Chapter 8 — Travelling waves and the Bergeron line

Every model in Modules 1 and 2 rested on a hidden assumption: when a voltage changes, it changes everywhere in the branch at once. That is what lumped means, and for a transmission line it is false. A disturbance launched at one end of a 50 km line does not reach the other end for roughly 170 microseconds, which is a long interval when the time step is 50 µs. In this chapter we take distance seriously, and the result is the most distinctive model in EMT simulation: a line represented not by an impedance but by a delay.

Learning objectives

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

• Explain when a line must be treated as distributed rather than lumped.
• Derive the wave equation for a lossless line and identify its surge impedance $Z_C$ and travel time $\tau$.
• Show that $v+Z_{C}i$ is unchanged along a travelling wave, and use that to build the two-terminal history-source model.
• Compute the reflection coefficient at a termination and predict voltage doubling at an open end.
• Build a distributed line in NumaSim from a Model plus two Terminals, and read the reflection staircase off the scope.

8.1 When distance starts to matter

The relevant question is how long a disturbance takes to cross the line. On an overhead line the propagation speed is close to the speed of light, so a 50 km span costs about $50/\mathrm{300,000}\approx 167\mu \text{s}$. That figure is only meaningful in comparison with the time step:

• If the crossing takes less than one step, the far end responds within a single step anyway, and a lumped model is honest.
• If the crossing takes many steps, a lumped model is a fiction. It reports the receiving end reacting before the disturbance could physically arrive.

The comparison between travel time and time step is the entire basis for choosing a line model, and Chapter 9 makes it quantitative. For the moment it is enough to observe that a line long enough to matter needs a model with memory of where things are, not merely when.

The physical picture is a chain of infinitesimal slices. Each slice of length $dx$ has series inductance $L^{\prime }dx$ and shunt capacitance $C^{\prime }dx$, where the primes mean "per unit length". Current entering a slice partly flows through and partly charges the slice's capacitance to ground. That handoff, repeated down the line, is what carries energy along at finite speed.

A differential slice of a lossless line. Series inductance $L^{\prime }dx$ opposes the change in current; shunt capacitance $C^{\prime }dx$ must be charged before the voltage can move on. Repeating the slice down the line is what gives the disturbance a finite speed.

8.2 The wave equation on a lossless line

Apply Kirchhoff to the slice. The voltage lost across the series inductance over the length $dx$ is

$$-\frac{\partial v}{\partial x}=L^{\prime }\frac{\partial i}{\partial t},$$

and the current diverted into the shunt capacitance is

$$-\frac{\partial i}{\partial x}=C^{\prime }\frac{\partial v}{\partial t}.$$

Differentiate the first with respect to $x$, the second with respect to $t$, and eliminate the mixed derivative:

$$\frac{{\partial }^{2}v}{\partial x^2}=L^{\prime }{C}^{\prime }\frac{{\partial }^{2}v}{\partial t^2}.$$

This is the wave equation, and its solutions are shapes that slide along the line without changing form. Writing the propagation velocity as

$$\nu =\frac{1}{\sqrt{L^{\prime }{C}^{\prime }}},$$

the general solution is any pair of such shapes, one moving in each direction:

$$v(x,t)=v^{+}(x-\nu t)+v^{-}(x+\nu t).$$

The argument $x-\nu t$ stays constant for an observer moving in the $+x$ direction at speed $\nu$; that observer sees $v^{+}$ frozen. The $v^{-}$ term is the same story travelling the other way. Nothing here is an approximation — for a lossless line this is the exact solution, and it is why the model we are about to build is unusually faithful.

8.3 Surge impedance and travel time

The companion current follows from the same two equations. For the forward wave alone, substituting $v^{+}$ back into $-\partial v/\partial x=L^{\prime }\partial i/\partial t$ gives a current proportional to the voltage, with the constant of proportionality

$$Z_C=\sqrt{\frac{L^{\prime }}{C^{\prime }}}.$$

$Z_C$ is the surge (or characteristic) impedance. Two properties of it are easily misread. It is not a resistance in the sense of dissipating energy, and it does not depend on the line's length. It is the ratio a travelling wave enforces between its own voltage and its own current, set purely by the line's geometry through $L^{\prime }$ and $C^{\prime }$. Overhead lines lie in the neighbourhood of a few hundred ohms; cables, whose $C^{\prime }$ is far larger, are an order of magnitude lower.

