SYMMETRICAL COMPONENTS


Symmetrical components allow phase quantities
of voltage and current to be replaced by three
separated balanced symmetrical components
Consider three phase balanced components
Ic1
Ib2
Ia1
Ib1
Ia2
Ic2

where
I a1 = I a1∠0o = I a1
I b1 = I a1∠240o = a 2 I a1
I c1 = I a1∠120o = aI a1
where a = 1∠120o , a 2 = 1∠240o , a 3 = 1∠360o
Energy Conversion Lab
Ia0
Ib0
Ic0
SYMMETRICAL COMPONENTS

Define the operator a


1+a+a2 = 0 where a=1∠120o
The order of phasor



abc: positive phase sequence
acb: negative phase sequence
abc (positive) sequence
acb (negative) sequence
I = I ∠0 = I
1
a

1
a
o
1
a
I a2 = I a2∠0o = I a2
I b1 = I a1∠240o = a 2 I a1
I b2 = I a2∠120o = aI a2
I c1 = I a1∠120o = aI a1
I c2 = I a2∠240o = a 2 I a2
zero sequence
I a0 = I b0 = I c0
Energy Conversion Lab
SYMMETRICAL COMPONENTS

Consider three phase unbalanced currents Ia, Ib, Ic

the symmetrical component of the currents
I a = I a0 + I a1 + I a2
I b = I b0 + I b1 + I b2 = I a0 + a 2 I a1 + aI a2
I c = I c0 + I c1 + I c2 = I a0 + aI a1 + a 2 I a2

the matrix form of the abc currents in term of
symmetrical component
0

I
1
1
1


I
 a
a
 I  = 1 a 2 a   I 1  ==> I abc = AI 012
a
 a 
 b 
 I c  1 a a 2   I a2 
 

the symmetrical component in term of the three phase
current 012
1 *
−1 abc
−1
Ia = A I
Energy Conversion Lab
where A =
3
A
SYMMETRICAL COMPONENTS

Consider three phase unbalanced currents Ia, Ib, Ic

the symmetrical component in term of the three phase
1
current I a0 = (I a + I b + I c )
3
1
I a1 = (I a + aI b + a 2 I c )
3
1
I a2 = (I a + a 2 I b + aI c )
3

Zero sequence current



one-third of the sum of the phase currents
in a three phase system with ungrounded neutral, the
zero sequence current can’t exist (KCL)
if neutral is grounded, zero-sequence current flow
between neutral and ground
Energy Conversion Lab
SYMMETRICAL COMPONENTS

Consider three phase unbalanced voltages Va, Vb, Vc

the symmetrical component of the currents
Va = Va0 + Va1 + Va2
Vb = Vb0 + Vb1 + Vb2 = Va0 + a 2Va1 + aVa2
Vc = Vc0 + Vc1 + Vc2 = Va0 + aVa1 + a 2Va2

the matrix form of the abc currents in term of
symmetrical component
0

V
1
1
1


V
 a
a 
V  = 1 a 2 a  V 1  ==> V abc = AV 012
a
 a 
 b 
Vc  1 a a 2  Va2 
 

the symmetrical component in term of the three phase
current
1 *
012
−1 abc
−1
Va
Energy Conversion Lab
=A V
where A =
3
A
Three Phase Transformations


Transformation is used to decouple variables
with time-varying coefficients and refer all
variables to a common reference frame
Transformation to decouple abc phase
variables



[f012]=[T012][fabc]
1 1 1 
[T012 ] = 1 1 a a 2 
3
1 a 2 a 


where a = e
j
2π
3
= 1∠120o
The symmetrical transformation is applicable
to steady-state vectors or instantaneous
quantities
SYMMETRICAL COMPONENTS

Consider three phase unbalanced currents Va, Vb, Vc

the symmetrical component in term of the three phase
1
Va0 = (Va + Vb + Vc )
current
3
1
Va1 = (Va + aVb + a 2Vc )
3
1
Va2 = (Va + a 2Vb + aVc )
3

Three phase complex power in terms of symmetrical
components




S3φ = VabcT Iabc* = (AVa012)T(AIa012)*
since AT=A, ATA*=3
S3φ = 3(Va012)T(Ia012)* = 3 Va0 Ia0* + 3 Va1 Ia1* + 3 Va2 Ia2*
total unbalanced power can be obtained from the sum of
the symmetrical component powers
Energy Conversion Lab
Sequence Impedances of Y-connected Loads

Consider a three phase balanced load with self and
mutual elements (Fig. 10.4 PSA-Saddat)
Vabc = Zabc Iabc
Va   Z s + Z n
V  =
Z + Z
n
 b  m
Vc   Z m + Z n
Zm + Zn
Zs + Zn
Zm + Zn
Zm + Zn   Ia 
Z m + Z n   I b 
Z s + Z n   I c 
Zabc
1 1 1 


