Chapter 10 — Multiconductor lines: sequence data and modal decomposition

Real transmission lines carry three phases on the same towers, close enough that each conductor's magnetic and electric field reaches the other two. Current in phase A induces voltage along phases B and C; charge on A raises the potential of B and C. The single-conductor model of Chapters 8 and 9 has nothing to say about any of it.

The remedy is one of the most elegant ideas in power-system analysis. Rather than build a coupled three-conductor wave model from first principles, we change coordinates: we find a set of combinations of the three phases that do not interact, solve each as an ordinary Chapter 8 line, and transform back. This chapter derives those coordinates, shows why the balanced case collapses to a familiar transformation, and connects the result to the three ways NumaSim accepts three-phase line data.

Learning objectives

By the end of this chapter you should be able to:

• Write the multiconductor line equations and identify what the coupling terms do.
• Explain how a modal transformation turns $n$ coupled lines into $n$ independent ones.
• Derive the balanced-line modes and relate them to Clarke components and to sequence quantities.
• Explain physically why the ground mode is slower and higher-impedance than the aerial modes.
• Choose among sequence, sequence-RLC, and balanced-matrix data entry, and convert between them.

10.1 Three conductors, one coupled problem

Repeat the slice argument of Chapter 8 with three conductors instead of one. The voltages and currents become vectors, and the per-unit-length parameters become matrices:

$$-\frac{\partial \mathbf{v}}{\partial x}={\mathbf{Z}}^{\prime }\mathbf{i},-\frac{\partial \mathbf{i}}{\partial x}={\mathbf{Y}}^{\prime }\mathbf{v},$$

where $\mathbf{v}=[v_a{v}_b{v}_c{]}^T$, and ${\mathbf{Z}}^{\prime }$ and ${\mathbf{Y}}^{\prime }$ are $3\times 3$. Their diagonals are the self impedance and self admittance already familiar from the single-conductor case. Their off-diagonals represent the coupling: $Z_{ab}^{\prime }$ is the voltage induced along conductor $a$ per unit of current in conductor $b$, and $Y_{ab}^{\prime }$ carries the charging current flowing between the two conductors.

Three conductors sharing a right of way. Each pair is coupled both magnetically (mutual inductance $L_m$) and electrostatically (mutual capacitance $C_m$), and each conductor also couples to earth. The coupling is what makes the three phases a single problem rather than three.

Repeating the derivation of Chapter 8, differentiation and substitution give

$$\frac{{\partial }^2\mathbf{v}}{\partial x^2}={\mathbf{Z}}^{\prime }{\mathbf{Y}}^{\prime }\mathbf{v},$$

which resembles the scalar wave equation with $L^{\prime }{C}^{\prime }$ replaced by the matrix product ${\mathbf{Z}}^{\prime }{\mathbf{Y}}^{\prime }$, and that resemblance is precisely the difficulty. In the scalar case $\sqrt{L^{\prime }{C}^{\prime }}$ gave a single travel time and $\sqrt{L^{\prime }/C^{\prime }}$ a single surge impedance. A matrix has no single square root of any physical meaning, and the three components of $\mathbf{v}$ remain mixed at every point along the line. No single travel time exists to delay by.

10.2 Changing coordinates: modal decomposition

The way forward is to abandon phase quantities as the working variables. The matrix ${\mathbf{Z}}^{\prime }{\mathbf{Y}}^{\prime }$ has eigenvectors, and along an eigenvector the matrix acts as a plain scalar multiplication. We therefore define new variables as combinations of the phase quantities, one per eigenvector:

$$\mathbf{v}=\mathbf{Q}{\mathbf{v}}_{\text{mode}},\mathbf{i}=\mathbf{Q}{\mathbf{i}}_{\text{mode}},$$

where the columns of $\mathbf{Q}$ are the eigenvectors. In these coordinates the wave equation becomes

$$\frac{{\partial }^2{\mathbf{v}}_{\text{mode}}}{\partial x^2}=\mathbf{\Lambda }{\mathbf{v}}_{\text{mode}},\mathbf{\Lambda }={\mathbf{Q}}^{-1}{\mathbf{Z}}^{\prime }{\mathbf{Y}}^{\prime }\mathbf{Q}=\mathrm{diag}({\lambda }_0,{\lambda }_1,{\lambda }_2),$$

and because $\mathbf{\Lambda }$ is diagonal, the three equations no longer mention one another. Each is a scalar wave equation — precisely Chapter 8 — with its own modal surge impedance and its own modal travel time.

