Dynamic Model of Induction Machine By Dr. Ungku Anisa Ungku Amirulddin Department of Electrical Power Engineering College of Engineering Dr. Ungku Anisa,

Dynamic Model of Induction MachineByDr. Ungku Anisa Ungku AmirulddinDepartment of Electrical Power EngineeringCollege of Engineering

Dr. Ungku Anisa, July 2008 1EEEB443 - Control & Drives

OutlineIntroductionThree-phase Dynamic ModelSpace Phasors of Motor VariablesThree-phase to Two-phase Transformation (Stationary)Two-phase (Stationary) Dynamic Model (in dsqs frame)Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq

frame)Voltage equationsTorque equation

Commonly-used Induction Motor ModelsStationary (Stator) Reference Frame ModelRotor Reference Frame ModelSynchronously Rotating Reference Frame Model

Equations in Flux LinkagesReferencesDr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 2

IntroductionPer phase equivalent circuit model only useful for analysing IM

performance in steady-stateall transients neglected during load and frequency variationsused in scalar control drives which do not require good transient

responseexample: drive systems for fans, blowers, compressors

Dynamic model used to observe dynamic (steady-state and transient) behaviour of IM since:Considers instantaneous effects of varying:

Voltages and currents Stator frequency Torque disturbances

machine is part of the feedback loop elements to control the dynamics of the drive system

High performance drive control schemes are based on dynamic model of IM

Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 3

IntroductionDynamic model – complex due to magnetic coupling

between stator phases and rotor phasesCoupling coefficients vary with rotor position and

rotor position vary with timeDynamic behavior of IM can be described by

differential equations with time varying coefficientsComplexity of dynamic model can be reduced by

employing space vector equations


Three-phase Dynamic ModelIM consists of three-phase windings spaced at 120 apartModel windings using simplified equivalent stator winding located on the magnetic axis of each phase.

5Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

a

b

b’c’

c

a’Simplified equivalent stator winding

ias

Magnetic axis of phase A

Magnetic axis of phase B

Magnetic axis of phase C

ics

ibs

Three-phase Dynamic Model

6Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Similar model is applied to represent the rotor ‘windings’

Rotor rotates at speed r

Rotor phase ‘a’ winding displaced from stator phase ‘a’ winding by angle r

stator, b

stator, c

stator, arotor, b

rotor, a

rotor, c

r

Three-phase Dynamic ModelVoltage equation for each stator phase:

Similarly, voltage equation for each rotor phase:

These equations can be written in a compact form.


dtdiRv

dtdiRv

dtdiRv

cscsscs

bsbssbs

asassas

dtdiRv

dtdiRv

dtdiRv

crcrrcr

brbrrbr

ararrar

'

'

'

Three-phase Dynamic ModelStator voltage equation (compact form):

Rotor voltage equation (compact form):

where:

abcs = stator flux linkage (flux linking stator windings )abcr = rotor flux linkage (flux that links rotor windings)


, , ,

cs

bs

as

cs

bs

as

cs

bs

as

i

i

i

v

v

v

abcsabcsabcs Ψiv

cr

br

ar

cr

br

ar

cr

br

ar

i

i

i

v

v

v

abcrabcrabcr Ψiv , ,

(1)

(2)

dtdRs abcsabcsabcs Ψiv

dtdRr abcrabcrabcr Ψiv '

Three-phase Dynamic ModelThe displacements between 3 phase stator (rotor) windings

are non-quadrature (i.e. not 90)Magnetic coupling exists between the 3 stator (or rotor)

phases, i.e. the flux linkage each stator (or rotor) phase is sum of:fluxes produced by the winding itselffluxes produced from the other two stator (or rotor) windingsfluxes produced by all three rotor (or stator) windings

Example: Flux linkage for stator phase ‘a’ is sum of:Fluxes produced by stator phase ‘a’ winding itselfFluxes produced by stator phase ‘b’ and stator phase ‘c’Fluxes produced by rotor phase ‘a’, ‘b’ and ‘c’


