Chapter 9 — Losses, lumped lines, and the time step

Chapter 8 built a line model that is exact for a line with no resistance, which is not a line that exists in practice. This chapter restores the losses and then confronts a question the travelling-wave model raises but cannot answer for itself: what is to be done with a line so short that a wave crosses it in less than one time step? Both answers hinge on the same comparison between the line's travel time and $\mathrm{\Delta }t$, and both lead to choosing the model and the step deliberately rather than by default.

Learning objectives

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

• Explain why line resistance is lumped rather than distributed, and where the lumps go.
• State the condition under which the lumped-loss model is trustworthy.
• Describe the nominal $\pi$ model and when it is the right choice.
• Predict which model NumaSim will use for a given line and time step, and recognize the automatic demotion to $\pi$.
• Quantify the error introduced by rounding the travel time to a whole number of steps, and pick a $\mathrm{\Delta }t$ that keeps it small.

9.1 Real lines dissipate

Add series resistance $R^{\prime }$ per unit length to the slice of Chapter 8 and the first transmission-line equation picks up a term:

$$-\frac{\partial v}{\partial x}=R^{\prime }i+L^{\prime }\frac{\partial i}{\partial t}.$$

The tidy consequences of Chapter 8 immediately come apart. The surge impedance is no longer the real constant $\sqrt{L^{\prime }/C^{\prime }}$; a travelling wave no longer keeps its shape, because different frequency components are attenuated by different amounts; and $v+Z_{C}i$ is no longer invariant along the wave, so the whole "read it here now, use it there later" mechanism loses its footing.

Doing this properly means treating $R^{\prime }$ and $L^{\prime }$ as functions of frequency and convolving in the time domain — the subject of Chapter 11, and not something NumaSim implements today. What it does implement is the compromise that EMT programs have used successfully for decades: keep the lossless travelling-wave core exactly as derived, and account for the resistance by inserting it as a handful of ordinary lumped resistors.

The compromise succeeds for a specific reason. On a transmission line the series resistance is small compared with the surge impedance: a 100 km line might have $3\mathrm{\Omega }$ of resistance against a $300\mathrm{\Omega }$ surge impedance. The resistance perturbs the propagation rather than governing it, so approximating where it acts costs far less accuracy than the crudeness of the approximation suggests.

9.2 Lumping the resistance

The question is where the lumps should go. Placing the whole resistance at one end is plainly asymmetric, and placing half at each end is better but still crude. The arrangement EMT programs settled on treats the line as two lossless half-lines, each of surge impedance $Z_C$ and travel time $\tau /2$, with the total resistance $R$ distributed as $R/4$ at each outer terminal and $R/2$ at the midpoint:

Lumped-loss line: two lossless half-lines of surge impedance $Z_C$, with the total series resistance $R$ split as $R/4$ at each end and $R/2$ in the middle. Eliminating the midpoint leaves a two-terminal model of exactly the same shape as the lossless one, but with terminal impedance $Z_C+R/4$.

The midpoint is of no interest to anyone, so it is eliminated algebraically. What survives is a two-terminal model with exactly the same structure as Chapter 8's — a conductance in parallel with a history source — but with the conductance now

$$G=\frac{1}{Z_C+R/4}$$

instead of $1/Z_C$. The extra $R/4$ is the quarter of the resistance that sits between the terminal and the first lossless section.

This model is trustworthy when

$$\frac{R}{4}\ll Z_C,$$

which is the formal statement that the resistance is a perturbation. The 10 km line in this chapter's lab has $R=0.5\mathrm{\Omega }$ and $Z_C\approx 300\mathrm{\Omega }$, giving $R/4=0.125\mathrm{\Omega }$, or four parts in ten thousand. A power line that violates the condition is difficult to construct. The condition does break down at high frequency, in studies such as power-line carrier, because the effective resistance climbs steeply with frequency and the lumps cease to be small.

