Representing unbalanced voltages by components

Fortescue's theorem states that three unbalanced phasors can be represented by three balanced sets of phasors: positive, negative and zero sequence. We show this by writing the phase voltages as sums of sequence components and proving the components can always be found uniquely.

Starting point: operator a

$$a = 1\angle120^\circ = -0.5 + j0.866, \quad a^2 = 1\angle240^\circ, \quad a^3 = 1, \quad 1 + a + a^2 = 0$$

Multiplying a phasor by $a$ rotates it by +120°.

Three balanced sets

 Positive (a-b-c)    Negative (a-c-b)    Zero
     Va1                 Va2           Va0 Vb0 Vc0
      ^                   ^             ^   ^   ^
     / \                 / \            |   |   |
  Vc1   Vb1          Vb2   Vc2        (all in phase)

• Positive: $V_{a1}$, $V_{b1} = a^2V_{a1}$, $V_{c1} = aV_{a1}$
• Negative: $V_{a2}$, $V_{b2} = aV_{a2}$, $V_{c2} = a^2V_{a2}$
• Zero: $V_{a0} = V_{b0} = V_{c0}$

Phase voltages as sums

$$\begin{aligned} V_a &= V_{a0} + V_{a1} + V_{a2} \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} \end{aligned} \qquad\Longleftrightarrow\qquad V_p = A V_s,\ \ A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$$

The components always exist

The determinant of $A$ is $3(a - a^2) = j3\sqrt3 \neq 0$, so $A$ is invertible, and for any given $V_a, V_b, V_c$ there is one unique set of components:

$$\begin{aligned} V_{a0} &= \tfrac13\left(V_a + V_b + V_c\right) \\ V_{a1} &= \tfrac13\left(V_a + aV_b + a^2V_c\right) \\ V_{a2} &= \tfrac13\left(V_a + a^2V_b + aV_c\right) \end{aligned} \qquad\Longleftrightarrow\qquad V_s = A^{-1}V_p$$

(obtained by adding the three equations, and by adding them after multiplying by $a, a^2$ or $a^2, a$, using $1+a+a^2=0$).

Hence a 3-phase unbalanced voltage system is exactly represented by its symmetrical components. A balanced positive sequence system gives $V_{a0} = V_{a2} = 0$; line-to-line voltages always have $V_0 = 0$ because they sum to zero.