1.90) into the above gives:

v1 = R1i1 +d

dt(L1i1) +

d

dt(Mi2) (1.93)

v2 = R2i2 +d

dt(L2i2) +

d

dt(Mi1) (1.94)

Because this system has saliency, then the inductances, L1 and L2 are func-tions of θ. However, because we are considering the machine to be linear, theinductances are not functions of current. Taking the appropriate derivatives


then the voltage equations become:

v1 = R1i1 + L1di1dt

+ i1dL1

dt+M

di2dt

+ i2dM

dt(1.95)

v2 = R2i2 + L2di2dt

+ i2dL2

dt+M

di1dt

+ i1dM

dt(1.96)

If (1.95) and (1.96) are respectively multiplied by i1 and i2, then one ob-tains the total instantaneous power for each of the two windings. If these twoexpressions are added together and integrated from time zero to time t, thenthe total energy input to the system can be calculated. This expression can beshown to have the form:

Ee =∫ t

0

(v1i1 + v2i2) dt

=∫ t

0

(R1i21 +R2i

22) dt+

∫(L1i1 di1 + L2i2 di2 + i1M di2 + i2M di1

+ 2i1i2 dM + i21 dL1 + i22 dL2) (1.97)

Comparing (1.97) to (1.44) allows the following relationships to be written:[Useful electricalenergy input

]=

∫ t

0

(v1i1 + v2i2) dt−∫ t

0

(R1i21 +R2i

22) dt (1.98)

Energy to

fieldstorage

+

Electricalenergy

to mechanicalenergy

=

∫(L1i1 di1 + L2i2 di2 + i1M di2

+ i2M di1 + 2i1i2 dM + i21 dL1 + i22 dL2) (1.99)Because the systemis linear we can in-terchange field en-ergy and field co-energy.

In order to get a value for the torque produced by this device one needsto separate the field energy from the mechanical energy in (1.99). This canbe achieved by locking the rotor of the transducer at some position, and thenenergizing the coils from zero current to some required value. Because the rotoris locked there can be no mechanical energy, therefore all the energy must befield energy. Note that this condition also implies that the dL1, dL2 and dMterms in (1.99) are zero. Therefore the stored field energy equation becomes:

Efe =∫ i1

0

L1i1 di1 +∫ i2

0

L2i2 di2 +∫ i2

0

i1M di2

=12L1i

21 +

12L2i

22 + i1i2M (1.100)

The integration in (1.100) implicitly relies on the fact that the energy storedin the field of an inductor depends on the instantaneous value of the current,and not on the time history of how the current attained this value. The reasonthat only the i1M di2 term is integrated in (1.100) is not immediately obvious.However it can be reasoned out using the following thought experiment. Assumethat i2 is initially held at zero by an open circuit. At the same time i1 is increasedfrom zero to some final value of I1. During this time the energy stored in the self


inductance is 12L1i

21. With the current in winding 1 held constant, the current

in winding 2 is then increased to I2. Similarly the energy stored in the winding2 self inductance is 1

2L2i22. However, due to the mutual coupling to winding

1 during this current increase there is an induced voltage in winding 1 whichattempts to produce a current to oppose the changing flux. The current sourcesupplying the I1 current produces a voltage to oppose this induced voltage sothat the current remains constant at I1. Therefore energy is being supplied towinding 1, and this energy must be going into the stored field since it cannotgo into mechanical energy. The energy supplied to winding 1 is:

EM =∫ T

0

Mdi2dt︸︷︷︸

Induced voltage

I1 dt

=∫ I2

0

MI1 di2 =MI1I2 (1.101)

Therefore the total energy is the self inductance energy terms plus (1.101). Amore formal proof can be obtained if the currents in both winding 1 and 2 areassumed to increase linearly from zero to their final values; i.e.:

i1(t) = K1ti2(t) = K2t

}for 0 ≤ t ≤ T (1.102)

If we resort to a form of the stored energy equation based on (1.97), we canwrite, using (1.102) the following for the stored field energy in the mutual flux:

EM =∫ T

0

(i2M

di1dt

+ i1Mdi2dt

)dt

=∫ T

0

2MK1K2t dt

=MK1K2T2 =Mi1(T )i2(T ) (1.103)

which is the same expression as (1.101).It is a simple matter to now calculate the torque produced by this system.

Because the system is linear we know that Efe = E′fe, therefore using (1.57) on

(1.100) one can write the following:

Te =∂Efe

∂θ


=12i21dL1

dθ+

12i22dL2

dθ+ i1i2

dM

dθ(1.104)

If the rotor is moving what happens with the stored field energy. Intuitiontells one that the field energy must be changing because the inductances arechanging. Clearly the change of the inductance with the angular movement isthe key to the production of torque. Therefore if torque is being produced thenmechanical energy is also being produced if there is movement. The questionto resolve is how much of the energy being input to the system during thismovement phase is going into mechanical energy and how much is going into


stored field energy. In order to analyze this situation differentiate (1.100) withrespect to time:

dEfe

dt= L1i1

di1dt

+12i1dL1

dt+ L2i2

di2dt

+12L2i2

dL2

dt+Mi2

di1dt

+Mi1di2dt

If (1.105) is integrated with respect to time then the total field energy can beobtained:

Efe =∫dEfe

=∫(L1i1 di1 +

12i21 dL1 + L2i2 di2 +

12i22 dL2 + i1i2 dM + i1M di2 + i2M di1)

(1.105)

The field energy related terms in this expression are obviously those used in(1.100), therefore the other terms are related to the mechanical energy, i.e. themechanical energy is:

Emo =∫(12i21 dL1 +

12i22 dL2 + i1i2 dM) (1.106)

If this expression is compared with (1.99), and eliminating the stored field energyterms one ends up with the following terms only:∫

(2i1i2 dM + i21 dL1 + i22 dL2) (1.107)

Clearly this is twice (1.106), therefore when the rotor is moved half the electricalinput energy associated with the movement is going into mechanical outputenergy, and the other half is being stored in field energy.

1.5.3 The Ellipse Diagram for co-energy

A classic case study for the use of co-energy is the synchronous reluctance ma-chine (SYNCREL). The development in the previous section of the basic torqueexpressions in terms of field energy and co-energy may appear to be very lowlevel. However, an understanding of this basic way of calculating torque is veryrelevant to the machines such as the SYNCREL as it exhibits a high degreeof saturation in normal operation. Co-energy torque calculation techniques arealso very useful for the switched reluctance machine for the same reason.

In this section we shall concentrate on the application of co-energy to calcu-late the torque of the SYNCREL. The advantage of using the co-energy approachis that the non-linearities caused by saturation and slotting effects can be ac-counted for, whereas with the conventional linearised models using sinusoidalapproximations these effects are ignored. Using co-energy in combination withfinite element analysis it is possible to get very accurate estimates of the torqueproduction and ripple for these machines at the design stage [5].

To demonstrate the application of co-energy to the reluctance machine weshall firstly consider a linear machine – i.e. no saturation or slotting effects.


Therefore the inductance expressions for the machine obey (A.55) and (A.56)derived previously. It will be shown in a future section that the dq-axis induc-tances can be related to the phase expressions in the following manner:

Ld = Lsl +32(L1 + L2) (1.108)

Lq = Lsl +32(L1 − L2) (1.109)

where Lsl � the self leakage of the phase windings. By manipulating these twoexpressions L1 and L2 can be obtained in terms of the dq-axes inductances:

L1 =Ld + Lq

3+

23Lsl (1.110)

L2 =Ld − Lq

3(1.111)

It is assumed that this machine is a three phase, four pole machine, and isexcited by currents of the form:

ia = Ipk cos(θpd + γ) (1.112)

ib = Ipk cos(θpd + γ − 2π3) (1.113)

ic = Ipk cos(θpd + γ +2π3) (1.114)

Notice that the currents are synchronized with the rotor position, and that theangle between the resultant mmf of the stator and the rotor d -axis is alwaysγ radians. For the purposes of this example we shall only consider γ = π/4radians.

From finite element analysis the dq parameters for an axially laminatedmotor with the phase current at 15 amps rms, and the current angle at π/4have been determined as [5]:

Ld = 77.3mHLq = 10.0mHLsl = 2.3mH

(1.115)

If (1.115) are substituted into (1.110) and (1.111) and these are substitutedinto (A.55) and (A.56), then the plots in Figure 1.12 can be obtained for theinductance variation with rotor angle.

To calculate the total flux linkage to the a-phase under these conditions thenthe following expressions have to be used:

ψaa = Laaia (1.116)ψab = Labib (1.117)ψac = Lacic (1.118)

and the total flux linkages with the a-phase are:

ψaf = ψaa + ψab + ψac (1.119)


Figure 1.12: Self and mutual inductance variation with rotor angle.

Clearly the expressions in (1.116), (1.117) and (1.118) involve multiplicativesinusoidal functions when the current expressions (1.112), (1.113) and (1.114)are substituted in. For example, (1.116) is an expression of the form:

ψaa = Ipk[L1 cos(θpd +π

4) + L2 cos 2θpd cos(θpd +

π

4)] (1.120)

and similarly for ψab and ψac. If (1.116), (1.117) and (1.118) are plotted thenFigure 1.13 results. Even though the self and mutual flux curves are nonsinu-soidal, the resultant total flux curve is sinusoidal. In [5] finite element (FE)based a-phase flux results have been compared to these ideal curves, and veryclose agreement is achieved [5].

In order to apply the co-energy concept to calculate the torque for the ma-chine a ψ vs i plot is required. This can be achieved by plotting the expression(1.119) versus the instantaneous a-phase current shown in (1.112). Figure 1.14shows this plot. Notice that the total flux linkage curve is an ellipse. The selfand mutual flux linkage terms and the aligned and unaligned linear magnetisa-tion lines are also shown. Note that the SYNCREL is not bound by these lines,as is the case for the switched reluctance machine. This is due to the effectof the mutual flux on the flux linking a single phase. Another major point ofdifference is that the ψ vs i plot traverses two quadrants, whereas the plot forthe switched reluctance machine stays in one quadrant. This occurs because ofthe bidirectional nature of the currents in the SYNCREL.

One can interpret this diagram as follows. If at t = 0 we have θpd = 0,then ia = Ipk/

√2. For the particular case in the plot Ipk = 22 Amp. The

starting point is marked on Figure 1.14. As θpd increases in a positive directionthe ia current will decrease in value. Therefore anticlockwise rotation of therotor corresponds to movement in an anticlockwise direction on the diagram (asshown by the arrow on the diagram).

Figure 1.14 shows a small movement in the rotor angle, from θpd = 0 to θpd =δ. The shaded area is the change in the co-energy for this movement, therefore


Figure 1.13: Flux linkage for the a-phase

T

Figure 1.14: Total, self and mutual flux versus current loci.


the instantaneous torque can be calculated using (1.57). Areas enclosed ina counter-clockwise direction correspond to positive (motoring) torque, whileareas in the clockwise direction give rise to negative (generating) torque.

Remark 8 Note that Figure 1.14 has been drawn for a linear system. Thesegment corresponding to the δ movement of the rotor is a triangle because ofthis linear assumption. In a real machine this area would have curved sidesdue to saturation at the higher currents when the rotor is aligned with a phase.Indeed in this situation the outside boundary of the diagram is no longer anellipse (this only occurs if the magnetic material is linear). In order to getaccurate instantaneous torques under this condition one would have to calculatethe area taking into account the non-linear shape of the segment [5].

One complete traverse of the ellipse corresponds to 360◦ electrical, and con-sequently to the passage of two rotor poles past the axis of the phase winding.Therefore, to conform with the normal convention applied to the switched re-luctance machine, the basic unit of energy conversion is half the area of theellipse, Eastroke = E2π/2, where E2π is the area for one complete revolution ofthe ellipse. The ellipse shown in Figure 1.14 is only for one phase, similar ellipsescan be drawn for the other two phases of a three phase machine. Therefore thetotal energy per stroke for the machine is:

Estroke = 3Eastroke =32E2π (1.121)

Assume that the machine in question has pp pole pairs; i.e. 2pp poles.Therefore in one mechanical revolution of the machine rotor 2pp poles will passeach phase axis. Clearly the total energy is the number of strokes multiplied bythe energy of a single stroke, i.e.:

ETstroke = 6pEastroke = 2ppEstroke (1.122)

The expression for the average torque for a revolution of the rotor is:

T ave =∆Efe

∆θ=ETstroke

2π=ppπEstroke (1.123)

For a general m phase, pp pole pair machine the average torque expressioncan easily be shown to be:

Tave =mppπEastroke

=ppπEstroke (1.124)

where Estroke = mEastroke .In order to calculate the torque using this technique one needs to calculate

the area of the ellipse. In general a numerical technique needs to be used to dothis (since the data for this curve in a real situation would come from a finiteelement analysis of the motor). One simple technique is the break the ellipse intoa number of sectors, each corresponding to the same small angular movementof the rotor. The area of each sector is then computed and summed to get thetotal area. Figure 1.15 shows the ellipse divided into a number of approximately


0

180

Figure 1.15: Segments used for ellipse area.

triangular sectors. The area of each of these areas can be approximated as atriangle, especially if the number of sectors is large. A general formula for thearea of a triangle is:

Atriag =12[x3(y1 − y2) + x2(y3 − y1) + x1(y2 − y3)] (1.125)

where the xi and yj (i, j = 1, 2, 3) are the vertices of the triangle in the xyplane.

If the triangles in Figure 1.15 are summed using (1.125) then the result forthis particular case is 45.4183 Nm. If the torque is calculated via the normal dqexpression for the torque of a three phase machine, i.e.:

T ave =32pp(Ld − Lq)idiq (1.126)

where pp � the pole pairs, then the result is 45.4275 Nm. Clearly as the numberof triangular segments is increased then the ellipse calculated torque approachesthe dq even more closely.

Remark 9 The above calculation is comparing the average torque using thedq expression and using the ellipse area assuming that the system is linear.Therefore any inaccuracy is due to the numerical inaccuracy of the triangles.It should be understood that the co-energy diagram for a real machine is notan ellipse, and one cannot use triangular segments to work out the accurateinstantaneous torques (as noted above). However even in the case of a saturatedmachine, triangles are quite satisfactory for the calculation of the average torquesince it is related to the total co-energy area, and not the area of the individualsegments.


Figure 1.16: Co-energy “ellipses” for a saturated SYNCREL

The main benefit of the ellipse is realised if one is dealing with data for a“real” machine – i.e. a machine with saturation and slotting effects. Saturationcharacteristics can be accounted for using the more traditional dq based formulaeby including saturation functions for the inductances, but slotting effects do notfit into the fundamental sinusoidal assumptions used to derived these types ofmodels. However, if a machine is designed using finite element (FE) modellingtechniques then the ψ versus i data is available from the FE package. SinceFE modelling uses the exact geometric dimensions of the machine stator androtor, as well as the characteristics of the magnetic materials, this data then willcontain the effects of the saturation and slots. Consequently, application of theellipse torque calculation technique to this data allows very precise prediction ofnot only the average torque from the machine, but also the torque ripple. Thisis an invaluable aid in designing a machine.

Reference [5] compares measured values of torque with torque data calcu-lated using the ellipse technique on the FE design data. The correspondencebetween the two sets of results is very good. One interesting point that is madein [5] is that the saturated “ellipse shapes” obtained from FE data are almostexactly the same as the ideal ones. The torque ripple present in a machinewith slotting manifests itself in irregular separation of the curved radial linesemanating from the centre of the ellipse. These lines each correspond to anequal angular movement of the rotor, and the co-energy enclosed in each smalltriangle is related to the instantaneous torque. In a slotted machine these areasbecome very irregular. These effects are shown schematically in Figure 1.16.Notice the curvature of the ellipses as the current level increases, and the differ-ent co-energy areas for the same angular movement of the rotor indicating thatthe machine has significant slotting ripple.

Chapter 2

The Kron PrimitiveMachine

2.1 Introduction

This chapter will develop some basic tools required for the transient analysis ofmachines. the modelling approach will be based on the Kron primitive machine.The Kron primitive is a generalisation of the DC commutator machine. It isuseful because almost all machines can be transformed into a Kron primitivemachine, allowing similar analysis of most machines within a common frame-work.

The develop the Kron machine we shall develop a model for the doublyfed machine, then convert this to a DC machine with a commutator, and thenfinally develop the basic primitive machine. The basic primitive model will thenbe used to analyse the DC machine, with particular emphasis on the separatelyexcited DC machine.

2.2 Model for the Doubly Fed Machine

The following development is with reference to Figure 1.11. This machine can berepresented schematically as in Figure 2.1. Note that even though this machineis shown with a salient rotor and stator, we shall be considering the situationwhere the either the rotor or the stator is cylindrical. The following voltageequations can be written for this system:

v1 = R1i1 + pψ1 (2.1)v2 = R2i2 + pψ2 (2.2)

where p denotes ddt .

These expressions can be expanded in terms of self and mutual inductancesto give:

v1 = R1i1 + p(L1i1 +M12i2) (2.3)v2 = R2i2 + p(L2i2 +M12i1) (2.4)

2.2 Model for the Doubly Fed Machine 37

�

v2

i2

i1

v1

F1

F2

�

( axis)d

Figure 2.1: Magnetic circuit conceptual diagram of a double fed machine

This in turn can be further expanded. Note that we shall assume that L12 andM12 are functions of θ (which they would be in the case of this doubly salientmachine), and in turn that θ is a function of time (i.e. the rotor is moving):

v1 = R1i1 + L1pi1 + i1pL1 +M12pi2 + i2pM12 (2.5)v2 = R2i2 + L2pi2 + i2pL2 +M12pi1 + i1pM12 (2.6)

Now considering the following term:

i1pL1 = i1dL1

dθ

dθ

dt

= i1ωdL1

dt(2.7)

where ω = dθ/dt and is the angular velocity of the rotor.Therefore (2.5) and (2.6) can be written as follows:

v1 = R1i1 + L1pi1 +M12pi2 + ωi1dL1

dθ+ ωi2

dM12

dθ(2.8)

v2 = R2i2 + L2pi2 +M21pi1 + ωi2dL2

dθ+ ωi1

dM12

dθ(2.9)

which can be written in matrix form as follows:

v = [R + Lp+ ωG] i (2.10)

where:

v =[v1v2

]i =

[i1i2

]


R =[R1 00 R2

]L =

[L1 M12

M12 L2

]

G =d

dθ

[L1 M12

M12 L2

]=dLdθ

Consider these equations for two separate cases: namely a cylindrical statorand rotor, and a salient pole rotor and a cylindrical rotor.

2.2.1 Zero saliency case

Because both the stator and the rotor are cylindrical then we have the followingsituation for the inductances of the machine: Note that the mu-

tual inductanceexpression is ap-proximately trueeven for a concen-trated or uniformdistributed windingin a DC machine.

L1 = constantL2 = constantM12 =Mm cos θ

Note that the above relationship for the mutual inductance will be assumedhere. Therefore under these conditions the voltage expression for the machinebecomes:[v1v2

]=

{[R1 00 R2

]+

[L1 Mm cos θ

Mm cos θ L2

]p+ ω

[0 −Mm sin θ

−Mm sin θ 0

]}[i1i2

](2.11)

This expression describes the complete electrical dynamics for this machinesystem. We have not considered the torque produced by the machine as yet,but when this is included then the mechanical dynamics can also be includedto give a full system description.

2.2.2 One degree of saliency case

In this case we consider the expressions for (2.10) with a salient stator and acylindrical rotor. It can be shown that the inductances for this situation are:

L1 = constantL2 = L′

2 + L′′2 cos 2θ

M12 =Mm cos θ

Therefore the voltage expressions become:[v1v2

]=

{[R1 00 R2

]+

[L1 Mm cos θ

Mm cos θ (L′1 + L′′

1 cos 2θ)

]p+ ω

[0 −Mm cos θ

−Mm cos θ −2L′′1 sin 2θ

]}[i1i2

](2.12)

2.2.3 Torque expression

The basic torque expression for this machine was found as (1.104), and is re-peated here for convenience:

Te =∂Efe

∂θ


=12i21dL1

dθ+

12i22dL2

dθ+ i1i2

dM12

dθ(2.13)


��

2�

�

2

�-�

Stator coil axis

Generation

Motoring

�i L2

2

2 2'' sin �

�i i Mm1 2 sin�

Figure 2.2: Torque plot for the double fed machine with DC rotor and statorcurrents.

By inspection this expression can be written in matrix form as follows:

Te =12[ i1 i2 ]

[dL1dθ

dM12dθ

dM12dθ

dL2dθ

] [i1i2

](2.14)

which can be written in matrix form as:

Te =12iTGi (2.15)

This equation turns out to be a general expression for virtually all machines. Forexample the torque for both the machine types presented above can be foundfor any particular stator and rotor currents by substituting for the inductanceterms in the G matrix.

Example: Let us consider a couple of examples of the use of the torqueexpression. Substitute into the G matrix the inductance expressions for the onedegree of saliency case, and we get:

Te =12[i1 i2

] [ 0 −Mm sin θ−Mm sin θ −2L′′

2 sin 2θ

] [i1i2

](2.16)

Consider that both the stator and the rotor are fed with DC currents. If thetorque expression components and their total are plotted then Figure 2.2 results.Note that the torque is pulsating, and that the average torque over a completecycle of rotation is zero. Another observation that one can make from this Note that torque is

positive in the anti-clockwise direction

diagram is that a net torque can be produced if the current in the either of thewindings (but not both) is reversed at the θ = 0 or θ = π angles.

2.3 Commutator Machines 40

++ + + + + + ++++++

+

+

F1F1

F2F2

d axis d axis

q axis q axis

�

�

i2

�

�

i1

i2

v1

v2

pseudo-stationary

Figure 2.3: A two pole commutator machine

2.3 Commutator Machines

An important group of AC and DC machines are those employing a commuta-tor. Whilst these machines are of considerable practical importance, they arealso important from a machine theory point of view as they form the basis ofthe generalised primitive machine. The concept of a pseudo stationary coil isfundamental to this, and this is the main reason that we are revisiting thesemachines. At the end of this section we shall have a full dynamic model of theDC motor.

