Multiphase systems – positive system, negative system and zero system

The chapter on three-phase current explained how voltages, currents and power are composed in various three-phase systems. In the special case of symmetrical consumers, it was shown that a network calculation can be carried out relatively easily. The simplified representation using single-phase equivalent circuits is particularly helpful for the calculation. These simplifications cannot be applied to unbalanced loads, which means that even for small networks, more comprehensive network calculations become confusing and time-consuming. Imbalances occur, for example, due to network faults such as a single-pole short circuit, switching operations or unbalanced loads on the individual phases. The relationships between triggering events and the results are difficult to understand based on numerical values alone. In order to be able to calculate comprehensive and unbalanced networks, the known system can be transformed into three new systems. The following three systems are examined in more detail for this purpose:

1 Positive-, negative-, and zero-sequence systems

The fundamental goal of the transformation is to convert an asymmetrical system of n vectors into n systems with a symmetrical vector arrangement. For a three-phase system, this means that three systems must be created. Each conductor, i.e. \(L_1\), \(L_2\) and \(L_3\), must be replaced by a system with a symmetrical vector arrangement. The transformation rule is based on the already known rotation operator \(\underline {a}\) and \(\underline {a}^2\) (see equation ??). The symmetry conditions for the current in the respective conductors can be established as follows (it should be noted here that everything that applies to the current also applies to the voltage!): \begin {align} \underline {I}_{\mathrm {L1}}&=\underline {I}_{0}+\underline {I}_{1}+\underline {I}_{2} \nonumber \\ \underline {I}_{\mathrm {L2}}&=\underline {I}_{0}+\underline {a}^2\cdot \underline {I}_{1}+\underline {a}\cdot \underline {I}_{2} \label {ZerlegungL2} \\ \underline {I}_{\mathrm {L3}}&=\underline {I}_{0}+\underline {a}\cdot \underline {I}_{1}+\underline {a}^2\cdot \underline {I}_{2} \nonumber \end {align}

The indices on the right-hand side of the equations represent the three systems: positive system (1), negative system (2) and zero system (0). If we look at the values of the positive system, the rotation operators result in clockwise phase currents. In the counter-system, the transformed phase currents are arranged in a left-rotating manner, and in the zero system, all components are aligned in the same direction. In principle, the zero system only occurs when the sum of all currents in the output system is not equal to zero.

In the next step, the so-called decomposition equations (equations ??) are rearranged to show the ratio of the current of the zero-sequence system to the conductor currents of the original system.

For the current in the zero-sequence system, the following applies by addition:

\begin {align} \underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}&=3\cdot \underline {I}_0+\underline {I}_1\cdot (1+\underline {a}^2+\underline {a})+\underline {I}_2\cdot (1+\underline {a}+\underline {a}^2) \notag \\ &=3\cdot \underline {I}_0 \\ \Leftrightarrow \underline {I}_0&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}) \end {align}

To determine the current of the secondary system, the equations are first multiplied so that the pre-factors before the secondary current add up to 1. hen the decomposition equations are added together again.

\(\underline {a}^3=1\) ; \(\underline {a}^4=\underline {a}\) ; \(\underline {a}^5=\underline {a}^2\) ; ...

\begin {align} \underline {I}_{\mathrm {L1}}&=\underline {I}_{0}+\underline {I}_{1}+\underline {I}_{2} \\ \underline {a}\cdot \underline {I}_{\mathrm {L2}}&=\underline {a}\cdot \underline {I}_{0}+\underline {a}^3\cdot \underline {I}_{1}+\underline {a}^2\cdot \underline {I}_{2} \\ \underline {a}^2\cdot \underline {I}_{\mathrm {L3}}&=\underline {a}^2\cdot \underline {I}_{0}+\underline {a}^3\cdot \underline {I}_{1}+\underline {a}\cdot \underline {I}_{2} \end {align}

\begin {align} \underline {I}_{\mathrm {L1}}+\underline {a}\cdot \underline {I}_{\mathrm {L2}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L3}}&=\underline {I}_{0}\cdot (1+\underline {a}+\underline {a}^2)+3\cdot \underline {I}_{1}+\underline {I}_{2}\cdot (1+\underline {a}^2+\underline {a}) \\ &=3\cdot \underline {I}_{1} \notag \\ \Leftrightarrow \underline {I}_{1}&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\underline {a}\cdot \underline {I}_{\mathrm {L2}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L3}}) \notag \end {align}

Finally, the current for the counter-system is calculated. The same principle is applied as for the system: first, the pre-factor is adjusted, then all decomposition equations are added:

\begin {align} \underline {I}_{\mathrm {\mathrm {L1}}}&=\underline {I}_{0}+\underline {I}_{1}+\underline {I}_{2} \\ \underline {a}^2\cdot \underline {I}_{\mathrm {\mathrm {L2}}}&=\underline {a}^2\cdot \underline {I}_{0}+\underline {a}^4\cdot \underline {I}_{1}+\underline {a}^3\cdot \underline {I}_{2} \\ \underline {a}\cdot \underline {I}_{\mathrm {\mathrm {L3}}}&=\underline {a}\cdot \underline {I}_{0}+\underline {a}^2\cdot \underline {I}_{1}+\underline {a}^3\cdot \underline {I}_{2} \end {align}

\begin {align} \underline {I}_{\mathrm {L1}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L2}}+\underline {a}\cdot \underline {I}_{\mathrm {L3}}&=(1+\underline {a}^2+\underline {a})\underline {I}_{0}+(1+\underline {a}+\underline {a}^2)\cdot \underline {I}_{1}+3\cdot \underline {I}_{2} \\ &=3\cdot \underline {I}_{2} \notag \\ \Leftrightarrow \underline {I}_{2}&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L2}}+\underline {a}\cdot \underline {I}_{\mathrm {L3}}) \notag \end {align}

This results in the following equations for the currents of all systems:

\begin {align} \underline {I}_0&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}) \\ \underline {I}_{1}&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\rlap {\underline {a}}\phantom {a^2}\cdot \underline {I}_{\mathrm {L2}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L3}}) \\ \underline {I}_{2}&=\frac {1}{3}\cdot (\underline {I}_{\mathrm {L1}}+\underline {a}^2\cdot \underline {I}_{\mathrm {L2}}+\rlap {\underline {a}}\phantom {a^2}\cdot \underline {I}_{\mathrm {L3}}) \end {align}

If you look closely at these three equations, you will see that conductor L1 does not have a rotation operator as a leading factor in any of the three equations. Therefore, conductor L1 should be selected as the reference conductor. This means that this is the conductor through which an imbalance occurs in the original system (e.g. due to a single-pole earth fault). The remaining symmetries in the original system can then be more easily considered via the other conductors Of course, any conductor can be taken as the reference conductor, since each conductor can be freely assigned an index (L1, L2, L3). However, this allows the initial requirement to be met: the symmetries of the real networks and equipment must be used to advantage!

As an example, an unbalanced current in a three-phase system and the transformation may look as follows:

Fig. 1 shows the phasor diagram of an unbalanced load. In the original system, the phasors show that both the angles and the amplitudes deviate from the symmetrical state. The transformation into the three systems results in symmetrical phasors, which are easier to calculate with. The rotation operators and the prefactor of \(\frac {1}{3}\) make it easy to see from the vectors how the transformed currents behave. To avoid having to write out the transformation equations in full every time, the equations can also be written in matrix notation:

\begin {align} \begin {bmatrix} \underline {I}_0 \\ \underline {I}_1 \\ \underline {I}_2 \end {bmatrix} = \frac {1}{3} \cdot \begin {bmatrix} 1 & 1 & 1 \\ 1 & \underline {a} & \underline {a}^2 \\ 1 & \underline {a}^2 & \underline {a} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1}} \\ \underline {I}_{\mathrm {L2}} \\ \underline {I}_{\mathrm {L3}} \end {bmatrix} \label {TraFoMatrix} \end {align}

The notation can be simplified further by combining the term with the rotation operators and the term with the leading factor into a matrix. This matrix is also called the symmetrisation matrix:

\begin {align} \begin {bmatrix} \underline {T} \end {bmatrix} =\frac {1}{3} \cdot \begin {bmatrix} 1 & 1 & 1 \\ 1 & \underline {a} & \underline {a}^2 \\ 1 & \underline {a}^2 & \underline {a} \end {bmatrix} \end {align}

In compact notation, this gives the following form:

\begin {align} \begin {bmatrix} \underline {I}_{012} \end {bmatrix} = \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix} \end {align}

According to the rules for calculating matrices, the inverse of \(\begin {bmatrix}\underline {T}\end {bmatrix}\) can be used to perform the inverse transformation:

\begin {align} \begin {bmatrix} \underline {I}_{\mathrm {L1}} \\ \underline {I}_{\mathrm {L2}} \\ \underline {I}_{\mathrm {L3}} \end {bmatrix} = \frac {1}{3} \cdot \begin {bmatrix} 1 & 1 & 1 \\ 1 & \underline {a}^2 & \underline {a} \\ 1 & \underline {a} & \underline {a}^2 \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_0 \\ \underline {I}_1\\ \underline {I}_2 \end {bmatrix} \end {align}

