Double-stub tuner for 75+j75 Ω on 50 Ω line

Double-stub tuner

A double-stub tuner is an impedance-matching network with two shunt stubs (usually short-circuited) at fixed positions on the line, a fixed distance apart (λ/8, 3λ/8 or λ/4). Only the stub lengths are adjusted. Stub 1 moves the admittance onto the rotated g = 1 circle. After travelling the spacing d this point lands on the g = 1 circle, and stub 2 cancels the remaining susceptance. It suits adjustable coaxial or waveguide tuners because nothing slides along the main line. Its limit is the forbidden region g > 1/sin²βd, which cannot be matched.

Design

Given: ZL = 75 + j75 Ω, Z0 = 50 Ω, spacing 3λ/8. Assumptions: short-circuited stubs, with the first stub at the load.

Step 1: Normalise and convert to admittance

zL = 1.500 + j1.500
yL = 1/zL = 0.333 − j0.333
ΓL = 0.542∠40.6°,  VSWR = 3.370

On the chart, zL is at 0.1936λ WTG and yL is diametrically opposite, at 0.4436λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.333 − j0.333.

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3333  → 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 = 0.333 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 = 0.5556, √ = 0.7454
Sol A: b1 = −1.4120
Sol B: b1 = +0.0787

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

Sol A: y after stub 1 = 0.333 − j1.745
   (0.3309λ) → after 0.375λ: y2 = 1.000 + j3.236
   b2 = −3.2361
Sol B: y after stub 1 = 0.333 − j0.255
   (0.4559λ) → after 0.375λ: y2 = 1.000 − j1.236
   b2 = +1.2361

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 = 0.0981λ, l2 = 0.0477λ
Sol B short: l1 = 0.2625λ, l2 = 0.3917λ

Answer: stub 1 at the load, stub 2 at 3λ/8 = 0.375λ from it. Solution A: l1 = 0.0981λ, l2 = 0.0477λ (shortest). Solution B: l1 = 0.2625λ, l2 = 0.3917λ.

 Z0=50Ω          3λ/8          
 o------+---------------+--------[75+j75Ω]
        |               |
     stub 2          stub 1
     l2=0.0477λ       l1=0.0981λ
       SC              SC

Check: y at stub 2 = 1.000 + j0.000, so VSWR drops from 3.37 to 1.