Figure 2.3 is a physical and conceptual diagram of a two pole commutatormachine. The coil on the rotor is said to be pseudo stationary because theconductors in the winding are moving (and therefore cutting flux) but the mmfproduced by the winding remains stationary in space. Of course it is the brushesthat allow this magic to occur. The conceptual diagram to the right of Figure 2.3shows the machine coils represented by concentrated coil equivalents. The coilon the rotor is now a pseudo-stationary coil.

The equations for this machine can now be written from (2.8) and (2.9)by realising that the d/dθ terms in (2.8) will be zero as the rotor coil from amagnetic point of view is space stationary with respect to the stator coil andthe rotor is round. Therefore the voltage equations expressions are:

v1 = R1i1 + L1pi1 +M12pi2 (2.17)

v2 = R2i2 + L2pi2 +M21pi1 + ωi2dL2

dθ+ ωi1

dM12

dθ(2.18)

Remark 10 Note that the expression for the rotor equation is the same as forthe non-commutator machine as the windings in the rotor are still moving atan angular velocity with respect to the stator field and consequently have thesame voltages induced in them due to the mutual coupling from the stator andthe variation in rotor magnetic circuit permeance due to the stator saliency. Itis an interesting property of pseudo stationary coils that they have rotationalvoltages induced in them from movement, but magnetically they are stationary

2.3 Commutator Machines 41

(hence the name). Note that θ based expressions for mutual and self flux do notchange with time and are only dependent on the brush angle.

The torque expression for this machine is a little different to that of (2.15)because of the disappearance of all the d/dθ terms in the first voltage equation.The torque expression can be calculated by looking at the total differential en-ergy input to the system and then assigning various components in the resultantexpression to losses, stored field energy and mechanical output energy. Now theincremental energy supplied in time dt is:

dEe = (v1i1 + v2i2)dt (2.19)

which can be expanded to:

dEe = (i21R1 + i22R2)dt︸︷︷︸incremental resistive loss

+ [12L1p(i21) +

12L2p(i22) +M12p(i1i2)]dt︸︷︷︸

incremental field storage

+ ω

(i1i2

dM12

dθ+ i22

dL2

dθ

)dt︸︷︷︸

incremental mechanical output energy

(2.20)

Using the relationship between torque and power – P = Tω, then one canwrite the following expression for the torque of the commutator machine withstator saliency:

Te = i1i2dM12

dθ+ i22

dL2

dθ

∴ Te = iTGi (2.21)

where:

i =[i1i2

]

G =[

0 0dM12dθ

dL2dθ

]Note that (2.21) is exactly twice the previous general expression (2.15) wherethe magnetic axis of the rotor winding was moving with respect to the stator.

Now let us write down the full expression for these equations for the statorsaliency commutator machine. Therefore the inductance values to be used are:

L1 = constant

L2 = L′2 + L

′′2 cos 2θ

dL2

dθ= −2L′′

2 sin 2θ

M12 =Mm cos θdM12

dθ= −Mm sin θ

and the voltage and torque expressions for a general brush displacement are:

Te =[i1 i2

] [ 0 0−Mm sin θ −2L′′

2 sin 2θ

] [i1i2

](2.22)

2.4 The Primitive Machine Concept 42

rotorcoil

Ra

Lr

�M Im f

+-

Ia

Va

Figure 2.4: Equivalent circuit of a DC machine in motoring mode

[v1v2

]=

[R1 00 R2

]+

[L1 Mm cos θ

Mm cos θ L2

]p

+ ω[

0 0−Mm sin θ −2L′′

2 sin 2θ

]

[i1i2

](2.23)

If one considers the situation where the brushes are at θ = −π/2 then theexpressions above become:

Te = i1i2Mm (2.24)v1 = i1R1 + L1pi1 (2.25)v2 = i2R2 + L2pi2 + ωMmi1 (2.26)

If one considers the steady state situation, then the normal steady state DCmachine equation can be obtained:

Te = If IaMm (2.27)Vf = IfRf (2.28)Va = IaRa + ωMmIf (2.29)

where f refers to field quantities, and a refers to the armature quantities. Theresultant equivalent circuit is shown in Figure 2.4.

We shall see in the next section that the DC commutator machine is thebasis of the Kron primitive machine

2.4 The Primitive Machine Concept

It will be shown in the subsequent chapters that most of the common machinetypes can be transformed into a form called the primitive machine or Kronprimitive machine form. The transformed machine has identical performancecharacteristics to the original machine. The reasons for carrying out these trans-formations are:


idr

iqr

iqs

idsvds

vdr

vqr

vqs

q axis

d axis

�

qs

qr

dr ds

pseudo stationary

Figure 2.5: Primitive dq machine

1. A variety of totally different machines can be analysed using the sameprimitive machine. The specifics of a machine are taken care of by thetransformations to and from the primitive machine.

2. There are a number of ways of doing the transformation, and some ofthe techniques result in significant simplifications in the complexity of themachine models.

3. It is often much easier to understand how a machine works in its primitiveform, as much of the complexity present in many machines become hiddenby the transformation process.

The primitive machine is essentially a generalization of the DC commuta-tor machine. Figure 2.5 shows a diagram of the general layout of a primitivemachine. Note that these machines are sometimes called dq machines as theyposition the coils on dq axes. This machine could contain an arbitrary num-ber of coils, but only two are shown in this case to keep the derivation simple.The nomenclature of the coil suffixes indicates the axis the coil lies on and thesupporting magnetic structure (s for stator and r for rotor).

By analogy with the commutator machine of the previous section one canproceed to write down the L and G matrices for a stator salient machine ofthe form of Figure 2.5. The coil pairs relative to the d axis coils are at 0 andπ/2 radians, and relative to the q axis coils are at 0 and −π/2 radians. Let usconsider the matrices for two coils as in (2.23):

L =[

L1 Mm cos θMm cos θ L2

]


Because the windings are separated by π/2 then the mutual inductance elementsof this matrix are non-zero only on the same magnetic axis (since cos θ = 0 forθ = π/2).

Now consider the G matrix:

G =[

0 0−Mm sin θ −2L′′

2 sin 2θ

]

The top row of this matrix is zero because the rotor coil is pseudo stationarywith respect to the stator coil, and consequently their is no relative movement ofthe rotor flux relative to the stator flux, and the stator circuit does no experienceany change in permeance as the rotor rotates. For θ = 0 and ±π/2, sin 2θ = 0and the only G term of concern is −Mm sin θ. Therefore the G matrix for theorthogonal coils becomes:

G =[

0 0∓Mm 0

]

Note that the sign can be deduced by applying Lenz’s law to the system. Terms in the Gmatrix are presentonly when coils arein space quadratureand at least one ispseudo stationary.

We shall now consider the four coil primitive machine case. Writing downthe voltage equations for this system in the most general form we get:

vds = Rdsids +d

dt(Ldsids) +

d

dt(Mdsdridr) +

d

dt(Mdsqriqr)

+d

dt(Mdsqsiqs) (2.30)

vdr = Rdridr +d

dt(Ldridr) +

d

dt(Mdrdsids) +

d

dt(Mdrqriqr)

+d

dt(Mdrqsiqs) (2.31)

vqs = Rqsiqs +d

dt(Lqsiqs) +

d

dt(Mqsqriqr) +

d

dt(Mqsdridr)

+d

dt(Mqsdsids) (2.32)

vqr = Rqriqr +d

dt(Lqriqr) +

d

dt(Mqrqsiqs) +

d

dt(Mqrdridr)

+d

dt(Mqrdsids) (2.33)

The above expressions can be significantly simplified using the observationsnoted above in relation to the terms that are relevant when the windings angles


are 0 or π/2 radians. Applying these principles the equations become:

vds = Rdsids + Ldsdidsdt

+Mdsdrdidrdt

(2.34)

vdr = Rdridr + Ldrdidrdt

+Mdrdsdidsdt

+ ω[iqrdMdrqr

dθ+ iqs

dMdrqs

dθ

](2.35)

vqs = Rqsiqs + Lqsdiqsdt

+Mqsqrdiqrdt

(2.36)

vqr = Rqriqr + Lqrdiqrdt

+Mqrqsdiqsdt

+ ω[idrdMqrdr

dθ+ ids

dMqrds

dθ

](2.37)

Note that the winding that is the source of the flux is taken as the referencewhen evaluating the angles between windings.

The above expression can be written in matrix form as follows:

vdsvdrvqsvqr

=

Rds 0 0 00 Rdr 0 00 0 Rqs 00 0 0 Rqr

+

Lds Mdsdr 0 0Mdrds Ldr 0 00 0 Lqs Mqsqr

0 0 Mqrqs Lqr

︸︷︷︸L matrix

p

+ ω

0 0 0 00 0 dMdrqs(θ)

dθdMdrqr(θ)

dθ0 0 0 0dMqrds(θ)

dθdMqrdr(θ)

dθ 0 0

︸︷︷︸G matrix

idsidriqsiqr

(2.38)

Substituting for the mutual inductance expressions one can write the aboveequation in matrix form as follows:

v = {R + Lp+ ωG}i (2.39)

where:

R =

Rds 0 0 00 Rdr 0 00 0 Rqs 00 0 0 Rqr

L =

Lds Mdsdr 0 0Mdrds Ldr 0 00 0 Lqs Mqsqr

0 0 Mqrqs Lqr

G =

0 0 0 00 0 Mdrqs Lqr

0 0 0 0−Mqrds −Ldr 0 0

v =

vdsvdrvqsvqr

i =

idsidriqsiqr

Let us consider the G matrix coefficients for a moment. Consider thedMdrqs(θ)

dθ term. Now Mdrds(θ) = Mdrds cos θ is the mutual inductance between


the dr and qs windings. In this part of the equation it is being used to givethe voltage induced in the dr winding due to flux generated in the qs winding.Therefore the angle between the winding is measured relative to the qs winding.Expanding:

dMdrqs(θ)dθ

= −Mdrqs sin θ (2.40)

whereMdrqs is the maximum mutual inductance between the d axis rotor wind-ing and the q axis stator winding. In this case the angle is −π/2 radians,therefore:

dMdrqs(θ)dθ

=Mdrqs (2.41)

The other term in the G matrix is the dMdrqr(θ)dθ type terms. These terms

relate to mutual inductance between two windings on the same magnetic struc-ture. Therefore if the windings are at an angle of θ = 0 then the dr and qrwindings are coincident, and assuming perfect couplings under theses condi-tions then the maximum mutual inductance would be equal to each windingsself inductance.

The torque expression for the primitive machine can be calculated using theexpression (2.21):

Te = [ ids idr iqs iqr ]

0 0 0 00 0 Mdrqs Lqr

0 0 0 0−Mqrds −Ldr 0 0

idsidriqsiqr

(2.42)

Expanding and collecting terms:

Te = (Mdrqsiqsidr −Mqrdsidsiqr)− (Ldr − Lqr)idriqr (2.43)

The second term in this expression is due at saliency (i.e. different permeancesin the d and q axis directions). In a non-salient machine this term woulddisappear. The internal mechanical power produced by the machine is easilycalculated from this expression using the expression:

Pe = Teω (2.44)

Often (2.39) is written in its expanded form as follows:vdsvdrvqsvqr

=

Rds + Ldsp Mdsdrp 0 0Mdrdsp Rdr + Ldrp ωMdrqs ωLqr

0 0 Rqs + Lqsp Mqsqrp−ωMqrds −ωLdr Mqrqsp Rqr + Lqrp

idsidriqsiqr

(2.45)

2.4.1 Use of the Primitive Machine - the DC Machine

In this section we shall consider an application of the primitive machine model.To keep things simple we shall look at a DC machine. As one might suspect theconversion of the DC machine model to the primitive machine is very straight


fa

c Compensatingwinding

d axis

q axis

�

(b) Primitive coil diagram

q axis

d axis

Compensating winding

(a) Physical arrangement

i fv f

ic

vc

ia

va

varm

Figure 2.6: Separately excited DC machine with a compensating winding

forward, since, as we have seen in the previous section the primitive machinemodel derivation was based on the DC machine.

Figure 2.6 shows a conceptual diagram of a separately excited DC machinewith a compensating winding or interpole.

In order to develop the primitive machine equations we can apply the rulesmentioned in the previous section. The machine is assumed to have a salientpole stator and a round rotor (which most machines of this type have). Clearlyfrom Figure 2.6 one can see that one of the windings present in the primitivemachine is not present in this machine, therefore it has one equation less thanthe general primitive machine equation. Let us construct the main inductancematrices. In order to understand where the elements go in these matrices weshall nominate the current vector to be:

i =

ificia

Firstly the L matrix. The only locations that elements appear are for mutualinductance terms for coils on the same axes and for self inductance terms:

L =

Lf 0 0

0 Lc Mca

0 Mac La

Now consider the G matrix. These terms occur in pseudo stationary coilsthat are orthogonal to a stationary coil or another pseudo stationary coil. Theterms take the general form of −M sin θ where the θ is measured relative tothe coil that is producing the flux (i.e. the seat of the flux). Therefore in thisparticular case the relevant coils are the armature coil (coil a) and the field coil(coil f) and the maximum mutual flux if the axes were lying on the same axis


would be Maf . Therefore we will have rotational voltage terms in the armatureand field equations. Therefore the G matrix becomes:

G =

0 0 0

0 0 0−Maf 0 0

Substituting the R, L and G matrices into the general expression (2.39)gives the following expression:

vfvcva

=

Rf 0 0

0 Rc 00 0 Ra

+

Lf 0 0

0 Lc Mca

0 Mac La

p

+ ω

0 0 0

0 0 0−Maf 0 0

ificia

(2.46)

which can be expanded giving: vfvcva

=

Rf + Lfp 0 0

0 Rc + Lcp Mcap−ωMaf Macp Ra + Lap

ificia

(2.47)

Now considering the currents we have in the connection shown in Figure 2.6:

ia = −ic and varm = va − vc (2.48)

Using this expression one can then further simplify (2.47) to give:[vfvarm

]=

[Rf + Lfp 0−ωMaf (Ra +Rc) + (La + Lc − 2Mac)p

] [ifia

](2.49)

which is the same expression as that for a separately excited DC machine withan armature inductance of (La + Lc − 2Mac). This inductance is small due tothe effect of the compensating winding since La ≈ Lc ≈Mac.

The torque expression for this machine can be obtained by using the generalexpression for the primitive machine and substituting for the specific G matrixfor this case:

Te = iTGi

= [ if ic ia ]

0 0 0

0 0 0−Maf 0 0

ificia

= −Maf iaif (2.50)

Note that the torqueis negative since ifia and if are posi-tive then the torqueis clockwise, whichis negative usingour convention.

The only remaining equation required to obtain a full dynamic model of aDC machine drive system is the load model. The standard model for a rotatingload is:

Jω + f(ω) + TF = Te (2.51)

2.5 Summary 49

where:

J � rotational moment of inertia

f(ω) � linear or non-linear friction coefficient

TF � fixed load torque

Equations (2.49), (2.50) and (2.51) constitute a complete dynamic model of thecompensated, separately excited DC machine. These equations can be used tocalculate the response of the system under various conditions – short circuitedgenerator, full voltage start etc. In addition they can be used for the design ofcontrollers for this machine.

Remark 11 If the if current is held constant, then the torque expression be-comes Te = Kia, where ia is governed by a first order linear differential equa-tion. One can see why the separately excited DC machine is simple to use inhigh performance control systems.

2.5 Summary

This chapter has shown how the primitive machine is derived from the basicDC commutator machine. Rules have been derived to determine the values ofthe various inductances used in the primitive machine model.

The primitive machine forms the basis for the analysis of a whole range ofmachines whose winding distributions can be modelled as a sinusoidal funda-mental without too much error. This assumption applies to common machinessuch as the induction machine, the synchronous machine, the synchronous re-luctance machine, and many brushless DC machines. However this modellingconcept cannot be applied to machines such as the switched reluctance machine.

Chapter 3

Frame transformations, DQand Space Vector Models

3.1 Introduction

This chapter develops the basic theory behind transformation of AC machinemodels into dq and space vector machine models. In the previous chapter wedeveloped the primitive machine model from the DC commutator machine. Thetransformations presented in this chapter allow an AC machine to be convertedinto a primitive machine form. This has the benefit of allowing all these machinetypes to be analysed using the same techniques. Another advantage is that thetransformed models are often considerably simpler than the non-transformedmodel, allowing simpler analysis and a more intuitive understanding of how themachine works.

The remainder of the chapter is organised as follows. The next major sectionwill present the basics of dq modelling, for both stationary frame and rotatingframe transformations. An example of how the transformations are applied isgiven for the synchronous reluctance machine. The final major section presentsthe other form of dynamic machine equations – space vector models, and showsthe relationship between this modelling technique and dq modelling. Once againthe SYNCREL is used as an example.

3.2 dq Models

Most electrical machines with sinusoidally distributed windings are modelledmathematically using a technique called dq modelling. It is not the purpose ofthis chapter to give an exhaustive derivation of dq modelling of machines, asthis could fill a whole text book in its own right. However, a brief overviewof the principles of dq modelling will be presented, and as an example the dqmodel for the SYNCREL will be derived.

The fundamental assumption used as the basis of dq modelling is that thewinding distribution in a machine is sinusoidal. In addition a number of othersecondary assumptions are made, which are similar to the assumptions used inAppendix A, namely:

3.2 dq Models 51

120

Figure 3.1: Three phase to two phase transformation

1. The machine does not exhibit stator or rotor slotting effects.

2. The machine iron is linear material, i.e. there is no saturation effects.

The combined effect of these assumption is that linear traditional circuitanalysis techniques can be used to analysis the electrical circuit of a machine.The sinusoidal assumption means that various spacial quantities in the machinecan be broken into orthogonal components. It is this that is used to carry outcoordinate transformations from the three phase axes of a machine to the twophase dq axes.

3.2.1 Stationary Frame Transformations

The general idea of a dq type of transformation can be obtained by consideringthe transformation of the three phase currents to their two phase equivalents.Consider Figure 3.1. This shows a conceptual diagram of a three phase ma-chine. The windings represented by the concentrated coils are actually spaciallysinusoidally distributed windings similar to that shown in Figure 1.1. The linesthrough the centre of the coils are the axes of the associated winding mmfs, andtherefore can be thought of a the vector that represents the sinusoidal quantities.

MMF transformations

Given that the space distribution of the mmfs for windings a, b and c can bemodelled similarly to (1.2) then the following expressions can be written for themmfs:

Fa(θp) = Fa cos θp (3.1)

Fb(θp) = Fb cos(θp − 2π

3

)(3.2)

Fc(θp) = Fc cos(θp +

2π3

)(3.3)

3.2 dq Models 52

where θp is as defined in Figure 3.1. The resultant mmf distribution for thethree phase machine is:

FT = Fa cos θp + Fb cos(θp − 2π

3

)+ Fc cos

(θp +

2π3

)(3.4)

Assuming that the three phase windings have identical turns, and they are beingdriven by three phase currents of the form:

ia = Ipk cosωt (3.5)

ib = Ipk cos(ωt− 2π

3

)(3.6)

ic = Ipk cos(ωt+

2π3

)(3.7)

then the peak mmfs for each of the phases are:

Fa = NIpk cosωt (3.8)

Fb = NIpk cos(ωt− 2π

3

)(3.9)

Fc = NIpk cos(ωt+

2π3

)(3.10)

where1 N = N/2.Therefore (3.4) can be written as:

FT = NIpk[cosωt cos θ + cos

(ωt− 2π

3

)cos

(θ − 2π

3

)

+ cos(ωt+

2π3

)cos

(θ +

2π3

)](3.11)


i.e. the resultant mmf is has a spacial sinusoidal distribution which is rotatingaround the machine at ωt electrical radians per second.

If the vectors associated with (3.1), (3.2) and (3.3) are resolved along twoorthogonal axes called the dq axes then the following expressions can be writtenfor the resultant dq axes mmfs:

Fsdq = TFabc (3.13)

i.e.[F sd

F sq

]=

[1 − 1

2 − 12

0√

32 −

√3

2

] FaFbFc

(3.14)

The ‘s’ superscript on these variables means that the dq axes are in a stationaryframe. The meaning of this will become clearer when we look at rotating frametransformations later. The ‘s’ subscript next to the ‘d ’ says that the d axis isfor the stator. An ‘r ’ subscript is used to refer to rotor quantities.

1Note that this definition for N results from the fact that the total mmf of the winding isexpended across two airgaps. Therefore the mmf per airgap is half the total mmf.

3.2 dq Models 53

Now let us consider what each of the mmfs for the axes are. Firstly expandingthe d axis expression we have:

F sds = NIpk

[cosωt− 1

2cos

(ωt− 2π

3

)− 1

2cos

(ωt+

2π3

)](3.15)

=32NIpk cosωt (3.16)

Similarly for the q axis we have:

F sqs =

√32NIpk

[cos(ωt− 2π

3)− cos(ωt+

2π3)]

(3.17)

=32NIpk sinωt (3.18)

Notice that the converted machine has time carrying sinusoidal mmfs on eachaxis that are 90◦ out of phase. Note that both the d and q axis windings are alsosinusoidally distributed as well – ie. F s

ds(θ) = F sds cos θ and F s

qs(θ) = F sqs sin θ.

In order to get the resultant space distribution from both these windings weadd together F s

ds(θ) and Fsqs(θ) similarly to the three phase case. Therefore we

get:

FT = F sds(θ) + F

sqs(θ)

=32NIpk(cosωt cos θ + sinωt sin θ)


Therefore the mmf distribution for the two phase machine is exactly the sameas the distribution for the three phase machine.

