d q Transormation

7/27/2019 d q Transormation

1/61

dq Theory for Synchronous Machinewithout Damper Winding


2/61

Relevance to Synchronous Machine

dq means direct and quadrature. Direct axis is aligned withthe rotors pole. Quadrature axis refers to the axis whoseelectrical angle is orthogonal to the electric angle of directaxis.

a axis

d axisq axis

b axis

c axis

m a

d

mq

22

2

memqr

mme

P

P min g

max,or r

min,or r max g

isr

d

a axis

m a

q axis

mq


3/61

Parks Transformation

Stator quantities ( S abc) of current, voltage, or flux can beconverted to quantities ( S dq0) referenced to the rotor.

This conversion comes through the K matrix.

01

0

dqabc

abcdq

SK S

KSS

13/2sin3/2cos

13/2sin3/2cos1sincos

2/12/12/1

3/2sin3/2sinsin3/2cos3/2coscos

3

2

1

meme

meme

meme

mememe

mememe

K

K

13/2cos3/2sin13/2cos3/2sin

1cossin

2/12/12/13/2cos3/2coscos

3/2sin3/2sinsin

32

1

r r

r r

r r

r r r

r r r

K

K

where

or

(MITs notation)

(Purdues notation)


4/61

Voltage Equations (1)

abcabcS abc dt d iR v

010101 dqdqS dq dt d

K iK R vK

010101 dqdqS dq dt d K K iK KR vK K

01

01

01

0 dqdqdqS dq dt d

dt d

K K KK iK KR v

01

000 dqdqdqS dq dt d

dt d

K K iR v

100

010001

sS RR

For stator windings

For field winding:

f f f f dt d

i Rv

Under motor reference convention for currents(i.e. the positive reference direction for currents is into the machine):


5/61


We derive the derivative of K-1:

Then, we get

000

00

001

r

r

dt d

K K

f f f

s

r d qq s

r qd d s

f

q

d

dt d i R

dt d

i R

dt d i R

dt d

i R

v

v

vv

000

03/2sin3/2cos


03/2cos3/2sin03/2cos3/2sin0cossin

1

r r

r r

r r

r

meme

meme

meme

me

dt

d K

dt d me

me

And for voltage, we get

2 P

dt d

mmer

r


6/61

Dynamical Equations for Flux Linkage

f f f

s

r d q sq

r qd sd

f

q

d

i Rvi Rv

i Rv

i Rv

dt

d

000

The derivations so far are valid for both linear and nonlinear models.

f

q

d

dqf

0

f f f

s

r d q sq

r qd sd

i Rv

i Rv

i Rv

i Rv

00

V

Let

we haveV

dt

d dqf


7/61

Flux Linkage vs. Current (1)

The next step is to relate current to flux linkage throughinductances. For salient pole rotor, the inductances canbe approximately expressed as

cf

bf

af

sf

L

L

LL

32

2cos

322cos

2cos

me B Alscc

me B Alsbb

me B Alsaa

L L L L

L L L L

L L L L

or:

32

2cos

322cos

2cos

r B Alscc

r B Alsbb

r B Alsaa

L L L L

L L L L

L L L L

2

mer

f T sf

sf ssabcf LL

LLL

cccbca

bcbbba

acabaa

ss

L L L

L L L

L L LL


8/61


32

2cos21

2cos21

32

2cos21

me B Acaac

me B Acbbc

me B Abaab

L L L L

L L L L

L L L L

Note: Higher order harmonics are neglected.

3

2cos

32

cos

cos

me sf fccf

me sf fbbf

me sf faaf

L L L

L L L

L L Lor:

32

2cos21

2cos21

32

2cos21

r B Acaac

r B Acbbc

r B Abaab

L L L L

L L L L

L L L L

32

sin

32

sin

sin

r sf fccf

r sf fbbf

r sf faaf

L L L

L L L

L L L

2

mer

Ref: 1. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,6 th Edition, pages 660-661.

