Single short-circuited stub design steps

Self-defined data: Z0 = 50 Ω, ZL = 100 + j80 Ω, single shunt short-circuited stub with the same Z0 = 50 Ω. Aim: find the stub position d (from the load) and length l so that the line sees Z0 (Γ = 0).

Principle

A shunt stub adds susceptance, so the work is done with admittances. Move from the load toward the generator until the real part of the line admittance is y = 1 + jb. At that point, connect a stub of susceptance −jb. The total is then y = 1.

Step 1: Normalise and plot

zL = (100 + j80)/50 = 2 + j1.6
|ΓL| = 0.555, VSWR = 3.49

Step 2: Convert to admittance

Move zL by 180° (λ/4) on the constant-VSWR circle:

yL = 1/zL = 0.305 − j0.244  (0.4584λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1 + j1.334  at 0.1718λ WTG
y2 = 1 − j1.334  at 0.3282λ WTG
d1 = 0.5 − 0.4584 + 0.1718 = 0.2134λ
d2 = 0.5 − 0.4584 + 0.3282 = 0.3697λ

Step 4: Required stub susceptance

Solution 1: bstub = −1.334
Solution 2: bstub = +1.334

Step 5: Stub length (short circuit at y = ∞, the right end of the chart, 0.25λ WTG)

For a short-circuited stub, y = −j cot βl. Start at the short-circuit point and move toward the generator to the required susceptance:

Sol. 1: −cot βl = −1.334 → l = 0.1024λ
Sol. 2: −cot βl = +1.334 → l = 0.3976λ

Result

Solution 1 is preferred because it uses the shorter line and the shorter stub, giving lower loss and wider bandwidth.

 Z0=50Ω           d = 0.2134λ
 o------------+----------------[ZL=100+j80]
              |
             stub l = 0.1024λ
              |
            short

Physical length example: at 2 GHz with an air line (λ = 15 cm), d = 3.20 cm and l = 1.54 cm.

Check: y at d1 = 1 + j1.334. The stub adds −j1.334, so y = 1 and the line is matched.