Derive symmetrical components of currents

Any unbalanced set of three-phase currents can be expressed as the sum of a zero sequence, a positive sequence and a negative sequence balanced set; the components of phase a are $I_{a0} = \frac13(I_a+I_b+I_c)$, $I_{a1} = \frac13(I_a+aI_b+a^2I_c)$ and $I_{a2} = \frac13(I_a+a^2I_b+aI_c)$.

Derivation for currents

Let $I_a, I_b, I_c$ be an unbalanced set of line currents. By Fortescue's theorem each is the sum of zero, positive and negative sequence components:

$$\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{b0} + I_{b1} + I_{b2} \\ I_c &= I_{c0} + I_{c1} + I_{c2} \end{aligned}$$

Using operator $a = 1\angle120^\circ$ (so $a^2 = 1\angle240^\circ$, $1 + a + a^2 = 0$, $a^3 = 1$), all components are written in terms of phase a:

• Zero sequence: $I_{b0} = I_{c0} = I_{a0}$
• Positive sequence (a-b-c): $I_{b1} = a^2 I_{a1}$, $I_{c1} = a I_{a1}$
• Negative sequence (a-c-b): $I_{b2} = a I_{a2}$, $I_{c2} = a^2 I_{a2}$

Therefore

$$\begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} \quad\Rightarrow\quad I_p = A I_s$$

Solving for the components:

1. Add the three equations: $I_a + I_b + I_c = 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} = 3I_{a0}$
2. Multiply the second by $a$ and the third by $a^2$, then add: $I_a + aI_b + a^2I_c = (1+a+a^2)I_{a0} + (1+a^3+a^3)I_{a1} + (1+a^2+a^4)I_{a2} = 3I_{a1}$
3. Multiply the second by $a^2$ and the third by $a$, then add: $I_a + a^2I_b + aI_c = 3I_{a2}$

$$\begin{aligned} I_{a0} &= \tfrac13\left(I_a + I_b + I_c\right) \\ I_{a1} &= \tfrac13\left(I_a + aI_b + a^2I_c\right) \\ I_{a2} &= \tfrac13\left(I_a + a^2I_b + aI_c\right) \end{aligned}$$

or in matrix form

$$I_s = A^{-1}I_p, \qquad A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$$

Components of phases b and c follow from those of phase a using the relations above.

Notes: $I_n = I_a + I_b + I_c = 3I_{a0}$, so zero sequence current exists only if there is a neutral or ground return. In a 3-wire or delta system $I_{a0} = 0$.