Three-phase Dynamic ModelIn general, the stator flux linkage vector:


rabcs,sabcs,abcs ΨΨΨ

cs

bs

as

csbcsacs

bcsbsabs

acsabsas

i

i

i

LLL

LLL

LLL

sabcs,Ψ

cr

br

ar

crcsbrcsarcs

crbsbrbsarbs

crasbrasaras

i

i

i

LLL

LLL

LLL

,,,

,,,

,,,

rabcs,Ψ

Flux linking stator winding due to stator currents

Flux linking stator winding due to rotor currents

(3)

Three-phase Dynamic ModelIn general, the rotor flux linkage vector:


Flux linking rotor winding due to rotor currents

Flux linking rotor winding due to stator currents

sabcr,rabcr,abcr ΨΨΨ

cr

br

ar

crbcracr

bcrbrabr

acrabrar

i

i

i

LLL

LLL

LLL

rabcr,Ψ

cs

bs

as

cscrbscrascr

csbrbsbrasbr

csarbsarasar

i

i

i

LLL

LLL

LLL

,,,

,,,

,,,

sabcr,Ψ

(4)

Three-phase Dynamic ModelSelf inductances in (3) and (4) consists of magnetising

inductance and leakage inductance:For stator:

For rotor:Due to symmetry in windings, mutual inductances between

stator phases in (3)(and rotor phases in (4)) can be written in terms of magnetising inductances:


lsmscsbsas LLLLL Stator leakage inductances

cs

bs

as

lsmsmsms

mslsms

ms

msmslsms

s,abcs

i

i

i

LL2

L

2

L2

LLL

2

L2

L

2

LLL

Note: Subsrcipt ‘s’ is replaced with ‘r’ for rotor phase leakage inductances, currents and flux linkage

lrmscrbrar LLLLL Rotor leakage inductances

Three-phase Dynamic ModelMutual inductances between the stator and rotor windings

depends on rotor position r:


cr

br

ar

rrr

rrr

rrr

mss

rr,abcs

iii

cos32cos3

2cos3

2coscos32cos

32cos3

2coscos

LNN

cs

bs

as

rrr

rrr

rrr

mss

rs,abcr

iii

cos32cos3

2cos3

2coscos32cos

32cos3

2coscos

LNN

Three-phase Dynamic ModelEquations (1) – (4):

completely describe dynamic characteristics of 3-phase IMconsists of 6 equations (3 for stator and 3 for rotor), i.e.

large number of equationsall equations are coupled to one another

Magnetic coupling complicates dynamic model in 3-phase!

Better to develop model based on space phasors:reduces number of equationseliminates magnetic coupling between phases


Space Phasors of Motor VariablesIf xa, xb, and xc are the 3-phase IM quantities, whereby:

The space phasor in the 3-phase system is obtained from the vectorial sum of the 3-phase quantities, i.e.:

(5)

x is called the space phasor or complex space vector

Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

cba

j

c

j

ba xaaxxexexx 23

4

3

2

3

2

3

2x

, where a = ej2/3

15

3

4cosˆ

32cosˆ

cosˆ

txx

txx

txx

c

b

a

Three-phase to Two-phase Transformation (Stationary)Any three-phase machine can be represented by an

equivalent two-phase machine using Park’s transformation

16

Three-phaseTwo-phase equivalent

There is magnetic coupling between phases

There is NO magnetic coupling between phases (due to 90 angle between phases)


easier way to obtain dynamics of IM

r

stator, qs

rotor,

stator, ds

rotatingr

Two-phase equivalent

rotor,

Three-phase to Two-phase Transformation (Stationary)Dynamic model of IM usually obtained using the two-phase

equivalent machinestator, b

rotor, b rotor, a

rotor, cstator, c

stator, a

r

rrotating

Three-phase


Three-phase to Two-phase Transformation (Stationary)All 3-phase system quantities have to be transformed

to 2-phase system quantitiesEquivalence between the two systems is based on

the equality of mmf produced and current magnitudes, i.e.:MMF produced by 2-phase system = MMF of 3-phase

systemcurrent magnitude of 2-phase system = current of 3-phase

systemThe use of space phasors enables the transformations

from 3-phase to 2-phase system.