This constitutes a complete algorithm, and it is the one NumaSim implements:

How a coupled three-phase line is solved. The phase quantities are projected onto modal coordinates, each mode is propagated as an independent Chapter 8 travelling-wave line with its own $Z$ and $\tau$, and the results are transformed back to phase quantities.

One caveat attaches to this, and it is the reason Chapter 11 exists. In general $\mathbf{Q}$ depends on frequency and its entries are complex, so the decomposition is only as clean as the assumption that ${\mathbf{Z}}^{\prime }$ and ${\mathbf{Y}}^{\prime }$ are constant. Everything that follows rests on that assumption.

10.3 The balanced line and Clarke components

For a line that is fully transposed — each conductor spends a third of the route in each physical position — all three conductors are electrically identical and all three pairs are coupled identically. Every matrix in the problem then has the same simple shape: equal diagonals and equal off-diagonals.

$${\mathbf{L}}^{\prime }=\left[\begin{array}{ccc}{L}_s& L_m& L_m\\ L_m& L_s& L_m\\ L_m& L_m& L_s\end{array}\right]$$

Matrices of this form have eigenvectors that can be written down without knowing $L_s$ or $L_m$ at all, which is what makes the balanced case so much more tractable than the general one. There are two distinct eigenvalues:

• The direction $[111{]}^T$, with eigenvalue $L_s+2L_m$. All three conductors carry the same current in the same direction.
• Everything orthogonal to it — a two-dimensional space, so a repeated eigenvalue $L_s-L_m$. The three currents sum to zero.

A convenient orthonormal choice of eigenvectors is Clarke's:

$$\mathbf{Q}=\left[\begin{array}{ccc}1/\sqrt{3}& 2/\sqrt{6}& 0\\ 1/\sqrt{3}& -1/\sqrt{6}& 1/\sqrt{2}\\ 1/\sqrt{3}& -1/\sqrt{6}& -1/\sqrt{2}\end{array}\right]$$

This matrix is real, constant, orthogonal (${\mathbf{Q}}^{-1}={\mathbf{Q}}^T$), and, most importantly, entirely independent of the line's parameter values. The same three columns diagonalize ${\mathbf{Z}}^{\prime }$, ${\mathbf{Y}}^{\prime }$, and their product, so one fixed transformation serves the whole model. This is the transformation NumaSim uses.

What the two mode types physically are

The distinction between the mode types is not a mathematical accident. It is a question of where the current returns.

The two mode types on a balanced line. In an aerial mode the currents sum to zero and return through the other phase conductors. In the ground mode all three currents flow the same way and must return through the earth, which is a poorer conductor and a much larger loop.

In an aerial mode the three currents sum to zero, so the outgoing and returning paths are conductors a few metres apart. The enclosed loop is small, making the inductance small, and the conductors are close, making the capacitance large. The consequence is a low surge impedance and propagation at nearly the speed of light.

In the ground mode all three currents flow in the same direction and return through the earth, tens of metres below and spread over a large, poorly conducting volume. The loop is very much larger, so the inductance is greater, and the effective separation is larger, so the capacitance is smaller. The consequence is

$$Z_0=\sqrt{\frac{L_0}{C_0}}\gg Z_1=\sqrt{\frac{L_1}{C_1}},{\tau }_0=d\sqrt{L_0{C}_0}>{\tau }_1=d\sqrt{L_1{C}_1}.$$

The ground mode is therefore slower and stiffer. On a typical overhead line $Z_0$ is roughly twice $Z_1$ and ${\tau }_0$ about one and a half times ${\tau }_1$. The difference is directly observable: a disturbance exciting both modes arrives at the far end twice, first with the aerial mode and again somewhat later with the ground mode.

