Chapter – 3

Reference Frame

Theory

Prof. B. S. Sree Shailan

• Imagination is more important than knowledge.

- Albert Einstein

• It is not work that kills men, it is worry. Work is

healthy; You can hardly put more on a man than

he can bear. But worry is rust upon the blade.

It is not movement that destroys the machine,

but friction.

- Henry Ward Beecher

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Topics to be covered• Real time model of a two phase

induction machine.

• Transformation to obtain constant

matrices.

• Three phase to two phase

transformation.

• Power equivalence.

Book: Electric Motor Drives by R. Krishnan

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Introduction• Power of Reference Frame Theory:

Eliminates Rotor Position Dependence –

Inductances and Capacitances

Transforms Nonlinear Systems to Linear

Systems for Certain Cases

Fundamental Tool For Rigorous Development of

Equivalent Circuits

Can Be Used to Make AC Quantities Become DC

Quantities

Framework of Most Controllers

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History of Reference Frame Theory

• 1929: Park’s Transformation

Synchronous Machine; Rotor Reference Frame

• 1938: Stanley

Induction Machine; Stationary Reference Frame

• 1951: Kron

Induction Machine; Synchronous Reference Frame

• 1957: Brereton

Induction Machine; Rotor Reference Frame

• 1965: Krause

Arbitrary Reference Frame

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Real-Time model of 2-ф Induction M/c

• Assumptions:

– Uniform air gap

– Balanced rotor and stator windings with

sinusoidal distributed mmf

– Inductance w. r. t. rotor position is sinusoidal

– Saturation and parameter changes are

neglected

• Windings are displaced in space by 900 electrical

and rotor winding α is at an angle θr from the

stator d-axis winding.

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• Number of turns per

phase in the stator and

rotor windings are

respectively T1 & T2.

Stator and rotor windings of a two-

phase induction motor

• Terminal voltages can

be expressed as the

sum of voltage drops in

resistances & rate of

change of flux linkages.

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• Under the assumption of uniform air gap, the

self inductances are independent of angular

positions & hence they are constants.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

• :

qs q qs qq qs qd ds q q

ds dq qs d ds dd ds d d

q qs d ds

q qs d ds

v R i L i L i L i L i

v L i R i L i L i L i

v L i L i R i L i L i

v L i L i L i R i

The equations are

L i

α α β β

α α β β

α α α α α αα α αβ β

β β β βα α β β ββ β

ρ ρ ρ ρ

ρ ρ ρ ρ

ρ ρ ρ ρ

ρ ρ ρ ρ

= + + + +

= + + + +

= + + + +

= + + + +

; rr dd qq sL L L L L Lαα ββ= = = =BSS

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• Mutual inductances between the stator

windings and between the rotor windings are

zero as they are displaced by 900. Thus,

Lαβ = Lβα = 0 and Ldq = Lqd = 0

• Mutual inductances between the stator and

rotor windings are a function of the rotor

position θr. Thus, with Lsr as the peak value of

the mutual inductance between a stator and a

rotor winding, we have,

cos ; sin

sin ; cos

d d sr r d d sr r

q q sr r q q sr r

L L L L L L

L L L L L L

α α β β

α α β β

θ θ

θ θ

= = = =

= = = = −

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• Solutions of the above equations are time-

consuming.

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

sin cos

co

•

s sin

sin cos

cos sin

:

qs s s qs sr r sr r

ds s s ds sr r sr r

sr qs r sr ds r rr rr

sr qs r sr ds r rr rr

The res

v

ulting equ

R L i L i L i

v R L i L i L i

v L i L i

ations are as follows

R L i

v L i L i R L i

α β

α β

α α

β β

ρ ρ θ ρ θ

ρ ρ θ ρ θ

ρ θ ρ θ ρ

ρ θ ρ θ ρ

= + + −

= + + +

= + + +

= − + + +

s q d rrwhere R R R and R R Rα β= = = =

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Transformation to Constant Matrices

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Transformation of

actual to fictitious

rotor variables

• Above transformation is valid for voltages,

currents and flux-linkages in a machine.

• Further, it is found that Tαβ = Tαβ-1 which is

required for inverse transformation.

• Tαβ is both orthogonal and symmetric.

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12 cos sin . .,

sin cos

drr r r

dqrr

qrr r r

i ii e i T i

i i

α

αβ αββ

θ θ

θ θ

= = −

• Fictitious rotor currents iqrr and idrr are the sum

of the projections of iα and iβ on the q and d

axis, respectively as given below:

• Rotor equations need to be referred to the

stator so that the stator and rotor d & q axes

windings gets physically connected.

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. .

. .

0 0

0 0

s s srqs qs

s s srds ds

qrr sr sr r rr rr rr r qrr

drr drrsr r sr rr r rr rr

R L Lv i

R L Lv i

v L L R L L i

v iL L L R L

ρ ρ

ρ ρ

ρ θ ρ θ

θ ρ θ ρ

+ + = − + − +

.

