Double-stub shunt tuner for admittance load

Given: Z0 = 50 Ω (Y0 = 0.02 ℧), YL = 0.00813 + j0.0065 ℧, spacing 3λ/8, first stub 0.01λ from the load. Short-circuited stubs are assumed (the usual choice; open-stub lengths are λ/4 different).

Step 1: Normalise and convert to admittance

yL = YL·Z0 = (0.00813 + j0.0065)×50 = 0.4065 + j0.325
zL = 1.501 − j1.200
yL = 1/zL = 0.406 + j0.325
ΓL = 0.469∠-41.7°,  VSWR = 2.765

On the chart, zL is at 0.3079λ WTG and yL is diametrically opposite, at 0.0579λ WTG.

Step 2: Move to the first stub (0.01λ from the load)

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

y1 = 0.425 + j0.385  (0.0679λ WTG)

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4250  → 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 = 0.425 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 = 0.6694, √ = 0.8182
Sol A: b1 = −2.2031
Sol B: b1 = −0.5667

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

Sol A: y after stub 1 = 0.425 − j1.818
   (0.3273λ) → after 0.375λ: y2 = 1.000 + j2.925
   b2 = −2.9250
Sol B: y after stub 1 = 0.425 − j0.182
   (0.4655λ) → after 0.375λ: y2 = 1.000 − j0.925
   b2 = +0.9250

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 = 0.0678λ, l2 = 0.0524λ
Sol B short: l1 = 0.1679λ, l2 = 0.3688λ

Answer (Smith chart values agree within reading accuracy): solution A, with l1 = 0.0678λ and l2 = 0.0524λ (shortest). Solution B: l1 = 0.1679λ, l2 = 0.3688λ.

 Z0=50Ω          3λ/8          0.01λ
 o------+---------------+--------[YL]
        |               |
     stub 2          stub 1
     l2=0.0524λ       l1=0.0678λ
       SC              SC

S-parameters of the matched network

Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.01λ) 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 = 0.469∠-41.7°, VSWR = 2.76
Tuner, solution A:
  S11 = 0.469∠106.4°   S12 = 0.883∠-15.9°
  S21 = 0.883∠-15.9°   S22 = 0.469∠41.7°
  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.