Open-circuited double stub for 60−j80 Ω

Given: ZL = 60 − j80 Ω, Z0 = 50 Ω, open-circuited shunt stubs, spacing d = 3λ/8, first stub 0.4λ from the load.

Step 1: Normalise and convert to admittance

zL = 1.200 − j1.600
yL = 1/zL = 0.300 + j0.400
ΓL = 0.593∠-46.8°,  VSWR = 3.911

On the chart, zL is at 0.3151λ WTG and yL is diametrically opposite, at 0.0651λ WTG.

Step 2: Move to the first stub (0.4λ from the load)

Rotate yL clockwise (toward the generator) on its VSWR circle:

y1 = 0.268 − j0.208  (0.4651λ WTG)

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2675  → 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 = 0.268 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 = 0.4635, √ = 0.6808
Sol A: b1 = −1.4730
Sol B: b1 = −0.1114

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

Sol A: y after stub 1 = 0.268 − j1.681
   (0.3341λ) → after 0.375λ: y2 = 1.000 + j3.545
   b2 = −3.5447
Sol B: y after stub 1 = 0.268 − j0.319
   (0.4477λ) → after 0.375λ: y2 = 1.000 − j1.545
   b2 = +1.5447

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

Step 6: Stub lengths

• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol A open : l1 = 0.3449λ, l2 = 0.2938λ
Sol B open : l1 = 0.4824λ, l2 = 0.1586λ

Answer (design): solution A, with stub 1 at 0.4λ from the load (open, l1 = 0.3449λ) and stub 2 at 0.775λ from the load (open, l2 = 0.2938λ). Solution B (l1 = 0.4824λ, l2 = 0.1586λ) is equally valid, with almost the same total length.

 Z0=50Ω          3λ/8          0.4λ
 o------+---------------+--------[60−j80Ω]
        |               |
     stub 2          stub 1
     l2=0.2938λ       l1=0.3449λ
       OC              OC

S-matrix of the design

Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.4λ) 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] ·[line d1], 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 = 0.593∠-46.8°, VSWR = 3.91
Tuner, solution A:
  S11 = 0.593∠143.6°   S12 = 0.805∠-174.8°
  S21 = 0.805∠-174.8°   S22 = 0.593∠46.8°
  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.