2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, page 52.


9/61


This matrix can be transformed into dq0 form and used tofind flux linkage.

abcf abcf abcf iL

f sf dq ssdq iLiK L K 01

01

dqf dqf dqf iL

f sf dq ssdq iKLiK KL 01

0

f T sf

sf ss

abcf LL

LLL

f

abc

abcf

f

abc

abcf i

ii

f sf abc ssabc iLiL

f f abcT sf f i L iL f f dq

T sf f i L 0

1iK L

f T sf

sf ssdqf

L1

1

K L

KLK KLL

f

dqdqf

0

f

dqdqf

i

0i

i

From with

where


10/61

Inductance Matrix in dq0 Frame

f sf

q

sf d

dqf

L L

L

L

L L

0023

000

000

00

0L

where

)(23

)(

2

3

B Amq

B Amd

L L L

L L L

ls

mqlsq

md lsd

L L

L L L

L L L

0

and

dqf dqf dqf iL

From d sf f f f

ls

qqq

f sf d d d

i Li L

i L

i Li Li L

23

00

Through derivations, we have

f T sf

sf ssdqf L1

1

K L

KLK KL

L


11/61

Dynamical Equation in terms of Current

V

dt

d dqf For linear model

from

VLi

1dqf

dqf

dt

d dynamical equationin terms of current

dqf dqf dqf iL

f f f

s

r d q sq

r qd sd

i Rv

i Rv

i Rv

i Rv

00

V

and

where

qqq

f sf d d d

i L

i Li L


12/61

Power

Electrical instantaneous Input Power on Stator can also beexpressed through dq0 theory.

011

0 )( dqT T

dqabcT abcccbbaain iviviv p iK K viv

00223 iviviv p qqd d in

200010

001

23

)(11

K K T


13/61

Torque

00223

iviviv p qqd d in

000

dt d

i Rdt

d i R

dt d

i R

v

v

v

s

r d qq s

r qd d s

q

d

From

we have

)(223

223

223 0

020

22d qqd m

q

qd

d qd sin ii P

dt d

idt

d idt

d iiii R p

Copper Loss Mechanical PowerMagnetic Power inWindings

Therefore, electromagnetic torque on rotor

)(223

d qqd m

meche ii P p

T

mech p


14/61

Equivalent Circuits (1)

f f f

s

r d qq s

r qd d s

f

q

d

dt d

i R

dt d

i Rdt

d i R

dt d

i R

v

v

v

v

000

d sf f f f

ls

qqq

f sf d d d

i Li L

i L

i L

i Li L

23

00

d axisdt

di L

dt di

Li Rv f sf d

d r qd sd


15/61


q axisdt

di Li Rv qqr d q sq

0 axisdt di

Li Rv s0

000

This circuit is not necessaryfor Y connected windingssince i 0=0.


16/61


Field windingdt di

Ldt

di Li Rv d sf

f f f f f 2

3


17/61

Combined Equivalent Circuit on d Axis (1)

dt

iid

Ldt

di

Li R

dt

di L

dt di

Li Rv

f d

md

d

lsr qd s

f sf

d d r qd sd

)( '

d axis equivalent circuit and field winding equivalent circuit can be combined:

md lsd L L L

mf lf f L L L

mf

sf

sf

md

f

a

L

L

L

L

N

N N

3

2

N

ii

L

Li f f

md

sf f

3

2'

f ad sf

f d mf

ad md

N N C L

N C L

N C L

23

2

2)

21(

82

0 g

avd P g

Dl C

From

(Details @ InductanceSM.ppt)

Let

a N f N and are effective number of turns of armature andfield windings.


