1
Chapter 1
INTRODUCTION AND MODELLING OF SIX PHASE
INDUCTION MOTOR
1.1 INTRODUCTION:
In the environment of industrial drives, the Variable
speed AC induction motors are replacing the DC motor based
drives because having the inbuilt disadvantage of mechanical
brushes and commutator, which undergo wear and tear with
time. In past decade the
technology opportunities in the area of
multi- phase Induction Motors undergo a significant evolution. But
soft control of torque and speed of these multi- phase induction
motors have forever been a confront to the engineers.
The Six phase AC motors are used for high power
industrial drive systems [1]-[8]. The important motivations and
the
list of significant inherent advantages
multi
level
inverter
fed
six
phase
behind the study of
induction
motors
are
“Enhanced system reliability, reducing the rotor harmonic current
losses, lowering the dc-link current harmonics, reduced torque
ripples, reduced harmonic power loss, better power distribution
per
phase,
improved
power
characteristics
and
improved
efficiency when compared to that of the three phase counterpart”
[3], [9]-[12]. The six phase induction motors have several applications
2
in “Traction, Hybrid electric vehicles, electric ship propulsion, and
the more electric air craft.”[13]- [15]
Multi phase (more than three phase) induction motors
are obtained by replacing the 3 -phase windings with more
number of phases [8]. Conventionally the three phase motors
were used for the drive applications since, easy availability of
three phase power sources. But in the present day with the
arrival
of
inverter
semiconductor
technology
switches,
generation
derived
of
from
power
multi phase power
sources is not a problem. Consequently control of multi phase
motors in high power industrial applications is now easily
achievable. In a three phase machine the conductors are
distributed in slots symmetrically for each phase group and the
conductors
belonging
to
each
individual
phase group are
connected in series. By subdividing each phase group of a usual
three phase motor into equal subgroups by disconnecting the
series connections of the conductors, more number of three
phase groups can be obtained from the same motor. In this
way the multi phase motors
such as
six phase, nine phase,
twelve, fifteen, eighteen phase motors can be produced from a
three phase motor by subdividing the phase groups into two, three
and four subgroups respectively [3].
“Six phase induction motors can be obtained by splitting
the individual three phase groups into two equal halves” [4]-[8].
3
In this approach the two set of three phases group with isolated
neutral are generated, which are 30 electrical degree phase
separated in space (Fig.1.1). The authors [16] have demonstrated
through analog simulation of six step voltage fed dual-three phase
induction motor drive. The performance of the d-q model for dual
three phase Induction motor is established [1]. Main advantage of six
phase Induction Motor is it can be started and run with some phases
open and performance in this faulted form can be almost similar when
all phases energized.
The pulse width modulation (PWM) inverters offer easy and
efficient control of voltage and frequency. The PWM inverter decreases
the amplitudes of lower order harmonics in the motor terminal voltage
by shifting the dominating harmonics towards higher frequencies. In
pulse width modulation control schemes for dual 2 level inverter
fed six phase induction motor, has effective phase voltages
consequent to the reference phase voltages are produced by switching
the inverter poles between voltage levels of +Vdc/2 and –Vdc/2, where
Vdc is the ‘DC-link voltage’ of the voltage source inverter
for
appropriate durations, during the switching period. If the pole
voltages of the inverter are permitted to vary between more
than two levels, subsequently a multilevel inverter system can
be accomplished [17]-[22]. Since the transition voltage levels in
each switching is lesser than that in the two-level inverters,
there is less switching losses in multi -level inverters. The
reduced
voltage
ripple
in
these
multi-level
inverters
also
4
reduces the current ripple in the inverter phases.
Multi-level
inverters are obtained from a number of minor discrete
voltage
sources,
and
they
produce
the
output
voltage
wave forms with more steps of smaller m agnitudes. Multi Level
Inverters have drawn marvelous interest in power industry. Significant
inherent advantages offered by the multi-level inverters compared to
their
two-level
inverters
counterparts
are
summarized
as
follows [23]-[27]:
1.
“It is achievable to obtain refined output voltage waveforms
moreover with reduced total harmonic distortion (THD) by
increased number of voltage levels of inverter (stepped output
voltage waveforms)
2.
It is feasible to reduce the electromagnetic interference (EMI)
problems by lowering the switching rate of change of voltage
(dv/dt)
3.
It
is
possible
to
realize
machine
currents
with
reduced