We now have a technique for going from a three phase machine mmf to anequivalent two phase machine mmf. However, in order for these transformationsto be very useful we have to have a technique to do the reverse. If we have aninverse transformation then we can relate values calculated in the two phasemachine back to the three phase machine. Therefore we need to make thetransformation of (3.13) invertible. In order to do this the T matrix and Fs

dq

vector are augmented as follows: F s

d

F sq

Fγs

=

1 − 1

2 − 12

0√

32 −

√3

21√2

1√2

1√2

FaFbFc

(3.20)

i.e. Fsdqγ = SFabc (3.21)

and Fabc = S−1Fsdqγ (3.22)

3.2 dq Models 54

where:

S−1 =

23 0

√2

3

− 13

1√3

√2

3

− 13 − 1√

3

√2

3

=23

1 0 1√2

− 12

√3

21√2

− 12 −

√3

21√2

i.e. S−1 =23S T (3.23)

The choice of the 1/√2 augmentation of T was made so that the property

in (3.23) was obtained. Note that the Fγ term is zero if the three phase mmfscontain no zero sequence components, else this term is not zero. Therefore, fora star connected machine Fγ always equals zero, since one cannot have zerosequence currents with this configuration.

Remark 12 Although the above analysis has been carried out assuming balancedsinusoidal currents the transformation expressions are valid for arbitrary currentwaveforms including DC.

Remark 13 The above dq mmf is mmf invariant with the three phase mmf.As we shall see this is not the preferred transformation.

Current Transformations

Given the mmf transformation in the previous section, it is a simple matter toconstruct the transformation for the three phase currents to their equivalent twophase currents. This transformation can be handled in two sensible ways. Thetransformation could be carried out in such a way that the transformed machineproduces the same total power as the original three phase machine. Such trans-formations are called power invariant transformations. Another transformationcan be implemented such that the transformed machine produces 2/3rds thepower of the three phase machine. This is one particular example of a powervariant transformation. Usually the power variant transformation is used, sinceit turns out that in steady state the two phase currents and voltages have ex-actly the same amplitude as the phase voltages and currents of the three phasemachine. If the magnitude of the two phase quantity is taken, and then pro-jected onto the relevant three phase axis, then the instantaneous value can befound for that phase. This transformation is commonly used in the literaturebecause of this property. Another advantage of this transformation is that theper phase inductance values found by the normal testing procedures can be ap-plied to each of the windings of the two phase machine. As we shall see shortlythis implies that end winding of the two phase machine has the same numberof turns as each individual winding of the three phase machine.

Consider the situation where we desire a power variant transformation – thetwo phase machine in this situation has 2/3rds the resultant mmf of the threephase machine. It can be seen from (3.19) that this means that the right hand

3.2 dq Models 55

side of (3.21) has to be multiplied by 2/3. Therefore (3.21) and (3.22) can bewritten as:

Fsdqy =

23SFabc (3.24)

Fabc =32S−1Fs

dqγ = STFsdqγ (3.25)

Now consider the mmf expressions expressed in terms of currents and windingturns:

Fsdqγ = N2φisdqγ (3.26)

Fabc = N3φiabc (3.27)

Using (3.24) one can write:

N2φisdqγ =23SN3φiabc (3.28)

where2:

N2φ = N2φ/2 � the half the number of turns for a windingof the two phase dqγ machine.

N3φ = N3φ/2 � the half the number of turns for each windingof the three phase machine.

Since the two phase dqγ machine is an artificial machine of our creation, weare free to choose the number of turns for each of the windings. Clearly if N2φ =N3φ, i.e. the two phase machine has the same number of turns on its windingsas the three phase machine, and the is relationship has the same form as themmf relationship above. Consequently the isdqγ vector has 2/3rds the magnitudeof the iabc resultant current vector. Therefore the current relationships betweenthe two machines is:

isdqγ =23Siabc (3.29)

iabc = ST isdqγ (3.30)

Voltage Transformations

Similarly, one can derive the relationship between the three phase and two phasevoltages. Consider the power relationships for the two machines:

P3φ = vsT

abciabc (3.31)

P2φ = vsT

dqγ idqγ (3.32)

We want P2φ = 2/3P3φ. Therefore substituting (3.31) and (3.32) into thisexpression and using (3.29) one can obtain:

vsdqγ =

23Svabc (3.33)

vabc = STvsdqγ (3.34)

2Note again that this definition arise from the mmf per airgap condition which is half thetotal mmf.

3.2 dq Models 56

Notice that this expression is in the same form as that for the current. Thereforeit has the same property that the magnitude of the voltage vector is 2/3rds thatof the voltage vector for the three phase machine. If one considers the casewhere the windings are excited by three phase currents of the form in (3.5),(3.6) and (3.7), then it is easy to show that:∣∣isdqγ∣∣ = Ipk (3.35)

i.e. the magnitude of the resultant dqγ vector is equal to the peak current in aphase in steady state. Similarly then we can write:∣∣vs

dqγ

∣∣ = Vpk (3.36)

where Vpk � the peak of three phase sinusoidal voltages supplying the abcwindings. Therefore the use of the 2/3rds power relationship has allowed one toeasily correlate the dqγ voltages and currents to the abc voltages and currents.

Impedance Transformations

Next we need to consider the transformation of the machine parameters be-tween the three phase and two phase machines. Consider the following generalexpressions for the two machines:

vabc = Zabciabc (3.37)vsdqγ = Zs

dqγ isdqγ (3.38)

Using (3.37) together with (3.30) and (3.34) one can write:

vsdqγ =

23SZabcST isdqγ (3.39)

Comparing this expression with (3.38) one can see that:

Zsdqγ =

23SZabcST (3.40)

and Zabc =23STZs

dqγS (3.41)

These general impedance transformations can be used to generate specifictransformations for the inductances and resistances for a three phase winding.For a three phase winding the impedance matrix can be written as:

Zabc =

Ra + Laap Labp Lacp

Lbap Rb + Lbbp LbcpLcap Lcbp Rc + Lccp

(3.42)

where p � d/dt.By inspection it can be seen that the resistive and inductive transformations

become:

Rsdqγ =

23SRabcST (3.43)

Rabc =23STRs

dqγS (3.44)

Lsdqγ =

23SLabcST (3.45)

Labc =23STLs

dqγS (3.46)

3.2 dq Models 57

To dqγs To abcFsdqy = 2

3SFabc Fabc = STFsdqγ

isdqγ = 23Siabc iabc = ST isdqγ

vsdqγ = 2

3Svabc vabc = STvsdqγ

Ψsdqγ = 2

3SΨabc Ψabc = STΨsdqγ

Lsdqγ = 2

3SLabcST Labc = 23S

TLsdqγS

Rsdqγ = 2

3SRabcST Rabc = 23S

TRsdqγS

Zsdqγ = 2

3SZabcST Zabc = 23S

TZsdqγS

Table 3.1: Summary of Stationary Frame Transformations

where:

Rabc =

Ra 0 0

0 Rb 00 0 Rc

Labc =

Laa Lab Lac

Lba Lbb Lbc

Lca Lcb Lcc

Flux Linkage Transformations

Now that we have the inductance and current transformations it is possible todevelop the transformations for the flux linkages. The flux linkage expressionsfor the three and two phase machines are:

Ψabc = Labciabc (3.47)Ψs

dqγ = Lsdqγ i

sdqγ (3.48)

If (3.45) and (3.30) are substituted into (3.48) then one gets:

Ψsdqγ =

23SLabc

23STSiabc

=23SLabciabc

∴ Ψsdqγ =

23SΨs

abc (3.49)

and Ψabc = STΨsdqγ (3.50)

The stationary frame transformations are summarized in Table 3.1.

3.2.2 Rotating Frame Transformations

The transformation in (3.21) allows the three phase windings to be representedby an equivalent set of two phase windings. These winding are stationary withrespect to the original three phase winding. It is then possible to project thestationary two phase winding onto two phase windings that are at some angleto the stationary winding axes and moving with respect to these axes.

The following discussion is with respect to Figure 3.2. This diagram shows arotating dq axes with respect to the stationary dq axes derived in the previous

3.2 dq Models 58

section. The angle θsr is defined with reference to the rotating axis (the subscriptbeing read as the angle of the stator axis (s) with respect to the rotating axis(r)) as this makes it easier to see the projections of the stationary quantitiesonto this axis. Using the normal convention for angle sign (anticlockwise ispositive angle), one can write the following expressions:

F rd1

= F sd cos θsr (3.51)

F rd2

= F sq cos(θsr +

π

2) = −F s

q sin θsr (3.52)

F rq1 = F s

q cos θsr (3.53)

F rq2 = F s

d sin θsr (3.54)

Clearly the total mmf then on each of the rotating axes is:

F rd = F r

d1+ F r

d2

= F sd cos θsr − F s

q sin θsr (3.55)

F rq = F r

q1 + Frq2

= F sq cos θsr + F s

d sin θsr (3.56)

This expression can be written more succinctly in matrix form:[F rd

F rq

]=

[cos θsr − sin θsrsin θsr cos θsr

] [F sd

F sq

](3.57)

The zero sequence component can be included by ensuring that it makes nocontribution to the projected vectors as follows:

F rd

F rq

F rγ

=

cos θsr − sin θsr 0

sin θsr cos θsr 00 0 1

F s

d

F sq

F sγ

(3.58)

and

F s

d

F sq

F sγ

=

cos θsr sin θsr 0

− sin θsr cos θsr 00 0 1

F r

d

F rq

F rγ

(3.59)

To make the θ definition consistent with the angle definition used to define theinductance expressions, use θsr = −θrs, where θrs is the angle of the rotatingaxis with respect to the stationary axis. Therefore the above can be written as:

F rd

F rq

F rγ

=

cos θrs sin θrs 0

− sin θrs cos θrs 00 0 1

F s

d

F sq

F sγ

(3.60)

and

F s

d

F sq

F sγ

=

cos θrs − sin θrs 0

sin θrs cos θrs 00 0 1

F r

d

F rq

F rγ

(3.61)

These relationships can be written in short form as:

Frdqγ = BFs

dqγ (3.62)

Fsdqγ = BTFr

dqγ (3.63)

3.2 dq Models 59

Figure 3.2: Two phase stationary to two phase rotating frame transformation

The stationary to rotating frame transformation can be combined with thethree phase to stationary two phase transformation to give the transformationfrom a three phase stationary frame to an arbitrary rotating frame. Clearly thetransformations for the mmf are (using (3.24) and (3.25)):

Frdqγ =

23CFabc (3.64)

Fabc = CTFrdqγ (3.65)

where:

C = BS =

cos θrs cos(θrs − 2π

3 ) cos(θrs + 2π3 )

− sin θrs − sin(θrs − 2π3 ) − sin(θrs + 2π

3 )1√2

1√2

1√2

(3.66)

CT = STBT =

cos θrs − sin θrs 1√

2

cos(θrs − 2π3 ) − sin(θrs − 2π

3 ) 1√2

cos(θrs + 2π3 ) − sin(θrs + 2π

3 ) 1√2

(3.67)

It can be shown that the all the transformations from the abc frame tothe dqγr frame have the same form as the stationary frame transformations ofTable 3.1, except that C and CT are substituted for S and ST respectively, andthe superscript on the variables becomes r.

From Faraday’s law it is possible to express the voltages in terms of rateof change of flux linkage. In the case of the rotating transformations, this rateof change can be from two causes; (a) the time rate of change of flux linkage

3.2 dq Models 60

To dqγr To abcFrdqy = 2

3CFabc Fabc = CTFrdqγ

irdqγ = 23Ciabc iabc = CT irdqγ

vrdqγ = 2

3Cvabc vabc = CTvrdqγ

Ψrdqγ = 2

3CΨabc Ψabc = CTΨrdqγ

Lrdqγ = 2

3CLabcCT Labc = 23C

TLrdqγC

Rrdqγ = 2

3CRabcCT Rabc = 23C

TRrdqγC

Zrdqγ = 2

3CZabcCT Zabc = 23C

TZrdqγC

Table 3.2: Summary of Rotating Frame Transformations

caused by the time rate of change of currents, and (b) the rate of change dueto the relative movement of the frames. The general Faraday relationship is:

vabc = pΨabc (3.68)

and Ψabc = CTΨrdqγ (3.69)

therefore vabc = pCTΨrdqγ (3.70)

As can be seen from (3.66), the C matrix is in general a time dependentmatrix, since θrs could be changing with respect to time. Therefore expanding(3.70) using the chain rule one gets:

vabc = {pCT }Ψrdqγ + CT {pΨr

dqγ} (3.71)

If one expands (3.71) by taking the appropriate derivatives, and then rearrangesthe result the following expression can be obtained:

vabc = CT

p

ψr

d

ψrq

ψrγ

+ ωrs

−ψr

q

ψrd

0

= CTvr

dqγ (3.72)

∴ vrdqγ = p

ψr

d

ψrq

ψrγ

+ ωrs

−ψr

q

ψrd

0

(3.73)

As we shall see in the next section, (3.73) is the form of the reluctance machinedq equations.

A summary of the transformations from a stationary frame to a rotatingframe are summarized in Table 3.2.

3.2.3 Example: SYNCREL Linear dq Model

As an example of the use of the above transformations we shall consider thesynchronous reluctance machine (SYNCREL). This machine was chosen partlyout of convenience (I already had the model developed), and also because itis a relatively simple machine that demonstrates saliency. The transformationprocess will be carried out in a two stage process. The reason for this is thatthe nature of the two phase stationary frame machine will be exposed, whereasif the direct transformation to the rotating frame is carried out then this modelis stepped over. The first step in process is to convert the three phase model

3.2 dq Models 61

Figure 3.3: Conceptual diagram of a three phase SYNCREL

of the machine to the two phase model of the machine. The following discus-sion is with reference to Figure 3.3. This diagram shows a three phase, twopole SYNCREL. The stationary dq frame is aligned with the d-axis along thea-phase mmf axis. The rotating d-axis is located along the high permeance axisof the rotor. Because the SYNCREL is a synchronous machine, the rotor hasto be synchronized with the rotating field in steady state to produce any usefultorque. Hence this frame is also synchronized with this field, and is known asa synchronously rotating reference frame. The synchronously rotating referenceframe has some very important properties that make it the frame that is mostuseful for control purposes. It will be seen in this frame that the angle depen-dence of the inductances disappears, and the currents and voltages become DCvalues in steady state.

The most complicated part of the three phase machine to two phase machineconversion is the inductance transformation, so we shall look at this in detail.The inductances for this model are calculated in Appendix A and appear in(A.55) and (A.56). These inductance expressions have to be transformed usingthe transformations in Table 3.1. Applying these transformations the inductancematrix in the stationary frame becomes:

Lsdqγ =

32

2

3Ll + L1 + L2 cos 2θ L2 sin 2θ 0L2 sin 2θ 2

3Ll + L1 − L2 cos 2θ 00 0 2

3Ll

(3.74)

Equation (3.74) can now be converted to the rotating frame by carrying outthe BLs

dqγBT transformation. After considerable manipulation one arrives at

3.2 dq Models 62

the following expression for the dq inductance matrix:

Lrdqγ =

Ll + 3

2 (L1 + L2) 0 00 Ll + 3

2 (L1 − L2) 00 0 Ll

(3.75)

If one assumes that the system has no zero sequence currents flowing (i.e.the machine has its winding Y connected for example) then the last column androw can be deleted from the above matrices. Therefore the relevant matrix forthe dq inductances is:

Lrdq =

[Ll + 3

2 (L1 + L2) 00 Ll + 3

2 (L1 − L2)

](3.76)

Notice in (3.76) that the θ dependent inductance values of the original threephase model have been converted to time invariant and θ independent induc-tances in the dq frame. This results from the fact that the dq reference frame istied to the rotor. If one were measuring the inductance whilst fixed to the rotor,the inductance will not change as the rotor is rotated (assuming a non-salientstator). In addition, the transformed windings that are fixed to this frame donot see any movement of the rotor from the moving d-axis, and therefore themutual inductance term to the orthogonal winding is zero. A consequence ofthis simplification of the inductances is that the dq frame dynamic equationsare much simpler than the three phase equations.

In (3.73) we calculated the generic form of the dq dynamic equations takinginto account only the voltage terms due to the flux linkages. If the three phaseconversion process is carried out for the resistance it can be shown that the dqvalues are identical to the three phase values. Therefore, the generic dq equationcan be rewritten in the following form if we include the resistive drop term anduse the fact that the dq inductances are time invariant:

vrdq =

[R 00 R

]irdq +

[Lrd 00 Lr

q

]pirdq + ωrs

[ −Lrq 0

0 Lrd

]irqd (3.77)

which can be written in scalar form as:

vrd = Rird + Lrddirddt − ωrsLr

qirq

vrq = Rirq + Lrqdirqdt + ωrsLr

dird

}(3.78)

where:

Lrd = Ll +

32(L1 + L2)

Lrq = Ll +

32(L1 − L2)

Equation (3.78) is shown in diagram form in Figure 3.4.It should be noted that the magnitude of the total flux linkage for the SYN-

CREL can be written in terms of the d and q-axis inductances as follows:

ψ =√(Lr

dird)2 + (Lr

qirq)2 (3.79)

3.2 dq Models 63

R

R

pd qr

qrL i

pd dr

drL i

+

+

-

-

Ldr

Lqr

vdr

vqr

d-axis

q-axis

idr

iqr

Figure 3.4: Model for the ideal dq equations

The other relevant part of the machine model is the torque. The torqueexpression (1.104) can be used as the basis for the development of the torquefor the linear reluctance machine. The expression developed in (1.104) was fora system where the rotor had a single excitation winding and the stator a singlewinding. However, the location of the second coil does not have to be on therotor, and it can be the q-axis coil instead. Obviously the self and mutualcoupling terms for the coils will now be the same as those in the stationaryframe dq model derived above.

The expression for the torque is (using (1.104)):

Te =12is

2

d

dLsd

dθrs+

12is

2

q

dLsq

dθrs+ isdi

sq

dLsdq

dθrs(3.80)

where the inductance terms are defined as in (3.74). Taking the derivatives inthis expression, and introducing the 3/2 factor to account for three phases, weget the following expression for the torque in terms of the stationary frame dqcurrents:

T se =

32

[(is

2

q − is2d )L2 sin 2θrs + 2L2isdi

sq cos 2θrs

](3.81)

Using the relationship:

isdqγ = BT irdqγ (3.82)

3.2 dq Models 64

one can substitute for isd and isq in (3.81) in terms of ird and irq, and obtain:

T re =

322L2i

rdi

rq

=32(Lr

d − Lrq)i

rdi

rq (3.83)

All of the analysis thus far has been for a single pole pair machine. A mul-tiple pole machine only requires a slight modification to the torque expression,and the ωrs term is in electrical radians per second in the dynamic equation.Therefore for a pp pole pair machine the torque expression is:

T re =

32pp(Lr

d − Lrq)i

rdi

rq (3.84)

The torque expression could also be found by applying the general torqueexpression derived for the primitive machine – (2.21), repeated here for conve-nience:

Te = iTGi (3.85)

where:

i =[irdirq

]and G =

[ −Lrq 0

0 Lrd

]Substituting for these vectors into the general torque expression:

Te = [ ird irq ][ −Lr

q 00 Lr

d

] [irdirq

]

= [ ird irq ][ −Lr

qird

Lrdi

rq

]= (Lr

d − Lrq)i

rdi

rq (3.86)

as was previously obtained. We can introduce the 32pp factor to give the same

torque as a pp pole pair three phase machine.The only other transformation of immediate interest that has not been ex-

plicitly carried out is the current transformation. It was eluded to in Sec-tion 3.2.1 that one property of the rotating transformations was that the mag-nitude of the current and voltage vectors was equal to that of a single phaseof the three phase machine in steady state. Another property that occurs isthat in steady state is that ird and irq have DC values if the dq-axes are rotatingsynchronously with the rotor. To formally show these properties consider theabc machine is being driven by currents of the form:

iaibic

=

Ipk cos(θrs + γ)Ipk cos(θrs + γ − 2π

3 )Ipk cos(θrs + γ + 2π

3 )

(3.87)

These currents are synchronized to the rotation of the rotor, and consequentlyso is the resultant current vector. Carrying out the transformation from the abcframe to the dq stationary frame we get:

isdqγ =

Ipk cos(θrs + γ)Ipk sin(θrs + γ)

0

(3.88)

3.3 Space Vector Model 65

and the further transformation to the dq rotating frame gives:

irdqγ =

Ipk cos γIpk sin γ

0

(3.89)

Notice that if the phase angle γ is zero then the q-axis current is zero, andall the current lies in the d-axis–i.e. along the high permeance axis of the rotor.If the peak value of the abc currents are constant then we have a constantamplitude D.C. value equal to the abc phase amplitude, in the dq-axes.

3.3 Space Vector Model

An alternative method for modelling machines that has become popular is thespace vector technique. This method of modelling is very similar to the dqmodelling technique, and in fact it is very simple to convert between the twodifferent types of models. The main reason for the popularity of the techniquehas been the growth in vector based control techniques, since this modellingmethod naturally fits this view of the machine. Its main advantage is that asimpler notation can be used for machine equations. For example, the electricaldynamics of an induction machine can represented by two equations (instead offour with a dq model). The form of the equations also evokes a resultant vectorway of thinking about the machine’s operation, as opposed to a componentvector approach with the dq modelling technique. A full discussion of spacevectors applied to the control of machines can be found in [6]. The applicationof space vectors to the reluctance machines has not been as pervasive as it haswith induction machines because the reluctance machine more naturally relatesto a component viewpoint. This is due to the presence two different permeanceaxes in the machine. However, in some situations space vectors are a usefultool for viewing this machine’s operation, and it is therefore justified to havea brief view of the space vector concepts applicable to reluctance machines. Itshould be emphasized that because the reluctance machine does not have anyrotor winding we have no need to develop rotor expressions, as is the case withthe induction machine.

