_06_ELC4340_Spring13_Symmetrical_Components.doc
Symmetrical Components
Transformation matrices and the decoupling that occurs in balanced three-phase systems.
Physical significance of zero sequence.
1.
Transformation Matrix
Fortescue's Theorem: An unbalanced set of N related phasors can be resolved into N systems of
phasors called the symmetrical components of the original phasors. For a three-phase system
(i.e. N = 3), the three sets are:
1. Positive Sequence - three phasors, equal in magnitude, 120o apart, with the same sequence
(a-b-c) as the original phasors.
o
2. Negative Sequence - three phasors, equal in magnitude, 120 apart, with the opposite
sequence (a-c-b) of the original phasors.
3. Zero Sequence - three identical phasors (i.e. equal in magnitude, with no relative phase
displacement).
The original set of phasors is written in terms of the symmetrical components as follows:
~
~
~
~
Va  Va0  Va1  Va2 ,
~
~
~
~
Vb  Vb0  Vb1  Vb2 ,
~
~
~
~
Vc  Vc0  Vc1  Vc2 ,
where 0 indicates zero sequence, 1 indicates positive sequence, and 2 indicates negative
sequence.
The relationship among the sequence components for a-b-c are
Positive Sequence
Negative Sequence
Zero Sequence
~
~
Vb1  Va1  1  120 o
~
~
Vb 2  Va 2  1  120 o
~
~
~
Va0  Vb0  Vc0
~
~
Vc1  Va1  1  120 o
~
~
Vc 2  Va 2  1  120 o
The symmetrical components of all a-b-c voltages are usually written in terms of the symmetrical
components of phase a by defining
a  1  120 o , so that a 2  1  240 o  1  120 o , and a 3  1  360 o  10 o .
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_06_ELC4340_Spring13_Symmetrical_Components.doc
~ ~ ~
Substituting into the previous equations for Va ,Vb ,Vc yields
~
~
~
~
Va  Va0  Va1  Va2 ,
~
~
~
~
Vb  Va 0  a 2Va1  aVa 2 ,
~
~
~
~
Vc  Va 0  aVa1  a 2Va 2 .
In matrix form, the above equations become
~
Va  1 1
~  
2
Vb   1 a
V~c  1 a
 
~
 Va 0 
 V~  ,
  a1 
2  ~ 
a  Va 2 
1
a
or in matrix form
~
~
~
~
Vabc  T  V012 , and V012  T 1  Vabc ,
where transformation matrix T is
1 1
T  1 a 2
1 a
1
1 1

1 1 
a  , and T  1 a
3
1 a 2
a 2 
1
a 2  .
a 
~
~
~
~
~
~
~
If Vabc represents a balanced set (i.e. Vb  Va  1  120 o  a 2Va , Vc  Va  1  120o  aVa ),
~
~
then substituting into V012  T 1  Vabc yields
~
Va 0 
1 1
~  1
Va1   3 1 a
V~a 2 
1 a 2


~
1   Va   0 
 ~ 
~
a 2  a 2Va   Va  .
~
a   aVa   0 
Hence, balanced voltages or currents have only positive sequence components, and the positive
sequence components equal the corresponding phase a voltages or currents.
~
~
~
~
~
~
If Vabc is an identical set (i.e. Va  Vb  Vc ), substituting into V012  T 1  Vabc yields
~
Va 0 
1 1
~  1
Va1   3 1 a
V~a 2 
1 a 2


~
~
1  Va  Va 
~   
a 2  Va    0  ,
~
a  Va   0 
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_06_ELC4340_Spring13_Symmetrical_Components.doc
which means that Va ,Vb ,Vc have only zero sequence components, and that these components are
identical and equal to Va .
~
~
Notice from the top row of V012  T 1  Vabc that V0 is one-third of the sum of the three phase
voltages or currents. Therefore, since the sum of three line-to-line voltages is identically zero
due to Kirchhoff's voltage law, line-to-line voltages can have no zero sequence components.
1.
Relationship Between Zero Sequence Currents and Neutral Current
Consider the relationship between zero sequence current and neutral current The zero sequence
current is


