---
title: "Double-stub for 80+j180 Ω at 3 GHz"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-80-j180-3ghz"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-80-j180-3ghz
site: BCT Engineering
---

# Double-stub for 80+j180 Ω at 3 GHz

RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines

- Asked in: 2070 Bhadra, 8+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-80-j180-3ghz
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

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.

## 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.

## Answer

**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
```text
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
```text
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:
```text
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
```text
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.
```text
Sol A short: l1 = 0.1143λ, l2 = 0.0395λ
Sol B short: l1 = 0.2615λ, l2 = 0.4246λ
```

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | at load | at load |
| Stub spacing | 0.375λ | 0.375λ |
| b1, b2 | -1.144, -3.950 | 0.072, 1.950 |
| Short l1, l2 | 0.1143λ, 0.0395λ | 0.2615λ, 0.4246λ |

Physical lengths (λ = 10.000 cm):
```text
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
```text
 Z0=100Ω          3.75 cm          
 o------+---------------+--------[80+j180Ω]
        |               |
     stub 2          stub 1
     l2=0.395 cm      l1=1.143 cm
       SC              SC
```
```text
Microstrip layout (dielectric: scale all lengths by 1/√εeff)
     l2            l1
     ||            ||
     ● via         ● via
     ||            ||
=====##== 3.75 ====##[ZL]
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At 3 GHz in an air line, what is the 3λ/8 stub spacing?
   A. 3.75 mm, since λ = 1 cm
   B. 3.75 cm, since λ = 10 cm
   C. 7.5 cm, since λ = 20 cm
   D. 2.5 cm, since λ = 6.7 cm
   Answer: B. λ = c/f = 3x10^8/3x10^9 = 10 cm, so 3λ/8 = 3.75 cm. (Key point: Wavelength and spacing)

2. Where is the first stub placed and what is its normalised admittance there?
   A. At 3λ/8 from the load, y1 = 0.8 + j1.8
   B. At the load, y1 = 0.8 + j1.8
   C. At λ/4 from the load, y1 = 1 - j1
   D. At the load, y1 = 0.206 - j0.464
   Answer: D. The first stub is at the load, so y1 = yL = 1/(0.8 + j1.8) = 0.206 - j0.464. (Key point: Stub 1 position)

3. After stub 1 and travel of 3λ/8, what must the admittance at stub 2 be before stub 2 is added?
   A. On the g = 1 circle, e.g. 1 + j3.950 for solution A
   B. On the real axis with b = 0
   C. Equal to the load admittance
   D. On the g = 0 circle
   Answer: A. Stub 2 can only cancel susceptance, so the admittance there must already have g = 1; stub 2 then supplies -b. (Key point: Stub 2 condition)

4. Using solution A (l1 = 0.1143λ, l2 = 0.0395λ), what is the final result?
   A. VSWR falls from 5.93 to 2.0
   B. VSWR remains 5.93 but the phase changes
   C. y at stub 2 = 1 + j0, so VSWR falls from 5.93 to 1
   D. y = 0 at stub 2, so all power is reflected
   Answer: C. With the stubs cancelling susceptance, the line sees y = 1, which is a perfect match (VSWR = 1). (Key point: Result of design)

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