Space vector modelling is based on the concept that the mmf of a threephase machine can be represented by a resultant vector that has a physicallocation in space. This stems from the fact that the individual windings of thephases are sinusoidally distributed, and the vector for each of the windings canbe considered to lie on the axis of the phases. It should be noted that the dqmodelling developed in the previous sections used similar assumptions, but themodelling approach was different.

3.3.1 Current Space Vectors

Stationary Frame Current Vectors

In a manner similar to (3.4) we can write the following expression for the resul-tant mmf in a three phase machine:

F sT = N3φ[ia(t)cosθ + ib(t)cos(θ − 2π

3) + ic(t)cos(θ − 4π

3)] (3.90)


where θ is the angle from the axis of the a-phase winding as previously.The notational simplicity of the space vector formulation is obtained by

introducing complex notation. In the following equations the “ ” is used todenote vectors in the complex form. Equation (3.90) can be written as:

F sT = N3φRe[ia(t)e−jθ + ib(t)ej(2π/3−θ) + ic(t)ej(4π/3−θ)]

=32N3φ

23Re

{[ia(t) + aib(t) + a2ic(t)]e−jθ

}(3.91)

where a = ej2π/3, and is a vector of unit length lying spatially along the axis ofthe b-phase. Similarly a2 = ej4π/3, and lies along the c-phase axis. Notice thatthis complex notation implicitly means that we have a set of pseudo “dq” axes,which now correspond to the real and imaginary axes.

Now consider the central part of the above mmf expression, namely:

ia(t) + aib(t) + a2ic(t) =[ia(t)− 1

2ib(t)− 1

2ic(t)

]+ j

√32

[ib(t)− ic(t)] (3.92)

which is the same expression that is obtained for the three phase to two phasetransformations for the dq model, where the imaginary axis expression corre-sponds to the q axis expression in the dq model.

Therefore let us define the current vector as follows:

is =23[ia(t) + aib(t) + a2ic(t)]

= |is| ejαs (3.93)

Using this definition of the current vector we can write the expression forthe mmf in the machine as:

F sT =

32N3φRe

{ise

−jθ}

(3.94)

Figure 3.5 shows pictorially (3.93) means. The |is| vector is the magnitude ofthe resultant current vector. Notice that the direction of this vector is spatiallythe same direction as the original mmf vector (since the two are related by ascalar). The αs angle is the angle of this vector with respect to the referencea-phase axis. If one were to add together the ia, ib, and ic current vectorsgraphically on this diagram, the resultant current vector would have the angleαs but be 3/2 times the magnitude. The 2/3rd term was introduced into (3.91),and then carried into (3.93), since the resultant current vector has the propertythat the vector can be directly projected back onto the three phase axes. Thisis the same situation as with dq modelling in that the 2/3 factor allows thecurrents in the two phase space vector representation to be directly correlatedwith the phase currents of the three phase machine. It should be noted thatimplicit in this projection is that there are not zero sequence currents flowing(i.e. ia + ib + ic = 0). It can also be shown that under this restriction thatthe space vector to three phase projections can be represented by the followingrelationships:

Re(is) = ia (3.95)

Re(a2is) = ib (3.96)Re(ais) = ic (3.97)


Figure 3.5: Resolving the current space vector onto the abc axes

In the particular case where the currents are of the form (3.5–3.7) then thespace vector can be written as follows:

is =23Ipk[cosωt+ (cos

2π3

+ j sin2π3) cos(ωt− 2π

3)

+ (cos4π3

+ j sin4π3) cos(ωt+

2π3)]

=23Ipk[cosωt− 1

2(−1

2cosωt+

√32

sinωt)

+ j√32

(−12cosωt+

√32

sinωt)− 12(−1

2cosωt−

√32

sinωt)

− j√32

(−12cosωt−

√32

sinωt)

= Ipk(cosωt+ j sinωt)

∴ is = Ipkejωt (3.98)

Therefore the resultant current vector has a constant magnitude and the angleαs is changing at the constant rate of ω, i.e. the vector is rotating around themachine at a constant angular frequency.

The space vector representation can be simply related back to the dq repre-


sentation. From (3.29) is can be seen that:

isd =23

[ia − 1

2ib − 1

2ic

](3.99)

isq =1√3[ib − ic] (3.100)

If one takes the real and imaginary components of (3.93) then one can write thefollowing:

Re(is) = Re[23(ia + aib + a2ic)

]

= Re[23(ia + (cos

2π3

+ j sin2π3)ib + (cos

4π3

+ j sin4π3)ic)

]

=23

[ia − 1

2ib − 1

2ic

]= isd (3.101)

Similarly:

Im(is) =1√3[ib − ic] = isq (3.102)

These projections can be seen in Figure 3.6. Note that the dq projections arenot as restrictive as the projections onto the abc axes. For example, if thereare zero sequence currents then (3.101) does not equal (3.95). Zero sequencecurrents require the presence of an additional space vector equation, as wasthe case with the dq equations. However, the discussion in this book shall befocussed on balanced (and usually Y connected) machines that do not have zerosequence current components.

Rotating Frame Current Vectors

Similar space vector expressions can be derived for frames that are not stationaryto the rotor. Consider the situation shown in Figure 3.7. Here we have theoriginal is vector as in Figure 3.6, as well as the same vector projected ontoanother frame which is possibly rotating.

The current vector can be written with reference to the rotating frame as:

ir = |ir| ejαr= |ir| ej(αs−θrs)

Now is = |is| ejαs = |ir| ej(θrs+αr) (as |is| = |ir| )= |ir| ejαrejθrs

∴ is = irejθrs = ire

−jθsr (3.103)

The sign of the angle in (3.103) is dependent on the reference axis for the an-gle difference between the two reference frames. The normal convention adoptedis that the old frame is taken as the reference, therefore the sign convention is:

xnew = xolde−jθnew−old (3.104)


Figure 3.6: Relationship between the dq axes and the current space vector.

Figure 3.7: Space vector rotating frame transformations


where:

θnew−old � the angle between the new and old axeswith reference to the old axis.

This relationship is general and can be applied to all space vector axis trans-formations.

In the case of the transformation in (3.103) we are transforming from therotating axis to the stationary axis. Applying the rule above, the angle fromthe old axis to the new axis is −θrs (or θsr). Therefore applying (3.103):

is = ire−jθsr (3.105)

3.3.2 Flux Linkage Space Vector

The total flux linking the phases in a three phase machine are:

ψa = Laaia + Labib + Lacic (3.106)ψb = Lbbib + Lbaia + Lbcic (3.107)ψc = Lccic + Lcaia + Lcbib (3.108)

Define the flux linkage space vector as follows:

ψs=

23(ψa + aψb + a2ψc) (3.109)

The justification for the definition of the space flux vector is that the funda-mental of the flux linkage to a single phase varies as a sinusoidal function of thecurrent angle to the axis of any particular phase. This was shown in Chapter 1.Therefore the flux linkage has similar sinusoidal properties to the mmf of themachine, and the same techniques can therefore be applied.

Substituting (3.106–3.108) into (3.109)and assuming that:

Lab = Lba

Lac = Lca

Lbc = Lcb

after a small amount of manipulation gives:

ψs=

23

[(ψa − 1

2ψb − 1

2ψc

)+ j

√32

(ψb − ψc)

](3.110)

which after further manipulation gives:

ψs=

23

[ 12

[(2Laa − Lab − Lac) ia + (2Lab − Lbb − Lbc) ib · · ·

+ (2Lac − Lcc − Lbc) ic]· · ·

+ j√32

[(Lbb − Lbc)ib + (Lab − Lac)ia + (Lbc − Lcc)ic

]](3.111)


This expression can be simplified greatly for a cylindrical rotor machine (i.e.the self inductances are equal, and the mutual inductances are equal) . Defining:

Laa = Lbb = Lcc = Ls

and

Lab = Lba = Lac = Lca = Lbc = Lcb =Mss

where Mss is the mutual inductance between 120◦ separated phases, then wecan write (3.111) as:

ψs= (Ls −Mss)

{23

[(ia − 1

2ib − 1

2ic) + j

√32

(ib − ic)]}

= Lsis (3.112)

where Ls = Ls − Mss. In the case of a normal three phase machine Mss =Mm cos 120◦ = − 1

2Mm. Note that ignoring leakage Mm = Ls. Therefore Ls =32Ls. This is the three phase equivalent inductance. We shall look at the

induction machinein detail in a fol-lowing chapter

Remark 14 Note that even though the inductance term is 32Ls in (3.112), the

flux linkage ψsis still 2/3rds that produced by the three phase machine as the

current vector is still 2/3rds the magnitude of the three phase current vector.

For the reluctance machine the expression is much more complicated. Sub-stituting the inductance expressions (A.55) and (A.56) into (3.111) letting θrs =θd, and after considerable manipulation one obtains the following expression forthe flux space vector in a stationary reference frame:

ψs= (L1 + L2 cos 2θrs)ia + (−L1

2+ L2 cos 2(θrs − π

3))ib · · ·

+ (−L1

2+ L2 cos 2(θrs +

π

3))ic · · ·

+ j1√3

[(√3L2 sin 2θrs)ia

+ (32L1 − L2(

32cos 2θrs +

√32

sin 2θrs))ib · · · (3.113)

+ (−32L1 + L2(

32cos 2θrs −

√32

sin 2θrs))ic]

(3.114)

The validity of this expression can be checked as follows. If (3.114) is calculatedfor θrs = 0 and ia = Ipk, ib = −Ipk/2, ic = −Ipk/2, (i.e. the mmf vector liescoincident with the a-phase) then the real part of the inductance is 3/2(L1+L2)as expected from the dq analysis. A similar result can be found for the imaginarycomponent for θrs = 90◦, ia = 0, ib =

√3/2Ipk, ic = −√

3/2Ipk (in this case themmf is at 90◦ and the rotor d-axis is also in this position).

If the currents are in the form of (3.87) then (3.114) can be simplified to thefollowing expression:

ψs

∣∣∣3φ currents

=32Ipk

[(L1 cos(θrs + γ) + L2 cos (θrs − γ)) · · ·

+ j (L1 sin(θrs + γ) + L2 sin(θrs − γ))]

=32Ipk

[L1e

j(θrs+γ) + L2ej(θrs−γ)

](3.115)


This special case for the currents has been chosen because it is the form of thecurrents that are applied to the machine when it is being vector controlled. Itcan be seen from that the resultant current is synchronized to the rotor positionsuch that the resultant current vector has an angle of γ radians with the rotorhigh permeance axis. Note that similar to the cylindrical rotor case the currentamplitude only appears as Ipk in (3.115), the 3/2 term being from the inductancepart of the expression. Therefore the magnitude of the flux linkage is 2/3rds ofthe flux linkage for the three phase machine, as was indicated from the definitionof the flux linkage expression (3.109).

3.3.3 Voltage Space Vector

In a manner analogous to the vector definitions of the current vector and fluxvector one can define the voltage vector:

vs =23(va + avb + a2vc) (3.116)

where va,vb,vc are the individual phase voltages. As with the current vectors,if there are no zero sequence voltages then the individual phase voltages canbe simply obtained from the voltage space vector using a similar expression to(3.95), (3.96) and (3.97):

Re(v) = va (3.117)

Re(a2v) = vb (3.118)Re(av) = vc (3.119)

The concept of the voltage space vector is quite abstract. However, itsexistence can be justified from the vectors already defined. The voltage ina machine is made up of two components; the resistive drop, and the inducedvoltage from changing flux linkages. We have already defined the current vector,and the resistive drop is simply this vector multiplied by the resistance (which isa scalar). The flux linkages have also been defined as a vector, and taking theirderivative in a vector sense also results in a vector. Therefore both componentsof the voltage are vectors, and consequently the voltage can be considered to be avector. It is easy to demonstrate that if one takes components of a voltage vectorfor a set of three phase windings one does get the individual abc componentsof the voltages. Note that this process requires that there are no zero sequencevoltages present.

Voltage in across a coil is usually related to two main things in an electricalcircuit – current through the coil resistance, or a changing linkage through thecoil (due either to time or spatial variations). For the concept of a voltage vectorto be useful it should give the correct results under these two conditions. Nowconsider the situation where we have the three coils of a three phase circuitcarrying DC currents. Therefore the individual currents in the windings areia, ib, and ic. Now if the voltage vector concept is to useful then:

v = iR (3.120)

Now writing down the voltage vector using its definition and the individual


phase voltages under this condition:

v =23(iaR+ aibR+ a2icR)

= R23(ia + aib + a2ic)

= iR (3.121)

Note that under this excitation the voltage vector and the current vector are inthe same spatial direction.

The other case to consider is when the coil is excited by a change in fluxlinkage. As mentioned previously there are two situations that occur – the coilis subjected to a time varying flux linkage which is stationary in space, andthe coil is subjected to a moving sinusoidal flux density waveform. We shallconsider each of these. Firstly the time varying flux linkage. Now, realisingthat θ is constant we can write:

dψs

dt=d

dt

[∣∣∣ψs

∣∣∣ ejθ]

=d∣∣∣ψ

s

∣∣∣dt

ejθ

= |vs| ejθ= vs (3.122)

Therefore the voltage vector is in the same direction as the flux vector underthis condition.

The other situation to consider is when we have a spatial sinusoidally dis-tributed flux distribution moving with respect to the coils. The best way toapproach this case is to again use the complex form of the flux linkage expres-sion:

ψs=

∣∣∣ψs

∣∣∣ ejθ(t) where θ(t) is the time varying vector angle

=∣∣∣ψ

s

∣∣∣ ejωt (3.123)

Therefore we can write:

vs =dψ

s

dt=

∣∣∣ψs

∣∣∣ ddtejωt

= jω∣∣∣ψ

s

∣∣∣ ejωt= jωψ

s(3.124)

Therefore the voltage vector is 90◦ out of phase with the flux linkage vector.As can be seen from the above expression the voltage vector “leads” the fluxlinkage vector by 90◦, which is the same result that is obtained in the temporaldomain for sinusoidal variation of flux linkages in an inductor.

3.3.4 Example: SYNCREL Space Vector Model

We have assembled enough of the space vector model machinery to constructthe space vector electrical model for the SYNCREL. This model is very simple,


in fact about as simple a model as one can get for a machine. The simplicityresults from the space vector notation. A SYNCREL has only one set of threephase windings on the stator, therefore the expression for the stator voltage inspace vector notation using stationary frame variables is:

vs = Ris +dψ

s

dt(3.125)

It is a straight forward process to verify this expression from the definitionsalready presented for the various space vectors.

Evaluation of the voltage from (3.125) is complex due to the nature of theflux linkage term in a stationary reference frame. A great simplification canbe achieved by converting this expression into a rotating frame synchronizedwith the rotor (as was done with the dq equations). Applying (3.104) to thevoltage, current and flux linkage vectors, one can write the following relationshipbetween the stationary and rotating reference frame vectors:

vs = vrejθrs

is = irejθrsψs= ψ

rejθrs

(3.126)

Substituting (3.126) into (3.125) gives:

vrejθrs = Rire

jθrs +d

dt

(ψrejθrs

)= Rire

jθrs +dψ

r

dtejθrs + ψ

rejθrsj

dθrsdt

∴ vr = Rir +dψ

r

dt+ jωrsψr

(3.127)

where ωrs =dθrsdt

θrs � angle of rotating frame wrt stationary frame (3.128)

Assuming that the currents being applied to the machine are of the form(3.87) then it is not difficult to show that:

is = Ipkej(θrs+γ) (3.129)

Therefore the rotating frame current space vector is, using (3.103):

ir = Ipkejγ (3.130)

Applying a similar transformation to (3.115), one obtains:

ψr=

32Ipk[(L1 + L2) cos γ + j(L1 − L2) sin γ]

=32Ipk

[L1e

jγ + L2e−jγ

](3.131)

3.3.5 Space Vector Power Expression

Now that we have the space vector representations for the dynamic equations weare in a position to calculate the input stator power for the machine in terms


of space vectors. Assuming that there are no zero sequence components thefollowing expression can be written for the three phase power of the machine:

P3φ = vaia + vbib + vcic= Re(vs)Re(is) + Re(a2vs)Re(a

2is) + Re(avs)Re(ais) (3.132)

This equation can also be written in a more compact form:

P3φ =32Re (vsi

∗s) (3.133)

where the “∗” means complex conjugate (ie. i∗s = 23 (ia+a

∗ib+a2∗ic) ). Because

the space vectors are closely related to the time domain phasors in steady statethe similarity of this expression with the time domain complex power expressionshould not be surprising. This expression can be confirmed by the followingexpansion:

Re (vsi∗s) = Re

((23

)2 (va + avb + a2vc

) (ia + a∗ib + a2∗

ic

))(3.134)

= Re((

23

)2 ( 32vaia +

32vbib +

32vcic · · ·

− j√32

(vaia + vbib + vcic)))

=23(vaia + vbib + vcic) (3.135)

As can be seen from (3.135) the space vector representation of the machine isabsorbing 2/3rds the power of the three phase machine. Hence the space vectortransformations we have developed are power variant transformations, as wasthe case for the dq transformations.

It should also be noted that power expressions are reference frame indepen-dent (as one would naturally expect if the reference frame concept was to beuseful). This can be shown as follows. Let use convert the voltage and currentvectors in the previous power expression to an arbitrary reference frame at someangle θ, with respect to the ie.

v′s = vse−jθ

i′∗s = i∗sejθ

Substituting these expressions into (3.133) we get:

Re(v′si′∗s ) = Re

[(23

)2 (va + avb + a2vc

)e−jθ

(ia + a∗ib + a2∗

ic

)ejθ

]

= Re

[(23

)2 (va + avb + a2vc

) (ia + a∗ib + a2∗

ic

)]

which is the same as (3.134). Therefore the power is invariant in differentreference frames.


3.3.6 Example: Space Vector Expression for SYNCRELTorque

The power expression developed above can be used as a means to calculate thetorque produced by the machine. A general expression for torque is:

T = Pω (3.136)

This expression can be used to develop the electro-magnetic torque for the spacevector model of the machine by utilizing the energy balance expressed in (1.44).If one can identify the loss and field storage terms then they can be subtractedfrom the total input energy to give the mechanical output energy. This can thenbe substituted into (3.136) to give the electromagnetic torque.

Consider the expression (3.125). The power expression for the machine canbe written using the relationship (3.133) as follows:

P3φ =32Re (vsi

∗s) =

32Re

[(Risi

∗s +

dψs

dti∗s

)](3.137)

Clearly the Risi∗s term is related to the power losses in the machine, therefore

thedψ

s

dt i∗s term must be related to stored field energy and mechanical output

power. Considering the last term for the special case of currents in the form(3.87), with Ipk constant with respect to time, and using (3.115) we can write:

dψs

dt=d

dt

(32Ipk

[L1e


])

= jωrs

(32Ipk

[L1e


])= jωrsψs

(3.138)

where ωrs =dθrsdt

Therefore the power expression under this steady state condition becomes:

P3φ =32Re

[(Risi

∗s + jωrsψs

i∗s)]

(3.139)

Clearly there is only one term related to the rotational power and that isjωrsψs

i∗s. Expanding this using i∗s = Ipke−j(θrs+γ) and (3.115) one gets:

P3φ =32Re

[32jωrsI

2pk

(L1 + L2e

−j2γ)]

=94ωrsI

2pkL2 sin 2γ (3.140)

∴ Te =94I2pkL2 sin 2γ (3.141)

If we remove the restriction that Ipk has to be constant, then we would endup with Ldi/dt type terms in (3.138). These terms are not related to ω in anyway, and result in change of stored field energy terms in (3.137). Therefore(3.140) is valid for transient conditions as well as for steady state.


The same expression can be obtained if the torque is calculated using therotating reference frame expression of (3.127). In this case the rotational powerterm is even more easily identified. Consider the power expression in this frame:

P3φ =32Re

(jωrsψr

i∗r)

=32Re

(jωrs

32Ipk

[L1e

jγ + L2e−jγ

]Ipke

−jγ

)=

94I2pk Re

(jωrsL1 + jωrsL2e

−j2γ)

=94ωrsI

2pkL2 sin 2γ (3.142)

which is the same as the power in the stationary frame case. Therefore the powerand torque produced is reference frame independent (as one should expect).

It is possible to develop a more general form of the mechanical power/torqueexpression. In the two examples above the key expression for the rotationalpower has the form:

Prot =32Re

(jωψi∗

)(3.143)

Regardless of whether the variables are in a rotating frame or a stationary framethe current and flux have the following form:

ψ =∣∣ψ∣∣ ejα and i = |i| ejβ

therefore:

Prot =32Re

(jω

∣∣ψ∣∣ |i| ej(α−β))

=32Re

(jω

∣∣ψ∣∣ |i| (cos(α− β) + j sin(α− β)))=

32Re(jω

∣∣ψ∣∣ |i| cos(α− β) − ω ∣∣ψ∣∣ |i| sin(α− β))

= −32ω∣∣ψ∣∣ |i| sin(α− β)

∴ Prot = −32ωψ × i (3.144)

and Trot = −32ψ × i (3.145)

Remark 15 The above general expressions for power and torque are the sameexpressions as rotational power and torque for the DC machine. In this machinethe torque produced is of the form:

Trot = Kψf ia (3.146)

where ia is the armature current and ψf is the flux linkage due to the field. Ina DC machine the physical arrangement with the commutator is such that thespatial angle between these two values is 90◦. Therefore the DC machine torqueexpression is the same at that for an AC machine with a right angle betweenthe current and flux vectors. As we shall see in the next chapter this torqueexpression is general for all AC machines satisfying sinusoidal assumptions.