1 ~ ~ ~
~
I a0  I a  I b  I c ,
3
and, from Kirchhoff's current law, the neutral current is


~
~ ~ ~
I n   I a  Ib  Ic .
Because the positive and negative sequence components of the a-b-c currents sum to zero, while
~
~
the zero sequence components are additive, then I n  3I a0 . Therefore, in a four-wire, threephase system, the neutral current is three-times the zero sequence current. In a three-wire, threephase system, there is no zero sequence current.
a
Ia
b
Ib
3 Phase,
Ic
4 Wire
c
n
In = 3Io
System
In = 3Io = Ia + Ib + Ic
Figure 1. Relationship Between Zero Sequence Currents and Neutral Current
(note – the neutral current is shown above as flowing out)
3.
Decoupling in Systems with Balanced Impedances
In a three-phase system with balanced impedances, the relationship among voltage, current, and
impedance has the form
~
Va   S
~  
Vb    M
V~c   M
 
M
S
M
~
M  I a 
~ 
~
~
M   I b  , or Vabc  Z abc  I abc ,
~
S   I c 
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_06_ELC4340_Spring13_Symmetrical_Components.doc
where S represents the self impedances of the phases, and M represents the mutual impedances.
~
~
This equation can be expressed in terms of sequence components by substituting Vabc  T  V012
~
~
and I abc  T  I 012 , yielding
~
~
T  V012  Z abc  T  I 012 .
Premultiplying by T 1 yields
~
~
~
V012  T 1Z abc  T  I 012  Z 012  I 012 ,
where Z 012  T 1Z abc  T . The symmetric form of Z abc given above yields
 S  2M
Z 012   0
 0
0
S M
0

0  ,
S  M 
0
which means that when working in sequence components, a circuit with symmetric impedances
is decoupled into three separate impedance networks with Z 0  S  2M , and Z1  Z 2  S  M
. Furthermore, if the voltages and currents are balanced, then only the positive sequence circuit
must be studied.
In summary, symmetrical components are useful when studying either of the following two
situations:
1. Symmetric networks with balanced voltages and currents. In that case, only the positive
sequence network must be studied, and that network is the "one-line" network.
2. Symmetric networks with unbalanced voltages and currents. In that case, decoupling applies,
and three separate networks must be studied (i.e. positive, negative, and zero sequences).
The sequence components of the voltages and currents can be transformed back to a-b-c by
using the T transformation matrix.
4.
Power
For a three-phase circuit, with voltages referenced to neutral,
~ ~
~ ~
~ ~
~T ~*
Pabc  Van I a*  Vbn I b*  Vcn I c*  Vabc
 I abc .
~
~
~
~
Substituting in Vabc  T  V012 and I abc  T  I 012 yields
~T
~*
Pabc  V012
 T T  T *  I 012
.
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_06_ELC4340_Spring13_Symmetrical_Components.doc
1 1
Since T  1 a 2

1 a
1
1 1

T
* 
a  , then T  T . Also, T  1 a
1 a 2
a 2 
1
a 2  . Therefore,
a 
 3 0 0
3 0 0
~T
~*


, or
T  T  0 3 0 , so that Pabc  V012  0 3 0  I 012
0 0 3
0 0 3
T
*


~ ~
~ ~
~ ~
Pabc  3 Va 0 I a*0  Va1 I a*1  Va 2 I a*2 .
Note the factor of three. If desired, the following power invariant transformation can be used to
avoid the factor of three:
1 1
1 
T
1 a2

3
1 a
1
1 1
1

1 a
a  , T 1 
3
1 a 2
a 2 
1
a 2  ,
a 
~ ~
~ ~
~ ~
so that Pabc  Va 0 I a*0  Va1 I a*1  Va 2 I a*2 .
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