Reflections and the reflection coefficient
What happens at an impedance mismatch? A 45 Ω load terminates a 50 Ω line — what fraction reflects?
| Level | ● screening |
| Track | SI & high-speed |
| Really testing | Γ in your head. |
Free full answer — one of 12 questions answered in full on this site.
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Say this first: Part of the wave reflects: Γ = (ZL − Z0)/(ZL + Z0). 45 Ω on 50 Ω: Γ = −5/95 = −5.3% — a small negative (downward) step. Open: +1 (the voltage doubles). Short: −1. Matched: 0.
Then explain: The transmitted wave = 1 + Γ. Small steps: Γ ≈ ΔZ/(2·Z0) — a 10% Z change reflects ~5%. Multiple reflections bounce between the source and load mismatches; a bounce (lattice) diagram tracks them.
Going deeper
Worked example — a connector section. A 60 Ω connector in a 50 Ω path: Γ = 10/110 ≈ +9.1% going in, −9.1% coming out. If the connector delay is short compared with the rise time, the two reflections nearly cancel and you see only a small blip.
In RF terms. Reflected power = Γ² ≈ 0.8%. Return loss = −20·log|Γ| ≈ 21 dB. The same number appears on a VNA plot as S11.
Quick mental math. Γ ≈ ΔZ/(2·Z0) for small steps: 10% impedance change → about 5% reflection. A ±10% fab tolerance on a 50 Ω line can therefore give up to about 5% reflection at each end of a section.
Where people go wrong
- Forgetting the sign: lower impedance than the line gives a negative (downward) step.
- Treating the transmitted wave as 1 − Γ; it is 1 + Γ.
More like this in the book
This is SI-02 from Hardware Engineer Interview Questions. The book has 16 questions on signal integrity & transmission lines and 141 in all, each answered the same way: what to say first, how to explain it, the numbers worth quoting and the follow-ups that come next.
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