Remark 16 In the power/torque expressions above thedψ

s

dt has a contributionto the rotational power because of the saliency of the rotor. The rotational termsresult because ψ

shas a spatial component due to the inductance variation with

rotor position. This situation does not occur with round rotor machines such asthe induction machine.

3.3.7 Relationship Between Space Vectors and dq Models

Clearly the space vector model and the dq model of a machine are very closelyrelated. The Real and Imaginary axes of the space vector model can be con-sidered to be the same as the dq axes. Therefore, by taking the componentsof the space vectors (i.e. taking the Re and Im parts) onto these axes one canobtain the dq representation of the variable or equation. For example, considerthe (3.115) representation for the flux linkage. Taking the Re and Im parts weobtain:

Re(ψs


)=

32Ipk[L1 cos(θ + γ) + L2 cos(θ − γ)] (3.147)

Im(ψs


)=

32Ipk[L1 sin(θ + γ) + L2 sin(θ − γ)] (3.148)

Calculating the flux linkage using (3.74) and (3.88) one gets the following:[ψsd

ψsq

]=

32Ipk

[L1 cos(θ + γ) + L2 (cos 2θ cos(θ + γ) + sin 2θ sin(θ + γ))L1 sin(θ + γ) + L2 (sin 2θ cos(θ + γ)− cos 2θ sin(θ + γ))

](3.149)

Since:

cos 2θ cos(θ + γ) + sin 2θ sin(θ + γ) = cos(θ − γ)sin 2θ cos(θ + γ)− cos 2θ sin(θ + γ) = sin(θ − γ)

then the Space Vector and dq expressions are equivalent. This equivalence ismore easily verified if the rotating versions of the two models are compared.Consider (3.131). If Re and Im parts are taken we have:

Re(ψr


)=

32Ipk [(L1 + L2) cos γ] = Ldid (3.150)

Im(ψr


)=

32Ipk [(L1 − L2) sinγ] = Lqiq (3.151)

since we know that Ld = 3/2(L1 + L2) and Lq = 3/2(L1 − L2), and id =Ipk cos γ, iq = Ipk sinγ from the dq model theory.

Finally it can be shown that the Space Vector and dq model theory give thesame torque and power expressions. Consider the following relationships:

I2pk sin 2γ = 2(Ipk cos γ)(Ipk sin γ)

= 2idiq (3.152)

2L2 =23

(32(L1 + L2)− 3

2(L1 − L2)

)=

23(Ld − Lq) (3.153)

Substituting these into (3.141) gives the normal dq torque expression (3.83).

3.4 Steady State Model 79

3.4 Steady State Model

Thus far this Chapter has been mainly considering the dynamic models. Thesteady state model of machines can be derived from the dq models, and asan example this section will develop the steady state voltages and currents forthe SYNCREL using the dynamic model as a starting point. The approach issimilar to that in [2].

Consider the ideal dq equation in a rotating reference frame as shown in(3.78). If the machine is in steady state then the derivative terms in this ex-pression will be zero. Hence the steady state form of (3.78) is:

vrd = Rird − ωrsLrqi

rq

vrq = Rird + ωrsLrdi

rd

}(3.154)

Using the standard transformation shown in Table 3.2 one can write:

vabc = CTvrdqγ (3.155)

Considering only phase a in this transformation it can be seen that (3.155) leadsto the following expression for the a phase voltage:

va = [

vrd︷︸︸︷(Rird − ωrsLr

qirq) cos θrs −

vrq︷︸︸︷(Rirq + ωrsL

rdi

rd) sin θrs] (3.156)

This equation can be written in the form:

va = Re{(vrd + jv

rq)e

jωrst}

(3.157)

We wish to express the voltage in terms of the phase voltage. Consider thefollowing form for the time domain expression for the voltage on phase a:

va = Vm cos(ωrst+ α) (3.158)

which may be written in complex notation as:

va = Re{√

2Vm√2ejαejωrst

}(3.159)

The phasor voltage for va may be written by inspection of (3.159) as:

Va =Vm√2ejα (3.160)

therefore (3.159) may be written as:

va = Re{√

2Vaejωrst}

(3.161)

Comparing (3.161) with (3.157) one can write the following:

Va =1√2

(vrd + jv

rq

)= V r

d + jV rq (3.162)

3.4 Steady State Model 80

Figure 3.8: Phasor diagram for a steady state SYNCREL

Similarly the phase current can be written in terms of the steady state dq cur-rents:

Ia =1√2

(ird + i

rq

)= Ird + jIrq (3.163)

Using the expressions for Va and Ia above one can write (3.162) as:

Va = RrdI

rd − ωrsLr

qIrq + j(Rr

qIrq + ωrsLr

dIrd) (3.164)

which allows the steady state phasor diagram of Figure 3.8 to be drawn. Noticefrom this Figure that the power factor for this machine can never be leading. Ifthe (3.164) is written as:

Va = RrdI

rd − ωrsL

rd

ξIrq + j(Rr

qIrq + ωrsLr

dIrd) (3.165)

where ξ = Ld/Lq then it can be seen that as ξ → ∞ and the d-axis flux in themachine stays constant, then the voltage and current vectors in Figure 3.8 willapproach each other, and the power factor will be unity.

Chapter 4

Vector Control of InductionMachines

4.1 Introduction

In the previous chapter we assembled much of the basic theory to tackle themodelling of the induction machine, and the derivation of vector control of theinduction machine. In this chapter we will extend the basic theory where neces-sary for the induction machine situation, and then apply this to the developmentof a number of different vector control strategies for this machine.

4.2 Vector Models for Induction Machines

The purpose of this section is to extend the theory developed for the singlewinding AC machine in the previous chapter to that of the induction machine.This involves considering modelling for a machine where one has windings onboth the rotor and the stator. Although the general principles are the same asfor the stator winding case the detailed expressions are different.

Figure 4.1 shows a conceptual diagram of a machine with three phase wind-ings on the rotor. Note that it this machine is an induction machine then therotor windings are usually short circuited. However for the moment the equa-tions will be developed as thought the windings are not shorted.

Similar expressions to those developed in the previous chapter for the statorwinding can be developed for the rotor winding. These will not be repeated here.These expressions are developed assuming that the rotor winding is stationary– in other words we are developing the expressions in a reference frame alignedwith the rotor.

4.2.1 Flux Linkage Expression

Some of the expressions developed previously have to be altered slightly inorder to account for the presence of the rotor. One of these is the flux linkageexpression as expressed in (3.111). With the rotor present the various total flux

4.2 Vector Models for Induction Machines 82

� rs

as

bs

cs

ar

br

cr

�rs

Figure 4.1: Conceptual diagram of an induction machine.

linkage expressions for the stator phases become:

ψas = Lsias +Mssibs +Mssics +Msriar cos θrs +

Msribr cos(θrs +2π3) +Msricr cos(θrs +

4π3) (4.1)

ψbs = Lsibs +Mssias +Mssics +Msriar cos(θrs +4π3) +

Msribr cos θrs +Msricr cos(θrs +2π3) (4.2)

ψcs = Lsics +Mssias +Mssibs +Msriar cos(θrs +2π3) +

Msribr cos(θrs +4π3) +Msricr cos θrs (4.3)

where:

Ls � the self inductance of the phase

Mss � the mutual inductance between stator phases

Msr � the maximum mutual inductance between a stator and rotor phase

Note the self inductance of a single stator winding can be written in termsof a leakage inductance and a magnetising inductance:

Ls = Lsl + Lsm (4.4)

where Lsl is a stator winding leakage inductance, and Lsm is a stator windingmagnetising inductance. If the stator has an effective number of turns equal


to Nse and the rotor has an effective number of turns Nre then the statormagnetising inductance is related to the mutual inductance between the rotorand the stator as follows: This is a standard

result from trans-former theoryLsm =

Nse

NreMsr (4.5)

If these expressions are substituted into the definition of the flux linkagevector (3.109), after considerable manipulation one can obtain the followingexpression:

ψs= Lsis + Lmire

jθrs

= Lsis + Lmi′r (4.6)

where the “′” means that the rotor current vector has been referenced to thestator reference frame. Notice that this expression has a marked resemblanceto the previous expression (3.112) – there is an additional term related to therotor. The various inductances in this expression are:

Ls = Ls − Lss = Lsl + Lsm −Mss and Mss = −12Mssm = −1

2Lsm

∴ Ls = Lsl +32Lsm (three phase stator inductance)

Lm =32Msr

Note that this expression can be further manipulated as follows:

ψs=

(Lsl +

32Lsm

)is +

32Nre

NseLsmi

′r (4.7)

Note that NreNsei′r is the rotor current referred to the stator. Therefore the ex-

pression can be written in terms of the stator magnetising current (which is thenormal form for a transformer) as follows: Note that referring

currents to the sta-tor using turns ra-tios is different toreferencing currentvectors to differentreference frames.

ψs= Lsis + Lmi

′rs (4.8)

where:

Lm =32Lsm

i′rs � the rotor current referenced and referred to the stator.

Remark 17 Note that the above expression is exactly analogous to an expres-sion for a transformer where the secondary current has been referred to thestator.

Similar analysis can be applied to the rotor giving the expression:

ψr= Lrir + Lmi

′s (4.9)


where:

Lr = Lr −Mrr = Lrl + Lrm and Mrr = −12Mrrm = −1

2Lrm

∴ Lr = Lrl +32Lrm (three phase rotor inductance)

Lm =32Mrs =

32Msr

Lrm � the rotor magnetising inductance =Nre

NseMrs

Remark 18 Note that if the effective rotor and stator turns are the same (i.e.Nse = Nre) then Lrm = Lsm = Msr. Therefore Lm = 3

2Lsm = 32Lrm and

both (4.6) and (4.9) are in the form of the conventional transformer equation.Referring currents to the stator or the rotor has no effect because the turns ratiois unity.

4.2.2 Magnetising Current

In a machine such as the induction machine there are two mmfs contributingto the total mmf in the machine – the stator mmf and the rotor mmf. Thetotal mmf in the machine can be calculated by adding together these two mmfs.Referring to (3.94) we can write:

FT = F sT + F r

T

=32Nse Re

{ise

−jθ}+

32NreRe

{i′re

−jθ}

=32Nse

[Re

{ise

−jθ}+Nre

NseRe

{i′re

−jθ}]

=32Nse Re

[(is +

Nre

Nsei′r)e

−jθ

](4.10)

Notice that the NreNsei′r term is the rotor current in a referenced to a stationary

frame referred to the stator (in the same way the secondary currents can bereferred to the primary in a transformer). The current expression in (4.10) isthe magnetising current as this is the current that produces the magnetisingcurrent in the machine:

im = is +Nre

Nsei′r = is + i

′rs (4.11)

4.2.3 Power and Torque Expressions

Other expressions that differ when we have a rotor with windings are those forpower and torque. Realising that the torque and power expressions are closelyrelated we will concentrate on the torque expression in this discussion. Themain difference between the evaluation of the torque expression here and thatcarried out in the previous chapter is that we must account for the power inthe rotor, since the equations for the induction machine allow for the doublefed machine case. Furthermore, if we don’t account for the rotor power thenwe would not have a torque expression at all, since it is the power in the rotor


that is contributing to the torque. As noted in the evaluation of the torque forthe SYNCREL we only needed to consider the stator in that case because theeffects of the rotor were reflected into the stator via the spatial variation of thestator inductances (which was caused by movement of the rotor). The rotorhad no electrical circuit, therefore was electrically passive.

Similarly to the expression (3.133) we can write the following expression forthe total power for the induction machine:

P3φ =32Re(usi

∗s + u

′ri

′∗r ) (4.12)

1We also know from conservation of energy arguments that equation (1.44)holds. Power is related to energy via the relation:

P3φ =dEe

dt⇒ dEe = P3φdt (4.13)

Therefore the power expression (4.12) can be broken into various componentsbased on (1.44). The losses section of this expression is normally broken invarious losses such as friction, iron losses etc., but we for simplicity shall onlyconsider the resistive loss component. Let us write the expression for the statorresistance loss:

PRsle =

32Re(uRsle i

∗s) (4.14)

=32Re(isRsi

∗s)

=32Re(|is|2Rs) (4.15)

where uRsle is the voltage across the stator resistance Rs.Similarly the losses in the rotor can be written as:

PRrle =

32Re(|i′r|2Rr) (4.16)

Using these two expressions we can write the increment energy loss due toresistance losses as:

dEle = PRle dt =

32Re

(|is|2Rs + |i′r|2Rr

)dt (4.17)

Now let us consider the energy that is being put into the field. The energybeing put into the field is the voltage across the inductive elements × the currentthrough the inductive elements. The voltage across the inductive elements isthe rate of change of flux linkage. The general expression for the field power is:

Pfe =32Re

(dψ

s

dti∗s +

dψr

dti∗r

)=

32Re

(dψ

s

dti∗s +

dψ′r

dti′∗r

)(4.18)

Note that this expression can also be written with the rotor variables referredto the stator reference frame and vice-versa since power expressions are invariant

1Note that the rotor variables are expressed in the stator reference frame.


under reference frame transformations. If (4.18) is integrate with respect to timethen the following expected expression for the field energy results:

Efe =32Re(ψ

si∗s + ψr

i∗r) =32Re(ψ

si∗s + ψ

′ri′∗r ) (4.19)

Remark 19 Note that it is implicit in our derivation of the field energy that therotor of the machine is stationary. If the rotor is not stationary then the dψ

r/dt

term will contain a rotationally related term. Such a term is not connected withfield energy but is connected with mechanical output power.

Rearranging (1.44) so that the mechanical energy term is the subject andwriting in terms of differentials gives:

dEme = dEe − dEle − dEfe (4.20)

Substituting for the terms in this expression, assuming that there is movementof the rotor, gives:

dEme =32Re(usi

∗s + u

′ri

′∗r )dt−

32Re

(|is|2Rs + |i′r|2Rr

)dt

− 32Re

dψ

s

dti∗s + i

′∗r

(dψ′

r

dt

∣∣∣∣∣θrs const

+ jωrsψ′r

)︸︷︷︸

Thedψ′r

dt term

dt (4.21)

Terms in this expression can be collected together so that it consists of compo-nents due to the stator and components due to the rotor:

dEme =32Re

[usi

∗s − |is|2Rs −

dψs

dti∗s

]dt︸︷︷︸

Stator mechanical energy components

+32Re

[u′ri

′∗r − |i′r|2Rr − i′∗r

(dψ′

r

dt


+ jωrsψ′r

)]︸︷︷︸

Rotor mechanical energy components

(4.22)

The stator section of this equation cannot contribute to mechanical outputpower, therefore all the terms in this section of the equation must add to bezero. Let us consider the rotor section of the equation – the terms:

u′ri′∗r − |i′r|2Rr − i′∗r

dψ′r

dt


= 0

for the same reason that the stator section terms equal zero – these terms do notinvolve anything related to mechanical motion and therefore cannot contributeto the mechanical output power. The last term on the other hand involvesωrs and therefore must have something to do with mechanical output power.


Therefore the mechanical output power is:

Prot = Pme =dEme

dt=

32Re(−jωrsψ′

ri′∗r ) (4.23)

∴ Trot = Tme =Pme

ωrs=

32Re(−jψ′

ri′∗r ) (4.24)

Note the similarity of these expressions with (3.143). Therefore in a mannersimilar to (3.144) and (3.145) we can write that the power and torque for theinduction machine is:

Prot = −32ωrsψ

′r× i′r (4.25)

Trot = −32ψ′r× i′r (4.26)

Remark 20 As can be seen from the torque and power expressions equations(3.144) and (3.145) are very general.

It is possible to derive a number of alternative expressions for the torqueof an induction machine using the flux expressions (4.6) and (4.9). Converting(4.9) to be relative to the stator reference frame and defining the magnetisingflux vector: Note that we have

assumed that thestator to rotoreffective turns areequal to get theexpression for ψ

m.

ψ′r= Lri

′r + Lmis (4.27)

ψ′m

= Lmi′r + Lmis = Lmim (4.28)

where ψ′m

refers to the magnetising flux referred to the stator reference frame(from its natural magnetising flux reference frame). Rearranging (4.28):

Lmis = ψ′m− Lmi

′r

and substituting into (4.27) gives:

ψ′r= Lri

′r + ψ

′m− Lmi

′r

Substituting this into (4.26) gives: Note that thetorque is relatedto the interactionof the magnetisingflux and the rotorcurrent. The rotorleakage flux doesnot contribute totorque production.

Trot = −32(Lri

′r + ψ

′m− Lmi

′r)× i′r

= −32ψ′m× i′r (4.29)

This expression can be further manipulated by using (4.28):

Trot = −32(Lm(i′r + is)× i′r (4.30)

= −32Lmis × i′r (4.31)

i.e. the torque is a function of the cross product of the stator and rotor current Assuming Nse =Nrespace vectors.


A different expression can be obtained from (4.26) as follows:

Trot = −32ψ′r× i′r

= −32(Lri

′r + Lmis)× i′r

= −32Lmis × i′r (as above)

Now realising that i′r × i′r = 0 allows one to write:

Trot = −32Lm

Ls(Lsis + Lmi

′r)× i′r

= −32Lm

Lsψs× i′r (4.32)

The torque can also be expressed totally in terms of stator quantities. Con-sider (4.31), by reversing the order of the cross product we can write:

Trot =32Lmi

′r × is (4.33)

Again using the fact that a vector crossed with itself is equal to zero we canwrite this expression as:

Trot =32(Lsis + Lmi

′r)× is

=32ψs× is (4.34)

This expression can be further developed. Assuming that the stator androtor effective turns are equal we can write:

ψs= ψ

sl+ ψ′

m(4.35)

where:

ψsl= Lslis

ψ′m

= Lmim = Lm(is + i′r)

Substituting for ψsin (4.35) we can write:

Trot =32(Lslis + Lmim)× is

=32Lmim × is

=32ψ′m× is (4.36)

Note that the sta-tor leakage flux doesnot contribute tothe torque.


xs

ys

xg

yg

xr

yr

�rs

�gs

� �gs rs�

� rs

� gs

Figure 4.2: Relationship between stationary and rotating frames and the generalreference frame for the induction machine

4.2.4 The Space Vector Model of the Induction Machine

In this section we develop the space vector model of the induction machine ina general reference frame. From this general reference frame it is then easyto generate models of the machine in the natural reference frames such as thestationary reference frame or the rotor reference frame.

Let us begin with the basic equations for the induction machine. These aresimple to deduce from the work we have already carried out on space vectormodelling. We simply write expressions of the form of (3.125) for both thestator and the rotor. Note that the equation for each of these has been writtenin its nature reference frame (i.e. stationary for the stator and rotating withthe rotor for the rotor):

vs = Rsis +dψ

s

dt(4.37)

vr = Rrir +dψ

r

dt(4.38)

We want to express these equations in a general reference frame rotating at anangular velocity of ωg rad/sec. Figure 4.2 shows the relationship between thestationary frame, the rotating frame and the general reference frame.

We can use the frame conversion factors to reference the stator expressionand the rotor expression to the general reference frame. Using (3.104) we canwrite for a generic vector in each of the natural reference frames:

xsg = xse−jθgs (4.39)

xrg = xre−j(θgs−θrs) = xre

j(θrs−θgs) (4.40)

Applying these conversion factors we can write the following:

vs = vsgejθgs ; is = isge

jθgs ; ψs= ψ

sgejθgs (4.41)


Substituting these expressions into (4.37) we can write:

vsgejθgs = Rsisge

jθgs +d

dt

(ψsgejθgs

)= Rsisge

jθgs + ψsgejθgs . j

dθgdt

+ ejθgsdψ

sg

dt

∣∣∣∣∣θg const

= Rsisgejθgs +

jωgψsg

+dψ

sg

dt

∣∣∣∣∣θg const

ejθgs

∴ vsg = Rsisg +dψ

sg

dt+ jωgsψsg

(4.42)

In a similar manner we can do the same with the rotor equation for themachine. In this particular case we have:

vrg = vrgej(θgs−θrs); ir = irge

j(θgs−θrs); ψr= ψ

rgej(θgs−θrs) (4.43)

Substituting these into (4.38) gives:

vrgej(θgs−θrs) = Rrirge

j(θgs−θrs) +d

dt

(ψrgej(θgs−θrs)

)= Rrirge

j(θgs−θrs) + ψrgej(θgs−θrs) . j

d(θgs − θrs)dt

+ ej(θgs−θrs)dψ

rg

dt

∣∣∣∣∣(θgs−θrs) const

= Rrirgej(θgs−θrs) +

j(ωgs − ωrs)ψrg

+dψ

rg

dt

∣∣∣∣∣(θgs−θrs) const

ej(θgs−θrs)

∴ vrg = Rrirg +dψ

rg

dt+ j(ωgs − ωrs)ψrg

(4.44)

Summarising, the induction machine equations in a general reference frameare:

vsg = Rsisg +dψ

sg

dt + jωgψsg

vrg = Rrirg +dψ

rg

dt + j(ωg − ωr)ψrg

(4.45)

Remark 21 With equation (4.45) one can easily generate the induction ma-chine equation in an arbitrary reference frame. This will be very handy whenwe consider the various forms of vector control which result from these expres-sions in particular reference frames.

