Double-stub 3λ/8 for 300+j300 Ω antenna

Given: antenna load ZL = 300 + j300 Ω, Z0 = 300 Ω, stub spacing 3λ/8. Assumptions: short-circuited 300 Ω stubs, with the first stub at the antenna terminals.

Step 1: Normalise and convert to admittance

zL = 1.000 + j1.000
yL = 1/zL = 0.500 − j0.500
ΓL = 0.447∠63.4°,  VSWR = 2.618

On the chart, zL is at 0.1619λ WTG and yL is diametrically opposite, at 0.4119λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.500 − j0.500.

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.5000  → 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 = 0.500 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 = 0.7500, √ = 0.8660
Sol A: b1 = −1.3660
Sol B: b1 = +0.3660

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

Sol A: y after stub 1 = 0.500 − j1.866
   (0.3247λ) → after 0.375λ: y2 = 1.000 + j2.732
   b2 = −2.7321
Sol B: y after stub 1 = 0.500 − j0.134
   (0.4721λ) → after 0.375λ: y2 = 1.000 − j0.732
   b2 = +0.7321

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 = 0.1006λ, l2 = 0.0558λ
Sol B short: l1 = 0.3058λ, l2 = 0.3506λ

Answer (design): solution A, with the first stub at the load (l1 = 0.1006λ), the second stub 0.375λ away (l2 = 0.0558λ).

 Z0=300Ω          3λ/8          
 o------+---------------+--------[300+j300Ω]
        |               |
     stub 2          stub 1
     l2=0.0558λ       l1=0.1006λ
       SC              SC

Justification

• Matchable: the forbidden region for 3λ/8 spacing is g > 2. The load conductance g = 0.5 lies outside it, so the design is possible.
• Perfect match: the admittance at stub 2 becomes 1 + j0. VSWR drops from 2.62 to 1, and no power is reflected (Γ goes from 0.447 to 0).
• Choice of solution A: both stubs are short (about 0.1λ and 0.06λ). This gives the least stub loss, the widest bandwidth and the smallest size.
• Short-circuited stubs are preferred: they do not radiate from their ends and are easy to tune with sliding shorts.
• 3λ/8 spacing gives the same forbidden region as λ/8 but more physical room between the junctions.
• Solution B (l1 = 0.3058λ, l2 = 0.3506λ) is also valid but longer.