Modes and sequence components

The resemblance between the aerial/ground split and positive- and zero-sequence impedances is not a coincidence, since both arise from diagonalizing the same balanced matrix. The $[111{]}^T$ direction is precisely what zero sequence means, and the two aerial modes span the same space as the positive- and negative-sequence directions. On a balanced line the positive and negative sequences are identical, which is why two of the three modes are degenerate. NumaSim uses the two vocabularies interchangeably: mode 0 is labelled zero-sequence, and the aerial modes carry the positive-sequence parameters.

10.4 Entering three-phase line data

Setting the Model's Conductors to 3 reveals the three-phase entry methods. All three ultimately produce the same six numbers — $R_1,L_1,C_1$ and $R_0,L_0,C_0$ — and differ only in the form in which the data is supplied.

Bergeron — Sequence Surge Impedance takes the modal quantities directly: Z1, τ1, Z0, τ0. It is the appropriate choice when the available data is in surge-impedance form, or when direct control over the arrival times is wanted. Internally,

$$L_1=Z_1{\tau }_1,C_1=\frac{{\tau }_1}{Z_1},L_0=Z_0{\tau }_0,C_0=\frac{{\tau }_0}{Z_0}.$$

Note that this method provides no way to specify resistance; it describes a lossless line.

Bergeron — Sequence RLC takes a Length plus per-kilometre $R_1,L_1,C_1$ and $R_0,L_0,C_0$. This is the form in which most line-constants programs and utility datasheets report their results, and it is the only three-phase entry permitting $R_0\ne R_1$. That distinction matters, because the earth return is genuinely more resistive.

Bergeron — Balanced Matrix takes a Length plus the self and mutual values directly: $R_s$, $L_s$, $L_m$, $C_s$, $C_m$. NumaSim applies the eigenvalues derived above:

$$L_1=L_s-L_m,L_0=L_s+2L_m,$$$$C_1=C_s+C_m,C_0=C_s-2C_m,R_1=R_0=R_s.$$

The capacitance formulas carry the opposite sign pattern to the inductance ones because the mutual entry in a nodal capacitance matrix is negative. $C_m$ is entered as a positive magnitude and the reduction supplies the sign. This is also why $C_1>C_0$ while $L_0>L_1$, and that pair of inequalities is what makes $Z_0>Z_1$.

The last equality deserves attention: mutual resistance is not modelled, so matrix entry forces $R_0=R_1$. Data with a distinctly larger zero-sequence resistance, as real earth-return data has, should be entered through the sequence-RLC method instead.

Each of these methods has a PI — counterpart for short lines, which follows Chapter 9's rules.

10.5 What the engine does with three phases

The three-phase travelling-wave line splits into two halves exactly as the single-phase one does, but each half now presents a $3\times 3$ Norton conductance rather than a scalar. Transforming $\mathrm{diag}(G_0,G_1,G_1)$ back to phase coordinates gives a matrix with equal diagonals and equal off-diagonals,

$$G_{\text{self}}=\frac{G_0+2G_1}{3},G_{\text{mutual}}=\frac{G_0-G_1}{3},$$

which NumaSim realizes without ever forming the matrix: a conductance $G_0$ from each phase to ground, plus a branch of value $(G_1-G_0)/3$ between each pair of phases. Three grounded conductances and three phase-to-phase branches reproduce the coupled Norton equivalent exactly.

The history side keeps the two mode families separate, each with its own integer delay from Chapter 9:

$$N_1=\mathrm{round}\left(\frac{{\tau }_1}{\mathrm{\Delta }t}\right),N_0=\mathrm{round}\left(\frac{{\tau }_0}{\mathrm{\Delta }t}\right).$$

The fallback test of Chapter 9 is all-or-nothing: the line is demoted to a $\pi$ if $min({\tau }_1,{\tau }_0)<\mathrm{\Delta }t$. Since the aerial mode is always the faster one, it is ${\tau }_1$ that decides.

Limits worth knowing

• One or three conductors. There is no six-conductor double-circuit model, and earth wires cannot be entered as extra conductors; they must be folded into the sequence data beforehand (Chapter 12).
• Balanced only. Matrix entry asks for a single $L_s$ and a single $L_m$, so an untransposed line, in which the three phases genuinely differ, cannot be represented. Most studies use transposed data in any case.
• No mutual resistance, as noted above.
• Constant parameters. $Z_0$, $Z_1$, ${\tau }_0$, ${\tau }_1$ do not vary with frequency. The consequences are Chapter 11's subject.

