You placed the series termination resistor at the receiver. It goes at the driver, in series with its output, and the reason is what the resistor is for: it is not damping the receiver, it is raising the source impedance until driver plus resistor equals the trace impedance. Put it at the far end and the driver still launches into a mismatch, the reflection still happens, and you have added a resistor that does nothing except drop a little voltage.

The other one worth undoing: you described ringing as a function of the 20 MHz clock. A 20 MHz clock with a 1 ns edge has spectral content past 300 MHz. The clock rate tells you how often the edge happens; the edge rate tells you whether the trace behaves as a transmission line at all.

A trace is a transmission line when the signal's rise time is short compared to the round-trip propagation delay. A useful threshold is that the trace is 'electrically long' when its one-way delay exceeds about a sixth of the rise time — at roughly 150 ps per inch on FR-4 inner layers, a 1 ns edge makes anything past about an inch worth thinking about.

At each impedance discontinuity, part of the wave reflects. The reflection coefficient is (Z2 − Z1) / (Z2 + Z1): an open circuit reflects the full amplitude in phase, a short reflects it inverted, and a matched load reflects nothing. An unterminated CMOS input is very nearly an open, so the edge arrives, reflects back to the driver, reflects again off the driver's low output impedance, and you see the sum of the launched and returning waves — a decaying oscillation whose period is the round-trip time.

Two measurements distinguish a reflection from everything else it might be. First, compare the ringing period against twice the trace's propagation delay: if they match, it is a reflection, and the match is usually unmistakable rather than approximate. Second, change the far end — add a load, or probe at a different point — and see whether the ringing amplitude changes as the reflection model predicts.

Before any of it, question the probe. A standard 10x probe with its ground lead is a few nanohenries in series with a few picofarads, which rings at a few hundred megahertz on its own. If the ringing changes character when you swap to a short ground spring, you were measuring your instrument.

Series termination at the driver: one resistor, no static current, works for a point-to-point link with the load at the far end. The launched edge is half amplitude until the reflection returns, so it is wrong for a multi-drop bus where receivers sit mid-line.

Parallel termination at the receiver: full amplitude everywhere, works for multi-drop, and burns static current continuously — which on a battery-powered board is often the end of the discussion. Thevenin and AC-coupled variants trade that current against component count and low-frequency behaviour.

Slowing the driver: many parts have a slew-rate control bit, and using it costs nothing. It is often the right answer, and it is the one candidates forget because it does not involve adding a component.

A 6-inch inner-layer trace on FR-4, driven by a part with roughly 20 Ω output impedance into a 50 Ω trace and an unloaded CMOS input.

Propagation delay is about 150 ps per inch, so one way is 900 ps and the round trip is 1.8 ns — expect a ringing period near 1.8 ns, about 550 MHz. The driver reflection coefficient is (20 − 50)/(20 + 50) = −0.43, and the open receiver reflects +1. So the edge arrives at nearly double amplitude, returns, inverts partially at the driver, and decays by roughly 43% per round trip: three or four visible cycles.

Series termination needs 50 − 20 = 30 Ω at the driver. Use 33 Ω, the nearest standard value, and check the result against the same round-trip prediction rather than against how the trace looks.