EE 426

DC Machine Modelling
dia Tm = Kψf ia
va = Eb + ia R a + La
dt ψf = Lf if
dif dω
vf = if R f + Lf Tm − Tl = J + Bω
dt dt
Eb = Kψf ω

Field Model

Armature Model

La
Electrical Time Constant (τa ) =
Ra
J
Mechanical Time Constant (τm ) =
B

This is further simplified to obtain the following block diagram.

Closed Loop Speed Control

• Inner control loop is provided to limit the converter and motor

current or motor torque below a safe limit.

• The PI Controller serves 3 purposes:

o Stabilizes the drive and adjusts the damping ratio at the desired
value.
o Makes the steady state error close to zero by integral action.
o Filters out noise by integral action.
Working of the Scheme
• Any positive speed error, caused by either an increase in the speed
command or an increase in the load torque, produces a higher
current reference Ia∗ .
• The motor accelerates due to an increase in Ia , to correct the speed
error and finally settles at a new Ia∗ which makes the motor torque
equal to the load torque and the speed error close to zero.

• For any large positive speed error, the current limiter saturates and
the current reference Ia∗ is limited to a value Iam
∗
, and the drive
current is not allowed to exceed the maximum permissible value.
• The speed error is corrected at the maximum permissible armature
current until the speed error becomes small and the current limiter
comes out of saturation.
• Now the speed error is corrected with Ia less than the permissible
value.

• A negative speed error will set the current reference Ia∗ at a negative
value.
• Since the motor current cannot reverse, a negative Ia∗ is of no use.
• It will however “charge” the PI controller.
• When the speed error becomes positive, the “charged” PI controller
will take a longer time to respond, causing unnecessary delay in the
control action.
• The current limiter is therefore arranged to set a zero – current
reference for negative speed errors.

Advantages of the Scheme

• Precise speed regulation even under varying load conditions.
• Improved torque control.
• Enhanced motor protection against overcurrent.
• Ability to maintain stable speed despite external disturbances.
Design of Current Controller

Ki K p K p (1 + sτi ) Kp
TF of PI Controller = K p + = Kp + = , τi =
s sτi sτi Ki

For simplifying calculations, we assume τi = τa .

Ia (s) 1 1
= =
Ia∗ (s) 1 + s τi R a 1 + sτcurrent
Kp
La Kp
⇒ Kp = , Ki =
τcurrent τa

Design of Speed Controller

Ki K p K p (1 + sτi ) Kp
TF of PI Controller = K p + = K p + = , τi =
s sτi sτi Ki

For simplifying calculations, we assume τi = τm .

ω(s) 1 1
= =
ω∗ (s) 1 + s τi B 1 + sτspeed
Kp
J Kp
⇒ Kp = , Ki =
τspeed τm

• Generally, τspeed = 10 × τcurrent (the speed controller is 10 times

slower than the current controller).

Braking Operation of Rectifier Controller Separately Excited DC Motor

• The direction of armature current cannot be reversed due to the

presence of fully controlled rectifier.

• Therefore, the only option available for the reversal of the flow of
power is to reverse both the rectifier output voltage Va and the
motor back emf E with respect to the rectifier terminals and make
|E| > |Va |.

• The rectifier output voltage can be reversed by making α > 90°.

• The condition |E| > |Va | can be satisfied for any motor speed by
choosing an appropriate value of α in the range 90° < α < 180°.

• The reversal of the motor emf with respect to the rectifier terminals
can be done by any of the following changes:
 o An active load coupled to the motor shaft may drive it in the
reverse direction. This gives reverse regeneration.
o The field current may be reversed, with the motor running in
the forward direction. This gives forward regeneration.
o The motor armature connections may be reversed with repsect
to the rectifier output terminals, with the motor still running in
the forward direction. This gives forward regeneration.

Controlled Rectifier Fed Separately Excited DC Motor

1) 1 – Phase Fully Controlled Rectifier

CCM:
1 π+α
Va = ∫ Vm sin(ωt) d(ωt)
π α
2Vm
= cos(α)
π
DCM:
β
1
Va = (∫ Vm sin(ωt) d(ωt)
π α
π+α
+∫ E d(ωt))
β
Vm (cos(α) − cos(β)) + E(π + α − β)
=
π

Powering Interval (α to β)
dia
va = vs = Vm sin(ωt) = E + ia R a + La
dt
Vm E t La
−τ
⇒ ia = sin(ωt − θ) − + Ae a , θ = tan−1 (ωτa ) , τa =
Z Ra Ra

Using the boundary condition ia (ωt = α) = 0, we obtain

E Vm
A=( − sin(α − θ)) eαcot (θ)
Ra Z

Vm E E Vm
⇒ ia = sin(ωt − θ) − +( − sin(α − θ)) e−(ωt−α)cot (θ)
Z Ra Ra Z

Using the boundary condition ia (ωt = β) = 0, we obtain

Vm E E Vm
sin(β − θ) − +( − sin(α − θ)) e−(β−α) cot(θ) = 0
Z Ra Ra Z

The boundary between CCM and DCM is given by β = π + α.

R a Vm e−π cot(θ) + 1
⇒ ωc = sin(α − θ) ( −π cot(θ) )
KZ e −1
• CCM/DCM can be determined by the motor speed/torque:
ω > ωc ⇒ DCM T < Tc ⇒ DCM
ω < ωc ⇒ CCM T > Tc ⇒ CCM

• The ideal no – load speed is obtained when Ia = 0.

• For 0 ≤ α ≤ π/2, Ia = 0 when E ≥ Vm .
• For π/2 ≤ α ≤ π, Ia = 0 when E ≥ Vm sin (α).

Vm π
, 0≤α≤
⇒ ωm0 ={ K 2
Vm sin(α) π
, ≤α≤π
K 2

2) 1 – Phase Half Controlled Rectifier

CCM:
1 π
Va = ∫ Vm sin(ωt) d(ωt)
π α
Vm
= (1 + cos(α))
π

DCM:
π
1
Va = (∫ Vm sin(ωt) d(ωt)
π α
π+α
+∫ E d(ωt))
β
Vm (1 + cos(α)) + E(π + α − β)
=
π

Powering Interval (α to π)
dia
va = vs = Vm sin(ωt) = E + ia R a + La
dt
Vm E E Vm
⇒ ia = sin(ωt − θ) − +( − sin(α − θ)) e−(ωt−α)cot (θ)
Z Ra Ra Z
⇒ ia (ωt = π) = iaπ

Freewheeling Interval (π to β)
dia
va = 0 = E + ia R a + La
dt
E
⇒ ia = − + Ae−ωtcot(θ)
Ra

Using the boundary condition ia (ωt = π) = iaπ , we obtain

 E π cot(θ)
A = (iaπ + )e
Ra

E E
⇒ ia = − + (iaπ + ) e−(ωt−π) cot(θ)
Ra Ra

Using the boundary condition ia (ωt = β) = 0, we obtain

E E
− + (iaπ + ) e−(β−π) cot(θ) = 0
Ra Ra

The boundary between CCM and DCM is given by β = π + α.

R a Vm sin(α − θ) e−πcot (θ) − sin(θ) e−αcot (θ)
⇒ ωc = ( )
KZ e−πcot (θ) − 1

The no – load speeds are given by:

Vm π
, 0≤α≤
ωm0 = { K 2
Vm sin(α) π
, ≤α≤π
K 2
3) 3 – Phase Fully Controlled Rectifier

1 α+2π/3 3Vm
Va = π ∫ Vm sin(ωt) d(ωt) = cos (α)
π
3 α+π/3
Multiquadrant Operation of Separately Excited DC Motor
1) Mechanical Switches

• The purpose of RS is to reverse the motor armature with respect to

the rectifier.

• One setting of switch RS gives operation in the 1st and 4th quadrants.
• The reversal of the armature connection then provides operation in
the 2nd and 3rd quadrants.

• The RS may consist of a relay operated contactor with two Normally

Closed (NC) and two Normally Opened (NO) contacts.
• The RS can also be realized using four thyristors.

• To avoid inductive voltage surges and to reduce the size of the

reversing switch, it is necessary to perform the switching at zero
armature current.

Speed Reversal
• The armature current ia is forced to zero quickly by setting the firing
angle at the highest permissible value.
• When the current value is zero, firing pulses are stopped.

• A dead time of 2 to 10 milliseconds is provided to ensure that the

current has in fact become zero.
• Now the armature is reversed by RS and the firing pulses are
released. The firing angle is already set at the highest value.
• The firing angle is gradually reduced.

• As the firing angle is reduced, the armature current tends to exceed

the safe limit but this is prevented by the current control loop.
• The motor decelerates to zero speed under regenerative braking and
then accelerates in the reverse direction with nearly maximum
torque under current control.
• When the speed reaches close to the steady-state value, the current
reduces and the motor settles at a speed in the reverse direction.

2) Dual Conveter
Non – Circulating Current Type

• Only one rectifier operates at any given time and the other is
blocked.

Speed Reversal
• The armature current ia is forced to zero by setting the firing angle of
rectifier 1 at the highest value.
• After zero current is sensed, a dead time of 2 to 10 ms is provided to
ensure the turn – off of all the thyristors of rectifier 1.
• Now the firing pulses are withdrawn from rectifier 1 and released to
rectifier 2.
• α2 is set to make the armature terminal voltage under continuous
conduction equal to the back emf.
• It is then reduced to make the maximum allowable current to flow.
• The motor torque builds up fast and the speed reversal is achieved at
the maximum torque under the influence of current control.

Circulating Current Type

• Both rectifiers operate simulataneously.

• The two rectifiers are controlled simultaneously in such a manner

that the sum of their average terminal voltages is zero so that no dc
current circulates in the loop formed by the two rectifiers.

⇒ Va1 + Va2 = 0
⇒ Vm cos(α1 ) + Vm cos(α2 ) = 0
⇒ cos(α1 ) = − cos(α2 )
⇒ α1 + α2 = 180°

• Since two rectifiers work in different modes, one in rectification and

another in inversion, their instantaneous voltages are not equal.
• This causes ac circulating current to flow in the loop formed by the
two rectifiers.
• The inductors L1 and L2 are connected to restrict the ac circulating
current.

Speed Reversal
• When operating in 1st quadrant, rectifier 1 will be rectifying
(0° < α1 < 90°) and rectifier 2 will be inverting (90° < α2 < 180°).
• α1 is increased and α2 is decreased.
• The motor back emf exceeds |Va1 | and |Va2 |.
• The armature current shifts to rectifier 2 and the motor operates in
the 2nd quadrant.
• As α2 is decreased gradually, the motor decelerates under
regenerative braking.

• When zero speed is reached α1 ≈ α2 ≈ 90°.

• Reduction of α2 below 90° will make it work as a rectifier and the
motor will accelerate to a speed in the reverse direction.

3) Field Current Reversal

• The field current can be reversed either by using a dual converter or

a single fully controlled rectifier with a reversing switch.

• Since the field current is much smaller compared to the armature

current, the ratings of the rectifiers of the field circuit will be much
smaller compared to those of the armature circuit.

• This makes this drive the cheapest 4 quadrant dc drive.

Speed Reversal
• The armature rectifier firing angle is set at the highest value to force
the armature current ia to zero.
• The firing angle of the rectifier supplying the field is now set for the
highest value.
• It inverts and the field current is forced to zero.

• After a suitable dead time, the second rectifier is activated at the

lowest firing angle.
• When the field current has nearly settled, and the motor back emf
has reversed, the armature rectifier is activated again.
• The firing angle is progressively changed to first brake and then
accelerate the machine in the reverse direction.

Chopper Control of Separately Excited DC Motor

Motoring

Ton
Va = δV, δ=
T
Duty Interval (0 ≤ t ≤ δT)
• Out of the total energy supplied, a part is absorbed by the armature
and converted into mechanical energy, a part is converted into heat
in resistance R a and the switch, and the remaining energy is stored in
the inductance La .

dia
V = E + ia R a + La
dt

Freewheeling Interval (δT ≤ t ≤ T)

• The magnetic energy stored in the inductance La is responsible for
maintaining the flow of the armature current.

dia
0 = E + ia R a + La
dt

Regenerative Braking

Toff
Va = δV, δ=
T
Energy Storage Interval (0 ≤ t ≤ (1 − δ)T)
• The motor terminal voltage remains zero and due to the back emf E
the armature current increases from ia1 to ia2 .
• The mechanical energy supplied by the load and the inertia of the
motor load system (only if the speed is changing) is converted by the
machine into the electrical energy.
• This energy is partly used in increasing the stored magnetic energy in
La , and the remainder is dissipated in R a and the switch.

dia
0 = E − ia R a − La
dt

Energy Transfer Interval ((1 − δ)T ≤ t ≤ T)

• The sum of the energy generated by the machine and the energy
stored in the inductor during the on period of the switch is partly
dissipated in resistance R a , and diode D and the remaining energy is
fed to the source, giving regenerative braking.

dia
V = E − ia R a − La
dt

1 δT
⇒ Regenerated Power (Prg ) = ∫ Via dt
T 0
Motoring and Regenerative Braking (2 Quadrant Chopper)
Ton
Va = δV, δ=
T
Motoring and Reverse Regenerative Braking (2 Quadrant Chopper)
Ton
Va = 2V(δ − 0.5), δ =
2T

Forward Motoring (0.5 < δ ≤ 1)

• S1 and S2 cannot be open simultaneously.
• Therefore, va ≥ 0.
Reverse Regenerative Braking (0 < δ < 0.5)
• S1 and S2 cannot be closed simultaneously.
• Therefore, va ≤ 0.

4 Quadrant Chopper
• If S2 is kept closed continuously and S1 and S4 are controlled, a 2
quadrant chopper is obtained, which can supply a variable positive
terminal voltage and the armature current in either direction.

• If S3 is kept closed continuously and S1 and S4 are controlled, a 2

quadrant chopper is obtained, which can supply a variable negative
terminal voltage and the armature current in either direction.

Rotating Magnetic Field in 3 – Phase Induction Motor

𝐀𝐂𝐁 Mechanical 𝐀𝐁𝐂 Electrical

vas = Vm cos(ωt) ias = Im cos(ωt − ϕ)

2π 2π
vbs = Vm cos (ωt − ) ibs = Im cos (ωt − ϕ − )
3 3
2π 2π
vcs = Vm cos (ωt + ) ics = Im cos (ωt − ϕ + )
3 3
Using the current phasors, it can be shown that MMF = NInet is a
rotating vector having a constant magnitude, i.e.
3NIm
|MMF| =
2
3 – Phase Induction Motor Modelling (𝐚𝐛𝐜 Frame)
Assumptions & Formulas
• ras = rbs = rcs = rs
• Ls = Lls + Lms
• Ms = Lms cos(δ)
• rar = rbr = rcr = rr
• Lr = Llr + Lmr
• Mr = Lmr cos(δ)
• P = d/dt