18/61


dt di

Ldt

di Li Rv d sf

f f f f f 2

3 '23

f f Nii

dt di

NLdt

di NLdt

di NLi NR Nv

d sf

f mf

f lf f f f 2

3

mf

sf

sf

md

L

L

L L

N 32

dt di

Ldt

di L

dt

di L N i R N Nv d md

f sf

f lf f f f

'2'2

23

23

dt

iid L

dt

di Li Rv f d md

f lf f f f

)( ''''''

lf lf

f f

f f

L N L

R N R

Nvv

2'

2'

'

2323


19/61


dt

iid L

dt di

Li Rv f d md d

lsr qd sd

)( '

dt iid

Ldt

di Li Rv f d md

f lf f f f

)( ''''''

From

f f Nvv'

'

23

f f Nii

we get

md md d ls

f sf d d d

i Li L

i Li L

' f d md iii


20/61

dq Theory for Permanent MagnetSynchronous Machine (PMSM)


21/61




01

0

dqabc

abcdq

SK S

KSS

13/2sin3/2cos


2/12/12/1


32

1

meme

meme

meme

mememe

mememe

K

K

13/2cos3/2sin

13/2cos3/2sin1cossin

2/12/12/13/2cos3/2coscos3/2sin3/2sinsin

32

1

r r

r r

r r

r r r

r r r

K

K

where

or

(MITs notation)

(Purdues notation)


22/61




K iK R vK

01

01

01

dqdqS dq dt d

K K iK KR vK K

01

01

01

0 dqdqdqS dq dt d

dt d

K K KK iK KR v

01

000 dqdqdqS dq dt d

dt d

K K iR v

100

010001

sS RR

For stator winding



23/61



Then, we get

000

00

001

r

r

dt d

K K

000

dt

d i R

dt d

i Rdt

d i R

v

vv

s

r d qq s

r qd d s

q

d

03/2sin3/2cos03/2sin3/2cos0sincos

03/2cos3/2sin


1

r r

r r

r r

r

meme

meme

meme

medt d

K

dt d me

me

And for voltage, we get

2 P

dt d

mmer

r


24/61


000 i Rv

i Rv

i Rv

dt

d

s

r d q sq

r qd sd

q

d


0

0

q

d

dq

00 i Rv

i Rvi Rv

s

r d q sq

r qd sd

V

Let

we haveV

dt

d dq 0


25/61



cccbca

bcbbba

acabaa

abc

L L L

L L L

L L LL

32

2cos

322cos

2cos

me B Alscc

me B Alsbb

me B Alsaa

L L L L

L L L L

L L L L


or:

32

2cos

322cos

2cos

r B Alscc

r B Alsbb

r B Alsaa

L L L L

L L L L

L L L L

2

mer


26/61


32

2cos21

2cos21

32

2cos21

me B Acaac

me B Acbbc

me B Abaab

L L L L

L L L L

L L L L


or:

32

2cos21

2cos21

32

2cos21

r B Acaac

r B Acbbc

r B Abaab

L L L L

L L L L

L L L L

2

mer


2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, page 52,

also pages 264-265.


27/61


This matrix can be transformed into dq0 form and used tofind flux linkage.

PMabcabcabcabc iL

PMabcdqabcf dq iK L K 0101

PMabcdqabcdq K iK KL K K 0101

0000 PMdqdqdqdq iL

PMabcdqabcdq K iK KL 01

0

where

)3/2cos(

)3/2cos(

)cos(

me

me

me

PM PMabc

or:

2

mer )3/2sin(

)3/2sin(

)sin(

r

r

r

PM PMabc


28/61

Inductance Matrix in dq0 Frame

Therefore, we get the following inductance matrix in dq0frame:

0

10

0000

00

L L

L

q

d

abcdq K KLL

where

)(23

)(23

B Amq

B Amd

L L L

L L L

ls

mqlsq

md lsd

L L

L L L

L L L

0

and

From00 i L

i L

i L

ls

qqq

PM d d d

0000 PMdqdqdqdq iL

000

PM

PMabc PMdq

K


29/61


V

dt

d dq 0For linear model from

VLi

1

0

0

dq

dq

dt

d dynamical equationin terms of current

00 i Rv

i Rv

i Rv

s

r d q sq

r qd sd

V

and

whereqqq

PM d d d

i L

i L

0000 PMdqdqdqdq iL

0

0

00

00

00

L

L

L

q

d

dqL

0000 /)(/)(

/)(

Li Rv Li Li Rv

Li Li Rv

ii

i

dt d

s

q PM r d d r q sq

d qqr d sd

q

d

For Y connected winding, since , only need to considerthe first two equations for i d and i q .