Using reasoning similar to that in the previous section we can write that therotational power is due to the true rotational motion in (4.45). Therefore the

4.3 A Heuristic Explanation of Vector Control 91

expression for the rotational power is:

Prot =32Re

{−jωrψrg

i∗rg}

= −32ωrψrg

× irg (4.46)

Therefore it is obvious that the rotational torque produced is:

Trot = −32ψrg

× irg (4.47)

This has the same form as the expression derived in the previous section, there-fore it can be seen that the various torque expressions derived in the previoussection carry over to the general reference frame except that the fluxes andcurrents have to be replaced by the general reference frame equivalents.. Theexpressions for the fluxes are:

ψsg

= Lsisg + Lmirg (4.48)

ψrg

= Lrirg + Lmisg (4.49)

One particular version of the torque expression that will be useful later isderived as follows. From (4.33) we know that we can write the following in thegeneral reference frame:

Trot =32Lmirg × isg (4.50)

This expression can be expanded as follows:

Trot =32Lm

Lr(Lrirg + Lmisg)× isg since isg × isg = 0

=32Lm

Lrψrg

× isg (4.51)

4.3 A Heuristic Explanation of Vector Control

In the previous section we developed some general space vector models of theinduction machine. These models are very useful in quantitative definitions ofthe fundamental equations for vector control. However, in this section we shallconsider vector control from a heuristic point of view.

The following discussion will be with reference to Figure 4.3. This diagramshows a squirrel cage induction machine with quadrature stator windings. Onecould consider the stator to be that of a DC machine where the q axis windingcorresponds to the compensation winding, and the d axis winding is the fieldwinding. For the sake of the following argument we will assume that the statorand rotor windings have the same number of effective turns. Let us now conducta thought experiment on this configuration.

Assume that a current iqs is suddenly injected into the sQ winding. ByLenz’s Law a current will flow in the rotor in such a way as to oppose thechange in flux caused by the increasing current in the sQ winding. The current


sD

sQ �m ' D

Q

Figure 4.3: Conceptual diagram of an induction machine with quadrature-phasestator windings.

directions are those marked in Figure 4.3. Therefore the current flowing at sometime t+o (where to is the time of application of the q axis current) is:

i′r = −iqs (4.52)

The effect of the injection of the current can be better scene from a space vectordiagram of the machine – see Figure 4.4. The resultant stator current is simplythe vector addition of the two stator currents. The induced rotor current directlyopposes the q axis stator current, and therefore there is no flux produced in theq axis. Therefore at time t+o the magnetising current is the d axis current, as itwould be in a compensated DC machine.

Assume that the rotor is held stationary. After the initial current inducedin the rotor the current will die away with the time constant being that of therotor circuit. Therefore at some time t > to the situation could be that inFigure 4.5. The length of the i′r vector will decrease in length until it is zero. Itis clear that under this condition that the magnetising current vector and fluxno longer coincide with the d axis of the machine. Therefore this situation isnow different from that of the DC machine.

What happens if we can move the stator so that the d axis remains alignedwith the magnetising current vector? That is, as the magnetising current vectormoves by δµm toward is we move the stator by the same amount. This move-ment of the stator can also be viewed as a movement of the rotor in a clockwisedirection if the stator is taken as the reference frame. If we have relative mo-tion between the stator and the rotor then we have the rotor bars cutting fluxand consequently there will be voltage produced in the bars. If one used the


D

Q

i s

i im ds'�

iqs i r '

�m

'

Figure 4.4: Space vectors in quadrature induction machine at time t+o .

D

Q

i s

ids

iqs

i r '

im '�m

Movement of im '

Figure 4.5: Space vectors in quadrature induction machine at t > to.

4.4 Special Reference Frames 94

D

Q

i s

i ir q' �

i im d' ��m sD

sQ

Figure 4.6: Position of the space vectors after the stator has been rotated.

F = qv×B expression the voltage induced for the relative motion in this exam-ple is such that the decreasing rotor current will be increased. Note that the Bfield would be slightly offset from the d axis because the iqs current is no longerfully compensated. This in turn changes the orientation of the induced voltagein the rotor and a consequent shift in the resultant rotor current so that com-pensation is reestablished. Figure 4.6 shows the situation where the stator hasbeen rotated so that the d axis still lies along the magnetising current vector.Note that this also implies that the rotor current, i′r is still orthogonal to themagnetising flux.

Remark 22 If one keeps the d axis winding aligned with the magnetising cur-rent vector, then the is vector will move by the same angular displacement as thed axis. Therefore implicitly the i′r vector must be orthogonal to the magnetisingcurrent vector.

In a real situation one would not move the stator winding, but the sameeffect could be achieved by controlling the stator d and q axis currents so thatthe resultant mmf (and hence current vector is) produced by the winding movesaround the machine by µm.

The movement of the stator current vector in relation to the magnetisingflux linkage phasor has important implications on the torque production of themachine. This will be investigated in the next section.

4.4 Special Reference Frames

4.4.1 The Magnetising Flux Linkage Reference Frame

In the previous section we heuristically discussed the basic concepts behindvector control. This discussion was with respect to the magnetising flux linkage.


i is sm,

�msD

sQ

x

y

�m

isx

isy

ids

iqs

� � �m mm mx'� �

Figure 4.7: Relationship between the dq frame and the special xy frame.

Now let us consider the implications of the magnetising flux reference frame onthe torque production of the machine.

The following discussion is with reference to Figure 4.7. As can be seen wehave defined a reference axes (the xy axes) that are rotating with the magnetis-ing current vector. Therefore the xy axes are rotating at:

ωm =dµmdt

(4.53)

One can also see that the relationship between the stator current in the mag-netising reference frame and the stator current in the stator frame is (using(3.104)):

ism = ise−jµm = isx + jiys (4.54)

From (4.36) we can deduce that the torque in a magnetising flux referenceframe is:

Trot =32ψm× ism

=32

∣∣∣ψm

∣∣∣ isy (4.55)

since there is no orthogonal component of the magnetising flux with this par-ticular orientation of the reference axes. Clearly the:

ψm

= Lm(ism + irm)

= Lmisx (4.56)= ψsm (4.57)


i is s r, �

�rsD

sQ

x

y

�mr

isx

isy

ids

iqs � r

�r

s

i imr mr�

��r r

Figure 4.8: Relationship between various space phasors in the stator and rotorflux linkage reference frames

therefore:

Trot =32ψsmisy

=32Lmisxisy

The basic principle of alignment with the magnetising axis forms the basisof magnetising flux vector orientated control. This shall be considered in moredetail later.

4.4.2 The Rotor Flux Linkage Reference Frame

Another reference frame that is very commonly used is the rotor flux referenceframe. In this reference frame the special axes are fixed to the rotor flux vector.Figure 4.8 is used in the following discussion.

We know from previous work that;

Trot =32Lm

Lrψrg

× isg (4.58)

Therefore if a reference frame, xy, is chosen so that it is coincident with therotor flux linkage vector then there by definition cannot be an orthogonal com-ponent of the rotor flux linkage in this frame of reference. Therefore the torqueexpression becomes:

Trot =32Lm

Lrψrψr

× isψr (4.59)

Now:

isψr = ise−jρr = isx + jisy (4.60)


If ψris the rotor flux linkage in a rotating frame, then:

ψrψr

= ψre−j(ρr−θr) = ψ

rejθre−jρr = ψ′

re−jρr =

∣∣∣ψr

∣∣∣ ejρre−jρr =∣∣∣ψ

r

∣∣∣ = ψrx

(4.61)

Substituting this into the torque expression (4.59) gives:

Trot =32Lm

Lrψrx × (isx + jisy)

=32Lm

Lrψrxisy (4.62)

This expression can be further manipulated into a form that is more familiar inthe vector control literature. This form of the expression is based in derivinga single current value related to the rotor flux linkage (the normal definition ofthe flux linkage involves the rotor and the stator fluxes):

imr =ψrψr

Lm=

(Lrirψr + Lmisψr)Lm

= isψr +Lr

Lmirψr

= isψr + (1 + σr)irψr; σr =Lrl

Lm(4.63)

This current vector lies along the x axis of the reference frame (since it isrelated to the rotor flux linkage vector via a scalar) as shown in Figure 4.8.Since ψ

rψr= ψrx then ψrx = Lm |imr| . Hence (4.62) becomes:

Trot =32L2m

Lr|imr| isy (4.64)

=32

Lm

(1 + σr)|imr| isy using Lr = (1 + σr)Lm (4.65)

As with the magnetising flux alignment case the torque can be controlledseparately from the flux in the machine if one can keep the |imr| value constantand use the isy current for the torque control.

4.4.3 Stator Flux Linkage Reference Frame

In a manner similar to rotor flux linkage vector alignment we can also align thespecial reference frame to the stator flux linkage vector. In a manner similar to(4.34), and realising that the reference frame is aligned with the stator flux, wecan write the torque as:

Trot =32ψsxisy (4.66)

Figure 4.9 shows the relationship between the special frame and the statorcurrent vector.

It is possible to put the equation for the torque in terms of currents. Thestator flux vector can be written as follows:

ψsψs

= Lsisψs + Lmirψs (4.67)


i is s s, �

i ims ms�

�s sD

sQ

x

y

s

isx

isy

iqs

ids

� � ��s s s sx, �

�ms

Figure 4.9: Relationship between the stationary reference frame and the specialreference frame fixed to the stator flux linkage space phasor.

where irψs is the rotor current referenced to the special stator flux referenceframe. Let us consider this current vector further:

irψs = irx + jiry = ire−j(ρs−θr) = ire

jθr = i′re−jρs (4.68)

Now let us define the stator magnetising current in the stator flux linkagereference frame:

ims =ψsψs

Lm=Ls

Lmisψs + irψs = (1 + σs)isψs + irψs (4.69)

where σs =Lsl

Lm(4.70)

Since the ψsψs

is coincident with the x axis of the reference frame then so isims. Therefore ψsx = Lmims. Substituting for ψsx in (4.66) we can write:

Trot =32Lm |ims| isy (4.71)

As with the previous special reference frames the torque is related to theinteraction of a current and a flux linkage vector that are in space quadrature.

Remark 23 In all these special reference frames one can see that the form ofthe torque expressions is identical to those for a separately excited DC machine.Furthermore the currents in these expressions are DC values in steady state.Therefore the choice of the reference frame is very important in simplifying acomplex model to that of a DC machine. The fact that the rotating referenceframes chosen create steady state DC currents has important implications onthe control strategies employed.

4.5 Derivation of Rotor Flux Oriented Vector Control 99

4.5 Derivation of Rotor Flux Oriented VectorControl

In the previous sections of this chapter we have derived a number of supportingexpressions that allow us to define the relevant expressions for vector controlin a number of different reference frames. In this section we shall derive theexpressions required for control in the rotor reference frame. This frame waschosen as it is one of the most popular control strategies used in practice. Thederivation of controllers in the other reference frames will be left as an exercise.

We begin with the equations for the induction machine in the general refer-ence frame (4.45), repeated here for convenience:

vsg = Rsisg +dψ

sg

dt+ jωgψsg

(4.72)

vrg = Rrirg +dψ

rg

dt+ j(ωg − ωr)ψrg

(4.73)

In a frame rotating with the rotor flux the speed of the frame is ωmr thereforethese expressions are modified as follows:

vsψr = Rsisψr +dψ

sψr

dt+ jωmrψsψr

(4.74)

vrψr = Rrirψr +dψ

rψr

dt+ j(ωmr − ωr)ψrψr

(4.75)

Let us firstly consider the stator voltage equation. We know that the statorflux can be expresses as follows:

ψsψr

= Lsisψr + Lmirψr (4.76)

and substituting this into (4.74) we get:

vsψr = Rsisψr +d

dt(Lsisψr + Lmirψr) + jωmr(Lsisψr + Lmirψr)

= Rsisψr + Ls

disψr

dt+ Lm

dirψr

dt+ jωmrLsisψr

+ jωmrLmirψr (4.77)

From the expression for the rotor magnetising current (4.63) we can write:

irψr =imr − isψr

(1 + σr)(4.78)

Because of the choice of the frame, we know that the frame lies along the imr

vector and consequently there is no orthogonal component to imr. Thereforethe expression for the rotor current can be rewritten as:

irψr =|imr| − isψr

(1 + σr)(4.79)


Substituting this for irψr in (4.77) we get:

vsψr = Rsisψr + Ls

disψr

dt+ Lm

d

dt

( |imr| − isψr

(1 + σr)

)

+ jωmrLsisψr + jωmrLm

( |imr| − isψr

(1 + σr)

)(4.80)

Now manipulating this equation to make the derivative of the stator current thesubject of the expression we get:

disψr

dt=vsψr

Ls− Rsisψr

Ls− Lm

Ls

d

dt

( |imr| − isψr

(1 + σr)

)

− jωmrisψr − jωmrLm

Ls

( |imr| − isψr

(1 + σr)

)(4.81)

Collecting terms we can write this expression as follows:[1− Lm

Ls(1 + σr)

]disψr

dt=vsψr

Ls− Rsisψr

Ls− Lm

Ls(1 + σr)d |imr|dt

− jωmrisψr

[1− Lm

Ls(1 + σr)

]

− jωmrLm |imr|Ls(1 + σr)

(4.82)

Realising that:

1− Lm

Ls(1 + σr)= 1− Lm

LsLrLm

= 1− L2m

LsLr

and:

σr = 1− k2r k is the coupling coefficient

Lm = k√LsLr ⇒ k =

Lm√LsLr

∴ σr = 1− L2m

LsLr

then we can write:

σrdisψr

dt=vsψr

Ls− Rsisψr

Ls− (1 − σr)d |imr|

dt− jωmrisψrσr − jωmr(1− σr) |imr|

(4.83)

which can be simplified to:

disψr

dt=vsψr

σrLs− Rsisψr

σrLs− (1 − σr)

σr

d |imr|dt

− jωmrisψr − jωmr(1− σr)σr

|imr|(4.84)


Breaking this into real and imaginary parts we get two coupled differentialequations for the stator circuit when rotor flux orientation is implemented:

disxdt

=vsxσrLs

− RsisxσrLs

− (1− σr)σr

d |imr|dt

+ ωmrisy (4.85)

disydt

=vsyσrLs

− RsisyσrLs

− ωmrisx − ωmr(1− σr)σr

|imr| (4.86)

Remark 24 Notice that the above expressions for isx and isy are cross coupledif we are attempting to control them through the voltages vsx and vsy. In additionthe rotor magnetising current has an influence on the expressions.

Remark 25 If we are using a voltage source inverter to control the above statorequations, we need a parameter dependent decoupler in order to gain decoupledcontrol of isx and isy.

Remark 26 If the stator is being fed from an ideal current source then thevalues of isx and isy are “impressed” on the stator. In this situation the currentsources are producing whatever voltage is required to make the currents in the xand y axes the current source value.

Following on from the last remark, if we do not have an ideal current sourcethen we can still achieve most of the advantages of the ideal current source. Wecan generate an current source by applying current feedback around a voltagesource inverter. Clearly this cannot be an ideal current source as it has a certainbandwidth in relation to reacting to changes in the load, and there is a finitevoltage that can be applied to the machine. However, even with these limitationsa reasonable current source can be obtained.

The main advantage of using current source feed to the inverter is thatthe complex dynamics of the stator are removed from the system. Let us forthe moment assume that the stator dynamics can be ignored due to an idealcurrent source, and then consider the rotor circuit. Realising that the rotor ofan induction machine is short circuited then we can write (4.75) as:

0 = Rrirψr +dψ

rψr

dt+ j(ωmr − ωr)ψrψr

(4.87)

Now for rotor orientation, then by definition we have:

ψrψr

= Lmimr = Lm |imr| (4.88)

therefore we can write:

0 = Rrirψr + Lmd |imr|dt

+ j(ωmr − ωr)Lm |imr| (4.89)

The rotor current can be expressed in terms of the rotor magnetising currentand the stator current using (4.79):

irψr =|imr| − isψr

(1 + σr)=Lm

Lr

(|imr| − isψr

)(4.90)


Substituting into (4.89) gives:

0 =RrLm

Lr|imr| −

RrLm

Lrisψr + Lm

d |imr|dt

+ j(ωmr − ωr)Lm |imr| (4.91)

Multiplying this by Lr/RrLm and rearranging we get:

Trd |imr|dt

+ j(ωmr − ωr)Tr |imr|+ |imr| − isψr = 0 (4.92)

where:

where Tr =Lr

Rr– the rotor time constant

Breaking this expression into real and imaginary parts:

Trd |imr|dt

+ |imr| = isx (4.93)

(ωmr − ωr)Tr |imr| − isy = 0 (4.94)

Rearranging the second of these equations and adding the torque expression(4.64) we end up with the conventional electrical rotor flux orientated equationsfor the induction machine:

Trd|imr|

dt + |imr| = isxωmr = ωr +

isy

Tr|imr|Trot = 3

2L2m

Lr|imr| isy

(4.95)

Remark 27 The first equation in (4.95) is the flux equation since imr is directlyrelated to ψ

rψrin the machine.

Remark 28 The second equation can be rearranged so that the last term is thesubject of the expression:

isyTr |imr|

= ωmr − ωr (4.96)

This is clearly in the form ωsl = ωe−ωr where ωsl is the slip frequency. There-fore isy

Tr|imr| is the slip frequency of the machine.

Remark 29 This expression is the same form as the dynamic equation of aseparately excited DC machine. The first equation has no coupling from thesecond equation. The second equation is related to the torque produced by themachine. This can be seen easily by rearranging the torque expression in termsof isy and substituting into the ωmr equation:

ωmr = ωr +23

Trot

Lm |imr|2(4.97)

Remark 30 One of the main control techniques used with these equations isthat |imr| is kept constant at the isx value, and the torque can be controlledseparately by the use of isy.

4.6 Structure of a Rotor Oriented Vector Drive 103

Remark 31 The above equations are fundamentally based on the fact that weare assuming that the position of the rotor flux is known. Therefore the isxand isy currents can be accurately applied at the correct position spatially in themachine.

Remark 32 If we have an ideal current source then the torque is algebraicallyrelated to the isy current . In practice a delay will occur in the rise of the torque,this being related to the leakage inductance of the machine.

Remark 33 There is no break away torque limit in rotor flux field orientedcontrol – the torque increases linearly with isy.

Torque control based on the equations in (4.95) are generally used in twocontrol philosophies – direct field oriented control, and indirect field orientedcontrol. In direct field oriented control the position of the rotor flux is deter-mined using flux feedback, the flux magnitude and position being obtained frommeasurements of the flux via Hall Effect sensors or search coils, or via a fluxmodel of the machine. Indirect field orientation on the other hand uses a feed-forward technique to calculate the position of the flux using the current rotorposition and the reference position of the slip frequency. The latter techniquehas proved to be very popular as it does not involve sensors or a flux model.Note that this technique implicitly has parameter sensitivity since the referenceposition is calculated using an expression that involves Tr.

4.6 Structure of a Rotor Oriented Vector Drive

4.6.1 Indirect Rotor flux Oriented Controller

The block diagram of Figure 4.10 shows the basic structure of a indirect vectorcontrolled drive based on the expressions developed in the previous section.Notice that all the control calculations are carried out in the rotating referenceframe.

The rotor flux reference angle is calculated on-line by integrating the refer-ence slip frequency:

ρr =∫ωslref dt+ θr

=∫

isyrefTrisxref

dt+ θr (4.98)

This integration clearly has to be implemented accurately in order for the rotorflux angle to be correct. In modern implementations this is carried out as adigital integration.

In the block diagram the current controller is implemented in the stationaryreference frame, however this is implementation dependent. For example, if thecurrent controllers are implemented using conventional PI controllers, then thecurrent control is carried out in the synchronously rotating rotor flux referenceframe. The reason for this is that PI controllers cannot provide accurate trackingof a sinusoidal reference, and the references in the stationary frame are sinusoidalin steady state. However in the rotating frame the steady state values of thecurrents are DC values, and PI controllers can track DC values with zero error.


IM

PWMmodulatorand currentcontroller

Inverter

ia

ib

ic

Current controlled inverter

�

�

�

�

�

�

dt

dTr 1

FieldWeakeningGenerator

rT

dt

d

�rref

rje

� 32 �

Rotating 2 phase tostationary 3 phase X'formation

r

m

L

L

2

3 2

�r

+-

+

+

-+

-+

-�r

�r

sxrefr iT

sxrefr

syref

slrefiT

i��

mrrefi

sxrefi

syrefi

dsrefi

qsrefi

SpeedController

rotrefT

r�

Mains

arefi

brefi

crefi

slref�

+

Figure 4.10: Block diagram of a indirect rotor flux vector oriented controlscheme

The field weakening block in the diagram generates the imrref value abovethe base speed. Therefore, this contains a function generator that drops therotor flux in a prescribed way so that the machine will operate in a constantpower mode above base speed. Below base speed the rotor flux is usually keptconstant at the maximum desired flux so that the transient performance of thedrive will be at its maximum.

The speed control block is usually implemented as a PI controller. Theoutput of the speed control block can be interpreted as the desired torque. Ifone wishes to implement position control then a further loop can be addedoutside the speed control loop. The control type is again usually a PI controller(although other more sophisticated control strategies can be used). The outputof this block is interpreted by the speed control loop as a desired speed.

4.6.2 Direct Rotor Flux Oriented Controller

This form of the rotor flux controller is probably a more obvious way of imple-menting a rotor flux oriented vector controller since it uses classical feedbackthat most engineers are familiar with. Figure 4.11 shows a block diagram ofthe controller. In this controller the rotor flux position is determined by a fluxmodel that processes current measurements from the machine. The torque andflux controllers would normally be conventional PI regulators.

The flux model is generated directly from the equations in (4.95). Figure 4.12is a block diagram of the flux model in a rotor flux reference frame. It is alsopossible to develop the flux model in a stationary reference frame.


IM

PWMmodulatorand currentcontroller

Inverter

ia

ib

ic

Current controlled inverter

�

�

�

�

�

�

FieldWeakeningGenerator

dt

d

�rref

rje

� 32 �

Rotating 2 phase tostationary 3 phase X'formation

r

m

L

L

2

3 2

�r

�r

+-

+

-+

-+

-�r

�r

mrrefi

sxrefi

syrefi

dsrefi

qsrefi

SpeedController

rotrefT

r�

Mains

arefi

brefi

crefi

mri

mri

-

+

-

+

Trot

FluxModel

isy

Fluxcontroller

TorqueController

Figure 4.11: Block diagram of a direct rotor flux field oriented vector controller.

