Double-stub matching 110+j110 Ω on 100 Ω line

Given: ZL = 110 + j110 Ω, Z0 = 100 Ω, stub spacing d = 3λ/8. Short-circuited shunt stubs are assumed. Assumption: the distance from the load to the first stub is not given, so the first stub is placed at the load (d1 = 0).

Step 1: Normalise and convert to admittance

zL = ZL/Z0 = 1.1 + j1.1
yL = 1/zL = (1.1 − j1.1)/(1.1² + 1.1²)
   = 0.4545 − j0.4545

On the Smith chart, zL is at 0.1706λ WTG (wavelengths toward generator). yL is diametrically opposite, at 0.4206λ WTG.

Step 2: Check the forbidden region

For spacing d the first-stub conductance must satisfy g ≤ 1/sin²(βd).

βd = 2π(3/8) = 135°,  1/sin²135° = 2
g = 0.4545 < 2  → matching is possible

Step 3: Rotated g = 1 circle

Rotate the g = 1 circle by 3λ/8 toward the load (270° anticlockwise). The first-stub point must lie on this rotated circle. Moving along the constant g = 0.4545 circle from yL, it meets the rotated circle at two points. Analytically (t = tan βd = −1):

b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.9091 − 0.2066 = 0.7025
√0.7025 = 0.8381
Solution A: b1 = 0.4545 − 0.1619 = +0.2927
Solution B: b1 = 0.4545 − 1.8381 = −1.3836

Step 4: Admittance after stub 1 and after moving 3λ/8

A: y1' = 0.4545 − j0.1619  (0.468λ WTG)
   move 3λ/8 → y2 = 1 − j0.8439 (0.343λ WTG)
B: y1' = 0.4545 − j1.8381  (0.326λ WTG)
   move 3λ/8 → y2 = 1 + j2.8439 (0.201λ WTG)

Both points lie on the g = 1 circle, as required.

Step 5: Second stub cancels the susceptance

A: b2 = +0.8439      B: b2 = −2.8439

Step 6: Stub lengths (short-circuited stub: y = −j cot βl)

Measure toward the generator from the short-circuit point up to the required jb. Because y is plotted on an ordinary Smith chart here, y = ∞ is the right end of the chart, which is 0.25λ on the WTG scale (not 0λ):

l = (1/2π)·cot⁻¹(−b), taken in 0 to λ/2

Solution A: b1 = +0.2927 → l1 = 0.2953λ
            b2 = +0.8439 → l2 = 0.3616λ
Solution B: b1 = −1.3836 → l1 = 0.0996λ
            b2 = −2.8439 → l2 = 0.0538λ

Result

Solution B uses shorter stubs, which gives less loss and wider bandwidth, so it is preferred.

 Z0=100Ω          3λ/8          ZL
 o----------+-------------+------[110+j110]
            |             |
         stub 2        stub 1
        0.0538λ        0.0996λ
          SC             SC

Check: with stub 1 = 0.0996λ, spacing 0.375λ and stub 2 = 0.0538λ, the input admittance at stub 2 is 1.000 + j0.000, so the load is perfectly matched.

Answer: first stub at the load, spacing 3λ/8 = 0.375λ, l1 = 0.0996λ, l2 = 0.0538λ (shortest set). The alternative is l1 = 0.2953λ, l2 = 0.3616λ.