Because the backward wave carries current in the opposite direction, the two solutions combine as

$$i(x,t)=\frac{1}{Z_C}(v^{+}(x-\nu t)-v^{-}(x+\nu t)).$$

Finally, if the line has length $d$, the time a wave needs to cross it is

$$\tau =\frac{d}{\nu }=d\sqrt{L^{\prime }{C}^{\prime }}.$$

Of these three quantities only two are independent. Given either $(Z_C,\tau )$ or $(L^{\prime },C^{\prime },d)$, NumaSim reconstructs the other pair from $L^{\prime }d=Z_C\tau$ and $C^{\prime }d=\tau /Z_C$.

8.4 The invariant that makes the model work

The step that turns this physics into an algorithm is to add $Z_C$ times the current equation to the voltage equation:

$$v(x,t)+Z_{C}i(x,t)=2v^{+}(x-\nu t).$$

The right-hand side depends on $x$ and $t$ only through the combination $x-\nu t$. So the quantity $v+Z_{C}i$ does not change for an observer riding along with the forward wave. Subtracting instead of adding gives the mirror statement: $v-Z_{C}i$ is unchanged along the backward wave.

That invariance is the entire model. Knowing $v+Z_{C}i$ at the sending end now is the same as knowing it at the receiving end one travel time from now, and nothing about the intervening line need be known, because nothing happening in between can overtake the wave.

8.5 A two-port with no instantaneous coupling

Label the ends $k$ and $m$, and define $i_{km}$ and $i_{mk}$ as the currents flowing into the line at each end. A wave leaving $k$ at time $t-\tau$ arrives at $m$ at time $t$. Riding along with it, $v+Z_{C}i$ is unchanged, where $i$ is measured in the direction of travel. At $k$ that direction is $i_{km}$; at $m$ the wave is leaving the line, so the current in the direction of travel is $-i_{mk}$. Equating the invariant at departure and arrival:

$$v_k(t-\tau )+Z_C{i}_{km}(t-\tau )=v_m(t)-Z_C{i}_{mk}(t).$$

Solve for the present current at $m$:

$$\boxed{{\displaystyle i_{mk}(t)=\frac{1}{Z_C}{v}_m(t)+I_m^{\text{hist}}(t-\tau )}}$$$$I_m^{\text{hist}}(t-\tau )=-\frac{1}{Z_C}{v}_k(t-\tau )-i_{km}(t-\tau ).$$

Repeating the argument for a wave travelling $m\to k$ gives the same form at the other end:

$$i_{km}(t)=\frac{1}{Z_C}{v}_k(t)+I_k^{\text{hist}}(t-\tau ),I_k^{\text{hist}}(t-\tau )=-\frac{1}{Z_C}{v}_m(t-\tau )-i_{mk}(t-\tau ).$$

Compared with the companion models of Chapter 2, the family resemblance is immediate: a conductance in parallel with a known current source. The conductance is $1/Z_C$, which is constant, real, and independent of $\mathrm{\Delta }t$. The current source is the history term, exactly as before.

One difference, however, is of the first importance. The history source at end $m$ depends only on conditions at end $k$, one travel time ago. No term links $v_k(t)$ to $v_m(t)$, so the two ends are not coupled at the present instant at all.

The travelling-wave line as two independent Norton equivalents. Each end sees a conductance $1/Z_C$ and a history current source; the only link between them is the delay line carrying conditions from the far end, $\tau$ seconds ago.

The consequence for the solver is substantial. Chapter 3 assembled one conductance matrix over the whole network; a travelling-wave line splits that matrix in two. NumaSim takes the decoupling literally. A Bergeron line is realized as two independent half-line components, each placed in its own electrical island with its own solve, exchanging nothing but delayed history through a ring buffer. A line in the middle of a circuit therefore does not enlarge the system to be solved; it divides it.

8.6 Reflections

A wave arriving at a termination generally cannot satisfy the termination's own $v$–$i$ relationship on its own, so a second wave is launched backwards to make up the difference.

Suppose a forward wave $v^{+}$ reaches an end terminated in a resistance $R_L$, and call the reflected wave $v^{-}$. The terminal voltage is the sum $v^{+}+v^{-}$; the current delivered to the load is $(v^{+}-v^{-})/Z_C$. Ohm's law at the termination ties them together:

$$v^{+}+v^{-}=R_L\frac{v^{+}-v^{-}}{Z_C}.$$

Solving for the ratio gives the reflection coefficient

$$\mathrm{\Gamma }=\frac{v^{-}}{v^{+}}=\frac{R_L-Z_C}{R_L+Z_C}.$$

Three cases carry all the intuition:

Termination	$\mathrm{\Gamma }$	Behaviour
Matched, $R_L=Z_C$	$0$	Nothing reflects. The line behaves as if it went on forever.
Open, $R_L\to \mathrm{\infty }$	$+1$	The reflected wave adds: the terminal voltage doubles.
Short, $R_L=0$	$-1$	The reflected wave cancels: the terminal voltage stays at zero.

Voltage doubling at an open end deserves particular attention, because it is not a numerical artifact but the reason energizing an unloaded line is a genuine insulation hazard. The arriving wave carries current that has nowhere to go, and the only way to satisfy the zero-current condition at the open end is to launch a backward wave whose current cancels the incident current. By the $v=Z_{C}i$ relation, that backward wave's voltage adds rather than cancels.

The same formula applies at the sending end, with $R_L$ replaced by the source's internal impedance. An ideal voltage source has zero impedance and therefore ${\mathrm{\Gamma }}_s=-1$: it reflects everything, with a sign flip. A wave that bounces off a mismatched load and returns to an ideal source is sent straight back out, inverted. Repeated round trips of this kind produce the reflection staircase examined in the lab.

8.7 Building a distributed line in NumaSim

Distributed lines are built from two palette components in the Transmission category rather than one, so the procedure is worth setting out in full.

1. Place a Transmission Line Model. It has no ports and takes part in no circuit — it is a named definition box. Give it a Model name (TL1 by default), leave Conductors at 1, and choose the Data entry method. Bergeron — Surge Impedance asks directly for Zc (Ω) and τ (travel time); Bergeron — RLC asks instead for Length (km) and per-kilometre $R$, $L$, $C$, from which the same two numbers are derived.
2. Place two Transmission Line Terminals, one at each end of the line in your circuit. On each, set Transmission Line Model to the model's name and give both the same Transmission Line name. Exactly two terminals may share a name — that pairing is what tells NumaSim which two points in the circuit are the ends of the same line.
3. Wire each terminal into the network like any other one-port.

When you press run, the model and its two terminals are fused into the pair of half-line components described in the previous section. Each terminal can optionally monitor its own current, with a selectable sign convention, which is how the sample below publishes Is and Ir.

8.8 Lab: watch a wave bounce

Lab 8A — the reflection staircase

Open the Bergeron line demo in simulator →

The circuit is a $100\text{V}$ ideal DC source at the sending end, a line with $Z_C=100\mathrm{\Omega }$ and $\tau =250\mu \text{s}$, and a selectable termination of $50$, $100$, or $200\mathrm{\Omega }$ at the receiving end. The run is $2\text{ms}$ at $\mathrm{\Delta }t=50\mu \text{s}$, so the travel time is exactly five steps and the window holds eight one-way transits. (Full description: Bergeron line demo.)

Begin with the selector on $100\mathrm{\Omega }$, for which ${\mathrm{\Gamma }}_L=0$. Nothing reflects, so Vr sits at zero for one travel time and then steps cleanly to $100\text{V}$ and stays there. That flat delay is the whole of the distributed model in a single trace; a lumped element would have moved Vr on the very first step.

Now switch the selector to $200\mathrm{\Omega }$, which mismatches the line:

$${\mathrm{\Gamma }}_L=\frac{200-100}{200+100}=\frac{1}{3}.$$

The first arrival overshoots to $100(1+\frac{1}{3})=133.3\text{V}$. The reflected third travels back, hits the ideal source, inverts (${\mathrm{\Gamma }}_s=-1$), and returns. Every round trip therefore multiplies the correction by ${\mathrm{\Gamma }}_s{\mathrm{\Gamma }}_L=-\frac{1}{3}$, so Vr updates only at odd multiples of $\tau$ and alternates about its final value:

Arrival	$t$	Vr
1st	$0.25\text{ms}$	$133.3\text{V}$
2nd	$0.75\text{ms}$	$88.9\text{V}$
3rd	$1.25\text{ms}$	$103.7\text{V}$
4th	$1.75\text{ms}$	$98.8\text{V}$

The sequence converges on $100\text{V}$, the only value a lossless line can settle at: once the transient has passed, the line is simply a conductor between an ideal source and a resistor.

