- 2078 Chaitra · 4+6 marks
Matching techniques and double-stub S-matrix
Short answer
Impedance matching sets Γ=0 using L-networks, quarter-wave transformers, single/double/triple stubs, tapers, and waveguide tuners. For a perfectly matched double-stub tuner referenced to the load, S=[[0,e^-jφ],[e^-jφ,0]], so S11=S22=0 and |S21|=1; with Z0 reference, lossless reciprocal matching gives S22=ΓL* and |S21|²=1-|ΓL|².
Illustrate different impedance matching techniques. What will be a resulting S-matrix, if you are asked to prepare a double-stub matching network of having perfect matching?
Impedance matching makes the load look like Z0 to the line, so that Γ = 0. This gives maximum power transfer, no standing waves, a better signal-to-noise ratio, and protection for the source.
Impedance matching techniques
- Lumped L-section (L-network): one series and one shunt reactance (L or C). It is simple, but works only up to a few GHz because real L and C have parasitics.
- Quarter-wave transformer: a λ/4 line with Z1 = √(Z0·RL) matches a real load. It is narrowband. Multisection (binomial or Chebyshev) versions give wider bandwidth.
- Single-stub tuning: one shunt (or series) short or open stub at a distance d from the load. Both d and l must be adjustable.
- Double-stub tuning: two shunt stubs at fixed positions (spacing usually λ/8, 3λ/8 or λ/4). Only the stub lengths are adjusted, which suits adjustable coaxial tuners. Some loads cannot be matched (forbidden region g > 1/sin²βd).
- Triple-stub tuner: removes the forbidden region.
- Tapered lines (exponential, triangular, Klopfenstein): very wideband.
- Slide-screw tuner, E-H tuner: used in waveguides.
L-section λ/4 transformer single stub
-[jX]-+- -[ Z1 , λ/4 ]- --+--d--[ZL]
| |
jB l (stub)
S-matrix of a perfectly matched double-stub network
For a two-port with both ports referenced to Z0, the S-parameters follow from the normalised ABCD matrix:
S11 = (A + B − C − D)/Δ S12 = 2(AD − BC)/Δ
S21 = 2/Δ S22 = (−A + B − C + D)/Δ
Δ = A + B + C + D
A double-stub tuner is a cascade of shunt stub 2, a line of length d, and shunt stub 1:
[ABCD] = |1 0| |cos βd j sin βd| |1 0|
|jb2 1| |j sin βd cos βd | |jb1 1|
Building blocks:
Shunt jb: S = 1/(2+jb) · | −jb 2 |
| 2 −jb |
Line βd: S = | 0 e^(−jβd) |
| e^(−jβd) 0 |
Condition for a perfect match: with the load ΓL on port 2,
Γin = S11 + S12 S21 ΓL/(1 − S22 ΓL) = 0
For a lossless, reciprocal tuner this holds when S22 = ΓL* (conjugate match at the load side). Unitarity then gives |S21|² = 1 − |ΓL|², and S12 = S21.
If port 2 is referenced to the load itself (the reference impedance is ZL), the matched network becomes an ideal matched, lossless, reciprocal two-port:
S = | 0 e^(−jφ) |
| e^(−jφ) 0 |
Here φ is the total phase delay. S11 = S22 = 0 (no reflection) and |S21| = 1 (all power reaches the load).
Numerical example (ZL = 110 + j110 Ω on a 100 Ω line, first stub at the load, d = 3λ/8, b1 = −1.384, b2 = −2.844, stub lengths 0.0996λ and 0.0538λ):
S11 = 0.466∠145.5° S12 = 0.885∠−45.8°
S21 = 0.885∠−45.8° S22 = 0.466∠−57.2°
ΓL = 0.466∠+57.2° → S22 = ΓL* ✓
|S11|² + |S21|² = 1 (lossless) ✓
Γin = 0 (VSWR = 1) ✓
Before matching, the line sees S11 = ΓL = 0.466∠57.2° (VSWR = 2.75). After matching, S11 at the input = 0.
✓ Answer checked by Claude Opus 5.5 (high effort)
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