---
title: "Double-stub with assumed load on 50 Ω line"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-arbitrary-load-50-ohm"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-arbitrary-load-50-ohm
site: BCT Engineering
---

# Double-stub with assumed load on 50 Ω line

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

- Asked in: 2071 Magh, 10 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-arbitrary-load-50-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

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.

## 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.

## Answer

**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
```text
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
```text
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:
```text
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
```text
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.
```text
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.

```text
 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.

## Check yourself

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

1. What choices does the answer make to design the two-stub system?
   A. ZL = 50 Ω matched, open stubs, spacing λ/2
   B. ZL = 100 + j100 Ω, short stubs, first stub at the load, spacing λ/8
   C. ZL = 100 + j100 Ω, open stubs, spacing λ/4 with stub 1 at 0.3λ
   D. Stubs in series, spacing 3λ/8, load purely resistive
   Answer: B. The load is assumed 100 + j100 Ω on a 50 Ω line, with short-circuited shunt stubs, the first at the load and spacing d = λ/8. (Key point: Assumptions)

2. For spacing λ/8 (βd = 45°), what is the largest load conductance g that can be matched and why does g1 = 0.25 qualify?
   A. g <= 1/sin(βd) = 1.41, and 0.25 is below it
   B. g >= 2 is required, so 0.25 fails
   C. Any g can be matched with two stubs
   D. g <= 1/sin^2(βd) = 2, and 0.25 is below 2
   Answer: D. Admittances with g > 1/sin^2(βd) cannot reach the rotated g = 1 circle; here the limit is 2 and g1 = 0.25 is inside it. (Key point: Forbidden region)

3. What is the role of the rotated g = 1 circle?
   A. Stub 1 must bring the admittance onto it, so after moving the spacing d it lands on g = 1
   B. Stub 2 is placed on it to read its length
   C. It marks where the load impedance is real
   D. It shows the VSWR circle after matching
   Answer: A. The g = 1 circle is rotated by the spacing toward the load; a point on it moves onto g = 1 after travelling d, where stub 2 cancels the remaining susceptance. (Key point: Rotated circle)

4. Which solution does the answer prefer and why?
   A. Solution A (l1 = 0.4233λ, l2 = 0.4574λ) because its stubs are longer
   B. Solution A because b1 is larger
   C. Solution B (l1 = 0.3347λ, l2 = 0.0869λ) because its total stub length is shorter
   D. Neither, both give VSWR above 1
   Answer: C. Solution B's stub total (about 0.42λ) is much less than A's (about 0.88λ), and both match the load. (Key point: Choosing solution)

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