The third position, $50\mathrm{\Omega }$, gives ${\mathrm{\Gamma }}_L=-\frac{1}{3}$. The first arrival now undershoots to $66.7\text{V}$, and because ${\mathrm{\Gamma }}_s{\mathrm{\Gamma }}_L=+\frac{1}{3}$ the staircase climbs monotonically — $66.7$, $88.9$, $96.3$, $98.8$ — rather than alternating. The line and the source are unchanged; the sign of a single coefficient has altered the entire character of the response.

Two further features of the run deserve attention:

1. Vs never moves. It is pinned at $100\text{V}$ throughout, because an ideal source has zero impedance, so all the visible activity is at the far end. Making the sending end respond requires source impedance: set the source's Impedance to R, give it a value, and Vs develops a staircase of its own.
2. Is starts at $1\text{A}$ regardless of the load. For the first $2\tau$ the source has no information about the far end and sees only the line, so the current is $100/Z_C=1\text{A}$. Only after the round trip does it begin working toward $100/R_L$. This is causality expressed directly in the waveform, and no lumped model reproduces it.

Lab 8B — the extremes

Open the open/short line demo in simulator →

Same line, same source, but the terminations are now pushed to the two ends of the range: $10\text{k}\mathrm{\Omega }$ standing in for an open circuit, $100\mathrm{\Omega }$ matched, and $1\mathrm{\Omega }$ standing in for a short. (Full description: Open and short line.)

The sample opens on the open end, where Vr reaches almost exactly $198\text{V}$ on the first arrival. This is the doubling of §8.6, blunted only because $10\text{k}\mathrm{\Omega }$ is not quite infinite: ${\mathrm{\Gamma }}_L=9900/10100=0.980$. Since ${\mathrm{\Gamma }}_s{\mathrm{\Gamma }}_L=-0.980$, the ringing is barely damped, and Vr swings between roughly $198\text{V}$ and near zero with a period of $4\tau =1\text{ms}$ for the whole run.

The $1\mathrm{\Omega }$ position inverts this behaviour completely. Here ${\mathrm{\Gamma }}_L=-0.980$, so the first arrival brings Vr to only $2\text{V}$, and because ${\mathrm{\Gamma }}_s{\mathrm{\Gamma }}_L$ is now $+0.980$ each round trip adds rather than alternates, so Vr creeps up in even increments of about $2\text{V}$. Plot Is alongside it: the current starts at the same $1\text{A}$ as always and then climbs in stages toward the $100\text{A}$ that a $1\mathrm{\Omega }$ load eventually draws.

The two runs use the same line and the same source. Only the sign of ${\mathrm{\Gamma }}_L$ differs, and it governs the entire character of the response.

8.9 Summary

• A line is distributed when a disturbance takes appreciably longer than one time step to cross it; the lumped assumption breaks because it lets effects outrun their causes.
• The lossless line obeys the wave equation, with propagation speed $\nu =1/\sqrt{L^{\prime }{C}^{\prime }}$, surge impedance $Z_C=\sqrt{L^{\prime }/C^{\prime }}$, and travel time $\tau =d\sqrt{L^{\prime }{C}^{\prime }}$.
• The quantity $v+Z_{C}i$ is unchanged along a forward-travelling wave. Applying that at the two ends yields a conductance $1/Z_C$ in parallel with a history source at each end — the same companion-model shape as Chapter 2, but with the history coming from the other end, $\tau$ seconds ago.
• Because there is no instantaneous path between the ends, the line decouples the network: NumaSim solves each end in its own island.
• A termination reflects a fraction $\mathrm{\Gamma }=(R_L-Z_C)/(R_L+Z_C)$ of the arriving wave. Open ends double the voltage, shorts cancel it, matched loads absorb it, and an ideal source reflects with ${\mathrm{\Gamma }}_s=-1$.

8.10 Problems

Problem 8.1. An overhead line has $L^{\prime }=1.0\text{mH/km}$ and $C^{\prime }=11.1\text{nF/km}$ and is $120\text{km}$ long. Find $Z_C$, the propagation speed, and the travel time.