9.3 What the loss does to the history term

The lossless history source of Chapter 8 depended only on the far end, one travel time ago. The lossy one is richer. Writing $Z=Z_C+R/4$ and $h=(Z_C-R/4)/Z$, the history current at end $k$ becomes a weighted sum of four delayed quantities:

$$I_k^{\text{hist}}=-\frac{Z_C}{Z^2}{v}_m(t-\tau )-\frac{Z_{C}h}{Z}{i}_{mk}(t-\tau )-\frac{R/4}{Z^2}{v}_k(t-\tau )-\frac{(R/4)h}{Z}{i}_{km}(t-\tau ).$$

The first two terms are the far-end contribution already familiar from Chapter 8. The last two are new, and they express something physical: a wave that left end $k$ one travel time ago has partly returned, because the $R/4$ lump at the far end is itself a small mismatch and reflects. Setting $R=0$ makes those two terms vanish identically, recovering Chapter 8 exactly.

NumaSim evaluates precisely these four coefficients at each line end. The result is exact at DC, where a steady current produces a drop of exactly $R$ ohms, and exact in the lossless limit. Between those extremes the loss is distributed approximately, which is the price of the lumping.

9.4 Short lines: the nominal $\pi$

The second problem is one of resolution. Consider a line 5 km long, which a wave crosses in about 17 µs, simulated at $\mathrm{\Delta }t=50\mu \text{s}$. The travelling-wave model must look up conditions from one travel time ago, and one travel time ago is less than one step. No such entry exists in the history buffer, and none ever will. The model is not merely inaccurate in this situation; it has nothing to read.

For such a line the delay is genuinely below the resolution of the study, so nothing is lost by ignoring it. The standard lumped representation is the nominal $\pi$: the whole series resistance and inductance in one branch, and the whole shunt capacitance split evenly between the two ends.

The nominal $\pi$ section: total series $R$ and $L$ in one branch, with the line's total capacitance $C$ split as $C/2$ at each terminal. NumaSim uses exactly one section — it does not cascade several.

Each element here is a Chapter 2 companion model, so the $\pi$ line adds nothing new to the solver: a series R-L branch and two shunt capacitors, all stamped into the same matrix as everything else. Notice that the two ends are now coupled through the series branch, which means a $\pi$ line does not split the network into islands the way a travelling-wave line does.

One limitation should be stated plainly: NumaSim uses a single $\pi$ section. Cascading several sections to approximate distributed behaviour, a common technique elsewhere, is not available from one component here. The desire to cascade sections is in any case a reliable indication that the travelling-wave model is the appropriate choice.

9.5 Choosing between the two models

The decision reduces to a single comparison.

Choosing a line model. The comparison is between the line's one-way travel time and the simulation time step; NumaSim applies the same test automatically and demotes a Bergeron line to a single nominal $\pi$ when the travel time falls below one step.

The Model's Data entry method — Bergeron — RLC or PI — RLC — expresses the modeller's intent, but that choice is not final. Before the run starts, NumaSim computes $\tau$ from the reduced parameters and applies the test itself. A line requested as Bergeron whose travel time proves shorter than the step is demoted to a nominal $\pi$; a line requested as $\pi$ remains $\pi$ regardless.

The demotion is silent. Nothing in the interface reports it, so it can be detected only by doing the arithmetic in advance or by noticing that the far end begins moving on the first step. It is best regarded as a safety net rather than a feature to rely on. When it fires unexpectedly, one of two things is true: the line is genuinely short and should have been declared $\pi$, or the time step is too coarse for the study at hand. The next two sections distinguish these cases.

9.6 The delay is a whole number of steps

The history buffer holds one entry per time step, so the delay a travelling-wave line can express is an integer multiple of $\mathrm{\Delta }t$. NumaSim rounds to the nearest:

$$N=\mathrm{round}\left(\frac{\tau }{\mathrm{\Delta }t}\right),{\tau }_{\text{eff}}=N\mathrm{\Delta }t.$$

There is no interpolation between buffer entries, so the line actually simulated has travel time ${\tau }_{\text{eff}}$ rather than $\tau$. The absolute error is at most half a step, which makes the relative error

$$\frac{|{\tau }_{\text{eff}}-\tau |}{\tau }\le \frac{\mathrm{\Delta }t}{2\tau }=\frac{1}{2N}.$$

The formula is a useful one, because $N$ is simply the number of steps per transit. Ten steps per transit gives 5 % accuracy in the travel time and fifty steps gives 1 %. Since travel time is the only quantity the lossless model takes from the data besides $Z_C$, an error in $\tau$ is an error in every reflection arrival for the remainder of the run. It does not average out; it accumulates as phase drift.