0)(3

10 cba iiii


30/61

Power


011

0 )( dqT T

dqabcT abcccbbaain iviviv p iK K viv


200010

001

23

)(11 K K T


31/61

Torque

00223

iviviv p qqd d in

000

dt d i R

dt d

i R

dt d

i R

v

v

v

s

r d qq s

r qd d s

q

d

From

we have

)(223

22

3

22

3 00

2

0

22

d qqd m

q

q

d

d qd sin ii

P

dt

d

idt

d

idt

d

iiii R p

Copper LossMechanical PowerMagnetic Power in

WindingsTherefore, electromagnetic torque on rotor

)(22

3d qqd

m

meche ii

P pT

mech p

qqq

PM d d d

i L

i L

qd qd q PM e ii L Li P T )(

22

3


32/61

Dynamical Equations of Motion

mm

damp Lem

dt d

T T T dt

d J

where

qT qd qd q PM e i K ii L Li P T )(22

3

For round rotor machine, qd L L q PM e i P T 43

mmdamp DT D m is combined damping coefficient of rotorand load.

d qd PM q

eT i L L

P iT

K )(4

3 torque constant

PM T P K 43


33/61

dq Theory for Synchronous Machinewith Damper Winding


34/61

Relevance to Synchronous Machine

dq means direct and quadrature. Direct axis is aligned withthe rotors pole. Quadrature axis refers to the axis whoseelectrical angle is orthogonal to the electric angle of direct

axis.

a axis

d axisq axis

b axis

c axis

m a

d

mq

22

2

memqr

mme

P

P min g

max,or r

min,or r max g

isr

d

a axis

m a

q axis

mq


35/61




01

0

dqabc

abcdq

SK S

KSS

13/2sin3/2cos


2/12/12/1


32

1

meme

meme

meme

mememe

mememe

K

K

13/2cos3/2sin


2/12/12/13/2cos3/2coscos3/2sin3/2sinsin

32

1

r r

r r

r r

r r r

r r r

K

K

where

or

(MITs notation)

(Purdues notation)


36/61




K iK R vK

01

01

01

dqdqS dq dt d

K K iK KR vK K

01

01

01

0 dqdqdqS dq dt d

dt d

K K KK iK KR v

01

000 dqdqdqS dq dt d

dt d

K K iR v

100

010

001

sS RR

For stator windings



37/61



Then, we get

00000

001

r

r

dt d

K K

000

dt d

i R

dt d

i Rdt

d i R

v

vv

s

r d qq s

r qd d s

q

d

03/2sin3/2cos


03/2cos3/2sin


1

r r

r r

r r

r

meme

meme

meme

medt d

K

dt d me

me

And for stator voltage, weget

2 P

dt d

mmer

r


38/61


For rotor windings:

We assume the rotor has field winding (magnetic field along d axis),one damper with magnetic field along d axis and one damper

with magnetic field along q axis.

qd qd qd k fk k fk r k fk dt d

iR v

q

d

k

k

f

r

R

R

R

00

00

00R

0

0 f

k fk

v

qd v

q

d qd

k

k

f

k fk


39/61


qqq

d d d

k k k

k k k

f f f

s

qr d q s

d r qd s

f

q

d

dt

d i R

dt d

i Rdt

d i R

dt d

i Rdt

d i R

dt d

i R

v

v

v

v

000

00

In summary:


40/61


qq

d d

q

d

k k

k k

f f f

s

r d q sq

r qd sd

k

k

f

q

d

i R

i Ri Rv

i Rv

i Rv

i Rv

dt

d 000


Let

we have V

dt

d dqf

q

d

k

k

f

q

d

dqf

0

qq

d d

k k

k k

f f f

s

r d q sq

r qd sd

i R

i R

i Rv

i Rv

i Rv

i Rv

00

V


41/61



32

2cos

322cos

2cos

me B Alscc

me B Alsbb

me B Alsaa

L L L L

L L L L

L L L L

or:

32

2cos

322cos

2cos

r B Alscc

r B Alsbb

r B Alsaa

L L L L

L L L L

L L L L

2

mer

rr T sr

sr ssabcf LL

LLL

cccbca

bcbbba

acabaa

ss

L L L

L L L

L L LL

qd

qd

qd

ck ck cf

bk bk bf

ak ak af

sr

L L L

L L L

L L LL

qd qq

qd d d

qd

k k k f k

k k k f k

fk fk f

rr

L L L

L L L

L L LL


42/61


32

2cos21

2cos21

32

2cos21

me B Acaac

me B Acbbc

me B Abaab

L L L L

L L L L

L L L L

32

cos

3

2cos

cos

me sf fccf

me sf fbbf

me sf faaf

L L L

L L L

L L Lor:

32

2cos21

2cos21

32

2cos21

r B Acaac

r B Acbbc

r B Abaab

L L L L

L L L L

L L L L

32

sin3

2sin

sin

r sf fccf

r sf fbbf

r sf faaf

L L L

L L L

L L L

2

mer


43/61


32

cos

32

cos

cos

me sk ck ck

me sk bk bk

me sk ak ak

d d d

d d d

d d d

L L L

L L L

L L L

or:

32

sin

32

sin

sin

r sk ck ck

r sk bk bk

r sk ak ak

d d d

d d d

d d d

L L L

L L L

L L L

2 mer

32

sin

32

sin

sin

me sk ck ck

me sk bk bk

me sk ak ak

qqq

qqq

qqq

L L L

L L L

L L L

32

cos

32

cos

cos

r sk ck ck

r sk bk bk

r sk ak ak

qqq

qqq

qqq

L L L

L L L

L L L


44/61


Note: Higher order harmonics are neglected in the above expressions.


2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, pages 52and 195.

qqq

d d d

mk lk k

mk lk k

mf lf f

L L L

L L L

L L L

0

0

d qqd

qq

d d

k k k k

f k fk

f k fk

L L

L L

L L

l i k C ( )


45/61


This matrix can be transformed into dq form and used tofind flux linkage.