\begin {align} \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix} = \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix} \end {align}

2 Three-phase power

How power is calculated has already been explained in the chapter on three-phase current. Here too, the equation for calculating power can be expressed in matrix notation for simplicity: \begin {align} \underline {S}_{\mathrm {L1L2L3}}&=\underline {U}_{\mathrm {L1}}\cdot \underline {I}_{\mathrm {L1}}^*+\underline {U}_{\mathrm {L2}}\cdot \underline {I}_{\mathrm {L2}}^*+\underline {U}_{\mathrm {L3}}\cdot \underline {I}_{\mathrm {L3}}^* \\ &= \begin {bmatrix} \underline {U}_{\mathrm {L1}} & \underline {U}_{\mathrm {L2}} & \underline {U}_{\mathrm {L3}} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1}}^* \\ \underline {I}_{\mathrm {L2}}^* \\ \underline {I}_{\mathrm {L3}}^* \end {bmatrix} = \begin {bmatrix} \underline {U}_{\mathrm {L1L2L3}} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix}_t ^* \notag \end {align}

With the given rotation operators, the currents and voltages in a symmetrical system can be related to one phase each. (\(\underline {U}_{\mathrm {L2}}=\underline {a}^2\cdot \underline {U}_{\mathrm {L1}}, \underline {U}_{\mathrm {L3}}=\underline {a}\cdot \underline {U}_{\mathrm {L1}}\))

This results in the familiar simplification for power: \begin {align} \underline {S}_{\mathrm {L1L2L3}}&=\underline {U}_{\mathrm {L1}}\cdot \underline {I}_{\mathrm {L1}}^*+\underline {U}_{\mathrm {L1}}\cdot \underline {I}_{\mathrm {L1}}^*+\underline {U}_{\mathrm {L1}}\cdot \underline {I}_{\mathrm {L1}}^* \\ &=3\cdot \underline {U}_{\mathrm {L1}}\cdot \underline {I}_{\mathrm {L1}}^* \end {align}

Now, using matrix notation, the transformation into the new systems can be calculated: \begin {align} \begin {bmatrix} \underline {U}_{\mathrm {L1L2L3}} \end {bmatrix} &= ( \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {U}_{012} \end {bmatrix} ) = \begin {bmatrix} \underline {U}_{012} \end {bmatrix} \cdot \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \\ \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix}^* &= ( \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix} )^* = \begin {bmatrix} \underline {T} \end {bmatrix}^{-1*} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix}^* \\ \underline {S}_{\mathrm {L1L2L3}} &= \begin {bmatrix} \underline {U}_{012} \end {bmatrix}_t \cdot \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {T} \end {bmatrix}^{-1*} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix}^* \\ &= \begin {bmatrix} \underline {U}_{012} \end {bmatrix}_t \cdot \begin {bmatrix} 1 & 1 & 1 & \\ 1 & \underline {a}^2 & \underline {a} \\ 1 & \underline {a} & \underline {a}^2 \end {bmatrix} \cdot \begin {bmatrix} 1 & 1 & 1 & \\ 1 & \underline {a}^2 & \underline {a} \\ 1 & \underline {a} & \underline {a}^2 \end {bmatrix}^* \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix}^* \notag \\ &= \begin {bmatrix} \underline {U}_{012} \end {bmatrix}_t \cdot \begin {bmatrix} 1 & 1 & 1 & \\ 1 & \underline {a}^2 & \underline {a} \\ 1 & \underline {a} & \underline {a}^2 \end {bmatrix} \cdot \begin {bmatrix} 1 & 1 & 1 & \\ 1 & \underline {a} & \underline {a}^2 \\ 1 & \underline {a}^2 & \underline {a} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix}^* \notag \\ &= \begin {bmatrix} \underline {U}_{012} \end {bmatrix} \cdot \begin {bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix}^* \notag \\ &=3\cdot (\underline {U}_0\cdot \underline {I}_o^*+\underline {U}_1\cdot \underline {I}_1^*+\underline {U}_2\cdot \underline {I}_2^*)=3\cdot \underline {S}_{012} \notag \end {align}

3 Equivalent circuit diagram

ECDs fundamentally serve to simplify the observation and understanding of components. We will now examine various operating resources and how the transformations behave in different network situations.

