Double stub advantages and matching steps

Advantages of double-stub over single-stub matching

1. The stub positions are fixed. Only the two stub lengths are varied, while a single stub needs both its position d and its length l to change.
2. It suits adjustable tuners (sliding shorts in coax or waveguide) for loads that vary. A single stub would need a movable stub along a slotted line, which is impractical.
3. It is easier to build in fixed microstrip or coax with a fixed junction. (Disadvantage: some loads cannot be matched, the forbidden region g > 1/sin²βd. A triple stub removes this.)

Steps for double-stub matching

1. Normalise ZL, plot zL, and convert it to yL (rotate 180°).
2. If the first stub is at d1 from the load, rotate yL toward the generator by d1 to get y1.
3. Draw the g = 1 circle rotated toward the load by the stub spacing d (chart angle 720°·d/λ).
4. From y1, move along its constant-g circle to the rotated g = 1 circle. This gives y1′ = g1 + jb′ and b1 = b′ − b(y1).
5. Rotate y1′ toward the generator by d. It lands on the g = 1 circle: y2 = 1 + jb2′.
6. Stub 2 must supply b2 = −b2′.
7. Convert b1 and b2 into stub lengths (from the short-circuit point for shorted stubs, or the open-circuit point for open stubs).
8. Check that y = 1 at stub 2.

Example

ZL = 60 − j80 Ω, Z0 = 50 Ω, short-circuited stubs, first stub at the load, spacing d = λ/8.

Step 1: Normalise and convert to admittance

zL = 1.200 − j1.600
yL = 1/zL = 0.300 + j0.400
ΓL = 0.593∠-46.8°,  VSWR = 3.911

On the chart, zL is at 0.3151λ WTG and yL is diametrically opposite, at 0.0651λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.300 + 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 = 0.3000  → 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 = 0.300 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 = 0.5100, √ = 0.7141
Sol A: b1 = +1.3141
Sol B: b1 = −0.1141

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

Sol A: y after stub 1 = 0.300 + j1.714
   (0.1675λ) → after 0.125λ: y2 = 1.000 − j3.380
   b2 = +3.3805
Sol B: y after stub 1 = 0.300 + j0.286
   (0.0481λ) → after 0.125λ: y2 = 1.000 + j1.380
   b2 = −1.3805

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).

Sol A short: l1 = 0.3965λ, l2 = 0.4542λ
Sol B short: l1 = 0.2319λ, l2 = 0.0998λ

Answer: solution B (l1 = 0.2319λ, l2 = 0.0998λ) gives the shorter total stub length.

 Z0=50Ω        λ/8
 o------+-------------+-------[60−j80 Ω]
        |             |
     stub 2        stub 1
     0.0998λ       0.2319λ
       SC            SC