10.6 Lab: two modes, two arrival times

Open the AC line fault demo in simulator →

A 230 kV source feeds a three-phase travelling-wave line into a $500\mathrm{\Omega }$ load, with a fault applied at the receiving end at $t=1.05\text{s}$ for 100 ms. The line is entered by sequence surge impedance:

$$Z_1=30\mathrm{\Omega },{\tau }_1=100\mu \text{s},Z_0=600\mathrm{\Omega },{\tau }_0=150\mu \text{s},$$

with $\mathrm{\Delta }t=20\mu \text{s}$. These values are deliberately exaggerated — a real 230 kV line would be nearer $Z_1=300\mathrm{\Omega }$, with $Z_0$ only about twice that — so that the two modes are easily distinguished. (Full description: AC line fault.)

What makes this lab work is the choice of observation point. The fault is at the receiving end, but the currents plotted, Ia, Ib and Ic, are measured at the source, a full line length away. Nothing at the source can react to the fault until a wave has crossed the line, so the sending end acts as a modal stopwatch from which the travel times can be read directly.

Consider first what that stopwatch should read. $N_1=\mathrm{round}(100/20)=5$ steps, so the aerial mode arrives exactly $100\mu \text{s}$ after the fault. $N_0=\mathrm{round}(150/20)=\mathrm{round}(7.5)=8$ steps, so the ground mode arrives at $8\times 20=160\mu \text{s}$ rather than at $150\mu \text{s}$. This is the quantization of Chapter 9, applied to each mode independently.

Run the sample as shipped, with the fault type set to ABC-G, a symmetrical three-phase fault. Zoom the plot tightly into the first $300\mu \text{s}$ after $t=1.05\text{s}$, far enough that individual time steps are visible rather than a millisecond-wide view. Three features stand out:

1. The currents follow their pre-fault trend undisturbed until the fifth step and then break sharply at $100\mu \text{s}$. Ib and Ic jump from under $0.5\text{kA}$ to about $11\text{kA}$ in that one step.
2. Nothing whatsoever happens at $160\mu \text{s}$. The trace runs straight through it. A balanced fault on a balanced line produces no zero-sequence current, so the ground mode is never excited — there is no second arrival to see.
3. $I_a+I_b+I_c=0$ at every step, to numerical precision. That is the same statement as point 2, read in the phase domain instead of the modal one.

To unbalance the fault, open the fault component, change Fault type to A-G (1φ to gnd), then re-run and zoom in the same way:

1. A second arrival appears at $160\mu \text{s}$. The first break remains at $100\mu \text{s}$, but the trace now kinks again sixty microseconds later. A single fault arrives twice, because the energy it launched has split between two modes travelling at different speeds. The second arrival lands at $160\mu \text{s}$ rather than $150$, so the predicted rounding is directly measurable.
2. Phases B and C respond, although the fault touched only phase A and nothing was ever connected to B or C. Their peaks grow by tens of percent over the cycle that follows. This is the coupling that the off-diagonal terms of ${\mathbf{Z}}^{\prime }$ and ${\mathbf{Y}}^{\prime }$ describe.
3. The fault current is much smaller, with a peak near $29\text{kA}$ against $291\text{kA}$ for the balanced case. A single-phase-to-ground fault must drive current through the zero-sequence path, and $Z_0=600\mathrm{\Omega }$ is twenty times $Z_1$. The ground mode is the bottleneck.

The two modes can be separated further by setting ${\tau }_0$ to $500\mu \text{s}$, leaving ${\tau }_1$ at $100\mu \text{s}$, and re-running the A-G case. The second arrival moves out to $500\mu \text{s}$ and becomes unmistakable: a clean $400\mu \text{s}$ of aerial-mode-only response, followed by the ground mode landing on top of it. As line data this is unphysical, since no real line has a ground mode five times slower than its aerial mode. It does, however, magnify what occurs at realistic ratios, and the two events are far easier to locate in the shipped $150\mu \text{s}$ case once they have been seen apart.

The balanced run repays one further observation. The sending-end currents do not jump to a final value and remain there; they climb in a staircase of roughly $11\text{kA}$ per tread, with a tread every $200\mu \text{s}$. That interval is $2{\tau }_1$, the round trip from source to fault and back. It is the reflection staircase of Chapter 8, operating in the aerial mode of a three-phase line.