Chapter 8's demonstration is deliberately clean in this respect: $\tau =250\mu \text{s}$ at $\mathrm{\Delta }t=50\mu \text{s}$ is exactly five steps, so ${\tau }_{\text{eff}}=\tau$ and the arrival times in the table are exact. Real line data rarely lands on a whole number of steps.

9.7 Choosing the time step

Taken together, the preceding sections constrain $\mathrm{\Delta }t$ from two directions.

The shortest line sets the ceiling. Every travelling-wave line in the network requires $\tau \ge \mathrm{\Delta }t$ merely to function, and requires $N$ comfortably above that to be accurate. The shortest line of interest therefore governs the step. A study containing a 100 km line and a 5 km stub is, as far as the time step is concerned, a study about the 5 km stub. Where the resulting step is unaffordable, the sound course is to model the stub as a $\pi$ deliberately rather than leave the fallback to do it unannounced.

Cost sets the floor. Halving $\mathrm{\Delta }t$ doubles the run time, and the travelling-wave line makes this slightly worse than usual because the history buffers grow: a line with $N$ steps of delay needs $N+1$ buffer slots per end.

A workable default for switching-surge work on transmission lines is a step in the tens of microseconds, which yields a few tens of steps per transit on a line of a hundred kilometres or so. Faster phenomena such as lightning and restrikes push toward the nanosecond range, at which point almost every line in the model is electrically long. This constraint sits alongside those established earlier: the step must still resolve the fastest switching action (Chapter 6) and the fastest ringing frequency (Chapter 3). The smallest of the three governs.

9.8 Lab: two ways to get a line wrong

Both parts of this lab use the same skeleton — a $100\text{V}$ ideal step, a line, and a selectable termination — so that the only quantities changing are the line data and the time step.

Lab 9A — watch the fallback fire

Open the short-line fallback demo in simulator →

This line is entered by the opposite route from Chapter 8's. Rather than $Z_C$ and $\tau$ directly, the model asks for a Length of $10\text{km}$ and per-kilometre $R^{\prime }=0.05\mathrm{\Omega }$, $L^{\prime }=1\text{mH}$, $C^{\prime }=0.0111\mu \text{F}$. The implied constants are worth deriving before running anything:

$$Z_C=\sqrt{\frac{L^{\prime }}{C^{\prime }}}=\sqrt{\frac{{10}^{-3}}{1.11\times {10}^{-8}}}\approx 300\mathrm{\Omega },\tau =d\sqrt{L^{\prime }{C}^{\prime }}\approx 33.3\mu \text{s}.$$

The run is set to $\mathrm{\Delta }t=50\mu \text{s}$, so $\tau <\mathrm{\Delta }t$: the travelling-wave request cannot be honoured, and the line runs as a nominal $\pi$. With the termination on $300\mathrm{\Omega }$, matched to $Z_C$, run the sample and examine Vr. It starts moving on the very first step, since a $\pi$ section has no delay to give. It then overshoots to about $112\text{V}$ and rings for a few steps before settling near $100\text{V}$; that overshoot becomes significant shortly. (Full description: Short-line fallback.)

Reducing the time step to $5\mu \text{s}$ and re-running changes two things at once:

1. The line is now travelling-wave, because $N=\mathrm{round}(33.3/5)=7\ge 1$. Vr sits at zero for $7\times 5=35\mu \text{s}$ and then steps to roughly $100\text{V}$ and remains there — a clean delay followed by a step, with no overshoot whatever, because the load is matched.
2. The overshoot in the $\pi$ result was never physical. It is the $\pi$ section's own $LC$ resonance, at a period of roughly $150\mu \text{s}$, which at $\mathrm{\Delta }t=50\mu \text{s}$ was being resolved with three steps per cycle.

The second observation is the important one. The fallback did not rescue the run from a poor time step; it exchanged one error for another and continued. A $\pi$ section is the correct model for a genuinely short line, not a means of tolerating an oversized step.