2
1 a a  = A
1 a a 2 


Z abc
 Zs + Zn
Z + Z
=
n
 m
 Z m + Z n
Zm + Zn
Zs + Zn
Zm + Zn
Zm + Zn 
Z m + Z n 
Z s + Z n 

Sequence Impedances of Y-connected Loads
Consider a three phase balanced load with self
and mutual elements in the above figure




voltage equation: Vabc = Zabc Iabc
use transformation: AVa012=ZabcAIa012
Va012=Z012Ia012, where Z012 = A-1ZabcA
Z012 in case of the above figure
Z 012



 Z s + 3Z n + 2 Z m
0
=


0
0 
Zs − Zm
0 

0
Z s − Z m 
0
1 1 1 


2
1
a
a
=A

1 a a 2 


impedances of nonzero terms appears in principle
diagonal
for a balanced load, three sequence impedances are
independent
current of each phase sequence produces voltage
drops of the same phase sequence only
Park Transformation
Park transformation to decouple
three-phase quantities into twophase variables (generator notation)



[fdq0]=[Tdq0(θd)][fabc]
generator notation, θq = θd + π/2

2π 

θ
θ
−
cos
cos


d
d

3



2π
2

Tdq 0 (θ d ) = − sin θ d - sin θ d −
3
3

1
1

2
2

[
]

2π 

cosθ d +

3 

2π


 - sin θ d +
3


1
2











cos θ d
- sinθ d

 
2π 
2π 

−1
Tdq 0 (θ d ) = cosθ d −

 - sin θ d −
3
3



 

2π 
2π 



 - sin θ d +
 cosθ d +
3
3





[
relationship between qd and abc
quantities,



]
b
positive d-axis is along with magnetic q
field winding axis
positive q-axis is along with internal
voltage ωLaf if
c
internal voltage leads magnetic field by
90 degree (generating)
d
θd

1


1


1

ω=ωs
a ω=0
Park Transformation

Park transformation (motor notation)




[fdq0]=[Tdq0(θd)][fabc]
motor notation, θq = θd - π/2
2π 
2π 



cos θ d cos θ d − 3  cos θ d + 3 





[Tdq 0 (θd )] = 23 sin θd sinθd − 23π  sinθd + 23π  



 

1
1

1

2
2
2


relationship between qd and abc
quantities,



positive d-axis is along with magnetic b
field winding axis
positive q-axis is along with negative of
the internal voltage ωLaf if (induced
voltage – motoring)
c
d-axis is referred from a-axis
d ω=ωs
θd
q
a ω=0
Park Transformation

Park transformation to decouple
abc phase variables


[fqd0]=[Tqd0(θq)][fabc]
generator notation, θq = θd + π/2
2π 
2π 



cos θ q cos θ q − 3  cos θ q + 3 





[Tqd 0 (θq )] = 23 sin θq sinθq − 23π  sinθq + 23π  





1
1

1

2
2
2




[T
qd 0
(θ q )
]
−1


cos θ q
- sinθ q

 
2π 
2π

= cosθ q −
 sin θ q −
3 
3

 

2π 
2π


 sin θ q +
 cosθ q +
3 
3










1


1


1

relationship between qd and abc
quantities,



q-axis is along with internal voltage
d-axis is along with the magnetic
field
q-axis is referred from a-axis
b
q
θq
c
d
ω=ωs
a ω=0
Transformation Between abc and qd0

Starting from positive sequence vector




i1  1 a
   2
i2  = 1 a
i   1 1
 0 
3 3
 
a  ia 
  
a  ib 

1  ic 
 
3
2
()
i2 = i1
*
()
3 *
= i
2

 

2
i 
1 a a  ia 
  * 2 
  
(i )  = 1 a 2 a  ib 
  3 
 i 
1
1
1
i
0 
 c 

 
2 2 2 
Let i = iqs − jids , the second row can be cancelled, the above
matrix can be reformed in terms of real part and imaginary part

 

2
iqs 
1 R(a) R( a )  ia 
 s  2 
  
2
id  = 1 - I(a ) - I(a)  ib 

  3  1 1


1
i
i0 
 c 

2 
2 2
1
1 

1


2
2  i 
iqs 
 a

 s  2 
3
3   
id  = 1  ib 
2
2
3

 
 i 

i0 
1
1
1
 c 

2 2
2 

[i ] = [T ][i ]
s
qd 0
s
qd 0
abc
Transformation Between abc and qd0

Balanced three-phase current in term of t


ia = I m cos(ωet + φ ), ib = I m cos(ωet −
2π
4π
+ φ ), ic = I m cos(ωet −
+φ)
3
3
Using the qd0 transformation, iqd0 becomes

iqs = I m cos(ωet + φ )
π
ids = − I m sin(ωet + φ ) = I m cos(ωet + φ + )
2
i0 = 0

