Events and Discontinuities

Why an event is not just another step

At an event the algebraic variables jump and the differential ones do not. A rotor angle cannot change discontinuously — it is the integral of a speed — but a bus voltage can and does, because the network equations carry no derivative and simply state what V must be right now, given x and the present topology.

So the integration rule must not be applied across the discontinuity. Doing so would average a pre-fault derivative with a post-fault one over a step in which neither held, which approximates nothing. The handling instead:

1. Step exactly onto the event time.
2. Apply the change and reassemble the Y-bus.
3. Hold x fixed and re-solve Y V − I_inj(x, V) = 0 for V alone.
4. Resume, with a couple of backward-Euler steps to damp the ringing the trapezoidal rule would otherwise show.

Step 3 is done by taking an ordinary step with h·a = 0: every differential row then reads x₁ − x₀ = 0, pinning x exactly, while the algebraic rows are untouched. One assembly, one code path, no separate pattern for the sub-block.

Both the pre-event and post-event points are recorded, at the same time value, so the discontinuity is visible in the trajectory rather than smoothed away by whoever plots it.

The states are pinned, not merely left alone

With h·a = 0 the Jacobian is block lower triangular, so in exact arithmetic the state update is zero. It is not zero in floating point: no backend does forward substitution in that order — each applies its own fill-reducing permutation and partial pivoting — and roundoff leaks across the block boundary, measured at ~1e-10 rad on a faulted network whose conditioning the fault admittance dominates.

Small, but wrong in kind and cumulative over a run with many events. A rotor angle is defined to be continuous across a discontinuity, so the algebraic re-solve declines to apply the update at all, and the gate asserts bit-identity rather than a tolerance.

The events

All of them are value-only. None changes the set of unknowns, so none forces a re-analysis, and one symbolic factorization serves a whole run. See the DAE chapter for how the topology-superset pattern buys that.

A "bolted" fault is a large admittance rather than an infinite one, because an infinite admittance is not a number. 1e6 per unit against a network whose entries are order 1 leaves |V| around 1e-6, four orders below any tolerance that matters. Going larger buys nothing and costs conditioning.

A LoadStep is applied as a change of admittance, not of power — Δy = −conj(Δs)/|V₀|² at the voltage the case was initialized at — because that is what the loads in this system are. It therefore delivers exactly Δs if the voltage happens to be back at V₀, and less, in proportion to |V|², if it is not.

A tripped unit is frozen, not removed. Nothing in the network can observe a disconnected machine's rotor, so integrating it would track a quantity no result depends on, and reconnecting it properly would need synchronization, which is not modelled.

Timing

Steps are truncated to land exactly on each event time, and the time is then snapped to the event's own value rather than accumulated — so an event time need not be a multiple of the step, and does not drift with the number of steps that preceded it.

No root-finding is needed for a scheduled event: it is already a time. A relay is different.

Relays: events triggered by a state

A relay watches a quantity — a bus voltage, a machine's speed, its angle excursion — and acts if it stays out of bounds for a stated delay.

The delay is the whole design. A relay that acted the instant a threshold were crossed would trip on every fault in the network rather than the ones it is meant to clear. And because the delay is measured from the crossing, an error in when the crossing happened is an error in when the relay acts — so the crossing is located, not rounded to whichever step happened to notice it. The step is shortened onto it with the same Illinois locator continuation uses to find a reactive limit along its curve, and from that instant the relay's action is an ordinary scheduled event.

A relay whose condition stops holding forgets. That is what separates it from a stopwatch, and it is what makes a fault cleared inside the delay trip nothing. It is gated as such: the same fault held for 0.2 s trips nothing and held for 0.9 s trips, against a 0.4 s relay.

One subtlety is worth naming, because getting it wrong is silent. A condition can begin holding at an event — a fault collapses a voltage within the instant — and then both ends of the following step are already inside the trip region, so a comparison local to that step sees no crossing at all and the relay never fires. What is compared is therefore the value at the last accepted point, remembered across events, rather than the step's own start.

Watching an algebraic quantity and watching a differential one differ in kind, and the gates cover both: a bus voltage moves discontinuously at an event, so its crossings tend to sit exactly on event times, while a rotor speed is integrated and crosses somewhere strictly inside a step. The located time reproduces the threshold to 1e-7 on a 20 ms step.

Backward-Euler damping, and what it costs

The trapezoidal rule is A-stable but not L-stable, so a discontinuity excites a numerical oscillation at the fastest mode that decays only as (−1)^k. Two backward-Euler steps after each event annihilate it.

The cost is exact and known: backward Euler evaluates δ̇ at the end of the step, so on a linearly growing speed each step overshoots by Ω_b·a·h²/2 with a = P_m/2H. Two steps leave a permanent offset of Ω_b·a·h² — 2.5e-5 rad at a 1 ms step on a typical machine — and it falls as h² despite backward Euler being first-order, because only a fixed number of steps ever use it.

Both of those numbers are gated against their closed forms, on a case where the exact trajectory is known: with P_e removed by a terminal fault the acceleration is constant, the speed is linear, and the angle is exactly quadratic — which the trapezoidal rule integrates with no error at all.

Islands

A switching event can leave a group of buses with no machine and no fixed-voltage bus in it. Such an island still solves — every bus has a path to ground through its own load — but the answer is a de-energized island, not a dynamic one, and reading it as a trajectory would be a mistake. The run reports it rather than presenting it.