10.7 Summary

• On a multiconductor line the per-unit-length parameters are matrices, and their off-diagonal terms couple the phases so tightly that no single surge impedance or travel time exists.
• Diagonalizing ${\mathbf{Z}}^{\prime }{\mathbf{Y}}^{\prime }$ produces modes — combinations of phase quantities that propagate independently, each as an ordinary Chapter 8 line with its own $Z$ and $\tau$.
• On a balanced line the eigenvectors are Clarke's, real and constant and independent of the parameter values, and the modes split into one ground mode ($[111]$, earth return) and two degenerate aerial modes (currents summing to zero).
• Physically, the earth return makes $L_0>L_1$ and $C_0<C_1$, hence $Z_0>Z_1$ and ${\tau }_0>{\tau }_1$: the ground mode is slower and higher-impedance, and it arrives second.
• NumaSim accepts sequence surge impedances, sequence RLC, or self/mutual matrix data, reducing all three to $R_1,L_1,C_1,R_0,L_0,C_0$. Matrix entry cannot express $R_0\ne R_1$; sequence RLC can.
• Each half-line stamps the coupled Norton equivalent as three grounded conductances plus three phase-to-phase branches, and delays the two mode families independently, $N_1$ and $N_0$ steps.

10.8 Problems

Problem 10.1. A balanced line has $L_s=1.5\mu \text{H/m}$, $L_m=0.57\mu \text{H/m}$, $C_s=11\text{pF/m}$, $C_m=1\text{pF/m}$. Find the modal parameters per metre, then $Z_1$, $Z_0$, and the two travel times for a 100 km line.

Solution 10.1

Apply the eigenvalue reductions:

$$L_1=L_s-L_m=1.5-0.57=0.93\mu \text{H/m},L_0=L_s+2L_m=1.5+1.14=2.64\mu \text{H/m},$$$$C_1=C_s+C_m=12\text{pF/m},C_0=C_s-2C_m=9\text{pF/m}.$$

Surge impedances:

$$Z_1=\sqrt{\frac{0.93\times {10}^{-6}}{12\times {10}^{-12}}}=\sqrt{7.75\times {10}^4}\approx 278\mathrm{\Omega },$$$$Z_0=\sqrt{\frac{2.64\times {10}^{-6}}{9\times {10}^{-12}}}=\sqrt{2.93\times {10}^5}\approx 542\mathrm{\Omega }.$$

Travel times over $d={10}^5\text{m}$:

$${\tau }_1=d\sqrt{L_1{C}_1}={10}^5\sqrt{1.116\times {10}^{-17}}={10}^5(3.34\times {10}^{-9})=334\mu \text{s},$$$${\tau }_0=d\sqrt{L_0{C}_0}={10}^5\sqrt{2.376\times {10}^{-17}}={10}^5(4.87\times {10}^{-9})=487\mu \text{s}.$$

Sanity checks: $Z_0/Z_1=1.95$ and ${\tau }_0/{\tau }_1=1.46$, both in the usual range, and the aerial propagation speed $1/3.34\times {10}^{-9}=3.0\times {10}^8\text{m/s}$ is the speed of light, as it should be for a mode that never enters the ground.

Problem 10.2. For the line of Problem 10.1, what is the largest time step at which it still runs as a travelling-wave line, and how many steps of delay does each mode get at $\mathrm{\Delta }t=50\mu \text{s}$?

Solution 10.2

The fallback test uses the smaller travel time, which is always the aerial mode's. So the line stays distributed as long as $\mathrm{\Delta }t\le {\tau }_1=334\mu \text{s}$. At anything larger, both modes are abandoned and the line becomes a single $\pi$ — even though the ground mode on its own would still have cleared the test.

At $\mathrm{\Delta }t=50\mu \text{s}$:

$$N_1=\mathrm{round}(334/50)=\mathrm{round}(6.68)=7\Rightarrow {\tau }_{1,\text{eff}}=350\mu \text{s},$$$$N_0=\mathrm{round}(487/50)=\mathrm{round}(9.74)=10\Rightarrow {\tau }_{0,\text{eff}}=500\mu \text{s}.$$

Both are within about 5 %, consistent with the $1/2N$ bound of Chapter 9.