abcf abcf abcf iL

qd k fk sr dq ssdqiLiK L K

01

01

dqf dqf dqf iL

qd k fk sr dq ssdqiKLiK KL

0

1

0

r T sr

sr ssabcf

LL

LLL

qd k fk

abc

abcf

qd k fk

abc

abcf i

ii

qd k fk sr abc ssabciLiL

qd qd k fk rr abcT sr k fk

iLiL qd qd k fk rr dq

T sr k fk

iLiK L 0

1

rr

T

sr

sr ssdqf LK L

KLK KLL

1

1

qd k fk

dqdqf

0

qd k fk

dqdqf i

ii

0

From with

where

I d M i i d F


46/61

Inductance Matrix in dq Frame

qq

d d d

d

q

d

k sk

k f k sk

fk f sf

sk q

sk sf d

dqf

L L

L L L

L L L L

L L

L L L

000230

00023

00023 00000

0000

000

0

L

where

)(23

)(23

B Amq

B Amd

L L L

L L L

mqlsq

md lsd

L L L

L L L

and

dqf dqf dqf iL From q sk k k k

f fk d sk k k k

k fk d sf f f f

k sk qqq

k sk f sf d d d

i Li L

i Li Li L

i Li Li L

i L

i Li L

i Li Li L

qqqq

d d d d d

d d

qq

d d

2323

23

000

Through derivations, we have

rr T sr

sr ss

dqf LK L

KLK KLL

1

1

ls L L0

D i l E ti i t f C t


47/61


V

dt

d dqf For linear model

from

VLi 1dqf

dqf

dt d dynamical equation

in terms of current

dqf dqf dqf iL

qq

d d

k k

k k

f f f

s

r d q sq

r qd sd

i R

i Ri Rv

i Rv

i Rvi Rv

00

V

and

where

qq

d d

k sk qqq

k sk f sf d d d

i Li L

i Li Li L

P


48/61

Power


0

11

0 )( dqT T

dqabc

T

abcccbbaain iviviv piK K viv


200

010

001

2

3)( 11 K K T

Torque


49/61

Torque

00223

iviviv p qqd d in

000

dt d i R

dt d

i R

dt d

i R

v

v

v

s

r d qq s

r qd d s

q

d

From

we have

)(22

32

2

32

2

3 00

20

22d qqd m

qq

d d qd sin ii

P

dt

d i

dt

d i

dt

d iiii R p

Copper LossMechanical PowerMagnetic Power inWindings

Therefore, electromagnetic torque on rotor

)(22

3d qqd

m

mech

e ii

P p

T

mech p

Equivalent Circuit on d Axis (1)


50/61


d axis of stator, field winding and d axis damper of rotor can form an equivalentcircuit.

Let md lsd L L L

mf lf f L L L

d d d mk lk k L L L

d d

d d

d d

k f d fk

k ad sk

k d mk

f ad sf

f d mf

ad md

N N C L

N N C L

N C L

N N C L

N C L

N C L

2

3

2

2

2

)21(8

20 g

avd P g

Dl C

From

(Details @ Inductance for SM.ppt)

, a N f N and are effective number of turns of armature, field and d axis

damper windings, respectively.

d k N



51/61


dt

iiid L

dt di

Li R

dt

di L

dt

di L

dt

di Li Rv

d

d

d

k f d md

d lsr qd s

k sk

f sf

d d r qd sd

)( ''

f a

f f

md

sf f i

N

N i

L

Li

32'Define

d

d

d

d

d k a

k k

md

sk k i

N

N i

L

Li

32'and

dt

di L

dt di

Ldt

di Li Rv d

d

k fk

d sf

f f f f f 2

3

dt di NL

dt di NL

dt di NL

dt di NLi NR Nv d d k fk d sf f mf f lf f f f 2

3

dt

di L

N

N

N

N

dt

di L

dt

di L

dt

di L N i R N Nv d

d

d

k fk

k

a

f

ad md

f sf

f lf f f f

''2'2

2

3

2

3

2

3

Define

f f Nvv'

a

f

N N

N and



52/61


dt

iiid L

dt

di Li Rv d k f d md

f lf f f f

)( '''''''

lf f

alf

f f

a

f

L N

N L

R N

N

R2

'

2

'

23

2

3

where

2

32

a

k f

md

fk

N N N

L L d d

dt

di

Ldt di

Ldt

di

Ldt

di

Li R

dt

di L

dt di

Ldt

di Li R

f

fk

d

sk

k

mk

k

lk k k

f fk

d sk

k k k k

d d

d

d

d

d d d

d d

d

d d d

23

23

0From

above

3

2

d d k

a

sk

md

N

N

L

L

next page



53/61


dt

di L

N N

N dt di

Ldt

di L

N

N dt

di L

N

N i R

N

N f fk

k f

ad md

k mk

k

ak lk

k

ak k

k

ad

d

d

d

d

d

d

d

d d

d

'2'2

'2

'

2

23

23

23

23

0

dt

iiid Ldt

di Li R

d d

d d d

k f d md

k lk k k

)(0

''''''

where

d

d

d

d

d

d

lk k

alk

k k

ak

L N

N L

R N

N R

2

'

2

'

