Sequence networks decoupled for static circuit

For a three-phase symmetrical static circuit (line, transformer, static load), the impedance matrix in sequence quantities is diagonal, so a current of one sequence produces voltage drops of that sequence only; this means the three sequence networks are decoupled.

Phase equations

Each phase has self impedance $Z_s$ and equal mutual impedance $Z_m$ to each other phase:

$$\begin{bmatrix} V_a\\V_b\\V_c \end{bmatrix} = \begin{bmatrix} Z_s&Z_m&Z_m\\Z_m&Z_s&Z_m\\Z_m&Z_m&Z_s \end{bmatrix}\begin{bmatrix} I_a\\I_b\\I_c \end{bmatrix}, \qquad V_p = Z_pI_p$$

Transformation

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

$$V_s = A^{-1}Z_pA\,I_s = Z_{seq}I_s$$

where

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

First find $Z_pA$, column by column (using $1 + a + a^2 = 0$):

• Column 1: each row $= Z_s + 2Z_m$, so column $= (Z_s+2Z_m)[1\ 1\ 1]^T$
• Column 2: rows $= Z_s + Z_m(a^2+a) = Z_s - Z_m$ times $[1\ a^2\ a]^T$
• Column 3: rows $= (Z_s - Z_m)$ times $[1\ a\ a^2]^T$

So $Z_pA = A\,\text{diag}(Z_s+2Z_m,\ Z_s-Z_m,\ Z_s-Z_m)$; the columns of $A$ are eigenvectors of $Z_p$. Therefore

$$Z_{seq} = A^{-1}Z_pA = \begin{bmatrix} Z_s+2Z_m & 0 & 0 \\ 0 & Z_s-Z_m & 0 \\ 0 & 0 & Z_s-Z_m \end{bmatrix}$$

Result

$$V_{a0} = Z_0I_{a0}, \qquad V_{a1} = Z_1I_{a1}, \qquad V_{a2} = Z_2I_{a2}$$

with $Z_0 = Z_s + 2Z_m$ and $Z_1 = Z_2 = Z_s - Z_m$.

All off-diagonal (mutual) terms between sequences are zero, so:

• Positive sequence current causes only positive sequence voltage drop, and similarly for negative and zero.
• Each sequence network can be drawn and solved independently as a single-phase circuit.
• The networks are connected to each other only at points of unbalance (fault points), which is the basis of unsymmetrical fault analysis.

If the circuit is not symmetrical (e.g. unequal mutual impedances in an untransposed line), off-diagonal terms appear and the sequence networks become coupled.