Double-stub short and open for inductive load

Assumptions: inductive load ZL = 50 + j100 Ω on a Z0 = 50 Ω line. Double shunt stubs, the first at the load, spacing d = λ/8. Both short- and open-circuited stubs are given.

Step 1: Normalise and convert to admittance

zL = 1.000 + j2.000
yL = 1/zL = 0.200 − j0.400
ΓL = 0.707∠45.0°,  VSWR = 5.828

On the chart, zL is at 0.1875λ WTG and yL is diametrically opposite, at 0.4375λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.200 − j0.400.

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

βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2000  → 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 = 0.200 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 = 0.3600, √ = 0.6000
Sol A: b1 = +2.0000
Sol B: b1 = +0.8000

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

Sol A: y after stub 1 = 0.200 + j1.600
   (0.1619λ) → after 0.125λ: y2 = 1.000 − j4.000
   b2 = +4.0000
Sol B: y after stub 1 = 0.200 + j0.400
   (0.0625λ) → after 0.125λ: y2 = 1.000 + j2.000
   b2 = −2.0000

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 toward the generator from the short-circuit point (y = ∞; with y plotted on an ordinary Smith chart, this is the right end, 0.25λ WTG).
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b). Measure toward the generator from the open-circuit point (y = 0; the left end, 0λ WTG).

Sol A short: l1 = 0.4262λ, l2 = 0.4610λ
Sol A open : l1 = 0.1762λ, l2 = 0.2110λ
Sol B short: l1 = 0.3574λ, l2 = 0.0738λ
Sol B open : l1 = 0.1074λ, l2 = 0.3238λ

Answer (choice): with short stubs, solution B (l1 = 0.3574λ, l2 = 0.0738λ). With open stubs, solution A (l1 = 0.1762λ, l2 = 0.2110λ, total 0.387λ) is slightly shorter than solution B (l1 = 0.1074λ, l2 = 0.3238λ, total 0.431λ).

Check: y at stub 2 = 1 + j0, so all standing waves on the main line are removed (VSWR goes from 5.83 to 1).

Electrical diagram

 Z0=50Ω          λ/8          
 o------+---------------+--------[50+j100Ω]
        |               |
     stub 2          stub 1
     l2=0.0738λ       l1=0.3574λ
       SC              SC

Physical connection (microstrip)

   stub 2 (l2)       stub 1 (l1)
      ||                ||
      ● via (short)     ○ open end
      ||                ||
======##===== λ/8 ======##====[ZL]
 in (50 Ω)                  load

A short stub ends in a via to ground. An open stub just ends (a small end-correction length is subtracted for fringing).