Lab 9B — quantize the delay

Open the Bergeron line demo in simulator →

This part returns to Chapter 8's line, with $Z_C=100\mathrm{\Omega }$ and $\tau =250\mu \text{s}$. Set the termination to $200\mathrm{\Omega }$ so that the reflection staircase is visible, and confirm the first arrival at $0.25\text{ms}$.

Changing the time step now moves the arrival:

$\mathrm{\Delta }t$	$N=\mathrm{round}(\tau /\mathrm{\Delta }t)$	${\tau }_{\text{eff}}$	Error
$50\mu \text{s}$	5	$250\mu \text{s}$	none
$60\mu \text{s}$	4	$240\mu \text{s}$	$-4\mathrm{\%}$
$80\mu \text{s}$	3	$240\mu \text{s}$	$-4\mathrm{\%}$
$150\mu \text{s}$	2	$300\mu \text{s}$	$+20\mathrm{\%}$
$200\mu \text{s}$	1	$200\mu \text{s}$	$-20\mathrm{\%}$

Two features of the table deserve comment. First, the error is not monotonic in $\mathrm{\Delta }t$: $60\mu \text{s}$ and $80\mu \text{s}$ give the same effective travel time, because they round to different values of $N$ that happen to yield the same product. Second, at $\mathrm{\Delta }t=200\mu \text{s}$ the fourth reflection has drifted a full transit from where it belongs. No warning accompanies this; the run completes normally and the waveform remains plausible. Only the arithmetic reveals that the arrival times are wrong.

9.9 Summary

• Line resistance is small compared with $Z_C$, so it is added as lumped resistors rather than distributed: $R/4$ at each terminal and $R/2$ at the midpoint, around two lossless half-lines.
• After eliminating the midpoint, the terminal conductance becomes $1/(Z_C+R/4)$, and the history source gains two extra terms drawn from the line's own end. The model is exact at DC and valid whenever $R/4\ll Z_C$.
• A line whose travel time is shorter than one time step cannot be represented as a travelling wave at all. The nominal $\pi$ — total $R$ and $L$ in series, $C/2$ shunt at each end — is the lumped alternative, and NumaSim demotes to it automatically and silently.
• The delay is stored as a whole number of steps, $N=\mathrm{round}(\tau /\mathrm{\Delta }t)$, so the simulated travel time is off by up to half a step: a relative error of $1/2N$.
• The shortest travelling-wave line in the network sets the ceiling on $\mathrm{\Delta }t$. Tens of steps per transit is the target, and choosing $\pi$ deliberately is preferable to letting the fallback choose it.

9.10 Problems

Problem 9.1. A 200 km line has a total series resistance of $6\mathrm{\Omega }$ and a surge impedance of $320\mathrm{\Omega }$. Compute the terminal conductance the lumped-loss model uses, and check the validity condition.

Solution 9.1

The lump at each terminal is $R/4=6/4=1.5\mathrm{\Omega }$, so

$$G=\frac{1}{Z_C+R/4}=\frac{1}{320+1.5}=\frac{1}{321.5}=3.11\times {10}^{-3}\text{S}.$$

The validity condition $R/4\ll Z_C$ is comfortably met: $1.5\mathrm{\Omega }$ against $320\mathrm{\Omega }$ is under half a percent, so the lumping perturbs the terminal impedance by less than 0.5 %.

Problem 9.2. A 4 km cable has $L^{\prime }=0.4\text{mH/km}$ and $C^{\prime }=0.25\mu \text{F/km}$, and the study runs at $\mathrm{\Delta }t=20\mu \text{s}$. Which model will NumaSim use, and why?

Solution 9.2

$$\tau =d\sqrt{L^{\prime }{C}^{\prime }}=4\sqrt{(0.4\times {10}^{-3})(0.25\times {10}^{-6})}=4\sqrt{{10}^{-10}}=4({10}^{-5})=40\mu \text{s}.$$

Since $\tau =40\mu \text{s}\ge \mathrm{\Delta }t=20\mu \text{s}$, the travelling-wave model is used, with $N=\mathrm{round}(40/20)=2$ steps of delay.