Problem 10.3. A single conductor of a three-phase line is energized while the other two are open. Which modes are excited, and what does the far end see?

Solution 10.3

Both. A phase-A-only excitation is $[100{]}^T$, which is not proportional to any single Clarke column. Projecting it gives a $1/\sqrt{3}$ component along the ground mode and the remainder along the aerial modes — so roughly a third of the energy travels in the ground mode and two thirds in the aerial modes.

Because the two mode families travel at different speeds, the far end sees the disturbance twice: an aerial-mode arrival at ${\tau }_1$, then a ground-mode arrival at ${\tau }_0$. Between the two arrivals the phase voltages are a partial reconstruction — the modal sum is incomplete, so the three phases show voltages that no steady-state analysis would predict, including on the two conductors that were never energized.

This is precisely why single-pole switching is a distinct study from three-pole switching, and why line energization is analyzed in the modal domain.

Problem 10.4. Your datasheet gives $Z_1=280\mathrm{\Omega }$, $Z_0=520\mathrm{\Omega }$, and propagation velocities of $2.98\times {10}^5$ and $2.10\times {10}^5\text{km/s}$ for a 160 km line. Fill in the four numbers the sequence surge-impedance entry wants.

Solution 10.4

The surge impedances transfer directly. The travel times come from length over velocity:

$${\tau }_1=\frac{160}{2.98\times {10}^5}=5.37\times {10}^{-4}\text{s}=537\mu \text{s},$$$${\tau }_0=\frac{160}{2.10\times {10}^5}=7.62\times {10}^{-4}\text{s}=762\mu \text{s}.$$

The entries are therefore $Z_1=280\mathrm{\Omega }$, ${\tau }_1=537\mu \text{s}$, $Z_0=520\mathrm{\Omega }$, ${\tau }_0=762\mu \text{s}$. Note that this method gives a lossless line; a study requiring attenuation must use the sequence-RLC form and supply $R_1$ and $R_0$.

Problem 10.5. Why can a fully transposed line be diagonalized by a real, constant matrix while a general untransposed line cannot?

Solution 10.5

Transposition makes all three conductors electrically interchangeable, so every matrix in the problem is symmetric and invariant under any permutation of the three phases. A matrix with that symmetry necessarily has the form $a\mathbf{I}+b\mathbf{J}$, and its eigenvectors are fixed by the symmetry alone — the $[111]$ direction and its orthogonal complement — regardless of what $a$ and $b$ are. Since $a$ and $b$ are the only place frequency could enter, the eigenvectors cannot depend on frequency either.

An untransposed line has three genuinely different conductor positions, so its matrices lose the permutation symmetry. The eigenvectors then depend on the actual entries, which are frequency-dependent and complex, so the transformation becomes complex and frequency-dependent too. Practical models handle this by fitting a constant real transformation over the frequency band of interest and accepting the error, which is why transposed data is so much easier to work with.

Problem 10.6. Matrix data is available for a line whose zero-sequence resistance is four times its positive-sequence resistance. How should it be entered?

Solution 10.6

Not through the balanced-matrix entry, which takes a single $R_s$ and assigns it to both sequences. The four-to-one ratio would be silently discarded and the zero-sequence transient left badly under-damped.

The data should be converted to sequence quantities by hand — $L_1=L_s-L_m$, $L_0=L_s+2L_m$, $C_1=C_s+C_m$, $C_0=C_s-2C_m$ — and entered through Bergeron — Sequence RLC, which accepts $R_1$ and $R_0$ independently. Nothing is lost in the conversion, because the reduction is exactly what the matrix entry would have performed; the difference is that control of the one parameter it cannot express is retained.

10.9 References

• E. Clarke, Circuit Analysis of A-C Power Systems, Wiley — the $\alpha \beta 0$ transformation used here for the balanced case.
• L. M. Wedepohl, "Application of matrix methods to the solution of travelling-wave phenomena in polyphase systems", Proc. IEE, 1963 — modal decomposition of multiconductor lines.
• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — modal travelling-wave line models.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — multiconductor lines and the eigenvalue/eigenvector formulation.

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