Double-stub on 75 Ω coax, load 109+j120 Ω

Given: Z0 = 75 Ω coax, ZL = 109 + j120 Ω, short-circuited 75 Ω coaxial stubs. Assumptions: the first stub is at the load, and the stub spacing is 3λ/8.

Step 1: Normalise and convert to admittance

zL = 1.453 + j1.600
yL = 1/zL = 0.311 − j0.342
ΓL = 0.568∠41.1°,  VSWR = 3.627

On the chart, zL is at 0.1930λ WTG and yL is diametrically opposite, at 0.4430λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.311 − j0.342.

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3111  → 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 = 0.311 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 = 0.5254, √ = 0.7248
Sol A: b1 = −1.3824
Sol B: b1 = +0.0673

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

Sol A: y after stub 1 = 0.311 − j1.725
   (0.3320λ) → after 0.375λ: y2 = 1.000 + j3.330
   b2 = −3.3302
Sol B: y after stub 1 = 0.311 − j0.275
   (0.4533λ) → after 0.375λ: y2 = 1.000 − j1.330
   b2 = +1.3302

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 = 0.0997λ, l2 = 0.0464λ
Sol B short: l1 = 0.2607λ, l2 = 0.3974λ

Answer (design): solution A, with l1 = 0.0997λ, l2 = 0.0464λ, and spacing 0.375λ.

 Z0=75Ω          3λ/8          
 o------+---------------+--------[109+j120Ω]
        |               |
     stub 2          stub 1
     l2=0.0464λ       l1=0.0997λ
       SC              SC

S-matrix

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 = 0.568∠41.1°, VSWR = 3.63
Tuner, solution A:
  S11 = 0.568∠145.3°   S12 = 0.823∠-37.9°
  S21 = 0.823∠-37.9°   S22 = 0.568∠-41.1°
  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.