Three-phase to Two-phase Transformation (Stationary)In the stationary 2-phase system, the space phasor is defined

as : (6)

The space phasor in the 2-phase system must equal that in the 3-phase system.

Hence, by comparing (5) and (6):


sq

sd xx jx

19

cbacba

sd xxxxaaxxx

2

1

2

1

3

2

3

2RexRe 2

cbcbasq xxxaaxxx

3

1

3

2ImxIm 2

Zero-sequence components, which may or may not be present.

Three-phase to Two-phase Transformation (Stationary)The abc dsqs transformation is given by:

(7)

Under balanced conditions, the zero-sequence component adds to zero, i.e.:


c

b

a

o

sq

sd

x

x

x

x

x

x

31

31

31

31

31

31

31

32

0

20

0 cbao xxxx

Three-phase to Two-phase Transformation (Stationary)Assuming balanced conditions, the abc dsqs transformation:

(8)The inverse transform (dsqs abc transformation) is given by:

(9) where:

(10)


abcabcsdq xTx

sdq

1abcabc xTx

3

13

1

00

0

1abcT

23

23

21

211

01

Tabc

Three-phase to Two-phase Transformation (Stationary)Transformation equations (8) and (9) apply to all 3-phase

quantities of the IM (i.e. voltages, current and flux linkages)Transformation matrices (Tabc and Tabc

-1) given by (10) causes the space phasor magnitude to be equal to peak value of the phase quantities, i.e.:

This is one of many abc dsqs and dsqs abc transformation matrices in literature, eg.: space phasor magnitude to be equal to 1.5 times peak value of the

phase quantities ( ) space phasor magnitude to be equal to rms value of the phase

quantities ( )


x̂x

x̂5.1x

2ˆx x

SPACE VECTORS


Space vector representation of the mmf distribution in an AC machine created by balanced positive-sequence three-phase sinusoidal currents. Each of the ABC (RGB) space vectors pulsates along its respective axis. The resultant vector (in black), of 1.5 magnitude, rotates at the excitation frequency.

Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html

SPACE VECTORSThis animation shows the motion of space vectors for the case of a balanced three-phase sinusoidal signal: fA = cos(ωt), fB = cos(ωt-α), fC = cos(ωt+α) where α = 2π/3.

The corresponding space vector is obtained from fR = (fA + γ fB + γ2 fC) = ejωt where γ = ejα.



Example 1 - Space Phasor & 3-2 phase transformationAn induction motor has the following parameters:


Parameter Symbol Value

Rated power Prat 30 hp (22.4 kW)

Stator connection Delta ()

No. of poles P 6

Rated stator phase voltage (rms)

Vs,rat 230 V

Rated stator phase current (rms)

Is,rat 39.5 A

Rated frequency frat 60 Hz

Rated speed nrat 1168 rpm

Example 1 - Space Phasor & 3-2 phase transformation (contd.)Assuming that the motor is operating under rated conditions,

with stator and rotor current phasors of:

Calculate the values the following stator current values at time t = 0:

ias, ibs and ics (3-phase stator phase currents) isd

s and isqs (2-phase stator phase currents)

iar, ibr and icr (3-phase rotor phase currents) ird

s and irqs (2-phase rotor phase currents)

Show that the magnitude of isds and isq

s is equal to the peak stator phase current


A/ph 5.265.39 sI

A/ph 6.1734.36 rI

Two-phase (Stationary) Dynamic Model (in dsqs frame)From the three-phase dynamic model (eq. 1 and 2):

Applying the transformation given by (8):

Note: dsqs – stator equivalent two-phase winding - rotor equivalent two-phase winding


r

stator, qs

rotor,

stator, ds

rotatingrrotor,

dtdRs abcsabcsabcs Ψiv dtdRr abcrabcrabcr Ψiv '

dtdRsssdq

ssdq

ssdq Ψiv

dtdRr rrr Ψiv '

Two-phase (Stationary) Dynamic Model (in dsqs frame)The rotor winding rotates at

a speed r

Hence, need to transform the rotor quantities from the to the stationary dsqs frame.


r

stator, qs

rotor,

stator, ds

rotatingrrotor,

r

r

xr

qs

dsBring all rotor and stator quantities to be on the same axis!