Note how much slower the cable is than an overhead line: $\nu =1/\sqrt{L^{\prime }{C}^{\prime }}={10}^5\text{km/s}$, a third of the speed of light, because the dielectric raises $C^{\prime }$ substantially. Its surge impedance is correspondingly low, $Z_C=\sqrt{0.4\times {10}^{-3}/0.25\times {10}^{-6}}=40\mathrm{\Omega }$. Two steps per transit is workable but crude, and the travel time is exact here only because 40 divides evenly by 20.

Problem 9.3. For the cable of Problem 9.2, how small must $\mathrm{\Delta }t$ be to keep the travel-time error below 2 %?

Solution 9.3

The bound is $1/2N$, so $1/2N\le 0.02$ requires $N\ge 25$ steps per transit. With $\tau =40\mu \text{s}$:

$$\mathrm{\Delta }t\le \frac{\tau }{25}=\frac{40}{25}=1.6\mu \text{s}.$$

That is a twelvefold reduction from the $20\mu \text{s}$ of Problem 9.2 and therefore twelve times the run time, which illustrates concretely why short cables are expensive to model as distributed lines and why the $\pi$ option exists.

Problem 9.4. A colleague reports that their line "isn't showing any propagation delay" even though they selected a Bergeron data-entry method. List the checks to be made, in order.

Solution 9.4

1. Compute the travel time from the data entered. For a per-kilometre entry, $\tau =d\sqrt{L^{\prime }{C}^{\prime }}$. A units mistake here — H/m in place of H/km, for instance — silently produces a $\tau$ wrong by three orders of magnitude.
2. Compare it with the time step. If $\tau <\mathrm{\Delta }t$ the line was demoted to a nominal $\pi$, which has no delay to show. This is by far the most likely cause, and nothing will have announced it: the demotion is silent, which is precisely why check 1 must be done by hand.
3. Establish whether the delay is simply too small to see. With $N=1$ the entire delay is one sample, invisible on a plot spanning milliseconds.
4. Confirm that the two terminals are paired — the same Transmission Line name on both, and both pointing at the same Model. An unpaired terminal is a different failure, but it presents in the same way, as a line that does not behave like a line.

Problem 9.5. Why does a travelling-wave line split the network into two independent solves while a $\pi$ line does not?

Solution 9.5

A travelling-wave line has no instantaneous path between its ends. Each terminal's model is a conductance to ground plus a current source whose value was fixed by conditions at the other end one travel time ago — a known number by the time the present step is solved. Neither end's equations contain the other end's present-step unknowns, so the conductance matrix has no entries linking them and the two sides can be factorized and solved separately.

A $\pi$ line has a series R-L branch tying the two terminals together within the same step. That branch stamps off-diagonal entries connecting the two nodes, so the ends appear in one another's equations and must be solved simultaneously.

Problem 9.6. A study contains a 150 km line and a 3 km line, both to be modelled as travelling-wave. Using $\nu =3\times {10}^5\text{km/s}$, what time step does each demand for 20 steps per transit, and what is the sensible course of action?

Solution 9.6

Travel times are ${\tau }_{150}=150/3\times {10}^5=500\mu \text{s}$ and ${\tau }_3=3/3\times {10}^5=10\mu \text{s}$. For 20 steps per transit:

$$\mathrm{\Delta }{t}_{150}\le \frac{500}{20}=25\mu \text{s},\mathrm{\Delta }{t}_3\le \frac{10}{20}=0.5\mu \text{s}.$$

The short line dominates by a factor of 50. Since one time step serves the whole study, everything would have to run at $0.5\mu \text{s}$ — fifty times the cost, almost all of it spent resolving a line 3 km long.

The practical course is to model the 3 km line as a nominal $\pi$ on purpose and run at $25\mu \text{s}$. At that step its 10 µs of delay falls below the resolution of the study in any case, so nothing observable is lost. What matters is that the decision is made explicitly rather than left to the fallback, which makes it without saying so.

9.11 References

• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — the lumped-resistance travelling-wave line and the nominal $\pi$ alternative.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — line-model selection, lumped losses, and the relationship between travel time and time step.
• A. Greenwood, Electrical Transients in Power Systems, 2nd ed., Wiley — physical treatment of attenuation and distortion on real lines.

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Previous: Chapter 8 — Travelling waves and the Bergeron line · Next: Chapter 10 — Multiconductor lines: sequence data and modal decomposition.