θ θr r

where is the time derivative of

• Applying the transformation to the α and β

rotor winding currents and voltages, we get,

2 2

1 1

2 2

;

•

; ;

;

/

/

:

r rr r rr qr qrr dr drr

qrr drrqr dr

w

w

R a R L a L v av v av

i ii i where

a a

k Tstator effective turns phasea

rotor effective turns ph

Steps involved

ase

a

k

r

T

e

= = = =

= =

= =

2

1 1 2; ;

• ,

: m sr m sr

Also Magnetizing and Mutual inductances

L T L TT ar L La e ∝ ∝ ∴ =

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• Note that the impedance matrix has constant

inductance terms and is no longer dependent

on the rotor position.

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. .

. .

0 0

0 0

s s mqs qs

s s mds ds

qr m m r r r r r qr

dr drm r m r r r r

system equation of induction m

R L Lv i

R L Lv i

v L L R L

ot

L i

v iL L L

o

R

r

L

ρ ρ

ρ ρ

ρ θ ρ θ

θ ρ θ ρ

+ + = − + − +

→

• Thus, the machine equations referred to the

stator are obtained as:

3-phase to 2-phase Transformation

• A dynamic model for the 3-phase inductionmachine can be derived from the two-phasemachine if the equivalence between three andtwo phases is established.

• Equivalence is based on the equality of the mmfproduced in the two-phase and three-phasewindings and equal current magnitude.

• Assuming that each of the 3-phase windingshas T1 turns/phase and equal currentmagnitudes, the 2-phase windings will have3T1/2 turns/phase for mmf equality.

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Two- and Three-phase stator windings

• Current i0 represents the imbalances in the a, b

and c phase currents and can recognized as the

zero-sequence component of the current.

0

• 0

2 2cos cos cos

3 3

2 2 2

sin sin sin3 3 3

1 1 1

2 2 2

:

c c c

qs as

ds c c c bs

cs

Relationship between dq an

i

d abc currents are

i

i i

i i

π πθ θ θ

π πθ θ θ

− +

= − +

[ ]0 , qd abc abc

i TT s ihu =

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• Transformation from two-phase currents tothree-phase currents can be obtained as

[ ]

[ ]

1

0

1

cos sin 1

2 2 cos sin 1

3 3

2 2cos sin

13 3

c c

abc c

abc abc q

c

c c

d

wher

i i

e

T

T

θ θ

π πθ θ

π πθ θ

−

−

= − −

+ +

=

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• This transformation can also be thought of as a

transformation from three (abc) axes to three

new (qd0) axes.

• Unbalances in the abc variables requires three

variables such as dq0.

• Balanced abc variables (equal phase

displacement and magnitude) requires two

independent dq variables and the third is a

dependent variable obtained as the negative

sum of two independent variables.BSS

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• Note that only leakage inductances and phase

resistances influence the zero-sequence

voltages and currents while the dq component

variables are influenced by the self and mutual

inductances and phase resistances.

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• Four system equations given by vqs, vds, vqr and

vdr for balanced conditions.

( ) ( )0 0 0 0 and ρ ρ= + = +s s ls s r r lr r

v R L i v R L i

• For unbalanced conditions, two more system

equations, one each for stator & rotor zero-

sequence voltages emerges. They are given by

Stanley’s Model

• Also known as stator-reference-frames model.

• Here, θc = 0 and hence, transformation matrix is

1 11

2 2

2 3 30 0

3 2 2

1 1 1

2 2 2

− −

= −

S

abcT for abc to qd

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An exampleAn induction motor has the following parameters:

5hp, 200V, 3-phase, 60Hz, 4-pole, star connected;

Rs = 0.277Ω; Rr = 0.183Ω; Lm = 0.0538H

Ls = 0.0553H; Lr = 0.056H

Effective stator to rotor turns ratio a = 3.

The motor is supplied with its rated and balanced

voltages. Find the q and d axes steady state

voltages, currents and phase currents Iqrr, Idrr, Iα &

Iβ when the rotor is locked. Use stator-reference-

frames model.

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• The applied phase voltages are as follows:

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2002 sin 163.3sin

3

2 2163.3sin & 163.3sin

3 3

as s s

bs s cs s

v t t

v t v t

ω ω

π πω ω

= × =

= − = +

0

• :

qs as

S

ds abc bs

cs

The d and q axes voltage

v v

v T vs ar

v v

e

=

( )2 1

, 3 2

qs as bs csHence v v v v

= − +

• For a balanced 3-phase input, vas + vbs + vcs = 0

• Substituting for vbs and vcs in terms of vas yields

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252 3

3 2qs as asv v v

= =

( ) 0

1 0

3• , ds cs bsSimilar v v v al ndy v= − =

( )

0

0

, 163.3sin 163.3 0 163.3

1 163.3cos

3

163.3 90 163.3

qs as s

ds cs bs s

Thus v v t V

and v v v t

j V

ω

ω

= = = =

= − =

= =

• The rotor is locked and hence

• For steady state evaluation,

ρ = jωs = j2πfs = j2π ×60 = j377 rad/sec

• The system equations in steady state are

• Note that the rotor windings are short-circuited

and hence rotor voltages are zero.