Two-phase (Stationary) Dynamic Model - dsqs frame transformOn the frame:

On the dsqs frame:

The angle between the two frames is r


rrr xx jxxr

srq

srd

srdq xx jxxr

xrds

r

r

xr

qs

ds

xrqs

ds

r

r

xr

qs

xrxr

Two-phase (Stationary) Dynamic Model - dsqs frame transformTherefore:

More elegantly :

(full derivation dss

frm trnsform)Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 30

r

r

rr

rrsrq

srd

x

x

x

x

cossin

sincos

xrds

r

r

xr

qs

ds

xrqs

r

r

xr

qs

xrxr

r

ds

rrx cossrdx rrx sin

rrx sinsrqx rrx cos

(11)

rjs

rdqre xx (12)

Two-phase (Stationary) Dynamic Model (in dsqs frame)Two-phase dynamic model:

To transform rotor quantities from the dsqs frame, from equation (12):

Substituting these into the rotor voltage equation above..


Expressed in rotating frame

dtdRsssdq

ssdq

ssdq Ψiv

dtdRr rrr Ψiv '

,srdq

jr

re vv

,srdq

jr

re ii

srdq

jr

re ΨΨ

Hence:

Therefore, the two-phase dynamic model in the stationary dsqs frame:

s

rdqrsrdq

srdqr

srdq

srdq

jsrdq

jr

srdq

j

rrrr

jdtdR

dtedeRe

dtdR

rrr

ΨΨiv

Ψiv

Ψiv

'

'

'

Two-phase (Stationary) Dynamic Model (in dsqs frame)


Expressed in stationary frame

Expressed in rotating frame

(13)

(14)

dtdRsssdq

ssdq

ssdq Ψiv

srdq

srdq

srdq

srdq ΨΨiv rr dtdR j'

Two-phase (Stationary) Dynamic Model (in dsqs frame)The flux linkages in (13) and (14) are given by:

where and Note that equations (13)-(16) each consists of two equations.

One from equating real quantitiesOne from equating imaginary quantities

Final dynamic equations in the stationary dsqs frame is given by substituting (15) and (16) into (13) and (14) and separating the real and imaginary equations.


srdq

ssdq

ssdq iiΨ ms LL

srdq

ssdq

srdq iiΨ '

rm LL (15)

(16)

lsms LLL ''lrmr LLL

Two-phase (Stationary) Dynamic Model (in dsqs frame)Final dynamic equations in the stationary dsqs frame:

Lm = mutual inductanceLr’ = rotor self inductances referred to statorRr

’ = rotor resistance referred to statorLs = stator self inductancevrd, vrq, ird, irq are the rotor voltages and currents referred to statorS = derivative operatorDr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 34

(17)

srq

srd

ssq

ssd

rrrrmmr

rrrrmrm

mss

mss

srq

srd

ssq

ssd

i

i

i

i

SLRLSLL

LSLRLSL

SLSLR

SLSLR

v

v

v

v

''

''

'

'

00

00

Example 2 – Dynamic Model of Induction Motor in dsqs frameThe induction motor from the Example 1 has the

following additional parameters:


Parameter Symbol Value

Rated torque Te,rat 183 Nm

Stator resistance Rs 0.294

Stator self inductance

Ls 0.0424 H

Referred rotor resistance

Rr’ 0.156

Referred rotor self inductance

Lr’ 0.0417 H

Mutual inductance Lm 0.041 H

Example 2 - Dynamic Model of Induction Motor in dsqs frame (contd.)Using the values of stator and rotor currents obtained in

the Example 1 , calculate the stator flux s and rotor flux r vectors at time t = 0.

Given that

Calculate the torque produced by the motor using:Stator flux s and stator current Is vector Rotor flux r and stator current Is vector Stator current Is and rotor current Ir vector


**'

* Im22

3Im

22

3Im

22

3rsmrs

r

msse iiL

Pi

L

LPi

PT


Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f).

