- 2071 Magh · 10 marks
Double-stub with assumed load on 50 Ω line
Short answer
Assume ZL=100+j100 Ω on 50 Ω line, short-circuited stubs, first stub at load, spacing λ/8=0.125λ. Solution B: l1=0.3347λ, l2=0.0869λ. This yields total shorter stub length and matches admittance to 1+j0.
By arbitrarily assuming a suitable load that connects to a 50-ohm transmission line find the lengths and spacing for a two-stub impedance matching system. Assume also a suitable separation between the stubs.
Assumptions: load ZL = 100 + j100 Ω on a Z0 = 50 Ω line. Short-circuited shunt stubs, the first stub at the load, and stub spacing d = λ/8.
Step 1: Normalise and convert to admittance
zL = 2.000 + j2.000
yL = 1/zL = 0.250 − j0.250
ΓL = 0.620∠29.7°, VSWR = 4.266
On the chart, zL is at 0.2087λ WTG and yL is diametrically opposite, at 0.4587λ WTG.
Step 2: First stub at the load
The first stub is placed at the load, so y1 = yL = 0.250 − j0.250.
Step 3: Forbidden-region check and rotated g = 1 circle
βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2500 → 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 = 0.250 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 = 0.4375, √ = 0.6614
Sol A: b1 = +1.9114
Sol B: b1 = +0.5886
Step 5: Move 0.125λ to stub 2 and find b2
Sol A: y after stub 1 = 0.250 + j1.661
(0.1649λ) → after 0.125λ: y2 = 1.000 − j3.646
b2 = +3.6458
Sol B: y after stub 1 = 0.250 + j0.339
(0.0548λ) → after 0.125λ: y2 = 1.000 + j1.646
b2 = −1.6458
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 = 0.4233λ, l2 = 0.4574λ
Sol B short: l1 = 0.3347λ, l2 = 0.0869λ
| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | at load | at load |
| Stub spacing | 0.125λ | 0.125λ |
| b1, b2 | 1.911, 3.646 | 0.589, -1.646 |
| Short l1, l2 | 0.4233λ, 0.4574λ | 0.3347λ, 0.0869λ |
Answer: first stub at the load, spacing λ/8 = 0.125λ. Solution B (l1 = 0.3347λ, l2 = 0.0869λ) has the shorter total stub length.
Z0=50Ω λ/8
o------+---------------+--------[100+j100Ω]
| |
stub 2 stub 1
l2=0.0869λ l1=0.3347λ
SC SC
Check: with these lengths the admittance at stub 2 is 1.000 + j0.000, so the VSWR on the main line drops from 4.27 to 1.
✓ Answer checked by Claude Opus 5.5 (high effort)
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