Notations
rs = Stator per phase resistance
Ls = Stator per phase self inductance
Lls = Stator per phase leakage inductance
Lms = Stator per phase magnetizing indutance
Ms = Mutual inductance b/w stator windings
Msr = Magnitude of mutual inductance b/w stator & rotor windings
Stator Flux Equations
2π
ψas = Ls ias + Ms ibs + Ms ics + Msr cos(θr ) iar + Msr cos (θr + )i
3 br
2π
+ Msr cos (θr − )i
3 cr
2π
ψbs = Ls ibs + Ms ics + Ms ias + Msr cos(θr ) ibr + Msr cos (θr + )i
3 cr
2π
+ Msr cos (θr − )i
3 ar
2π
ψcs = Ls ics + Ms ias + Ms ibs + Msr cos(θr ) icr + Msr cos (θr + )i
3 ar
2π
+ Msr cos (θr − )i
3 br

Rotor Flux Equations

2π
ψar = Lr iar + Mr ibr + Mr icr + Msr cos(θr ) ias + Msr cos (θr − )i
3 bs
2π
+ Msr cos (θr + )i
3 cs
2π
ψbr = Lr ibr + Mr icr + Mr iar + Msr cos(θr ) ibs + Msr cos (θr − )i
3 cs
2π
+ Msr cos (θr + )i
3 as
2π
ψcr = Lr icr + Mr iar + Mr ibr + Msr cos(θr ) ics + Msr cos (θr − )i
3 as
2π
+ Msr cos (θr + )i
3 bs

Stator Voltage Equations Rotor Voltage Equations

vas = rs ias + Pψas var = rr iar + Pψar
vbs = rs ibs + Pψbs vbr = rr ibr + Pψbr
vcs = rs ics + Pψcs vcr = rr icr + Pψcr

We can use matrices to represent the voltage equations.

Voltage Equations in Matrix Form
abc abc
Vabc = RIabc + PLabc abc
abc Iabc

abc
Vabc = [vas vbs vcs var vbr vcr ]T
abc ibs ics iar ibr icr ]T
Iabc = [ias

rs 0 0 0 0 0
0 rs 0 0 0 0
0 0 rs 0 0 0 L11 M12
R= Labc
abc = [ T ]
0 0 0 rr 0 0 M12 L22
0 0 0 0 rr 0
[0 0 0 0 0 rr ]

Ls Ms Ms Lr Mr Mr
L11 = [Ms Ls Ms ] L22 = [Mr Lr Mr ]
Ms Ms Ls Mr Mr Lr

2π 2π
cos(θr ) cos (θr + ) cos (θr − )
3 3
2π 2π
M12 = Msr cos (θr − ) cos(θr ) cos (θr + )
3 3
2π 2π
[ cos (θ r + ) cos (θr − ) cos(θr ) ]
3 3

Torque Equation
dωm
Tm − Tl = J
dt
1 abc T dL abc
Tm = Iabc I
2 dθr abc
2θe
θr =
P
3 – Phase to 2 – Phase Conversion

2π 2π
MMFabc = Ns (ia cos(ϕ) + ib cos ( − ϕ) + ic cos ( + ϕ))
3 3
ib ic √3
⇒ MMFabc = Ns ((ia − − ) cos(ϕ) + (i − ic ) sin(ϕ))
2 2 2 b

π
MMFαβ = Nαβ (iα cos(ϕ) + iβ cos ( − ϕ))
2
⇒ MMFαβ = Nαβ (iα cos(ϕ) + iβ sin(ϕ))

Comparing MMFabc with MMFαβ , we obtain

1 1
Ns 1 − − ia
iα 2 2
[i ] = [ib ]
β Nαβ √3 √3 i
[0 2 − 2 ]
c

We consider α, β, 0 as Positive, Negative, and Zero Sequence

respectively.
Let us assume that
Ns
i0 = k ( ) (i + ib + ic )
Nαβ a

1 1
iα 1 − −
Ns 2 2 ia Ns
⇒ [iβ ] = √3 √3 [ib ] ⇒ Iαβ0 =( ) MIabc
Nαβ 0 − Nαβ
i0 2 2 ic
[k k k ]