harmonics, resulting into compact torque pulsations in the drive
system; resulting in fewer stress on the insulation of machine
phase windings
4.
The Inverter is able to operate with lower switching frequency and
therefore the switching losses are reduced.
5.
Lesser amplitudes of alternating Common Mode Voltage and
hence lesser bearing currents”.
Thus, although it was initially designed to reduce the
5
harmonic contents in the output voltage waveforms, the multi-level
inverters have very quickly established themselves as a preferred
option for realizing high-voltage high-power drives for industrial,
marine, utility and traction applications, using power devices of lower
voltage ratings. “Today, multi-level inverters are extensively used in
high-power drive applications for laminators, mills, conveyors,
pumps, fans, blowers, compressors, etc.” [11]. Reduction in Harmonic
losses are prime requirement in the application of high power drive.
Multilevel inverters and inverter fed multiphase motors are very
promising in this respect. Multi-level inverters has materialized as a
favored solution for high-voltage high-power industrial induction
motor drive applications. Multi-level inverters have much reduced
harmonic content in output voltage waveforms and facilitate them to
be operated at much smaller frequencies resulting into lesser
switching losses. A simplified SVPWM method for three level inverters
has been developed [28].The Space- vector pulse width modulation
(SVPM) procedure for dual three phase drive has been developed in
[5]. Because having limited range of switches the traditional 2-level
inverter may not be functional in high voltages. Thus, in the recent
years
Multi
level
inverters
are
widely
used
for
high
voltage
applications.
Hence in this thesis, different dual multi level inverters are used
independently to analyze the performance of six phase induction
motor.
6
1.2 ASSERTION OF PROBLEM
In this thesis it is proposed to analyze the Six Phase Induction
motor using Space- vector PWM of dual two level inverter and different
dual multi level inverters independently with MATLAB/Simulink
software. The investigations are carried out by following steps.
 Development of MATLAB/Simulink model of Six Phase Induction
motor.
 Development of MATLAB/Simulink based algorithm for dual 2-level
Inverter fed Six phase Induction motor
 MATLAB/Simulink based algorithm for dual 3-level Inverter fed Six
phase Induction motor
 Development of MATLAB/Simulink based algorithm for dual 4-level
Inverter fed Six phase Induction motor
 Development of MATLAB/Simulink based algorithm for dual 5-level
Inverter fed Six phase Induction motor
 Development of MATLAB/Simulink based algorithm for dual 6-level
Inverter fed Six phase Induction motor
 Development of MATLAB/Simulink based algorithm for dual 7-level
Inverter fed Six phase Induction motor
1.3
METHODOLOGY
 Survey, Study and analysis of pertinent Literature.
7
 In this thesis a modular, SIMULINK Six phase induction motor
model is developed. In the modular scheme every block solves
one of the modal equations. The inputs given to Six phase
induction motor are the load torque, the dual three-phase
voltages and fundamental frequency. The outputs are the
electrical torque, the speed of rotor and six phase currents.
 In order to study the performance of the six phase induction
motor modal the Space Vector Pulse Width Modulation scheme
of dual 2,3,4,5,6&7 inverter levels are used independently for
six phase induction motor.
1.4 OVERVIEW OF THESIS
Six-phase induction motors are realized by splitting the
stator windings of three phase induction motor into two
identical halves, which has a phase displacement of 30
electrical degrees between them.
Fig.1.1 The winding disposition of six phase induction motor
The dual three phase groups(Fig.1.1) named as A, B,C phase
8
group and M,N,O phase group, have an angular separation of 30
electrical
degrees
between
them. Fig.1.1
exhibits
the
winding
disposition of six phase induction motor. A,B,C phases are fed from
inverter(X) of the six phases of the stator winding and M,N,O
phases are fed from inverter(Y) with isolated power supplies(Fig.1.2).
A
M
Vdc
C
N
B
Inverter (X)
Vdc
O
SPIM
Inverter (Y)
Fig.1.2 Implementation of dual Inverter fed Six Phase Induction
Motor with isolated power supplies.
Similarly the following investigations are carried out and
simulated to analyze the performance of Six Phase Induction Motor
using multi level inverters.
 Inverter(X) and Inverter (Y) (Fig.1.2) are replaced with two
identical three level SVPWM Inverters and fed to Six Phase
Induction Motor.
 Inverter(X) and Inverter (Y) (Fig.1.2) are replaced with two
identical four level SVPWM Inverter and another four level
9
SVPWM Inverter(dual four level inverter) respectively and fed to
Six Phase Induction Motor.
 Inverter(X) and Inverter (Y) (Fig.1.2) are replaced with five level
SVPWM Inverter and another five level SVPWM Inverter(dual five
level inverter) respectively and fed to Six Phase Induction Motor.
 Inverter(X) and Inverter (Y) (Fig.1.2) are replaced with six level
SVPWM Inverter and another six level SVPWM Inverter(dual six
level inverter) respectively and fed to Six Phase Induction Motor.
 Inverter(X) and Inverter (Y) (Fig.1.2) are replaced with seven level
SVPWM Inverter and another seven level SVPWM Inverter (dual
seven level inverter) respectively and fed to Six Phase Induction
Motor.
The Thesis is organized into 5 chapters.
Chapter-1 elucidates the introduction,
Assertion of problem,
Methodology of Research, Overview of thesis and illustrates the
Modelling of Six phase induction motor .
In chapter-2 discusses Simulation of Six phase induction motor
by adopting both dual two level SVPWM inverter and dual three
level SVPWM inverter independently and these simulated results
are compared and presented. Initially the conventional ‘2-level’
SVPWM Inverter X and another ‘2-level’ SVPWM inverter Y
are fed to both ends of Six phase Induction motor. Finally, the
dual 2-level SVPWM inverters are replaced by the proposed dual
10
3-level SVPWM inverters to analyze the Six Phase Induction
Motor.
In chapter-3 deals with dual 4-level SVPWM inverter and
dual 5–level inverters which are fed to six phase induction motor
independently and
simulation results of Six phase Induction
motor are presented.
The dual 6-level SVPWM fed Six phase Induction motor
dual 7–level SVPWM
fed Six phase Induction motor
and
simulated
results are presented and compared in chapter-4.
The investigations are concluded in the chapter-5.
1.5 MODELLING OF SIX PHASE INDUCTION MOTOR (SPIM)
“The Six-phase induction motor can be comprehend by
splitting the phase belt of a conventional Three phase Induction
motor with an angular displacement of 30 electrical degrees”.
Fig.1.1 exhibits the Winding disposition of Six phase induction
motor. As exposed in the Fig.1.1, the six phases of the stator
winding can be grouped into two separate three phase winding
groups, and appropriately connecting the phasor groups. The two
groups named A,B,C group and M,N,O group, have an angular
displacement of 30 electrical degrees between them. while harmonic
currents of the order 6x + 1 (x = 1,3,5,7,9 ....etc.,) are present in the
stator windings the flux vector generated by one phase group (A,
B, C) cancels the flux vector produced by the other (M,N,O) three
phase group, thereby creating zero air gap flux due to these
11
harmonic currents[7]. while harmonic currents belonging to the
12x±1, where x = 0, 1,2,3,4,5....etc., harmonic group which
include
the
fundamental
component,
flow
in
the
stator
windings the flux add each other to create a resultant air gap
flux. Since the neutrals of the phase groups are isolated (see
fig.1.1), no triplen harmonic currents will flow in the motor phases.
Zhao Yifan and Lipo Thomas. A, have derived a model of
six
pha se
induction
motor[7]
where,
set
of
harmonic
components belonging to the 6x +1 (x = 1,3,5,7,9 ....etc.,)
order and the set of harmonic components belonging to the
12x ± 1w h ere x= 0,1,2, 3, 4, 5 . . . . etc. , or der a r e or thogona lly
decomposed
into
two
subspaces.
This
orthogonal
decomposition of the different vector groups is accomplished by
defining a general vector SK (wt ) as shown in equation (1.1)
assumptions [7] have been made in deriving the dual three-phase
induction machine model:
sk (wt )  [cos(wt ) cos(wt   ) cos(wt  4 ) cos(wt  5 ) cos(wt  8 ) cos(wt  9 )] ...(1.1)
Where in equation 1 .1, ‘  ’ is equal to 3 0°, which
correspond to the space phase separation between the two
three phase groups and k=0,1,3,5,7,9,11.... etc., correspond
to the fundamental and all harmonic order. The three sets of
harmonics, consisting of the positive sequence 12x ± 1,
where
x= 0,1,2,3,4,5 ....etc., and the zero sequences z1 6x ± 1, where
x = 1,3,5 ....etc., and z2 (triplen components) span subspaces defined
12
by equation1.2. The positive sequence 12x ± 1,where x = 0,1,2,3,4,5
....etc., span the subspace S1, the zero sequence component zl defined
by 6x + 1, where x= 1,3,5,7,9,11 ....etc., span the subspace S2 and the
zero sequence
component
z2
defined
by
the
triples
order
harmonics span the subspace S3 [7]. The two terms in each
subspace represent the analogous orthogonal components in
that
subspace.
These
two
orthogonal
components
in
each
subspace are obtained by substituting wt equal to 0 and π/2
correspondingly in equation (1.I). Hence a six dimensional matrix
can be achieved by arranging the three subspaces as represented in
equation
It can be prominent that the harmonics spanning the subspace
S1, S2 and S3 are orthogonal to each other as represented by
equation (1.2b)
S1● S2T = S1●S3T = S3●S2T = 0
………………….. (1.2b)
ref. [7]
A transformation matrix C can be defined by multiplying
equation (1.2) with a constant 1/√3 and the entire harmonic
component voltages of the six phase machine can be decomposed into
13
their respective orthogonal subspaces by using this transformation
matrix C as shown in equation (1.3).
 V 
1 cos 
V 
0 sin 
 

 V1 
1 cos 5
   1/ 3 
 V2 
0 sin 5
Vo1 
1
0
 

1
0
Vo 2 
cos 4
cos 5
cos 8
sin 4
cos 8
sin 5
cos 
sin 8
cos 4
sin 8
1
sin 
0
sin 4
1
0
1
0
cos 9  Vas 
sin 9  Vms 
cos 9  Vbs 
   .......... (1.3)
sin 9  Vns 
0  Vcs 
 
1  Vos 
Vas , Vbs , Vcs are the phase voltages of A, B, C three phase group.
Vms , Vns , Vos are the phase voltages of M,N,O three phase group.
V , V are the two orthogonal voltage components of the 12x +1,where
x = 0, 1, 2, 3,4,5,6 ... etc., order harmonic set spanning subspace S1.
V1, V2 are the two orthogonal voltage components of the 6x+1,
where x =1,3,5,7,9,…..etc., order harmonic set spanning subspace S2
Vo1, Vo2 are two orthogonal voltages of the triplen harmonic set
spanning subspace S3.
The equation (1.4) shows the transformation matrix used for the
orthogonal decomposition.
1 cos 
0 sin 

1 cos 5
C   (1 / 3 )
0 sin 5
1
0

1
0
cos 4
cos 5
cos 8
sin 4
cos 8
sin 5
cos 
sin 8
cos 4
sin 8
1
sin 
0
sin 4
1
0
1
0
cos 9  Vas 
sin 9  Vms 
cos 9  Vbs 
   …….(1.4)
sin 9  Vns 
0  Vcs 
 
1  Vos 
The constant 1/√3 is multiplied with eqn.(1.2) to obtain the
transformation matrix (C) (eqn.1.4), so that (C)-1 equals to (C)T. Now
14
from the phasor diagram of the winding disposition of Fig.1.1,
equation for stator voltage in Actual six dimensional space can be
written as
VS   RS iS  
d
(LSS iS   LSr ir )
dt
………………………(1.5)
Where in this equation (1.5)
VS  , iS 
and ir  are the ‘input voltage’ , ‘stator current’ & ‘rotor
current vectors’ correspondingly defined by equations (1.6), (1.7) and
(1.8) correspondingly.
T
VS   Vas