Scaled current space vector

 s
i = iq − jids = I m {cos(ωet + φ ) + j sin(ωet + φ )}
~ jω e t
j (ωe t +φ )
jφ jω e t
= I me
= I me e = 2I ae
1
~
where I a =
I m e jφ , which is the phasor quantity
2
Transformation Between abc and qd0

Scaled current space vector
 s
i = iq − jids = I m e j (ωet +φ )


clearly for balanced three-phase current, iqs and ids
are orthogonal and they have the same peak value
as the abc phase current
Ids peaks 90o ahead of iqs and the resultant current
I rotates counter-clockwise at a speed of ωe from
initial position of φ to the a phase axis at t=0
b
q
c
d
a
 s
i = iq − jids
Transformation Between qd0 to arbitrary
reference frame

New rotating qd axes with stationery qd axes


s
iq  cos θ − sin θ  iq 
 =
 s 
sin
cos
θ
θ
i

 id 
 d
t
θ (t ) = ∫ ω (t )dt + θ (0)
0
qd component space vector form
(
)
iq − jid = iq − jid e − jθ


s
s
the above equation implies rotating stationery qd components
backward by angle θ
synchronous rotating frame w.r.t. stationery frame
(
)
iq − jid = iq − jid e − j (ωet +θe ( 0 ))
e
e
s
s
= I m cos(φ − θ e (0)) + jI m sin(φ − θ e (0))



quantities in synchronous frame are constant
relationship between syn. frame and peak value phasor of a phase
current
~
i = iqs − jids = (iqe − jide )e jωet = 2 I a e jωet
~
∴ (iqe − jide ) = 2 I a
syn. frame quantities and peak value phasor quantities of phase a
current are the same
Transformation Between abc and qd0

Full transformation from stationery qd frame to
arbitrary qd rotating frame

full transformation form
iq  cos θ − sin θ
  
id  = sin θ cos θ
i0  0
0
 

s
0 iq 
 s

0 id 

1 i0 
 
In matrix notation, [iqd0] in terms of original abc currents [iabc]
[iqd 0 ] = [Tθ ][iqds 0 ] = [Tθ ][Tqds 0 ][iabc ] = [Tqd 0 ][iabc ]

Total instantaneous power into three phase circuit in
arbitrary qd0 frame


Pabc = va ia + vbib + vcic =
3
(vqiq + vd id ) + 1 v0i0
2
3
no restriction on abc currents, could be balanced or
unbalanced
Project 4-1
Complex quantities in transformation


Transform the instantaneous three-phase ac
current to space vectors in positive and
negative-sequence in the spatial domain
 The abc currents are of the form
ia=10cos(2πt)
ib=10cos(2πt-2π/3)
ic=10cos(2πt+2π/3)
Using the following dq0 transformation matrix

2π 

θ
θ
cos
cos
−


d
d

3



2π
2

Tdq 0 (θ d ) = − sin θ d - sin θ d −
3
3

1
1

2
2
[
]
2π 

cosθ d +

3 

2π


 - sin θ d +
3


1
2









Project 4-1
Complex quantities in transformation

Show the two rotating space vectors id and iq
components corresponding to sinusoidal and
complex phase currents

run the dq0 sequence component in





stationary frame ωe=0 frame
rotating frame ωe
-ωe frame
2 ωe frame
5 ωe frame
Sequence Current Space Vector

Sequence space vector



i1 = (i2 ) *
Balanced three-phase current in term of t
2π
4π
 i = I cos(ω t ), i = I cos(ω t −
ω
),
i
=
I
cos(
t
−
)
a
m
e
b
m
e
c
m
e
3
3
Sequence current space vector

3
I m e jωet
2
*
3
− jωe t
2
= i1
i2 = ia + a ib + aic = I m e
2
i1 = ia + aib + a 2ic =
()

Resultant airgap mmf
(
)


i1  1 a
   2
i2  = 1 a
i   1 1
 0 
3 3
 
a  ia 
  
a  ib 

1  ic 
 
3
2
N sin
 N sin  3 
jθ a
− jθ a
=
Fs =
i2e + i1e
 I m  cos(θ a − ωet )
4
 2  2 
for balanced 3 phase currents, Fs is a rotating space
vector which has a sinusoidal spatial distribution
around the airgap with speed of ωe