�

Lm

�

rT

r�

�r

+

+

rr ��

rmr ��

isy

isx

rje

��23 � dt

dTr 1

1

rT

ia

ib

ic

�r

mri

Figure 4.12: Flux model in a rotor flux reference frame

4.7 Magnetising Flux Orientation 106

4.7 Magnetising Flux Orientation

We shall not derive the equations for magnetising flux orientation in these notes.However, it is useful to look at the rotor equations in this frame and comparethem with those for the rotor flux oriented frame:

disxdt

+isxTrl

− ωslisy =

(|imm|+ Tr d|imm|

dt

)Trl

(4.99)

disydt

+isyTrl

= ωsl

(TrTrl

|imm| − isx)

(4.100)

where:

Trl =Lrl

Rr

ωsl = ωm − ωrimm = irm + ism where the m subscript means magnetising frame

Notice that the cross coupling in these equations is much more complex thanin the case of rotor flux orientation. Therefore to gain decoupled control of thetorque and flux using this frame we require decoupling equations.

Remark 34 Clearly rotor flux orientation gives the classical DC machine equa-tions for the control without any decoupling equations when the system is beingcurrent fed.

Appendix A

Calculation of Inductancesfor Salient Pole Machines

A.1 Calculation of Inductances

One of the fundamental parameters of any machine model is the inductance ofthe armature windings of the machine under all operating conditions. Later inthis chapter we shall that the variation of inductances with respect to the me-chanical position of the rotor is directly connected with electromagnetic energyconversion in all machines, and hence with the production of torque. Thereforethe calculation of the SYNCREL inductances is fundamental to understandingthe machines operation.

In the case of the SYNCREL, the armature is on the stator, since the rotordoes not have any windings. We will find that the inductance of a particularwinding varies depending on the position of the rotor in relation to the winding,and the degree of magnetic saturation of the stator and the rotor iron.

This section will determine the self and mutual inductances for the statorwindings of the SYNCREL. The derivation of these inductances will be carriedout in a detailed and formal manner using a traditional approach [2]. A differentapproach, and in many ways a simpler and more elegant one using the conceptof winding functions is shown in Appendix B

Remember!! Modify the winding function stuff for the SYNCREL.

The following standard assumptions are made in the following analysis:

1. The stator windings are sinusoidally distributed. When excited with cur-rent a sinusoidal spatial distribution of mmf is produced.

2. The machine does not exhibit any stator or rotor slotting effects.

3. The machine iron is a linear material, i.e. it is not subject to magnetic sat-uration effects. The permeability of the material is very large in compar-ison to air. Therefore the permeance of the magnetic paths is dominatedby the air gaps.

4. The air gap flux density waveforms can be adequately represented by theirfundamental component.

A.1 Calculation of Inductances 108

Figure A.1: Two pole three phase SYNCREL – conceptual diagram

5. The stator turns are all full pitched (i.e. they cover π electrical radians).

6. There is no leakage flux – i.e. there is perfect coupling between the wind-ings.

Figure A.1 is a conceptual schematic of a two pole, three phase SYNCREL.Note that the rotor shape does not represent a realistic rotor, but is drawn inthis manner to accentuate the variable reluctance in the d and q axes. The axisof the rotor which offers the minimum reluctance to the passage of flux acrossthe air gap from the stator to the rotor is called the d-axis. The maximumreluctance path is denoted as the q-axis. Note that following development willuse the concept of dq axes before the concept has been rigorously developedin a more general framework. In the following development the dq axes areclosely associated with the physical configuration of the rotor, therefore thegeneral development can be left to later without having too many problemsunderstanding this material.

A few preliminary conjectures, based on heuristics, can be made about thevariation of the winding inductance with respect to the angular rotor position:

Conjecture 35 The winding self inductance will be a maximum when the d-axis of the rotor is aligned with the axis of the winding.

Remark 36 This conjecture concurs with ones intuitive understanding of fluxinteracting with iron. The presence of iron in a coil will result in more flux perunit of current. When the d-axis is aligned with the axis of a coil then therewill be more iron in the flux path for the coil.


Conjecture 37 The winding self inductance will be a minimum when the q-axis of the rotor is aligned with the axis of the winding.

Remark 38 If there is less iron in the coils flux path then it is harder to produceflux for a given amount of current in the coil. Clearly if the q-axis is alignedwith the axis of the coil then there is a larger air path and less iron for the fluxto travel through.

Conjecture 39 As the rotor is rotated between these two positions the self in-ductance varies. The period of the phase inductance variation is half the periodof the mmf variation for the phase winding.

Remark 40 This is fairly obvious since the phase inductance is a maximumwhen a d-axis rotor pole aligns with the phase axis, and this occurs when therotor has rotated through π electrical radians.

Conjecture 41 There is mutual inductance between the three phase stator wind-ings that is a function of the rotor position.

Remark 42 Clearly as the rotor is rotated the amount of iron in the paths thatwould be taken by the mutual flux will vary, and hence the amount of flux linkingthe windings will vary.

A complete analysis of this situation involves computing of all the harmonicsof the flux density waveform and then calculating the total flux linkage withthe winding. One then obtains inductance expressions containing a numberof harmonic terms [2]. The harmonic term amplitudes decrease rapidly withincreasing harmonic number, allowing the approximation of considering only thefundamental to be made. The constant reluctance path approximation madein the following analysis is essentially the same approximation. If the windingfunction technique is use to calculate the inductances then the harmonic effectsare sometimes more readily included. However, the accuracy of this techniqueis critically dependent on the accuracy of the inverse air gap function.

A.1.1 Self Inductances

Firstly consider the self inductance of the a-phase sinusoidally distributed wind-ing. A useful technique to calculate inductances in situations like this is toconsider that the stator mmf can be broken into two sinusoidally distributedcomponents which can be considered to be acting along the d-axis and the q-axis of the rotor.(this is possible because of the assumed sinusoidal nature ofthe mmf, which implicitly allows components to be taken). Let us consider afew simple cases of the application of this concept. Figure A.1 can be used asan aid to visualise the situation. If, for example. the rotor d-axis is aligned withthe axis of the a-phase winding then the total a-phase mmf acts on the d-axispermeance, and there is no component acting on the q-axis permeance. Sincethe stator mmf is spatially sinusoidally distributed, then this means that the airgap flux density waveform would be sinusoidally distributed. Similarly if therotor q-axis is aligned with the a-phase axis, then the total a-phase mmf actson the q-axis permeance. Between these two rotor positions the permeance seenby the winding is, in general, a complex function of the rotor angular position.


Consequently, the air gap flux density distribution is also a complex function ofthe rotor angle.

The sinusoidally distributed mmf on the stator of the machine can be brokeninto two sinusoidal components which are centred on the d and q-axes respec-tively, regardless of the position of the rotor. These component mmfs are thenacting on the d and q-axis permeances, Pd and Pq. Since these permeances areconstant, this is equivalent to saying that the component mmfs are acting on twoconstant air gaps, gd and gq, for the d and q-axes [2]. Therefore the resultantcomponent air gap flux densities should be spatially sinusoidal, and consequentlythe resultant total air gap flux density should also be sinusoidal. This contra-dicts the statements made in the previous paragraph about the complex natureof the air gap flux density. However, the fundamental of the actual air gap fluxdensity is, in practice, very close to that obtained using this approximation, andmeasured inductances for real machines are in reasonable agreement with thecalculated values based on the approximation. The reason for this is that sinu-soidally distributed windings will only link to the components on a flux densitywaveform that have the same pole number as the winding, as was previouslyshown in Section 1.2. Therefore, for an ideal sinusoidally distributed windingonly the fundamental component of the flux density can link to the winding,and consequently harmonic flux densities only contribute to leakages.

Remark 43 An ideal sinusoidally distributed winding cannot be constructed –all true windings have winding space harmonics. These winding harmonics cantherefore link to harmonic flux densities of the same poll number. This can leadto the generation of harmonic voltages, and more complex inductance variationswith rotor position.

Addition Could add a section here examining the assumption that the d andq-axes air gaps can be modelled as constant air gaps. Could consider anideal 2 pole axially laminated machine, looking at the effective air gapseen by the mmf in both the axes.

The following is with reference to Figure A.2, which is a laid out diagram ofFigure A.1. This diagram shows the two fictitious air gaps, with the componentmmfs acting on the d and q-axes respectively. The resultant air gap flux densitydistributions are shown for the two axis waveforms. Notice that the resultantair gap flux density waveform is distorted away from the d-axis of the rotor bythe q-axis flux waveform, the degree of distortion being related to the differencebetween the air gap lengths and the mmf applied in the axes.

In order to calculate the self inductance of the a-phase winding the totalself flux linkage must be calculated for the winding. This self flux linkage hasseparate components contributed by both the d and q-axis fluxes.

Using the approach in [2] we calculate the flux due to one of the componentmmfs acting on one of the air gaps by proceeding in the following manner:

1. Calculate the flux in an incremental area at some angular position in themachine accounting for the spatial distribution of the mmf.

2. One then integrates up these incremental fluxes for a total span of a singlecoil. This gives the total flux linking one coil.


Figure A.2: Developed diagram of a SYNCREL.

3. Calculate the flux linking all the coils that have their axes at some angularposition around the machine. This is achieved by multiplying the valueobtained in point 2 by the number of turns that lie in the same positionas the single coil.

4. Finally integrate up the previous value over the coil span accounting forthe change in the number of turns with angular spatial variation.

5. Once the flux linkage for each air gap is found then the total flux linkageto the a-phase is found by adding together the linkages due to the d andq axes.

Consider the d-axis, as shown in Figure A.3. The expression for the fluxover a 180◦ electrical span of the d-axis mmf can be found as follows. Considerthe incremental permeance over an angle of dβ:

dPd =µodA

gd(A.1)


Figure A.3: d axis developed diagram for SYNCREL

where:

dA � the incremental area.= lmr dβ

β � machine periphery angle relative to the d axis.

lm � the length of the machine.

r � the radius of the machine at the centre of the air gap.

µo � the permeability of free space.

Therefore the incremental flux can be written as:

dφd = dPd(Fd cosβ

)=Fdµolmr

gdcosβ dβ (A.2)

where Fd cosβ is the d-axis component mmf.


To find the total flux linking a single coil whose most clockwise coil side startsat α radians relative to the d-axis position, we integrate the d-axis incrementalfluxes dφd for the dA elements using the following integration:

φd =∫ α+π

α

dφd

=∫ α+π

α

[Fdµolmr

gdcosβ

]dβ

=−2Fdµolmr

gdsinα (A.3)

where:

Fd = Fa cos θpd the component mmf at θpd, and (A.4)

Fa � the peak mmf of the a-phase.

θpd � the angle of the d-axis around the machine periphery (elec-rad) (A.5)

Remark 44 Note that the above definition of the flux linkage per turn impliesthat the normal vector for the coil area is at the angle α+π/2 radians. Realisingthis is important in getting the correct sign for the total flux linkage of the coil.

Clearly the maximum flux of 2Fdµolmr/gd is obtained when the coil sideα = −π/2 – this means that the coil axis is a 0 radians and hence aligns withthe component mmf axis. Equation (A.3) can be written in terms of the totald-axis permeance by utilising the following expression:

Pd =∫ π

2

−π2

dPd

=∫ π

2

−π2

µolmr

gddβ

=µolmrπ

gd(A.6)

therefore (A.3) can be written as:

φd(α) =−2FdPdπ

sinα (A.7)

If a coil side starts at some angle α with respect to the d-axis then the coilaxis is at α+ π/2. Define:

αa � angle of the coil axis relative to the d-axis

and hence:

αa = α+π

2(A.8)

and consequently:

α = αa − π

2(A.9)


Substituting this into (A.7) we can write the flux for a single turn whose axisis at αa with respect to the d-axis as:

φd(αa) =−2FdPdπ

sin(αa − π

2)

=2FdPdπ

cosαa (A.10)

This expression can be further manipulated so that the flux is a function of theangle of the d-axis and the coil axis with respect to the axis of the a-phase. Let:

θa � the angle of the coil axis with respect to the a-phase

therefore:

αa = θa − θpd (A.11)

Substituting this into (A.10) we can write the following:

φd(θa, θpd) =2FdPdπ

cos(θa − θpd) (A.12)

We are now in a position to calculate the flux linkage to the turns of a-phaseat some particular coil axis angle θa for some constant d-axis. The number ofturns that have their coil axis at angle θa can be deduced from the turns densityfunction (1.1) as:

nta(θa) = na cos θa (A.13)

Remark 45 Clearly nta(θa) can be negative. The concept of a negative numberof turns/radian at a particular coil axis angle is related to the concept of a nega-tive number of conductors around the periphery of the machine (the sign in thiscase arising from the direction of current in the conductors at that point).Theturns density function expressed in θa is essentially the mmf/ampere for thewinding at a particular position. This is also known as a winding function.Therefore the negative sign indicates that the flux produced is in the oppositedirection across the air gap (i.e. from the stator to the rotor instead of fromthe rotor to the stator).

Therefore the total flux linkage for the number of turns at θa is:

ψd(θa) =2FdPdna

πcos θa cos(θa − θpd) (A.14)

We are now in the position to calculate the total flux linkage of the d-axisflux to the a-phase by integrating the flux linkage ψd(θa) at each position θa forthe coil span of the winding. Therefore the total flux linkage is:

ψad(θpd) =2FdPdna

π

∫ γ+π

γ

cos θa cos(θa − θpd)dθa (A.15)

Carrying out this integration and simplifying the result we obtain:

ψad(θpd) = FdPdna cos θpd (A.16)


In a similar fashion, the flux linkage of the q-axis flux with the a-phase canbe found. The procedure is identical to the above so it will not be presentedin detail. Instead we will simply state the results of the intermediate steps andthen present the final flux linkage result.

The incremental permeance for the q-axis is:

dPq =µolmr dβ

gq(A.17)

and therefore the total permeance of over a coil span is:

Pq =∫ π

0

dPq

=µolmrπ

gq(A.18)

The q-axis is at an angle of π/2 radians with respect to the d-axis. Thereforethe variation of the q-axis mmf is:

Fq = Fq cos(β − π

2) (A.19)

Therefore the q-axis incremental flux linkage is:

dφq = dPq Fq cos(β − π

2) (A.20)

Since cos(β − π2 ) = sinβ, and substituting for dPq in (A.20) gives:

dφq =µoFqlmr

gqsinβ dβ =

FqPqπ

sinβ dβ (A.21)

Consequently the expression for the flux linkage for a single coil can be writtenas:

φq(α) =∫ α+π

α

dφq

=FqPqπ

∫ α+π

α

sinβ dβ

=2FqPqπ

cosα (A.22)

where α � an angle relative to the d-axis.Carrying out the angle conversion to the coil axes relative to the a-phase as

was done in the d-axis case we can write:

φq(θa) = −2FqPqπ

sin(θa − θpd) (A.23)

The total flux linkage of the q-axis flux to the a-phase can therefore bewritten as:

ψaq(θpd) =∫ γ+π

γ

nta(θa)φq(θa) dθa

=−2FqPqna

π

∫ γ+π

γ

cos θa sin(θa − θpd) dθa

∴ ψaq(θpd) = −FqPqna sin θpd (A.24)


In the above expressions the peak values of the d and q-axes mmfs are foundby taking components of the a-phase mmf onto the d and q-axes respectivelyas follows:

Fd = Fa cos(−θpd) = Fa cos θpd (A.25)

Fq = Fa sin(−θpd) = −Fa sin θpd (A.26)

where Fa � the peak of the a-phase mmf = naia (from (1.2)). Note that thenegative sign in front of the θpd terms results from the fact that the angle ismeasured relative to the d-axis, and not the a-phase axis,since we are projectingthe a-phase mmf onto the d and q axes.

The total flux linkage to the a-phase can now be calculated by using super-position and adding the components linking to it from the d and q-axes. Using(A.16) and (A.24) we get:

ψaa(θd) = ψad(θd) + ψaq(θd)

= naFa(Pd cos2 θpd + Pq sin2 θpd)

= n2aia(Pd cos

2 θpd + Pq sin2 θpd)

=n2aia2

[(Pd + Pq) + (Pd − Pq) cos 2θpd] (A.27)

The rotor self inductance can therefore be calculated as a function of thed-axis position as:

Laa =ψaa

ia= L1 + L2 cos 2θpd (A.28)

where:

L1 =N2

8(Pd + Pq)

L2 =N2

8(Pd − Pq)

N � total number of turns in sinusoidal winding= 2na

Figure A.4 shows a plot of (A.28). Notice that the inductance varies as afunction of cos 2θpd with a constant offset as mentioned in conjecture 39.

The self inductances for the other two phases can be found similarly as:

Lbb = L1 + L2 cos 2(θpd − 2π

3

)(A.29)

Lcc = L1 + L2 cos 2(θpd +

2π3

)(A.30)

Addition Perhaps a remark about the fact that this analysis gives accurateinductance results since only the fundamental components of the flux den-sity distribution link to the sinusoidal winding, as proved in a previoussection.


Figure A.4: ‘a’ phase inductance plot.

A.1.2 Mutual Inductances

In addition to the self inductance of the winding there is also mutual inductancebetween the a, b, and c-phases. These inductances are also a function of theposition of the rotor, since its position clearly changes the reluctance of the fluxpaths between the windings. The process of calculating the general expressionsfor these inductances is very similar to that for the self inductances. We shallwork out in detail the mutual inductance between two windings and then simplystate the relationships between the other windings.

Let us consider the mutual inductance between the a-phase and the b-phase.The spatial sequence of the phases is as shown in Figure 3.3. The windingconductor density distribution for the b-phase is:

nb(θp) = nb sin(θp − 2π3) (A.31)

Therefore the number of coils with their axes at some angle θa with respect tothe a-phase (i.e. the winding function) is:

ntb(θ) = nb cos(θa − 2π3) (A.32)

As with the self inductance we shall work out the flux linkage for the d andq axes separately, and then use superposition to calculate the total flux linkage.We can write the expression for the flux linkage for a single turn using theexpression (A.10) calculated for the self inductance case:

φd(αa) =2FdPdπ

cosαa (A.33)


Again we can say:

αa = θa − θpd (A.34)

allowing us to again write the flux expression as:

φd(θa, θpd) =2FdPdπ

cos(θa − θpd) (A.35)

We can now write the flux expression for the coils that have their axis at θa as:

ψd(θa, θpd) =2FdPdnb

πcos(θa − 2π

3) cos(θa − θpd) (A.36)

Finally we now find the total linkage of the d -axis flux by integrating over a coilspan of the b-phase:

ψdba =2FdPdnb

π

∫ γ+π

γ

cos(θa − 2π3) cos(θa − θpd)dθa (A.37)

After considerable manipulation this expression can be written as:

ψdba = FdPdnb cos(θpd − 2π3) (A.38)

Using (A.25) the expression becomes:

ψdba = FaPdnb cos θpd cos(θpd − 2π3) (A.39)

Now let us consider the q-axis contribution to the b-phase flux. Using (A.10)we can again write an expression for the q-axis flux linking a single turn centredat the angle αaq relative to the q-axis:

φq(αaq) =2FqPqπ

cosαaq (A.40)

The αaq angle con be converted to angle relative to the a-phase:

θa = θpd + αaq +π

2(A.41)

and therefore:

αaq = θa − (θpd +π

2) (A.42)

Hence φq can be written as:

φq(θa, θpd) =2FqPqπ

cos(θa − θpd − π

2) (A.43)

=2FqPqπ

sin(θa − θpd) (A.44)

Now using the winding function we can write:

ψq(θa, θpd) =2FqPqnb

πcos(θa − 2π

3) sin(θa − θpd) (A.45)


Integrating over the coil span:

ψqba =2FqPqnb

π

∫ γ+π

γ

cos(θa − 2π3) sin(θa − θpd)dθa (A.46)

After considerable manipulation we arrive at the expression for the fluxlinkage from the q-axis to the b-phase:

ψqba = FqPqnb cos(θpd − π

6) (A.47)

Using (A.26) this expression can be written as:

ψqba = −FaPqnb sin θpd cos(θpd − π

6) (A.48)

We are now in a position to calculate the total mutual flux linkage to theb-phase from the a-phase as follows:

ψba = ψdba + ψqba

= Fanb[Pd cos θpd cos(θpd − 2π3)− Pq sin θpd cos(θpd − π

6)]

= Fanb

{Pd

[−14(1 + cos 2θpd) +

√34

sin 2θpd

]+

Pq

[√34

sin 2θpd +14(1− cos 2θpd)

]}(A.49)

After manipulation we get the following expression for this mutual inductance:

ψba =nanbia

2

[−(Pd + Pq)2

+ (Pd − Pq) cos(2θpd − 2π3)]

(A.50)

For a balanced machine na = nb, therefore the term in front of this expressionis n2

aia/2. Therefore this expression is the same as that for the self inductancesand hence we can write the mutual inductance in the same form as that for theself inductances:

ψba =N2ia8

[−(Pd + Pq)2

+ (Pd − Pq) cos(2θpd − 2π3)]

(A.51)

Dividing (A.51) by ia gives the inductance expression:

Lba = Lab = −L1

2+ L2 cos 2(θpd − π

3) (A.52)

where L1 and L2 are as defined in (A.28).By a similar process it can be shown that the other mutual inductances are:

Lca = Lac = −L1

2+ L2 cos 2(θpd +

π

3) (A.53)

Lcb = Lbc = −L1

2+ L2 cos 2θpd (A.54)


A.1.3 Summary

Assuming that the mmf for each phase varies sinusoidally around the machine,and that the resultant mmf in the machine acts on two different air gaps for thelow and high permeance axes, then the self and mutual inductances of a phasewinding vary as follows with θpd (the angle of the d-axis with the a-phase). Inthe above derivations we did not take into account the leakage inductance termin each of the self inductances. If we assume that the leakage does not changewith rotor position (which may not be a valid assumption) then the leakage canbe included by the addition of the term Ll as shown below:

Self Inductances

Laa = Ll + L1 + L2 cos 2θpdLbb = Ll + L1 + L2 cos 2

(θpd − 2π

3

)Lcc = Ll + L1 + L2 cos 2

(θpd + 2π

3

) (A.55)

Mutual Inductances

Lba = Lab = −L12 + L2 cos 2(θpd − π

3 )Lcb = Lbc = −L1

2 + L2 cos 2θpdLca = Lac = −L1

2 + L2 cos 2(θpd + π3 )

(A.56)

where:

L1 =N2

8(Pd + Pq)

L2 =N2

8(Pd − Pq)

N � total number of turns in sinusoidal winding= 2na

Ll � the leakage inductance of each phase

Appendix B

Winding Functions

B.1 Introduction

The computation of self and mutual inductances for machines is crucial whenone is trying to develop a set of dynamic equations for the performance of amachine. A technique using the concept of winding functions has been shownto be useful for developing inductance expressions for complex machine andwinding configurations [1, 7]. Because of the difficulty of obtaining the originalsource for this technique [3], this appendix will attempt to develop the basis ofthe technique for some very simple winding configurations, and then show thatthe basic idea can be extended to more complex geometries.