23

2

3

d d

d d

k f

a

fk

md

k

a

mk

md

N N

N L L

N

N L L

23

23

2

2



54/61


From

we get

md md d ls

k sk f sf d d d

i Li L

i Li Li Ld d

''d k f d md iiii

dt iiid L

dt di Li Rv d k f d md

d lsr qd sd

)(''

dt

iiid L

dt

di Li Rv d k f d md

f lf f f f

)( '''''''

dt iiid L

dt di Li R d d

d d d

k f d md

k lk k k )(0

'''

'''

'd k

i

Equivalent Circuit on q Axis (1)


55/61


q axis equivalent circuit and q axis damper equivalent circuitcan be combined:

Letmqlsq L L L

qqq mk lk k L L L

qq

qq

k aq sk

k qmk

aqmq

N N C L

N C L

N C L

23

2

2)

21(8 2

0

P g Dl C av

qFrom

(Details @ Inductance for SM.ppt)

s N and are effective number of turns of

stator and q axis damperwindings, respectively.

d k N



56/61


dt

iid L

dt

di Li R

dt

di L

dt

di L

dt

di Li R

dt

di L

dt

di Li Rv

q

q

q

q

q

k q

mqq

lqr d q s

k

sk q

mqq

lqr d q s

k

sk q

qr d q sq

)(

'

q

q

q

q

d k

a

k

k

md

sk

k i N

N i

L

Li

3

2'where

dt

di

Ldt

di

Ldt

di

Li R

dt

di L

dt

di Li R

q

sk

k

mk

k

lk k k

q sk

k

k k k

q

q

q

q

qqq

q

q

qqq

23

23

0From

above

3

2

qq k

a

sk

mq

N

N

L

L

next page



57/61

q q ( )

dt

di L

dt

di L

N

N dt

di L

N

N i R

N

N qmq

k

mk k

ak

lk k

ak k

k

a q

q

q

q

q

q

qq

q

'2'2'

2

23

23

23

0

dt

iid Ldt

di Li R

qq

qqq

k q

mq

k

lk k k

)(0

'''''

where

q

q

q

q

q

q

lk k

alk

k k

ak

L N

N L

R N

N R

2

'

2

'

23

2

3

qq

qq

k f

a

fk

mq

k

a

mk

mq

N N

N L

L

N

N

L

L

23

2

3

2

2



58/61

q q ( )

From

we get

mqmqqls

k sk qqq

i Li L

i Li Lqq

'qk qmq

iii

dt iid L

dt di Li Rv q

k qmq

qlqr d q sq

)('

dt

iid L

dt

di Li R qq

qqq

k q

mq

k

lk k k

)(0

'''''

'qk

i

Equivalent Circuit on 0 Axis


59/61

q

0 axisdt di

Li Rv s0

000

This circuit is not necessaryfor Y connected windingssince i 0=0.

Dynamical Equations fromEquivalent Circuits (1)


60/61


dt di

Li Rv

dt di L

dt di Li Rv

dt

di L

dt

di Li R

dt di L

dt di Li R

dt

di L

dt

di Li Rv

dt di L

dt di Li Rv

ls s

md md

f lf f f f

mqmq

k

lk k k

md md

k lk k k

mqmq

qlsqd r q sqq

md md

d lsd qr d sd d

q

qqq

d

d d d

00000

'''''

''''

''''

0

0

where

'

''

q

d

k qmq

k f d md

iii

iiii

Dynamical Equations fromEquivalent Circuits (2)


61/61

Equivalent Circuits (2)The equations can be written in matrix form as:

VI

Ldt d

where

mqlk mq

md lk md md

md md lf md

ls

mqmqlsq

md md md lsd

L L L

L L L L

L L L L

L

L L L

L L L L

q

d '

'

'0

0000000

000

000000000

000

L

'

'

'0

q

d

k

k

f

q

d

i

i

ii

i

i

I

''

''

'''000

qq

d d

k k

k k

f f f

s

d r q sqq

qr d sd d

i R

i R

i Rvi Rv

i Rv

i Rv

V

VLI 1

dt d or

d q Transormation

Documents

Transcript of d q Transormation