Relation Between Space Vector And
Phase Quantity

Current space vector and phase currents

current space vector

peak value of phase current
i1 =
3
I m e jωet
2
i = I m e jωet

relations between current space vector and phase expression
()
2
i = i1
3

()
3
or i1 = i
2
Balanced sequence current space vector

3
i1 = ia + aib + a ic = I m e jωet
2
3
i2 = ia + a 2ib + aic = I m e − jωet = i1
2
1
1
i0 = (ia + ib + ic ) = I m ∗ 0 = 0
3
3
2
()
*
qd0 Transformation to Series RL

Consider a three phase balanced RL transmission line
with self and mutual elements
[Vs ] − [VR ] = [ R ][i ] + p[ L][i ]
Where
vasgs 


abc
[VS ] = vbsgs 

v
csgs


varg r 


[VRabc ] = vbrgr 
v 
 crgr 
ra + rg rg
rg 


[ R abc ] =  rg rb + rg rg 

 r
r
r
r
+
g
g
c
g


 Laa + Lgg − 2 Lag

[ Labc ] =  Lab + Lgg − Lag − Lbg
L + L − L − L
gg
ag
cg
 ac
Lac + Lgg − Lcg − Lag 

Lbb + Lgg − 2 Lbg
Lbc + Lgg − Lcg − Lbg 
Lbc + Lgg − 2 Lbg − Lcg Lcc + Lgg − 2 Lcg 
Lab + Lgg − Lbg − Lag
qd0 Transformation to Series RL

Consider a three phase balanced line with self and mutual
elements in Fig. 5.17







voltage equation: Δ[Vabc]= [Rabc][iabc]+p [Labc][iabc]
use transformation:
[Tqd0(θ)] -1Δ Vaqd0= [Rabc] [Tqd0(θ)]-1 [iqd0]+ [Labc] p [Tqd0(θ)]-1 [iqd0]
[Tqd0(θ)] [Tqd0(θ)]-1 Δ Vqd0= [Tqd0(θ)] [Rabc] [Tqd0(θ)]-1 [iqd0]+ [Tqd0(θ)]
[Labc] p ( [Tqd0(θ)]-1 [iqd0] )
Δ Vqd0= [Rqd0] [iqd0]+ [Lqd0] p[iqd0]+ [Tqd0(θ)] [Labc] [iqd0] p [Tqd0(θ)]-1
[Rqd0]=[Tqd0(θ)] [Rabc] [Tqd0(θ)]-1,
[Lqd0]= [Tqd0(θ)] [Labc] [Tqd0(θ)]-1
Rqd0 and Lqd0 in case of Fig. 5.17
R qd 0


=


rs − rm
0
0
rs − rm
0
0

0 

rs + 2rm 
0
Lqd 0

=


Ls − Lm
0
0

0 
Ls − Lm

0
Ls + 2 Lm 
0
impedances of nonzero terms appears in principle diagonal
0
qd0 Transformation to Series RL

Transform from abc to qd0 equivalent circuit
speed voltage
speed voltage
∆vq = (rs − rm )iq + ( Ls − Lm )
diq
+ ( Ls − Lm )id
dθ q
dt
dt
dθ q
did
∆vd = (rs − rm )id + ( Ls − Lm )
− ( Ls − Lm )iq
dt
dt
di
∆v0 = (rs + 2rm )i0 + ( Ls + 2 Lm ) 0
dt
Space Vector and Transformations

Air gap mmf due to current ia(t)


Fa1=(Nsine/2) ia(t) cos(θa), Fa1 is centered about aphase winding axis
space vector notation
Fa1

N sin
=
ia ,
2
ia =i a (t ) cos θ a
where
Resultant airgap mmf by currents flowing into
all three windings
 


e +e
cos θ a =
Fs = Fa1 + Fb1 + Fc1
(
jθ a
− jθ a
i1  1 a
   2
i2  = 1 a
i   1 1
 0 
3 3
2
)
N sin
N
ia + ib + ic = sin (ia cos θ a + ib cos θ b + ic cos θ c )
2
2
2π
4π
2π
4π
j
j
−j
−j



N sin  jθ a 
θ
j
−
3
3 
3
3 
a


+ e  ia + ib e + ic e 
=
+ ic e
e  ia + ib e

4  



N
N
= sin e jθ a ia + ib a 2 + ic a + e − jθ a ia + ib a + ic a 2 = sin i2 e jθ a + i1e − jθ a
4
4
=
{ (
)
(
)}
(
)
a 2  ia 
  
a  ib 

1  ic 
 
3
Project. 4-2 qd0 Transformation

Given a one-line diagram of a three-phase
system as shown below, sketch the inputoutput relations between the qd0 component


generator is represented by an equivalent voltage
source E behind a source inductance Lg
since it is a three-wire system, no zero sequence
component. zero-sequence circuit is omitted