3.1 Symmetrical voltage source

Symmetrical voltage sources can be found, for example, in mains supply and also in synchronous machines. The first step is to look at the output system. The relationships between the star voltage and the individual phases are already known from the chapter ???. Using the formula for the transformation (cf. ??), it can now be shown mathematically that in such a symmetrical case, the voltages of the counter and zero systems are 0.

Only in the neutral system does a driving voltage occur, equal to the star voltage: \begin {align} \begin {bmatrix} \underline {U}_0 \\ \underline {U}_1 \\ \underline {U}_2 \end {bmatrix} &= \frac {1}{3}\cdot \begin {bmatrix} 1 & 1 & 1 & \\ 1 & \underline {a} & \underline {a}^2 \\ 1 & \underline {a}^2 & \underline {a} \end {bmatrix} \cdot \begin {bmatrix} \underline {U}_{\mathrm {L1}} \\ \underline {U}_{\mathrm {L2}} \\ \underline {U}_{\mathrm {L3}} \end {bmatrix} \\ \begin {bmatrix} \underline {U}_{012} \end {bmatrix} &= \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {U}_{Y } \\ \underline {a}^2\cdot \underline {U}_{Y } \\ \underline {a}\cdot \underline {U}_{Y } \end {bmatrix} \end {align}

Solving the equations for the transformed systems yields the following stress results: \begin {align} \underline {U}_0&=\frac {1}{3}\cdot {U}_{Y }\cdot (1+\underline {a}^2+\underline {a})=0 \\ \underline {U}_1&=\frac {1}{3}\cdot {U}_{Y }\cdot (1+\underline {a}\cdot \underline {a}^2+\underline {a}^2\cdot \underline {a})=U_{Y } \\ \underline {U}_2&=\frac {1}{3}\cdot {U}_{Y }\cdot (1+\underline {a}^2\cdot \underline {a}^2+\underline {a}\cdot \underline {a})=0 \end {align}

With the results of the counter and zero systems, it can be said figuratively that these are short-circuited. Thus, the respective ECDs can be created for the three transformed systems:

In Figure 2, it should be clearly emphasised once again that, on the one hand, three individual systems are formed and, furthermore, that in the special case of symmetrical sources, a voltage only occurs in one of the systems.

3.2 Symmetrical load

A star connection with a return conductor should be selected for the consideration of the loads. An impedance Z is assumed in each conductor, including the return conductor. For the symmetrical case, the impedances are assumed to be equal; only the impedance in the return conductor may differ from the others and is given its own index.

For the original system, the equation for the voltage is set up as follows:

\begin {align} \underline {U}_{\mathrm {L1}}=\underline {Z}\cdot \underline {I}_{\mathrm {L1}}+\underline {Z}_\mathrm {E}\cdot (\underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}) \\ \underline {U}_{\mathrm {L2}}=\underline {Z}\cdot \underline {I}_{\mathrm {L2}}+\underline {Z}_\mathrm {E}\cdot (\underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}) \\ \underline {U}_{\mathrm {L3}}=\underline {Z}\cdot \underline {I}_{\mathrm {L3}}+\underline {Z}_\mathrm {E}\cdot (\underline {I}_{\mathrm {L1}}+\underline {I}_{\mathrm {L2}}+\underline {I}_{\mathrm {L3}}) \end {align}

Here, too, matrix notation can be used to simplify the expression: \begin {align} \begin {bmatrix} \underline {U}_{\mathrm {L1}} \\ \underline {U}_{\mathrm {L2}} \\ \underline {U}_{\mathrm {L3}} \end {bmatrix} = \begin {bmatrix} \underline {Z}+\underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} \\ \underline {Z}_\mathrm {E} & \underline {Z}+\underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} \\ \underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} & \underline {Z}+\underline {Z}_\mathrm {E} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1}} \\ \underline {I}_{\mathrm {L2}} \\ \underline {I}_{\mathrm {L3}} \end {bmatrix} \end {align}

With...

\begin {align} \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} = \begin {bmatrix} \underline {Z}+\underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} \\ \underline {Z}_\mathrm {E} & \underline {Z}+\underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} \\ \underline {Z}_\mathrm {E} & \underline {Z}_\mathrm {E} & \underline {Z}+\underline {Z}_\mathrm {E} \end {bmatrix} \end {align}

...results in

\begin {align} \begin {bmatrix} \underline {U}_{\mathrm {L1L2L3}} \end {bmatrix} = \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix} \end {align}

If transformed using the transformation equations, the following applies: \begin {align} \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {U}_{012} \end {bmatrix} = \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix} \end {align}