The winding function based expression for the mutual inductance betweentwo arbitrary windings , i and j, in a machine is:

Lij = µ0rl

∫ 2π

0

g−1(θ, θrm)Ni(θ, θrm)Nj(θ, θrm) dθ (B.1)

where:

r � radius of the circle of integration

l � length of the stack of the machine

g(θ, θrm) � airgap function

Ni(θ, θrm) � ‘i’ phase winding function

Nj(θ, θrm) � ‘j’ phase winding function

θ � mechanical angle around the machine

θrm � mechanical angle of the rotor

µ0 � permeability of free space

Remark 46 θrm is the angle of the rotor with respect to a reference. Thisangle is only relevant in relation to the calculation of inductance if the rotor hassaliency – that is the air gap function is not a constant with respect to θ.

Remark 47 The radius r is not obvious when we are dealing with singly ordoubly salient structures. The value must be chosen in such a way as to ensure

B.2 Ideal Sinusoidal Winding 122

n na a( ) sin

n nb b( ) sin( )

(Rad)

Figure B.1: Two pole sinusoidal winding layout

that the flux linking into the winding is correct, since the radius is a crucialcomponent in determining the dA areas when computing incremental fluxes.

The following sections will, by examples, develop an understanding of theterms in this equation and how they are derived for a particular machine.

B.2 Ideal Sinusoidal Winding

We shall begin by considering the standard sinusoidally distributed windings.The windings we will consider will be idealized ones, but the same technique canbe used for real windings that are distributed in slots. The situation that weare considering is the mutual inductance between two sinusoidally distributedwindings with the same pole number that are at some phase angle to each other.For simplicity we shall assume that the machine has a uniform air gap g and theiron of the machine has infinite relative permeability. This in turn implies thatone can consider that the mmf is all expended in driving flux across the machineair gaps. Figure B.1 shows the layout of the windings. We are assuming thatwe have a two pole configuration.

B.2.1 Conventional inductance calculation

The plan of attack is to firstly look at calculating the mutual inductance betweenthe windings using a conventional technique. Then we shall use a slightly differ-ent way of calculating the inductance, which will lead to the winding functionformulation.


In order to calculate the mutual inductance between the two windings wehave to calculate the flux that links the two windings. In this example we shallassume that winding ‘a’ has a current of ia amps and we are interested in theflux that links to winding ‘b’.

Figure B.2 shows the ‘a’ phase winding sitting on the stator of a machine.The mmf for winding ‘a’ can be computed by carrying out the Ampere’s lawline integral:

i =∮

H.dl (B.2)

around the path shown in Figure B.2, and realising that H = F/l, where F isthe mmf.

For some small increment dδ the amount of enclosed current is:

di = naia sin δ dδ (B.3)

where δ is an angle around the machine circumference.If we carry out integrate over the coil span then we are enclosing the current

inside the path of integration . To calculate this current we need to add togetherthe incremental currents around the periphery of the machine for the length ofthe enclosed path. This gives the total enclosed ampere turns of the winding atsome position θ. Therefore the expression is:

iT (θ) =∫ θ+π

θ

di

=∫ θ+π

θ

naia sin δ dδ

= 2naia cos θ

= 2Fa cos θ (B.4)

For a typical machine we assume that the magnetic materials have infinitepermeability. This in turn means that no magnetic field intensity is requiredto force flux through the iron. Consequently all of the H, and therefore F , isexpended to force flux across the airgap.

The mmf for the ‘a’ phase can be found from iT (θ) by realising that theintegration path in Figure B.2 crossing two air gaps, therefore half the mmfcalculated is used to cross each air gap. Therefore the mmf expression for the‘a’ phase becomes:

Fa(θ) = Fa cos θ (B.5)

where Fa(θ) = naia.Now that we have an expression for the mmf of the ‘a’ phase we can compute

the flux density at some arbitrary angle θ around the periphery of the machineusing the basic fact that:

B = µoH (B.6)


Figure B.2: Calculation of the MMF for a sinusoidal winding.

and H = F/l we can write:

Ba(θ) =µogFa(θ)

=µoFag

cos θ (B.7)

In order to find the total flux linkage to winding ‘b’ we need to add up allthe incremental fluxes over a complete coil span of winding ‘b’.

The incremental flux linking any incremental area of the machine is at someangle θ is:

dφ = Ba(θ)dA (B.8)

Now if the radius of the area in question is r then:

dA = rl dθ (B.9)

where dθ is an incremental angle. Therefore:

dφ = Ba(θ)rl dθ (B.10)

To get the total flux linking a single turn of the ‘b’ phase whose coil side startsat γ we need to integrate up the incremental fluxes for the total coil span of the


coil. Stated mathematically this is:

φab(γ) =∫ γ+π

γ

dφ

=µorlFag

∫ γ+π

γ

cos θ dθ

= −2µorlFag

sinγ (B.11)

In order to find the total flux linkage for the whole phase we need to integrateup the flux for each coil of the phase.

The number of turns whose coil sides start in the ‘b’ phase at some particularangle θ around the machine is:

nb(θ) dθ = nb sin(θ − α) dθ (B.12)

Therefore the flux linking the coils starting at θ is:

nb(θ)φab(θ) dθ (B.13)

Finally in order to get the total flux linkage for the entire phase we need tointegrate up the flux linking the coils for each value of θ for the entire phasespan. Therefore the expression for the flux linking from the ‘a’ phase to the ‘b’phase is:

λab =∫ α+π

α

nb(θ)φ(θ) dθ

= −2µorlFanbg

∫ α+π

α

sin(θ − α) sin θ dθ

∴ λab = −µorlFanbπg

cosα (B.14)

B.2.2 Alternative inductance calculation

In this section we shall carry out the same inductance calculation as above, butin this case using the turns function for the ‘b’ phase. One of the main differ-ences that arise from this approach is that the integration is carried out over2π mechanical radians. The only real difference between these two techniquesis that we calculate the total flux linkage to winding ‘b’ for an incrementalarea dA at some angle of θ, and then add up all the areas for the whole wind-ing. At any particular θ there are in general a number of turns that are beinglinked. In the previous method we calculated the flux linking a whole singlecoil of phase ‘b’ with its coil side starting at some angle θ, and then addedup the flux for all the coils. The difference between the two techniques can beseen in Figure B.3, where the first technique finds the flux linking a group ofcoils whose coil sides start at θ1, and the second technique finds the incremen-tal flux linking all the coils at position θ2. This is represented in the diagramas B(θ2)

(∫ θ2o n sin δ dδ

)rl dθ where the

∫ θ2o n sin δ dδ term corresponds to the

cumulative number of turns to the θ2 point along the winding.


Figure B.3: Different methods of calculating the flux linking a coil.

Therefore general cumulative coil number expression for any arbitrary angleθ is:

C(θ) =∫ θ

α

n sin(θ − α) dθ= n [1− cos(θ − α)] (B.15)

This function is plotted in Figure B.4. Notice that the maximum number ofcumulative turns occurs at α + π (i.e. after the pole pitch of the winding) asintuition would tell us.

In order to calculate the total mutual flux between the ‘a’ and ‘b’ phasesusing this approach we have to add up the incremental mutual flux linkagesfor the cumulative number of ‘b’ phase coils over the 2π radian span of the ‘b’phase. Using (B.15) the ‘b’ phase cumulative number of coils at any position θis:

Cb(θ) = nb [1− cos(θ − α)] (B.16)

Clearly the total flux linking the ‘b’ winding is:

λab =∫ 2π

0

dλ

=∫ 2π

0

Cb(θ) dφ (B.17)

Using (B.7) and (B.8) we can expand the dφ expression so that we get:

λab =µ0rl

g

∫ 2π

0

Cb(θ)Fa(θ) dθ (B.18)


2

2n

Figure B.4: Cumulative number of turns for a sinusoidally distributed winding.

This can be further expanded by substituting for Cb(θ) and Fa(θ) to give:

λab =µ0rlFanb

g

∫ 2π

0

[1− cos(θ − α)] cos θ dθ

=µ0rlFanb

g

{∫ 2π

0

cos θ dθ −∫ 2π

0

cos(θ − α) cos θ dθ}

= −µ0rlFanbg

∫ 2π

0

cos(θ − α) cos θ dθ

= −µ0rliag

∫ 2π

0

nb cos(θ − α)na cos θ dθ (B.19)

Notice in (B.19) that the terms inside the integral are Fb(θ)/ib and Fa(θ)/ia.These terms are called the winding functions for the winding and are usuallygiven the notation below:

Na(θ) � Fa(θ)/ia (B.20)

Nb(θ) � Fb(θ)/ia (B.21)

allowing the above expression to be written as:

λab = −µ0rliag

∫ 2π

0

Nb(θ)Na(θ) dθ (B.22)

As a test we can substitute the appropriate expressions for our examplesystem into the above and we get:

λab = −µ0rliag

∫ 2π

0

nb cos(θ − α)na cos θ dθ

= −µ0rliaπnanbg

cosα

= −µ0rlFanbπ

gcosα (B.23)

B.3 Non-sinusoidal winding 128

which is the same as (B.14).From (B.22) expression it is simple to see that the mutual inductance be-

tween these two windings is:

Lab =|λab|ia

= µ0rl

∫ 2π

0

g−1Nb(θ)Na(θ) dθ (B.24)

The situation used for the above development is very simple. Consequently(B.24) is a simpler function form compared to what can happen in more complexmachines. For example, one of the windings can be located on a rotor, andtherefore the winding function for this winding can become a function of therotor position as well as θ. Also in the above development the air gap g has beenassumed to be constant. However, if saliency is present in a machine structurethen g will be a function of θ. The situation is even more complex if the saliencyis on the rotor, as g then also becomes a function of the rotor angle as well.Therefore, the general form of the mutual inductance expression becomes:

Lab = µ0rl

∫ 2π

0

g−1(θ, θrm)Na(θ, θrm)Nb(θ, θrm) dθ (B.25)

B.3 Non-sinusoidal winding

As an example of the utility of this technique for determining the mutual induc-tances we shall calculate the mutual inductance between two windings that havedifferent pole pitches. Furthermore the distribution of the windings around themechanical periphery of the machine is not symmetric. The case is simplified byconsidering the two windings to be concentrated windings and to be mounted oncylindrical magnetic structures. The usual infinite iron permeability assumptionis used, and the air gap is assumed to be constant.

The approach taken will be to firstly calculate the mutual inductance us-ing basic principles, and then to calculate the inductances using the windingfunction technique.

B.3.1 Inductance Using Basic Principles

Figure B.5 is a developed diagram of the winding arrangement for the machine.The plots show the turns function for the ‘b’ winding and the mmf as a functionof θ for the ‘a’ winding. Figure B.6 shows the physical arrangement of thewindings.

Remark 48 The span of the coil defined in Figure B.5 is arbitrary. There aretwo possible definitions for coil span for both coils, and it does not matter whichis chosen.

The mmf diagram for the ‘a’ winding can be determined by using the mag-netic circuit concept. The current in the winding produces a flux φ across an airgap of length g an area rlτa, where r is the circumference of the machine, andl is the axial length as in previous derivations. This same flux has to cross thereturn air gap, which is the same length but has an area of rl(2π− τa). Clearlythe first air gap and the second air gaps have different reluctances due to the


b

2

2

nb

Fa ( )

a

2

2a

a an i

aa an i

2

Figure B.5: Turns function and mmf distribution for two fractional pitch wind-ings.

a

b

Figure B.6: Physical layout of the non-sinusoidal winding.


different areas of the air gaps. This can be clearly seen from the definition ofreluctance:

R =g

µ0A(B.26)

where A is the area orthogonal to the flux.Using these ideas we can write the following expression for the flux produced

by the ‘a’ winding:

φ =naia

Rτa +Rτa

(B.27)

where Rτa and Rτa correspond to the reluctance of the τa and 2π − τa air gapsthat the flux has to cross. Since magnetic circuits obey the analogous relation-ships as conventional electrical circuits, we can write the following expressionsfor the mmf required for the flux to bridge each of the air gaps:

F τaa =

Rτa

Rτa +Rτa

naia (B.28)

F τaa =

Rτa

Rτa +Rτa

naia (B.29)

Now:

Rτa =g

µ0rlτa(B.30)

Rτa =g

µ0rl(2π − τa) (B.31)

therefore:

F τaa =

gµ0rlτa

gµ0rlτa

+ gµ0rl(2π−τa)

naia

=(2π − τa

2π

)naia (B.32)

and

F τaa =

gµ0rl(2π−τa)

gµ0rlτa

+ gµ0rl(2π−τa)

naia

=τa2πnaia (B.33)

Clearly Fa = F τaa + F τa

a = naia. The negative sign results in Figure B.5 due tothe convention that flux flowing out of the rotor is due to positive mmf.

Given this information we can now write down the flux density produced bythe ‘a’ winding in the air gap corresponding to τa:

Bτaa = µ0

F τaa

g

=µ0

g

(2π − τa

2π

)naia (B.34)


This flux density links the ‘b’ phase winding from α to τa, therefore the fluxthrough the winding due to this flux density is:

φτaab =∫ τa

α

µ0

g

(2π − τa

2π

)naiarl dθ

=µ0

g

(2π − τa

2π

)naiarl(τa − α) (B.35)

Similarly the remaining section of the ‘b’ phase is linked by the return fluxfrom the ‘a’ winding. The return flux density is:

Bτaa = µ0

F τaa

g

= −µ0

g

( τa2π

)naia (B.36)

Therefore the flux component linking to the ‘b’ winding is:

φτaab =∫ τb+α

τa

−µ0

g

( τa2π

)naiarl dθ

= −µ0

g

( τa2π

)naiarl(τb + α− τa) (B.37)

The total flux linkage from winding ‘a’ to ‘b’ is therefore:

λab = nb(φτaab + φτaab)

=µ0nbnaiarl

g

[(2π − τa

2π

)(τa − α)−

( τa2π

)(τb + α− τa)

]

=µ0nbnaiarl

g

[2π(τa − α)− τaτb

2π

](B.38)

and consequently the mutual inductance is:

Lab =µ0nbnarl

g

[2π(τa − α) − τaτb

2π

](B.39)

B.3.2 Inductance Using Winding Functions

We shall now compute the inductance using (B.24). The first step is to findthe winding functions for the windings. This is very simple as we have alreadycalculated the mmf waveform for the ‘a’ winding. Its winding function is sim-ply this divided by ia. The winding function for the ‘b’ phase can be writtenby inspection because of the similarity to the ‘a’ winding. Therefore the twowinding functions are:

Na(θ) ={ (

2π−τa2π

)na for 0 ≤ θ ≤ τa

− (τa2π

)na for 0 ≥ θ ≥ τa (B.40)

Nb(θ) ={ (

2π−τb2π

)nb for α ≤ θ ≤ τb + α

− (τb2π

)nb for α ≥ θ ≥ τb + α (B.41)

B.4 Flux Linkage Expression 132

Breaking (B.24) into a piecewise continuous integral and substituting theappropriate values for Na(θ) and Nb(θ) in each of the integrals we can write:

λab =µ0nanbrl

g

{∫ α

0

(2π − τa

2π

)(− τb2π

)dθ +

∫ τa

α

(2π − τa

2π

)(2π − τb

2π

)dθ

+∫ τb+α

τa

(− τa2π

)(2π − τb

2π

)dθ +

∫ 2π

τb+α

(− τa2π

)(− τb2π

)dθ

}(B.42)

∴ λab =µ0nanbrl

g

{(2π − τa

2π

)(− τb2π

)α+

(2π − τa

2π

)(2π − τb

2π

)(τa − α)

+(− τa2π

)(2π − τb

2π

)(τb + α− τa) +

(− τa2π

)(− τb2π

)(2π − τb − τa)

}(B.43)

After some simplification and dividing by ia we obtain the following expres-sion for the mutual inductance:

Lab =µ0nbnarl

g

[2π(τa − α) − τaτb

2π

](B.44)

which is exactly the same as that derived using the basic principles approach.

B.4 Flux Linkage Expression

There is another useful expression that is related to the general inductanceexpression, and that is the general flux linkage expression. This expression canbe simply derived from (B.1) by using the following relations:

B(θ, θrm) = µ0H and (B.45)

F (θ, θrm) =∮

H.dl = Hg(θ, θrm) (B.46)

therefore (B.46) can be written as:

F (θ, θrm) =g(θ, θrm)B(θ, θrm)

µ0(B.47)

If (B.1) is multiplied by ii (or ij) then we and up with an expression of theform:

Lijii = λij = µ0rl

∫ 2π

0

g−1(θ, θrm)Fi(θ, θrm)Nj(θ, θrm) dθ (B.48)

and substituting (B.47) for Fi(θ, θrm) in this we can then write the followingfor the general mutual flux linkage expression:

λij = rl∫ 2π

0

Bi(θ, θrm)Nj(θ, θrm) dθ (B.49)

B.5 A Note on Winding Functions for Multi-pole Machines 133

B.5 A Note on Winding Functions for Multi-pole Machines

Consider a generic winding with the following conductor distribution:

n(θ) = npk sin ppθ (B.50)

where:

npk � the peak conductor density [conductors/rad]

θ � periphery angle around machine [mech-rad]

pp � pole pairs of the winding.

Assuming that the poles are series connected, the total number of turns inthe winding is equal to the number of turns in a pole pair multiplied by thenumber of pole pairs. We can find the number of turns in a pole pair by addingup the number of conductors in π electrical radians of the winding,which is π/pmechanical radians. We can therefore write the total turns for a phase of thewinding as follows:

N = p∫ π

pp

0

npk sin ppθdθ

= p[−npkpp

cos ppθ]πp

0

= 2npk (B.51)

Now let use consider the mmf for a winding with the distribution as in (B.50).Assume that the winding is carrying a current of i Amp in each conductor.Therefore we can see that for an angle dθ we have the following total current inan element of the winding at some angle θ :

diT = npki sin ppθdθ (B.52)

To find the mmf produced by the winding we employ Ampere’s Law and inte-grate to get the total current enclosed by a coil span at some angle γ :

FT =∫ γ+π

p

γ

npki sin ppθdθ

=2npkipp

cos ppθ

=Ni

ppcos pθ

= N ′i cospθ (B.53)

where:

N ′ � the turns/phase/pole pair = N/pp

B.6 Conclusion 134

FT is the total mmf, which is expended across two air gaps in a machine.Therefore the mmf/air gap is:

F =FT2

=N ′i2

cos ppθ (B.54)

=Ni

2ppcos ppθ =

2npki2pp

cos ppθ

=npki

ppcos ppθ (B.55)

The winding function is defined as the mmf/amp, therefore the windingfunction for this multi-pole winding becomes:

NA(θ) =N ′

2cos ppθ (B.56)

=npkpp

cos ppθ (B.57)

Therefore we can express the amplitude of the winding function in terms of theturns/phase/pole pair, or alternatively in the peak conductors/rad/pole pair.

B.6 Conclusion

This appendix has attempted to show the basis for the use of winding functionsfor the calculation of the mutual inductance between windings. Two exampleshave been used to achieve this, one involving idealised sinusoidal windings, andthe other an unusual set of windings. In a real machine one does not havepure sinusoidally distributed windings. In these situations one can computethe winding functions by simply evaluating the mmf waveforms that are pro-duced by the real windings. These mmf waveforms will contain all the windingharmonics, therefore the inductances calculated will be accurate subject to theinfinite permeability assumption and the modelling of the air gap function.

Bibliography

[1] F. Liang, L. Xu, and T. Lipo. D-q analysis of a variable speed doubly ACexcited reluctance motor. Electric Machines and Power Systems, 19(2):125–138, March 1991.

[2] D. O’Kelly and S. Simmons. Introduction to Generalized Electrical MachineTheory. McGraw Hill, England, 1968.

[3] N. Schmidt and D. Novotny. Introductory Electro-Mechanics. Ronald Press:New York, 1965.

[4] G. Slemon. Electric Machines and Drives. Addison-Wesley, 1992.

[5] D. Staton, W. Soong, and T. Miller. Unified theory of torque productionin switched reluctance and synchronous reluctance motors. IEEE Trans. onIndustry Applications, IA-31(2):329–337, 1995.

[6] P. Vas. Vector Control of AC Machines. Oxford University Press, 1990.

[7] L. Xu, F. Liang, and T. Lipo. Transient model of a doubly excited reluctancemotor. IEEE Trans. on Energy Conversion, 6(1):126–133, March 1991.

intro to vector control of induction mc for electrical engg

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