Double-stub for 10 GHz antenna, Γ = 0.45∠60°

Given: f = 10 GHz, ΓL = 0.45∠60°, Z0 = 100 Ω. Assumptions: short-circuited stubs, the first stub at the antenna, spacing 3λ/8. Microstrip with effective permittivity εeff ≈ 1.9 (for example a 100 Ω line on RT/Duroid 5880), so λg = λ0/√εeff = 3 cm/√1.9 = 2.176 cm.

Load impedance

zL = (1 + Γ)/(1 − Γ) = 1.060 + j1.036
ZL = 105.98 + j103.58 Ω

Step 1: Normalise and convert to admittance

zL = 1.060 + j1.036
yL = 1/zL = 0.483 − j0.472
ΓL = 0.450∠60.0°,  VSWR = 2.636

On the chart, zL is at 0.1667λ WTG and yL is diametrically opposite, at 0.4167λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.483 − j0.472.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4826  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.483 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.7323, √ = 0.8557
Sol A: b1 = −1.3841
Sol B: b1 = +0.3274

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.483 − j1.856
   (0.3253λ) → after 0.375λ: y2 = 1.000 + j2.773
   b2 = −2.7732
Sol B: y after stub 1 = 0.483 − j0.144
   (0.4706λ) → after 0.375λ: y2 = 1.000 − j0.773
   b2 = +0.7732

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0996λ, l2 = 0.0551λ
Sol B short: l1 = 0.3004λ, l2 = 0.3548λ

Physical lengths (λ = 2.176 cm):

d1 = 0.000 cm, spacing = 0.816 cm
Sol A short: l1 = 0.217 cm, l2 = 0.120 cm
Sol B short: l1 = 0.654 cm, l2 = 0.772 cm

Answer (design): solution A, with l1 = 0.0996λg = 0.217 cm, l2 = 0.0551λg = 0.120 cm, and spacing 0.375λg = 0.816 cm.

Physical diagram (microstrip)

   stub 2              stub 1
  l2=1.20 mm          l1=2.17 mm
     ||                  ||
     ● via               ● via
     ||                  ||
=====##=== 8.16 mm ======##[antenna]
 100 Ω microstrip feed

S-matrices before and after the design

Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.450∠60.0°, VSWR = 2.64
Tuner, solution A:
  S11 = 0.450∠145.9°   S12 = 0.893∠-47.1°
  S21 = 0.893∠-47.1°   S22 = 0.450∠-60.0°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.