1
1 0
2k
2 1 √3 1
⇒ M −1 = −
3 2 2 2k
1 √3 1
[− 2 −
2 2k]

We choose k such that

2
M −1 = MT
3

1 1
1 − −
2 2
1 √3 √3
⇒k= ⇒ M= 0 −
√2 2 2
1 1 1
[√2 √2 √2 ]

1) Total Power Invariant (Clarke’s Transformation)

T
Pabc = va ia + vb ib + vc ic = Iabc Vabc

T
Pαβ0 = vα iα + vβ iβ + v0 i0 = Iαβ0 Vαβ0
Comparing Pabc with Pαβ0 , we obtain
T T T Ns T
Iabc Vabc = Iαβ0 Vαβ0 ⇒ Iabc Vabc = ( ) Iabc MT Vαβ0
Nαβ
Ns
⇒ Vabc = ( ) MT Vαβ0
Nαβ

We know that
Nαβ
Iabc = ( ) M −1 Iαβ0
Ns

We choose the ratio of N such that

Ns Nαβ
( ) MT = ( ) M −1
Nαβ Ns

Ns 2
⇒ =√
Nαβ 3

2 2
⇒ Iαβ0 = √ MIabc and Iabc = √ MT Iαβ0
3 3

2) Per Phase Power Invariant

va ia + vb ib + vc ic 1 T
pabc = = Iabc Vabc
3 3

vα iα + vβ iβ + v0 i0 1 T
pαβ0 = = Iαβ0 Vαβ0
2 2

Comparing pabc with pαβ0 and perfoming similar steps, we obtain

Ns 2
=
Nαβ 3
3 – Phase Induction Motor Modelling (𝛂𝛃 Frame)
Formulas
• L̅s = Lls + 3/2Lms
• L̅r = Llr + 3/2Lmr
• ̅̅̅̅̅
Msr = 3/2Msr

Notations
• Subscript is used to represent stator variables.
• Superscript is used to represent rotor variables.

Stator Flux Equations

π
ψα = L̅s iα + ̅̅̅̅̅
Msr cos(θr ) iα + ̅̅̅̅̅
Msr cos ( + θr ) iβ
2
π
ψβ = L̅s iβ + ̅̅̅̅̅
Msr cos(θr ) iβ + ̅̅̅̅̅
Msr cos ( − θr ) iα
2
ψ0 = Lls i0

Rotor Flux Equations

π
ψα = L̅r iα + ̅̅̅̅̅
Msr cos(θr ) iα + ̅̅̅̅̅
Msr cos ( − θr ) iβ
2
π
ψβ = L̅r iβ + ̅̅̅̅̅
Msr cos(θr ) iβ + ̅̅̅̅̅
Msr cos ( + θr ) iα
2
0 0
ψ = Llr i
Stator Voltage Equations
vα = rs iα + Pψα
dθr α
⇒ vα = rs iα + L̅s Piα + ̅̅̅̅̅
Msr cos(θr ) Piα − ̅̅̅̅̅
Msr sin(θr ) i
dt
dθr β
− ̅̅̅̅̅
Msr sin(θr ) Piβ − ̅̅̅̅̅
Msr cos(θr ) i
dt

vβ = rs iβ + Pψβ
dθr β
⇒ vβ = rs iβ + L̅s Piβ + ̅̅̅̅̅
Msr cos(θr ) Piβ − ̅̅̅̅̅
Msr sin(θr ) i
dt
dθr α
+ ̅̅̅̅̅
Msr sin(θr ) Piα + ̅̅̅̅̅
Msr cos(θr ) i
dt

v0 = rs i0 + Pψ0
⇒ v0 = rs i0 + Lls Pi0

Rotor Voltage Equations

v α = rr iα + Pψα
dθr
⇒ v α = rr iα + L̅r Piα + ̅̅̅̅̅
Msr cos(θr ) Piα − ̅̅̅̅̅
Msr sin(θr ) i
dt α
dθr
+ ̅̅̅̅̅
Msr sin(θr ) Piβ + ̅̅̅̅̅
Msr cos(θr ) i
dt β

v β = rr iβ + Pψβ
dθr
⇒ v β = rr iβ + L̅r Piβ + ̅̅̅̅̅
Msr cos(θr ) Piβ − ̅̅̅̅̅
Msr sin(θr ) i
dt β
dθr
− ̅̅̅̅̅
Msr sin(θr ) Piα − ̅̅̅̅̅
Msr cos(θr ) i
dt α

v 0 = rr i0 + Pψ0
⇒ v 0 = rr i0 + Llr Pi0

We can use matrices to represent the voltage equations.

Voltage Equations in Matrix Form
αβ0 αβ0 αβ0 αβ0 αβ0 dθr αβ0
Vαβ0 = RIαβ0 + Lαβ0 PIαβ0 + Gαβ0 I
dt αβ0

αβ0 T
Vαβ0 = [vα vβ v0 vα vβ v0]

αβ0 T
Iαβ0 = [iα iβ i0 iα iβ i0 ]

rs 0 0 0 0 0
0 rs 0 0 0 0
0 0 rs 0 0 0
R=
0 0 0 rr 0 0
0 0 0 0 rr 0
[0 0 0 0 0 rr ]

L̅s 0 0 ̅̅̅̅̅
Msr cos(θr ) −M̅̅̅̅̅
sr sin(θr ) 0
0 L̅s 0 ̅̅̅̅̅
Msr sin(θr ) ̅̅̅̅̅
Msr cos(θr ) 0
αβ0 0 0 Lls 0 0 0
Lαβ0 =
̅̅̅̅̅
Msr cos(θr ) ̅̅̅̅̅
Msr sin(θr ) 0 L̅r 0 0
−M ̅̅̅̅̅ ̅̅̅̅̅
sr sin(θr ) Msr cos(θr ) 0 0 L̅r 0
[ 0 0 0 0 0 Llr ]

0 0 0 −M̅̅̅̅̅ ̅̅̅̅̅
sr sin(θr ) −Msr cos(θr ) 0
0 0 0 ̅̅̅̅̅
Msr cos(θr ) −M ̅̅̅̅̅
sr sin(θr ) 0
αβ0
Gαβ0 = 0 0 0 0 0 0
̅̅̅̅̅
−M sr sin(θr )
̅̅̅̅̅
Msr cos(θr ) 0 0 0 0
̅̅̅̅̅
−M ̅̅̅̅̅
sr cos(θr ) −Msr sin(θr ) 0 0 0 0
[ 0 0 0 0 0 0]

Alternative Method to Obtain the Voltage Equations

We know that
2
Vαβ0 = √ MVabc
3
 αβ0 C 0 2
⇒ Vαβ0 = [ 11 abc
] Vabc abc
= CVabc C11 = √ M
0 C11 3
abc αβ0
⇒ Vabc = C −1 Vαβ0

The voltage equations in the abc frame are given by

abc abc
Vabc = RIabc + PLabc abc
abc Iabc

αβ0 αβ0 αβ0

⇒ C −1 Vαβ0 = RC −1 Iαβ0 + PLabc −1
abc C Iαβ0

Pre – multiplying both sides by C, we obtain

αβ0 αβ0 −1 αβ0
Vαβ0 = CRC −1 Iαβ0 + CPLabc
abc C Iαβ0

αβ0 αβ0 αβ0 αβ0

⇒ Vαβ0 = CRC −1 Iαβ0 + CLabc −1 abc −1
abc C PIαβ0 + C(PLabc )C Iαβ0

Simplifying each of the terms, we obtain

αβ0 αβ0 αβ0 αβ0 αβ0 dθr αβ0
Vαβ0 = RIαβ0 + Lαβ0 PIαβ0 + Gαβ0 I
dt αβ0