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Double-stub for 80+j180 Ω at 3 GHz

Short answer

For ZL=80+j180 Ω, Z0=100 Ω at 3 GHz, spacing 3λ/8=3.75 cm. Short-circuited stubs with first at load: l1=1.143 cm (0.1143λ), l2=0.395 cm (0.0395λ). Physical diagram: microstrip with via grounds, stubs separated by 3.75 cm.

Design a double-stub impedance matching network for a given load of 80 + j180 Ohm connected to a 100-Ohm transmission line at 3 GHz with a three-eighths wavelength separation between the stubs. Illustrate necessary diagrams to show physical connections.

Given: ZL = 80 + j180 Ω, Z0 = 100 Ω, f = 3 GHz, spacing 3λ/8. Assumptions: short-circuited stubs, the first stub at the load, and an air line (λ = c/f = 3×10⁸/3×10⁹ = 10 cm).

Step 1: Normalise and convert to admittance

zL = 0.800 + j1.800
yL = 1/zL = 0.206 − j0.464
ΓL = 0.711∠51.3°,  VSWR = 5.931

On the chart, zL is at 0.1787λ WTG and yL is diametrically opposite, at 0.4287λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.206 − j0.464.

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

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2062  → 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 = 0.206 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 = 0.3699, √ = 0.6082
Sol A: b1 = −1.1442
Sol B: b1 = +0.0721

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

Sol A: y after stub 1 = 0.206 − j1.608
   (0.3377λ) → after 0.375λ: y2 = 1.000 + j3.950
   b2 = −3.9496
Sol B: y after stub 1 = 0.206 − j0.392
   (0.4385λ) → after 0.375λ: y2 = 1.000 − j1.950
   b2 = +1.9496

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 = 0.1143λ, l2 = 0.0395λ
Sol B short: l1 = 0.2615λ, l2 = 0.4246λ
QuantitySolution ASolution B
Stub 1 from loadat loadat load
Stub spacing0.375λ0.375λ
b1, b2-1.144, -3.9500.072, 1.950
Short l1, l20.1143λ, 0.0395λ0.2615λ, 0.4246λ

Physical lengths (λ = 10.000 cm):

d1 = 0.000 cm, spacing = 3.750 cm
Sol A short: l1 = 1.143 cm, l2 = 0.395 cm
Sol B short: l1 = 2.615 cm, l2 = 4.246 cm

Answer (design): solution A, with the stubs 3.75 cm apart, l1 = 0.1143λ = 1.143 cm, l2 = 0.0395λ = 0.395 cm.

Check: y at stub 2 = 1 + j0, so VSWR goes from 5.93 to 1.

Physical connection

 Z0=100Ω          3.75 cm          
 o------+---------------+--------[80+j180Ω]
        |               |
     stub 2          stub 1
     l2=0.395 cm      l1=1.143 cm
       SC              SC
Microstrip layout (dielectric: scale all lengths by 1/√εeff)
     l2            l1
     ||            ||
     ● via         ● via
     ||            ||
=====##== 3.75 ====##[ZL]

✓ Answer checked by Claude Opus 5.5 (high effort)

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