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.

0rθ =

0 0

0 0

0 00

0 00

ρ ρ

ρ ρ

ρ ρ

ρ ρ

+ + = +

+

s s m qsqs

s s m dsds

m r r qr

m r r dr

R L L iv

R L L iv

L R L i

L R L i

• Numerical values for the parameters and variablesare substituted to solve for currents.

• Note that the stator and rotor currents aredisplaced by 900 among themselves as expected ina 2-phase machine.

• The zero-sequence currents are zero because zerosequence voltages are nonexistent with balancedsupply voltages.

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0

qs

0

ds

0

qr

0

dr

i =35.37-j108.18=113.81 -71.9

i =108.18+j35.37=113.8118.1

i =-34.88+j103.63=109.34108.6

i =-103.63-j34.88=109.34 -161

.

:

4

The currents are

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0

0

0

0

1 0 1 113.8 71.9

1/ 2 3 / 2 1 113.8168.1

113.8 48.11/ 2 3 / 2 1

• :

as qs

bs ds

cs

i i

i i

i

The phase currents are

i

− = − − = −

0

0

i 328.02108.6

i 328.0

•

2 161 4

:

.

qrr qr

drr dr

Rotor currents are ai

ai

= =

= = −

• The α and β currents, assuming θr = 0 are

0

0

cos sin 328.02 161.4

sin cos 328.02108.6

drrr r

qrrr r

ii

ii

α

β

θ θ

θ θ

− = = − −

2) Derive the steady-state equivalent circuit from

the dynamic equations of the induction motor.

• This can be derived by substituting for the

d- and q-axes voltages in system equations.

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2 2sin ; sin & sin

3 3as m s bs m s cs m sv V t v V t v V t

π πω ω ω

= = − = +

0

0

0

sin 0

cos 90

0 0

,

qs as m s m

S

ds abc bs m s m

cs

v v V t V

v T v V t V

v v

Thus

ω

ω

= = =

In steady State, =j 0s qr drand v vρ ω = =

• Substituting these into system equations yields,

• The input voltages are in quadrature so the

currents have to be in quadrature because the

system in steady state is linear and they can be

represented as ids = jiqs and idr = jiqr

• Substituting these equations into the above

equation and considering only one stator and

rotor equation with rms values yields,

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0 0

0 0

0

0

ω ω

ω ω

ω ω ω ω

ω ω ω ω

+ + = − + −

+

s s s s m qsm

s s s s m dsm

s m r m r s r r r qr

r m s m r r r s r dr

R j L j L iV

R j L j L ijV

j L L R j L L i

L j L L R j L i

• Rearrange the rotor equation with the aid of

ωsl = ωs– ωr = Sωs. Thus, rotor equation is

• Rotor and stator equations, when combined

give the equivalent circuit with the magnetizing

current as the sum of stator and rotor currents.

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( )

( ) ( )( )0

s s s s s s m r

m s r s r s r r r

V R j L I j L I

jL I R j L I

ω ω

ω ω ω ω

= + +

= − + + −

0 rm s s s r r

RjL I j L I

Sω ω

= + +

, s m ls r m lrA L L L and L Ll o Ls = + = +

Power Equivalence

• 3

t

i abc abc as as bs bs cs cs

phase instantaneous p

P v i v i v i

ower inp i

i

ut s

v= +

−

= +

[ ]( ) [ ]

[ ]( ) [ ]

( )( ) ( )

1 1

0 0

1 1

0 0

0 0

3

• , ,

2 12

− −

− −

−

=

=

= + + →

t

i abc qd abc qd

tt

qd abc abc qd

qs qs ds ds

P T v T i

v T T

On tran

i

v i v

sforming to q a

i v

nd

i

d axes we get

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• Zero-sequence current does not exist for a

balanced three-phase machine.

• The model development has so far kept the

d- and q-axes stationary with respect to the

stator. These axes or frames are known as

reference frames.

• The speed of the reference frames is arbitrary

in an arbitrary reference frame.

• The input power given by equation (1) remains

valid for all occasions.

( )3,

2= +

i qs qs ds dsThus P v i v i

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Assignment1) Discuss the suitability of stationary reference

frames for the induction motor in power

system studies.

2) Explain the method of obtaining the constant

matrices for transformation on to the fictitious

q and d axes for a 2-phase induction motor.

3) Explain the need for 3-phase to 2-phase

transformation and the method of carrying out

such transformation with an example.

4) Derive the power equivalence in dq0 variables.

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