In the stationary dsqs (αβ in the diagram) frame:• dsqs components are time varying sinusoidal signals at stator frequency (2f)



If we want the Induction Motor to behave like a DC motor, the two-phase components must be constant values.

This can be achieved by having a two-phase frame that rotates together with the space vector.

A rotating dq frame!Source:http://www.ece.umn.edu/users/riaz/animations/listanimations.html

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)The dq referance frame is rotated

at an arbitrary speed g

On the dsqs frame:

On the dq frame:

The angle between the two frames is g where:


tgg

g

g

d

q

x

qs

ds

xds

xqs

sq

sd

sdq xx jxx

qddq xx jxx

Two-phase Dynamic Modelin Arbitrary Rotating Frame(dsqs dq frame transform)

Therefore:

More elegantly :

(full derivation dsqs

d frm trnsform)Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 40

g

g

d

q

x

qs

ds

xds

xqs

xqxd

g

gsdx cosdx g

sqx sin

gsdx sinqx g

sqx cos

sq

sd

gg

gg

q

d

x

x

x

x

cossin

sincos

sdq

jdq xex g

(18)

(19)

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Hence, space vectors in the

stationary dsqs frame will have to be transformed into the rotating dq frame using:

Equation (20) will have to be employed onto equations (13) –(16) to obtain the IM dynamic model in the rotating dq frame.


dqjs

dq xex g (20) g

g

d

q

x

qs

ds

) :(Note

cossin

sincos

tggsdq

jdq

sq

sd

gg

gg

q

d

ge

x

x

x

x

xx

xds

xqs

rdqrgrdqrdqrrdq

rdqj

rrdqj

sdqj

rrdqj

jdtdR

ejdtedeRe gggg

ΨΨiv

ΨΨiv

'

'

Subst. (20) into (13), stator voltage equation:

Subst. (20) into (14), rotor voltage equation:

Expressed in stationary frame dtdR ssdq

ssdqs

ssdq Ψiv

srdqr

srdq

ssdqr

srdq jdtdR ΨΨiv '

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)


Expressed in arbitrary rotating frame

Expressed in stationary frame

Expressed in arbitrary rotating frame

sdqgsdqsdqssdq

sdqj

sdqj

ssdqj

jdtdR

dtedeRe ggg

ΨΨiv

Ψiv

(21)

(22)

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Subst. (20) into the flux linkages equations of (15) and (16):

Stator flux linkage:

Rotor flux linkage:

Airgap flux linkage:

where and


lsms LLL ''lrmr LLL

rdqsdqsdq iiΨ ms LL

rdqsdqrdq iiΨ 'rm LL

rdqsdqodq iiΨ mm LL

(23)

(24)

(25)

rq

rd

sq

sd

rrrrgmmrg

rrgrrmrgm

mmgsssg

mgmsgss

rq

rd

sq

sd

i

i

i

i

SLRLSLL

LSLRLSL

SLLSLRL

LSLLSLR

v

v

v

v

''

''

')()(

)(')(

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Note that equations (21)-(25) each consists of two equations.

One from equating real quantitiesOne from equating imaginary quantities

Final dynamic equations in the arbitrary rotating dq frame is given by substituting (23)-(24) into the (21)-(22) and separating the real and imaginary equations.

Final dynamic equations in the arbitrary dq frame:

Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives 44(26)

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)Torque equation in the arbitrary dq frame:

Product of voltage and current conjugate space vectors:

It can be shown that for ias + ibs + ics = 0,

Input power to the IM:


csbs2

ascs2

bsas*ss aiiai

32

vaavv32

iv

cscsbsbsasas*ss iviviv

32

ivRe

*ssivRe

2

3 cscsbsbsasasin ivivivP


Input power to the IM:

If and then:


*ssivRe

2

3 cscsbsbsasasin ivivivP

qqddqdqd*ssin iviv

23

)jii)(jvv(Re23

ivRe23

P

q

d

v

vv

q

d

i

ii vi

23

P tin


The IM equation given by (26) can be written as:

Note that the matrices:[R] = consists of resistive elements[L] = consists of coefficients of the derivative operator S[G] = consists of coefficients of the electrical rotor speed r