Vms
Vbs
Vns
Vcs
Vos
iS   ias
ibs
ins
ics
ios


T
ims
ir   iar
ibr
inr
icr
ior


T
imr




………………….. (1.6)
………………… (1.7)
…………….(1.8)
In equation (1.5) the stator resistance matrix [Rs], the stator self
inductance [Lss] and the stator to rotor mutual inductance matrixes
[Lsr] are defined by equations (1.9), (1.10) and (1.11) respectively.
 rs

0
0
[ RS ]  
0
0


0
0
rs
0
0
0
rs
0
0
0
0
0
0
0
0
0
rs
0
0
0
0
0
0
rs
0
0

0
0  ………………………………..….(1.9)

0
0

rs 

15
 Lls
0

0
LSS   
0
0

 0
1

 cos( )

cos(4 )
M
cos(5 )
cos(8 )

cos(9 )
0
0
0
0
Lls
0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
0
Lls
0
0
0
0
0 
0 
0 

0 
0 

Lls 
cos( )
1
cos(4 )
cos(3 )
cos(5 )
cos(4 )
cos(8 )
cos(7 )
cos(3 )
1
cos( )
cos(4 )
cos(4 )
cos(7 )
cos( )
cos(4 )
1
cos(3 )
cos(3 )
1
cos(8 )
cos(5 )
cos(4 )
cos( )
cos(9 ) 
cos(8 ) 
cos(5 )  ….(1.10)

cos(4 )
cos( ) 

1

Where, Lls is the stator leakage inductance.
M is the mutual inductance between any two windings with zero
angular separation between them.
θ is the phase angle between the (A, B, C) and (M,N,O) set of
windings and is equal to π/6
Lsr  
 cos( r )
 cos(   )
r

cos( r  4 )
M
cos( r  5 )
 cos( r  8 )

cos( r  9 )
cos( r   )
cos( r )
cos( r  4 ) cos( r  5 ) cos( r  8 )
cos( r  3 ) cos( r  4 ) cos( r  7 )
cos( r  3 )
cos( r )
cos( r   ) cos( r  4 )
cos( r  4 ) cos( r   )
cos( r )
cos( r  3 )
cos( r  7 ) cos( r  4 ) cos( r  3 )
cos( r )
cos( r  8 ) cos( r  5 ) cos( r  4 )
cos( r   )
cos( r  9 ) 
cos( r  8 ) 
cos( r  5 ) 

cos( r  4 )
cos( r   ) 

cos( r ) 
.... (1.11)
Where,
θr is the relative position between the rotor and the stator.
By applying the orthogonal transformation to equation (1.5) the stator
voltage equation may be written as
[C ]V S   [C ]RS [C ] 1 [C ]i S  
Where,
[C ' ][C ]1  [C ][C ]T  I


d
[C ]LSS [C ] 1 [C ]i S   [C ]LSr [C ] 1 [C ]i r 
dt
………(1.12)
16
i.e.,
[C ]1  [C ]T
…………………………………(1.13)
 1
 cos 

cos 4
1
C   (1 / 3 ) 
cos 5
 cos 8

cos 9
0
sin 
sin 4
sin 5
sin 8
sin 9
1
0
1 0
cos 5 sin 5 0 1
cos 8 sin 8 1 0 ………..………..……….. (1.14)

cos  sin  0 1
cos 4 sin 4 1 0

cos 9 sin 9 0 1
Since [Rs] is a diagonal matrix with each diagonal element equal to Rs,
the term [C][Rs][C]-1 = [Rs]…………………………………………………. (1.15)
Now
from
eqn.(1.12),
by
applying
both
left
and
right
transformation to the self inductance matrix of equation (1.10),
equation (1.16) is obtained.
C LSS C 1
 Lls
0

0
 C 
0
0

 0
1

 cos( )

cos(4 )
[C ] M 
cos(5 )
cos(8 )

cos(9 )
0
0
0
0
Lls
0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
0
Lls
0
0
0
0
cos( )
1
cos(3 )
cos(4 )
cos(7 )
cos(8 )
cos(4 )
cos(3 )
0
0 
0  1
C  
0
0

Lls 
cos(5 )
cos(4 )
cos(8 ) cos(9 ) 
cos(7 ) cos(8 ) 

