As frequency rises, lumped L and C become awkward to use. The values get so small that parasitics swamp them, and part-to-part spread turns directly into spread in performance. Stub matching gets around this by using the transmission line itself as the element.
Only two parts
With distributed elements there are really only two tools:
- Line length — rotates the point clockwise along a constant-
|Γ|circle - Stub — adds susceptance at that position
If the line's characteristic impedance equals the chart reference, lengthening the line leaves |Γ| unchanged. The point simply swings around, keeping its distance from the center. This is the decisive difference from lumped elements: think of a line as something that rotates the match rather than advancing it, and nothing gets confusing.
Three facts worth memorizing
| Line | What it does |
|---|---|
| λ/4 open stub | Becomes a short |
| λ/4 shorted stub | Becomes an open — the same as adding nothing |
| λ/2 line | Whatever you connect comes back out unchanged |
Open and short swap at λ/4 because that is half a turn (180°) around the chart. At λ/2 you go all the way round and come back. One full turn of the chart corresponds to λ/2 — a conversion you will use from here on.
Single-stub matching, step by step
At 1 GHz, match a 100 Ω load to 50 Ω. Both the line and the stub are 50 Ω.
The load's reflection coefficient is
Γ_L = (100 − 50) / (100 + 50) = 1/3The point will now travel around the |Γ| = 1/3 circle.
There are two steps:
- Lengthen the line to carry the point to where the normalized conductance is 1
- Put a shunt stub there to cancel the remaining susceptance
Why aim for g = 1? Because a shunt stub can only add susceptance. It cannot move the conductance, so the line has to bring the point onto the g = 1 circle first.
The line length works out to
θ = acos(−1/3) / 2 = 0.9553 rad = 54.74° = 0.1520 λAt that position the impedance and normalized admittance are
Z = 33.333 − j23.570 Ω
y = 1.0000 + j0.7071The conductance is exactly 1. All that remains is a susceptance of +j0.7071.
Add a stub to cancel it. An open stub's admittance is y = j tan θ, so choose the length for which tan θ = −0.7071:
θ = 144.74° = 0.4020 λAdding them up:
y = 1.0000 + j0.7071 − j0.7071 = 1.0000 + j0
Zin = 50.000 + j0 ΩWe are in the center.
There are two solutions — which one to take
A constant-|Γ| circle meets the g = 1 circle at two points, so there are always two solutions.
| Line length | Open stub | Shorted stub | |
|---|---|---|---|
| Solution A | 0.1520 λ | 0.4020 λ | 0.3480 λ |
| Solution B | 0.3480 λ | 0.0980 λ | 0.1520 λ |
Both match perfectly at the design frequency. But their bandwidths differ. The fractional bandwidth where |S11| stays below −15 dB is
Solution A (stub 0.4020 λ) → 16.7 %
Solution B (stub 0.0980 λ) → 19.9 %The shorter stub is the wider-band one. The reason is straightforward: a stub's electrical length scales with frequency, so the longer the stub, the more its susceptance swings for the same frequency offset — and the sooner the match falls apart.
The shorter one wins in implementation too. It takes less board area, has lower loss and is less sensitive to fabrication error. When in doubt, take the shorter stub — because all three point the same way.
Open stubs and shorted stubs
The same job can be done with an open stub or a shorted stub; the lengths differ by λ/4.
- Microstrip — open stubs are the natural choice. A short needs a via, and its inductance becomes an error
- Waveguide and coax — a short can be easier to build
- When DC has to pass — a shorted stub is grounded at DC, so it can double as the bias path
That last point is used all the time. Using one shorted stub for both matching and bias feed is standard practice in power amplifiers.
Note also that an open stub leaks fields from its tip. Even if it is ideally a perfect open, it looks slightly longer in reality (the end effect), so it is normal to make it a touch shorter than the design value.
The quarter-wave transformer is a special case
Sometimes a single line does the job without a stub. If the load is purely resistive, a λ/4 line alone gives a match:
Zc = √(Z0 × RL)To match 200 Ω to 50 Ω, Zc = √(50 × 200) = 100 Ω. If you are free to choose the line impedance, this is the simplest solution there is.
But the load must be purely resistive. If there is reactance on it, you first have to lengthen the line to bring it onto the real axis, or cancel it with a stub. Real loads are usually complex, so single-stub matching gets used more often.
A quarter-wave transformer also gains bandwidth with more sections, on the same principle as multi-stage lumped matching.
Try it
Lines and stubs are elements whose behavior is easy to see on the chart.
Smith Match — Matching Network Designer
Add a transmission line and change its length, and the point moves on a circle around the center. Add a stub where it crosses the g = 1 circle and you have retraced the procedure above. Put Solution A and Solution B side by side in the sweep view and the bandwidth difference is plain to see.