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Double-stub with assumed load on 50 Ω line

Short answer

Assume ZL=100+j100 Ω on 50 Ω line, short-circuited stubs, first stub at load, spacing λ/8=0.125λ. Solution B: l1=0.3347λ, l2=0.0869λ. This yields total shorter stub length and matches admittance to 1+j0.

By arbitrarily assuming a suitable load that connects to a 50-ohm transmission line find the lengths and spacing for a two-stub impedance matching system. Assume also a suitable separation between the stubs.

Assumptions: load ZL = 100 + j100 Ω on a Z0 = 50 Ω line. Short-circuited shunt stubs, the first stub at the load, and stub spacing d = λ/8.

Step 1: Normalise and convert to admittance

zL = 2.000 + j2.000
yL = 1/zL = 0.250 − j0.250
ΓL = 0.620∠29.7°,  VSWR = 4.266

On the chart, zL is at 0.2087λ WTG and yL is diametrically opposite, at 0.4587λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.250 − j0.250.

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

βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2500  → matchable

Rotate the g = 1 circle by 0.125λ (90° 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 = 0.250 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 45° = +1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.4375, √ = 0.6614
Sol A: b1 = +1.9114
Sol B: b1 = +0.5886

Step 5: Move 0.125λ to stub 2 and find b2

Sol A: y after stub 1 = 0.250 + j1.661
   (0.1649λ) → after 0.125λ: y2 = 1.000 − j3.646
   b2 = +3.6458
Sol B: y after stub 1 = 0.250 + j0.339
   (0.0548λ) → after 0.125λ: y2 = 1.000 + j1.646
   b2 = −1.6458

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 = 0.4233λ, l2 = 0.4574λ
Sol B short: l1 = 0.3347λ, l2 = 0.0869λ
QuantitySolution ASolution B
Stub 1 from loadat loadat load
Stub spacing0.125λ0.125λ
b1, b21.911, 3.6460.589, -1.646
Short l1, l20.4233λ, 0.4574λ0.3347λ, 0.0869λ

Answer: first stub at the load, spacing λ/8 = 0.125λ. Solution B (l1 = 0.3347λ, l2 = 0.0869λ) has the shorter total stub length.

 Z0=50Ω          λ/8          
 o------+---------------+--------[100+j100Ω]
        |               |
     stub 2          stub 1
     l2=0.0869λ       l1=0.3347λ
       SC              SC

Check: with these lengths the admittance at stub 2 is 1.000 + j0.000, so the VSWR on the main line drops from 4.27 to 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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