---
title: "Double-stub matching 110+j110 Ω on 100 Ω line"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-110-j110"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-110-j110
site: BCT Engineering
---

# Double-stub matching 110+j110 Ω on 100 Ω line

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

- Asked in: 2080 Chaitra, 12 marks
- Video: [Double shunt stub matching network problem solving using smith chart by Dr. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=AuO61UXmMlw) — RF Design Basics
- 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-110-j110
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

Assuming a load of 110 + j110 ohm is connected to a 100-ohm transmission line, find the lengths and spacing for a double-stub impedance matching system using three-eighths wavelength separation between the stubs.

## Short answer

For ZL=110+j110 ohm, Z0=100 ohm, spacing 3lambda/8 and first stub at the load, normalize z=1.1+j1.1 giving y=0.4545-j0.4545. Preferred solution: stub lengths l1=0.0996lambda and l2=0.0538lambda; alternative l1=0.2953lambda and l2=0.3616lambda.

## Answer

**Given:** ZL = 110 + j110 Ω, Z0 = 100 Ω, stub spacing d = 3λ/8. Short-circuited shunt stubs are assumed.
**Assumption:** the distance from the load to the first stub is not given, so the first stub is placed **at the load** (d1 = 0).

#### Step 1: Normalise and convert to admittance
```text
zL = ZL/Z0 = 1.1 + j1.1
yL = 1/zL = (1.1 − j1.1)/(1.1² + 1.1²)
   = 0.4545 − j0.4545
```
On the Smith chart, zL is at 0.1706λ WTG (wavelengths toward generator). yL is diametrically opposite, at 0.4206λ WTG.

#### Step 2: Check the forbidden region
For spacing d the first-stub conductance must satisfy g ≤ 1/sin²(βd).
```text
βd = 2π(3/8) = 135°,  1/sin²135° = 2
g = 0.4545 < 2  → matching is possible
```

#### Step 3: Rotated g = 1 circle
Rotate the g = 1 circle by 3λ/8 toward the load (270° anticlockwise). The first-stub point must lie on this rotated circle. Moving along the constant g = 0.4545 circle from yL, it meets the rotated circle at two points. Analytically (t = tan βd = −1):
```text
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.9091 − 0.2066 = 0.7025
√0.7025 = 0.8381
Solution A: b1 = 0.4545 − 0.1619 = +0.2927
Solution B: b1 = 0.4545 − 1.8381 = −1.3836
```

#### Step 4: Admittance after stub 1 and after moving 3λ/8
```text
A: y1' = 0.4545 − j0.1619  (0.468λ WTG)
   move 3λ/8 → y2 = 1 − j0.8439 (0.343λ WTG)
B: y1' = 0.4545 − j1.8381  (0.326λ WTG)
   move 3λ/8 → y2 = 1 + j2.8439 (0.201λ WTG)
```
Both points lie on the g = 1 circle, as required.

#### Step 5: Second stub cancels the susceptance
```text
A: b2 = +0.8439      B: b2 = −2.8439
```

#### Step 6: Stub lengths (short-circuited stub: y = −j cot βl)
Measure toward the generator from the short-circuit point up to the required jb. Because y is plotted on an ordinary Smith chart here, y = ∞ is the right end of the chart, which is 0.25λ on the WTG scale (not 0λ):
```text
l = (1/2π)·cot⁻¹(−b), taken in 0 to λ/2

Solution A: b1 = +0.2927 → l1 = 0.2953λ
            b2 = +0.8439 → l2 = 0.3616λ
Solution B: b1 = −1.3836 → l1 = 0.0996λ
            b2 = −2.8439 → l2 = 0.0538λ
```

#### Result

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 position | at load | at load |
| Stub spacing | 0.375λ | 0.375λ |
| Stub 1 length | 0.2953λ | 0.0996λ |
| Stub 2 length | 0.3616λ | 0.0538λ |

**Solution B** uses shorter stubs, which gives less loss and wider bandwidth, so it is preferred.

```text
 Z0=100Ω          3λ/8          ZL
 o----------+-------------+------[110+j110]
            |             |
         stub 2        stub 1
        0.0538λ        0.0996λ
          SC             SC
```

**Check:** with stub 1 = 0.0996λ, spacing 0.375λ and stub 2 = 0.0538λ, the input admittance at stub 2 is 1.000 + j0.000, so the load is perfectly matched.

**Answer:** first stub at the load, spacing 3λ/8 = 0.375λ, l1 = 0.0996λ, l2 = 0.0538λ (shortest set). The alternative is l1 = 0.2953λ, l2 = 0.3616λ.

## Check yourself

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

1. For a double-stub tuner with spacing d, which condition on the first-stub conductance g must hold for matching to be possible?
   A. g <= 1/sin^2(βd)
   B. g >= 1/sin^2(βd)
   C. g = 1 exactly
   D. g <= sin^2(βd)
   Answer: A. The first-stub point must be able to reach the rotated g = 1 circle, which requires g <= 1/sin^2(βd). Here βd = 135° so the limit is 2 and g = 0.4545 passes. (Key point: Forbidden region check)

2. Why is the g = 1 circle rotated by the stub spacing (3λ/8) toward the load?
   A. Because the load admittance must be rotated toward the generator by 3λ/8 to reach stub 1
   B. Because the point just after stub 1 must land on g = 1 after moving 3λ/8 toward the generator to stub 2
   C. Because the second stub is placed at the load
   D. Because short-circuited stubs add 270° of phase
   Answer: B. After stub 1 the admittance moves 3λ/8 toward the generator and must then have g = 1. So before moving, it must lie on the g = 1 circle rotated back toward the load. (Key point: Rotated g = 1 circle)

3. In the design, what does stub 1 change and what does stub 2 do?
   A. Stub 1 changes the conductance; stub 2 changes the spacing
   B. Both stubs add conductance so that g = 1
   C. Stub 1 adds susceptance only (g unchanged) to reach the rotated circle; stub 2 cancels the susceptance left at g = 1
   D. Stub 1 cancels the load reactance fully; stub 2 is only for bandwidth
   Answer: C. A shunt stub adds only susceptance, so g stays 0.4545 at stub 1. Stub 2 then cancels the jb of y2 = 1 + jb, leaving y = 1. (Key point: Role of stub 1 and stub 2)

4. Which solution gives the final preferred answer, and why?
   A. Solution A (0.2953λ, 0.3616λ), because it has the larger susceptance
   B. Solution A, because longer stubs are more stable
   C. Neither; the load lies in the forbidden region
   D. Solution B (0.0996λ, 0.0538λ), because shorter stubs give less loss and wider bandwidth
   Answer: D. Solution B uses the shorter stubs (l1 = 0.0996λ, l2 = 0.0538λ), which reduces loss and widens bandwidth, so it is preferred. (Key point: Preferred solution)

## Close variants set in other exams

- [Double-stub tuner for 75+j75 Ω on 50 Ω line](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Adouble-stub-tuner-75-j75) — RF and Microwave Engineering, chapter 2, asked 2×

## Related in RF and Microwave Engineering

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- [Double-stub for 80+j180 Ω at 3 GHz](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Adouble-stub-80-j180-3ghz) — RF and Microwave Engineering, chapter 2
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