[F] = consists of coefficients of the reference frame speed g


iiiiv gr FGSLR


Hence, the input power is given by:


iiiiiiiivi gt

rtttt

in FGSLRP 2

3

2

3

PowerLosses in winding resistance

Rate of change of stored magnetic energy

Mechanical power

Power associated with g – upon expansion gives zero


The mechanical power is most important:

By observing equation (26), [G] consists of terms associated with r :


ii rt

emmech GTP 2

3

rq

rd

sq

sd

rrrrgmmrg

rrgrrmrgm

mss

mss

rq

rd

sq

sd

i

i

i

i

SLRLSLL

LSLRLSL

SLSLR

SLSLR

v

v

v

v

''

''

')()(

)(')(

00

00


Therefore, mechanical power:


ii rt

emmech GTP 2

3

r

rdrsdm

rqrsqm

t

rq

rd

sq

sd

em

iLiL

iLiL

i

i

i

i

T

'

'

0

0

2

3

i G


Since m = r / (P/2), hence the electromagnetic torque:

(27)


r

rdrsdm

rqrsqm

t

rq

rd

sq

sd

emmech

iLiL

iLiL

i

i

i

i

TP

'

'

0

0

2

3

rqsdrdsqme iiiiLP

T 22

3

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame) - SummaryFinal dynamic equations in the arbitrary dq frame:

Torque equation in the arbitrary dq frame:


rqsdrdsqme iiiiLP

T 22

3

rq

rd

sq

sd

rrrrgmmrg

rrgrrmrgm

mmgsssg

mgmsgss

rq

rd

sq

sd

i

i

i

i

SLRLSLL

LSLRLSL

SLLSLRL

LSLLSLR

v

v

v

v

''

''

')()(

)(')(

(26)

(27)

Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame Speed of reference frame: Dynamic model in the stationary reference frame:

Torque equation in the stationary reference frame:


srqssd

srd

ssqme iiiiL

PT

22

3

0g

srq

srd

ssq

ssd

rrrrmmr

rrrrmrm

mss

mss

srq

srd

ssq

ssd

i

i

i

i

SLRLSLL

LSLRLSL

SLSLR

SLSLR

v

v

v

v

''

''

'

'

00

00

(28)

(29)

(30)

Assuming balanced conditions, vsds, vsq

s, isds, isq

s can be obtained from:

(31)The inverse transform is given by:

(32) where:

and (33)

vrds, vrq

s, irds, irq

s obtained from transforming vabcr vr vrdqs

using equations (31), (32), (33) and (11)

Three-phase to Two-phase Transformation (Stationary)


abcabcsdq xTx

sdqabcabc xTx 1

3

13

1

00

0

1abcT

23

23

21

211

01

Tabc

Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame This model is used when:

stator variables are required to be actual (i.e. same as in the actual machine stator)

rotor variables can be fictitiousAllows elegant simulation of stator-controlled induction motor

drivesphase-controlled and inverter-controlled IM drives (i.e. this IM

model is used for variable voltage control at constant frequency)Input variables are well defined and can be used to find vsd and

vsq easily Reduce computations leading to real-time control applications


Commonly-used Induction Motor Models – Rotor Reference Frame Speed of reference frame: Dynamic model in the rotor reference frame:

Torque equation in the rotor reference frame:


rg

rrq

rrd

rsq

rsd

rrm

rrm

mmrsssr

mrmsrss

rrq

rrd

rsq

rsd

i

i

i

i

SLRSL

SLRSL

SLLSLRL

LSLLSLR

v

v

v

v

'00

0'0

(34)

(35)

(36) rrqrsd

rrd

rsqme iiiiL

PT

22

3

Two-phase Transformation (from stationary stator dsqs frame to rotor drqr frame)Assuming balanced conditions, vsd

r, vsqr, isd

r, isqr can be obtained from

vsds, vsq

s, isds, isq

s using:

(37)

The inverse transform is given by:

(38)