1
cos( ) cos(4 ) cos(5 )  1 …….. (1.16)
C 
cos( )
1
cos(3 ) cos(4 )
cos(4 ) cos(3 )
1
cos( ) 

cos(5 ) cos(4 ) cos( )
1 
By applying the left and right transformation to the first term
and doing the right transformation to the second term, equation (1.16)
can be summarized to equation (1.17)
17
C LS S C 1
 Lls
 0

 0
 
 0
 0

 0

0
0
0
0
Lls
0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
0
Lls
0
0
0
0
3
3 cos( / 2)

 3 cos( )
3 cos( / 2   )

3 cos(4 ) 3 cos( / 2  4 )
(1 / 3 )[C ]M 
3 cos(5 ) 3 cos( / 2  5 )
3 cos(8 ) 3 cos( / 2  8 )

3 cos(9 ) 3 cos( / 2  9 )
0 
0 

0 

0 
0 

Lls 

0 0 0 0
0 0 0 0
0 0 0 0 …………………… (1.17)

0 0 0 0
0 0 0 0

0 0 0 0
By simplification, the term C LSS C 1 can be written as
C LS S C 1
3

3 cos( / 2)


0
M
0


0

0

The
 Lls
 0

 0

 0
 0


 0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
3 cos( / 2)
3
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
transformed
self
0
0
0

0
0

0
inductance
0 
0 

0 

0 
0 

Lls 

………………… (1.18)
matrix
can
be
further
simplified and represented as
 Lls  3M
 0

 0
1
C LSS C   
 0
 0

 0
0
Lls  3M
0
0
0
0
0
0
0
0
Lls
0
0
Lls
0
0
0
0
0
0
0
0
Lls
0
0
0 
0

0
0

Lls 
…………………….. (1.19)
Likewise the rotor mutual inductance matrix of equation. (1.11)
can be reduced and represented as shown in equation. (1.20).
18
C LSS C 1
3M cos( r )  3M sin( r )
 3M sin( ) 3M cos( )
r
r


0
0

0
0


0
0

0
0

0 0 0 0
0 0 0 0
0 0 0 0
 ……………………….. (1.20)
0 0 0 0
0 0 0 0

0 0 0 0
The neutrals of both the three phase groups are isolated and there are
no
triplen
components
in
the
phase
currents,
the
subspace
corresponding to the triplen components can be removed from
equation(1.12) and the resulting equation simplified as
 Lls  3M
V 
is 
 0
V 
i 
   [ Rr ]4 x 4  s   d 
V1
 i1s  dt  0

 
 
V 2 
 is 
 0
3M cos( r )  3M sin( r )

d  3M sin( r ) 3M cos( r )
0
0
dt 

0
0

0
0
Lls  3M
0
0
Lls
0
0
0  is 
0  is 

0   i1s 
 
Lls  i2 s 
0 0 ir 
0 0 ir  ……………………………………………… (1.21)
0 0  i1r 
 
0 0 i2 r 
where,
is , is are the two orthogonal components of stator currents spanning
subspace S1 .
i1s , i2 s are the two orthogonal components of stator currents spanning
subspace S 2 .
ir , ir are the two orthogonal components of stator currents spanning
subspace S1 .
Ilr, i2s are the two orthogonal components of rotor currents spanning
subspace S2.
19
Likewise to the equation (1.5), the rotor voltage equation can be
written as,
0=[Rr][ir]+d/dt([Lrr][ir]+[Lrs][is])
 rs 0 0 0 0 0 


 0 rs 0 0 0 0 
 0 0 rs 0 0 0 
[ Rr ]  