Solution 8.1

$$Z_C=\sqrt{\frac{L^{\prime }}{C^{\prime }}}=\sqrt{\frac{1.0\times {10}^{-3}}{11.1\times {10}^{-9}}}=\sqrt{9.01\times {10}^4}\approx 300\mathrm{\Omega }.$$$$\nu =\frac{1}{\sqrt{L^{\prime }{C}^{\prime }}}=\frac{1}{\sqrt{(1.0\times {10}^{-3})(11.1\times {10}^{-9})}}=\frac{1}{3.33\times {10}^{-6}}\approx 3.0\times {10}^5\text{km/s},$$

which is essentially the speed of light — the expected answer for an overhead line in air.

$$\tau =\frac{d}{\nu }=\frac{120}{3.0\times {10}^5}=4.0\times {10}^{-4}\text{s}=400\mu \text{s}.$$

Problem 8.2. The line of Problem 8.1 is to be entered using the Bergeron — Surge Impedance method. Which two numbers are required? Suppose a colleague then enters $Z_C=300\mathrm{\Omega }$ and $\tau =400\mu \text{s}$ for a line they claim is $240\text{km}$ long. Is that consistent?

Solution 8.2

The required entries are $Z_C=300\mathrm{\Omega }$ and $\tau =400\mu \text{s}$. The length and the per-kilometre parameters are not asked for, because $Z_C$ and $\tau$ already contain everything the model needs.

The colleague's numbers are not consistent with $240\text{km}$ unless the line is very different from Problem 8.1's. Travel time scales with length at fixed $L^{\prime }$ and $C^{\prime }$, so doubling the length to $240\text{km}$ would double $\tau$ to $800\mu \text{s}$. Getting $400\mu \text{s}$ over $240\text{km}$ implies $\nu =6\times {10}^5\text{km/s}$ — twice the speed of light, which is impossible. Note that $Z_C$ alone would not have caught the error: it does not depend on length at all.

Problem 8.3. A $400\mathrm{\Omega }$ line is terminated in $1200\mathrm{\Omega }$. A $60\text{kV}$ step is launched from the sending end. What is the receiving-end voltage immediately after the wave arrives?

Solution 8.3

$$\mathrm{\Gamma }=\frac{1200-400}{1200+400}=\frac{800}{1600}=0.5.$$

The terminal voltage is the incident wave plus the reflected wave:

$$v_m=v^{+}(1+\mathrm{\Gamma })=60(1.5)=90\text{kV}.$$

The result is half again the applied voltage on a line that is only lightly loaded, which is why energizing a long, weakly terminated line is treated as a genuine overvoltage study rather than a formality.

Problem 8.4. Explain, without algebra, why the voltage doubles at an open end but the current does not.

Solution 8.4

An open end forces the current to be zero there. The arriving wave brings current with it, so the only way to satisfy that constraint is to launch a backward wave carrying an equal and opposite current — the two currents cancel, giving zero, as required.

Voltage and current in a travelling wave are, however, locked together by $v=Z_{C}i$ with a sign that depends on the direction of travel. A backward wave whose current is the negative of the forward wave's has a voltage equal to the positive of the forward wave's. The same reflection that cancels the current therefore adds the voltage: the boundary condition is satisfied in current at the cost of doubling the volts.

Problem 8.5. In the lab, Is held at exactly $1\text{A}$ for the first two travel times whichever termination was selected. Why $2\tau$ rather than $\tau$, and why exactly $1\text{A}$?

Solution 8.5

The $1\text{A}$ is $V/Z_C=100/100$. At the instant of energization the source has no information about the far end, so the only quantity available to it is the line's own surge impedance: the line accepts current at the rate $Z_C$ dictates, exactly as if it extended forever.

The duration is $2\tau$ rather than $\tau$ because the information must make the round trip — $\tau$ for the wave to reach the termination and a further $\tau$ for the reflection it provokes to return. Until that returning wave arrives at $0.5\text{ms}$, the sending end cannot distinguish a short from an open or a matched load.

Problem 8.6. A line is terminated in its own surge impedance at both ends and energized by a step. How many reflections occur, and what does the receiving-end waveform look like?

Solution 8.6

None. With $R_L=Z_C$ the reflection coefficient is zero at both terminations, so the incident wave is fully absorbed and nothing is sent back.

The receiving-end voltage is therefore flat at zero for one travel time, then steps once to its final value and remains there. This is the cleanest available demonstration of the delay itself, with no reflection staircase superimposed on it, which is why the lab begins from the matched case before mismatching the load.

8.11 References

• L. Bergeron, Water Hammer in Hydraulics and Wave Surges in Electricity, ASME Committee / Wiley — the graphical travelling-wave method, borrowed from hydraulics, that the line model is named after.
• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — the distributed-parameter line as a pair of Norton equivalents coupled only by delay.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — travelling-wave line models and the two-port history-source formulation.

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