Note: vsds, vsq

s, isds, isq

s can be obtained from vabcs and iabcs using (31), (32) and (33)


ssq

ssd

rr

rrrsq

rsd

x

x

x

x

cossin

sincos

rsq

rsd

rr

rrssq

ssd

x

x

x

x

cossin

sincos

trr :Note

Commonly-used Induction Motor Models - Rotor Reference Frame vrd

r, vrqr, ird

r, irqr obtained from transforming vabcr vrdq

r

using equations (31), (32) and (33)This model is used when:

Switching elements and power are controlled on the rotor side

Example: for simulations of slip-energy recovery scheme


Commonly-used Induction Motor Models – Synchronously Rotating Reference Frame Speed of reference frame: Dynamic model in the synchronously rotating frame:

Torque equation in the synchronously rotating frame:


sg (39)

(40)

(41) rqsdrdsqme iiiiLP

T 22

3

rq

rd

sq

sd

rrrrsmmrs

rrsrrmrsm

mmgsssg

msmssss

rq

rd

sq

sd

i

i

i

i

SLRLSLL

LSLRLSL

SLLSLRL

LSLLSLR

v

v

v

v

')()(

)(')(

Two-phase Transformation (from stationary stator dsqs frame to synchronously rotating dq frame)Assuming balanced conditions, vsd, vsq, isd, isq can be obtained from

vsds, vsq

s, isds, isq

s using:

(42)

The inverse transform is given by:

(43)

Note: vsds, vsq

s, isds, isq

s can be obtained from vabcs and iabcs using (31), (32) and (33)


ssq

ssd

ss

ss

sq

sd

x

x

x

x

cossin

sincos

sq

sd

ss

ssssq

ssd

x

x

x

x

cossin

sincos

tss :Note

Commonly-used Induction Motor Models - Synchronously Rotating Reference Framevrd, vrq, ird, irq obtained from transforming vabcr vr

vrdqs vrdq using equations (31), (32), (33), (11) and

(42)Synchronous reference frame:

transforms sinusoidal inputs into dc signalsprovides decoupled torque and flux channels

Hence, IM control similar to separately excited DC motor achieved by employing vector control schemes



Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f).

In the stationary αβ (dsqs) frame:• αβ (dsqs) components are time varying sinusoidal signals at stator frequency (2f)

In the rotating synchronous dq frame:• dq components are constant • values depend on the orientation of the space vectors with respect to the dq axes.


Equations in Flux LinkagesAll dynamic equations presented are consists of 4 variables,

i.e. vsdq, vrdq, isdq and irdq

Note that in squirrel-cage IM: vrdq = 0 at all timesIf the equations are required to contain flux linkages (i.e.

either sdq, rdq or odq), the dynamic model can be obtained by substituting irdq using the following equations respectively:


rdqsdqsdq iiΨ ms LL

rdqsdqrdq iiΨ 'rm LL

rdqsdqodq iiΨ mm LL

(23)

(24)

(25)

ReferencesKrishnan, R., Electric Motor Drives: Modeling, Analysis and

Control, Prentice-Hall, New Jersey, 2001.Bose, B. K., Modern Power Electronics and AC drives, Prentice-

Hall, New Jersey, 2002.

Dr. Ungku Anisa, July 2008 64EEEB443 - Control & Drives

dsqs frame transform


xrds

r

r

xr

qs

xrxr

r

ds

rrx cossrdx rrx sin

rrx sinsrqx rrx cos

rj

rsrdq

rrrr

rrrr

rrrr

srq

srd

exx

xx

xx

xx

xx

sinjcosj

cossinj

sincos

jx r

xrqs

Back

dsqs dq frame transform


g

g

d

q

x

qs

ds

xds

xqs

te

xx

xx

xx

xx

ggjs

dqdq

ggsq

sd

gsqg

sd

gsqg

sd

qd

g

where

sinjcosj

cossinj

sincos

jx

xx

xqxd

g

gsdx cosdx g

sqx sin

gsdx sinqx g

sqx cos

Back

Documents

Dynamic Model of Induction Machine By Dr. Ungku Anisa Ungku Amirulddin Department of Electrical Power Engineering College of Engineering Dr. Ungku Anisa,