Symmetrical components and complex power

Symmetrical components

Symmetrical components (Fortescue's theorem, 1918) state that any set of three unbalanced phasors can be replaced by the sum of three balanced sets:

• Positive sequence: three equal phasors 120° apart with the same phase sequence as the original (a-b-c).
• Negative sequence: three equal phasors 120° apart with opposite sequence (a-c-b).
• Zero sequence: three equal phasors in phase with each other.

With operator $a = 1\angle120^\circ$:

$$\begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} V_{a0} \\ V_{a1} \\ V_{a2} \end{bmatrix}, \quad\text{i.e. } V_p = A V_s$$

The same relation holds for currents, $I_p = A I_s$. This lets an unbalanced problem be solved as three balanced, independent problems.

Complex power in terms of symmetrical components

Total complex power in phase quantities:

$$S = V_a I_a^* + V_b I_b^* + V_c I_c^* = V_p^T I_p^*$$

Substituting $V_p = AV_s$, $I_p = AI_s$:

$$S = (AV_s)^T (AI_s)^* = V_s^T A^T A^* I_s^*$$

$A$ is symmetric ($A^T = A$), and since $a^* = a^2$:

$$A^* = \begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$$ $$A^T A^* = \begin{bmatrix} 3 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3&0&0\\0&3&0\\0&0&3 \end{bmatrix}$$

using $1+a+a^2 = 0$, $a^3 = 1$, $a^4 = a$. Hence

$$S = 3V_s^T I_s^* = 3\left(V_{a0}I_{a0}^* + V_{a1}I_{a1}^* + V_{a2}I_{a2}^*\right)$$

So total power is three times the sum of the sequence powers of phase a, with no cross products between different sequences. For a balanced system only the positive sequence term remains: $S = 3V_{a1}I_{a1}^*$.