For symmetrical voltage sources, the voltage could be calculated from the transformation depending on the original system. The same should be done for the impedances for the consumers. To do this, the equations must be transformed in such a way that a ratio of transformed impedances and impedances of the original system is obtained. To achieve this, both sides are first expanded with \(\begin {bmatrix}\underline {T}\end {bmatrix}\) and then rearranged based on Ohm’s law. \begin {align} \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {U}_{012} \end {bmatrix} &= \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix} \\ \begin {bmatrix} \underline {U}_{012} \end {bmatrix} &= \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \cdot \begin {bmatrix} \underline {I}_{012} \end {bmatrix} \\ \begin {bmatrix} \underline {Z}_{012} \end {bmatrix} &= \begin {bmatrix} \underline {T} \end {bmatrix} \cdot \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \begin {bmatrix} \underline {T} \end {bmatrix}^{-1} \end {align}

Performing the matrix operation of the last equation yields the desired relationship between the original and transformation systems. \begin {align} \begin {bmatrix} \underline {Z}_{012} \end {bmatrix} = \begin {bmatrix} \underline {Z}+3\underline {Z}_E & 0 & 0 \\ 0 & \underline {Z} & 0 \\ 0 & 0 & \underline {Z} \end {bmatrix} \end {align}

Now, the respective ECDs of the transformed systems can also be established for the symmetrical loads:

In Fig. 3, the same impedances occur in the positive and negative systems. If the consumers are connected in a delta configuration, the impedances would have to be converted to star values using the known ratios. For zero impedances, currents only flow if there is actually a connection to a star. Then the impedances in the return conductor occur with triple the value.

3.3 Symmetrical lines

The line is the connection between the voltage source and the consumers. Until now, considerations regarding three-phase current have only distinguished between generation and consumers. With the lines, an additional component is now added that must be considered separately. Lines generally have their own impedances, coupling impedances and conductor-earth impedances. However, the conductor-earth impedances can be neglected. Each conductor has an intrinsic impedance and two coupling impedances, i.e. to the other conductors. The voltage for this calculation is defined as the difference between the voltages at the beginning and end of the line. This allows the matrix equation for a line system to be set up:

\begin {align} \begin {bmatrix} \underline {U}_{\mathrm {L1A}}-\underline {U}_{\mathrm {L1B}} \\ \underline {U}_{\mathrm {L2A}}-\underline {U}_{\mathrm {L2B}} \\ \underline {U}_{\mathrm {L3A}}-\underline {U}_{\mathrm {L3B}} \end {bmatrix} = \begin {bmatrix} \underline {Z}_\mathrm {S} & \underline {Z}_\mathrm {K} & \underline {Z}_\mathrm {K} \\ \underline {Z}_\mathrm {K} & \underline {Z}_\mathrm {S} & \underline {Z}_\mathrm {K} \\ \underline {Z}_\mathrm {K} & \underline {Z}_\mathrm {K} & \underline {Z}_\mathrm {S} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1}} \\ \underline {I}_{\mathrm {L2}} \\ \underline {I}_{\mathrm {L3}} \end {bmatrix} \end {align}

From this, a compact notation can be derived (with \(U_{\mathrm {L1A}}-U_{\mathrm {L1B}}=U_{\mathrm {L1}}\), ...): \begin {align} \begin {bmatrix} \underline {U}_{\mathrm {L1L2L3}} \end {bmatrix} = \begin {bmatrix} \underline {Z}_{\mathrm {L1L2L3}} \end {bmatrix} \cdot \begin {bmatrix} \underline {I}_{\mathrm {L1L2L3}} \end {bmatrix} \end {align}

We now have an equation that is very similar to the one for symmetrical consumers. Here, too, the equation must be transformed so that there is a direct dependency between the original system and the transformation system. Since the equation has the same basic structure, the same extensions and transformations can be carried out as for symmetrical consumers: \begin {align} \begin {bmatrix} \underline {Z}_{012} \end {bmatrix} = \begin {bmatrix} \underline {Z}_\mathrm {S}+2\cdot \underline {Z}_\mathrm {K} & 0 & 0 \\ 0 & \underline {Z}_\mathrm {S}-\underline {Z}_\mathrm {K} & 0 \\ 0 & 0 & \underline {Z}_\mathrm {S}-\underline {Z}_\mathrm {K} \end {bmatrix} \end {align} When transforming, it is always important that the transformed impedance matrix consists only of values on the main diagonal so that there are only direct connections between the individual conductors and the transformation systems.

Fig. 4 shows the ESB of the three systems for symmetrical lines. The dashed line is intended to reflect the neutral point treatment on the generation and consumption side for reference to the star voltages.