 0 0 0 rs 0 0 
0 0 0 0 r 0
s


 0 0 0 0 0 rs 


 Llr
0

0
Lrr   
0
0

 0
0
0
0
0
Llr
0
0
Llr
0
0
0
0
0
0
0
0
Llr
0
0
Llr
0
0
0
0
………………………… (1.22)
… (1.23)
 1
 cos( )
0

cos(4 )
0 
M
0

cos(5 )
0
cos(8 )
0


Llr 
cos(9 )
cos( )
1
cos(4 )
cos(3 )
cos(3 )
cos(4 )
cos(7 )
cos(8 )
cos(5 )
cos(4 )
cos(8 ) cos(9 ) 
cos(7 ) cos(8 ) 
1
cos( ) cos(4 ) cos(5 )  …

cos( )
1
cos(3 ) cos(4 )
cos(4 ) cos(3 )
1
cos( ) 

cos(5 ) cos(4 ) cos( )
1 
(1.24)
 cos( r )
 cos(   )
r

cos( r  4 )
Lrs   M cos(  5 )
r

cos( r  8 )

cos( r  9 )
cos( r   )
cos( r )
cos( r  3 )
cos( r  4 )
cos( r  7 )
cos( r  8 )
cos( r  4 ) cos( r  5 ) cos( r  8 ) cos( r  9 ) 
cos( r  3 ) cos( r  4 ) cos( r  7 ) cos( r  8 ) 

cos( r )
cos( r   ) cos( r  4 ) cos( r  5 )  (1.25)
cos( r   )
cos( r )
cos( r  3 ) cos( r  4 )

cos( r  4 ) cos( r  3 )
cos( r )
cos( r   ) 

cos( r  5 ) cos( r  4 ) cos( r   )
cos( r ) 
Where [Rr], [Lrr] and [Lrs] mentioned in equation (1.22) are named
rotor resistance matrix, rotor self inductance matrix and rotor to
stator
mutual
inductance
matrix
respectively,
defined
by
the
equations (1.23),(1.24) and (1.25).
The orthogonal transformation can be applied to the rotor
equation (1.22) and the resultant equation can be written as,
0  [C ]Rr [C ]1[C ]ir  


d
[C ]Lrr [C ]1[C ]ir   [C ]Lrs [C ]1[C ]is  ……………........…. (1.26)
dt
Substituting equation (1.23) to equation (1.25) in equation (1.26) and
performing orthogonal transformation, the final rotor equations can be
represented as shown in equation (1.27).
20
i.e.,
 Llr
ir 
0 

i 
0 
   [ Rr ]4 x 4  r   d 
 i1r  dt 
0 

 
 
0 
i2 r 

 3M cos( r ) 3M sin( r )

d  3M sin( r ) 3M cos( r )
dt 


0
0
0
0
 3M
0
0
0
Llr  3M
0
0
0
Llr
0
0
0
0  ir 
0  ir 

0   i1r 
 
Llr  i2 r 
0 0 is 
0 0 i s  ………………………………………….…. (1.27)
0 0  i1s 
 
0 0 i 2 s 
From equation (1.21) and equation (1.27) the corresponding
voltage equations spanning subspaces S1 and S2 can be separated
out as shown by equation.(1.28.1) and equation (1.28.2).
 RS
V 

V 
    0
 0
0

 
0
 
 0
0
0
RS
0
0
Rr
0
0
0  is 
i 
0
  s  
0  ir 
 
Rr  
ir 

0
3M cos( r )  3M sin( r ) is 
 Lls  3M
 

0
Lls  3M 3M sin( r ) 3M cos( r )  i s 
d
 ir 
0
dt  3M cos( r ) 3M sin( r ) Llr  3M

 
0
Llr  3M  i r 
 3M sin( r ) 3M cos( r )
………………… (1.28.1)
Where equation (1.28.1) represents for the subspace S1 and the
equations for the subspace S2 are represented by equation (1.28.2)
V1   RS
V   0
 2  
0  0
  
0  0
Equation
0
0
RS
0
0
Rr
0
0
(1.28.1)
0   i1s 
 Lls



0  i2 s  d  0

0   i1r  dt  0

 
Rr   i1r 
0
represents
0
0
Lls
0
0
Llr
0
0
that
machine model is converted to
the
an
0   i1s 
0  i2 s 
…. …..(1.28.2)
0   i1r 
 
Llr   i1r 
complete
equivalent
six phase
two
phase
induction machine.
To analyze the performance of the
Six phase Induction Motor
controlled by the Space-vector PWM scheme with the dual Two Level
21
and dual Three-level Inverters independently fed six phase Induction
Motor the investigations are carried out by using MATLAB/Simulink
software which is explained in the chapter-2.