Single-stub tuner for antenna with S-matrix

Assumptions: the antenna is a half-wave dipole with ZL = 73 + j42.5 Ω. It is fed by a lossless Z0 = 50 Ω line at f = 1 GHz (air line, λ = 30 cm). A single shunt stub is used. Both short and open stubs are given.

A single-stub tuner places a shunt stub at a distance d from the load, where the line admittance is y = 1 + jb. The stub supplies −jb, so the total admittance becomes 1 (matched).

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.460 + j0.850
ΓL = (zL − 1)/(zL + 1) = 0.371∠42.5°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 2.181

Draw the constant-VSWR circle through zL (zL is at 0.1909λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.512 − j0.298   (0.4409λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j0.800  at 0.1553λ WTG
y2 = 1.000 − j0.800  at 0.3447λ WTG
d1 = 0.2143λ from the load
d2 = 0.4038λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b). Measure toward the generator from the short-circuit point (y = ∞; with y plotted on an ordinary Smith chart, this is the right end, 0.25λ WTG).
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b). Measure toward the generator from the open-circuit point (y = 0; the left end, 0λ WTG).

Sol 1: bstub = −0.7999
   short stub l = 0.1426λ
   open stub  l = 0.3926λ
Sol 2: bstub = +0.7999
   short stub l = 0.3574λ
   open stub  l = 0.1074λ

Physical lengths (λ = 30.000 cm):

Sol 1: d = 6.430 cm, short l = 4.279 cm, open l = 11.779 cm
Sol 2: d = 12.113 cm, short l = 10.721 cm, open l = 3.221 cm

Answer (choice): solution 1 with a short-circuited stub (d = 0.2143λ, l = 0.1426λ) uses the shortest lengths. It has the lowest loss and the widest bandwidth.

 Z0=50Ω        d = 0.2143λ (6.43 cm)
 o----------+------------------[Dipole 73+j42.5Ω]
            |
          stub l = 0.1426λ (4.28 cm)
            |
          short

Check: y(d) = 1 + j0.800. The stub adds −j0.800, so y = 1 and Γ = 0.

S-matrix

Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching (line sees the load directly):
  S11 = ΓL = 0.371∠42.5°,  VSWR = 2.18
Matching network (solution 1):
  S11 = 0.371∠111.8°   S12 = 0.928∠-55.4°
  S21 = 0.928∠-55.4°   S22 = 0.371∠-42.5°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred.