---
title: "Single short-circuited stub design steps"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:single-short-circuited-stub-design"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/single-short-circuited-stub-design
site: BCT Engineering
---

# Single short-circuited stub design steps

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

- Asked in: 2079 Chaitra, 10 marks
- Video: [Single shunt stub matching network using smith chart by Dr. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=r78seP5upKA) — 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/single-short-circuited-stub-design
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

With self-defined line and load impedances and mentioning the steps, design a single short-circuited matching stub.

## Short answer

For Z0=50 ohm, ZL=100+j80 ohm, use a single shunt short-circuited stub. Normalize z=2+j1.6, convert to y=0.305-j0.244; the g=1 circle is met at 1+j1.334 or 1-j1.334. Preferred solution: stub position d=0.2134lambda and length l=0.1024lambda; alternative d=0.3697lambda, l=0.3976lambda.

## Answer

**Self-defined data:** Z0 = 50 Ω, ZL = 100 + j80 Ω, single **shunt short-circuited** stub with the same Z0 = 50 Ω.
**Aim:** find the stub position d (from the load) and length l so that the line sees Z0 (Γ = 0).

#### Principle
A shunt stub adds susceptance, so the work is done with admittances. Move from the load toward the generator until the real part of the line admittance is y = 1 + jb. At that point, connect a stub of susceptance −jb. The total is then y = 1.

#### Step 1: Normalise and plot
```text
zL = (100 + j80)/50 = 2 + j1.6
|ΓL| = 0.555, VSWR = 3.49
```

#### Step 2: Convert to admittance
Move zL by 180° (λ/4) on the constant-VSWR circle:
```text
yL = 1/zL = 0.305 − j0.244  (0.4584λ WTG)
```

#### Step 3: Move toward the generator to the g = 1 circle
The VSWR circle cuts the g = 1 circle at two points:
```text
y1 = 1 + j1.334  at 0.1718λ WTG
y2 = 1 − j1.334  at 0.3282λ WTG
d1 = 0.5 − 0.4584 + 0.1718 = 0.2134λ
d2 = 0.5 − 0.4584 + 0.3282 = 0.3697λ
```

#### Step 4: Required stub susceptance
```text
Solution 1: bstub = −1.334
Solution 2: bstub = +1.334
```

#### Step 5: Stub length (short circuit at y = ∞, the right end of the chart, 0.25λ WTG)
For a short-circuited stub, y = −j cot βl. Start at the short-circuit point and move toward the generator to the required susceptance:
```text
Sol. 1: −cot βl = −1.334 → l = 0.1024λ
Sol. 2: −cot βl = +1.334 → l = 0.3976λ
```

#### Result

| Solution | Stub position d | Stub length l |
|---|---|---|
| 1 | 0.2134λ | 0.1024λ |
| 2 | 0.3697λ | 0.3976λ |

Solution 1 is preferred because it uses the shorter line and the shorter stub, giving lower loss and wider bandwidth.

```text
 Z0=50Ω           d = 0.2134λ
 o------------+----------------[ZL=100+j80]
              |
             stub l = 0.1024λ
              |
            short
```

**Physical length example:** at 2 GHz with an air line (λ = 15 cm), d = 3.20 cm and l = 1.54 cm.

**Check:** y at d1 = 1 + j1.334. The stub adds −j1.334, so y = 1 and the line is matched.

## Check yourself

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

1. Why is the single shunt-stub design carried out with admittances?
   A. Shunt elements add susceptance directly in parallel, so admittances simply add
   B. The Smith chart only works for admittance
   C. Admittance makes the VSWR circle larger
   D. Short-circuited stubs have zero admittance
   Answer: A. A shunt stub is in parallel with the line, so its susceptance adds to the line admittance. (Key point: Why admittance)

2. How is the stub position d found?
   A. Move from zL toward the load until x = 0
   B. Move from yL toward the generator along the VSWR circle until y = 1 + jb (g = 1)
   C. Move 90° from the load regardless of the VSWR
   D. Move until the susceptance is zero
   Answer: B. At the point where g = 1 the line admittance is 1 + jb, and the stub then needs to supply -jb. Two such points exist on the circle. (Key point: Stub position)

3. At y1 = 1 + j1.334, what susceptance must the stub supply, and what is the resulting total?
   A. +1.334, total 1 + j2.668
   B. -1.334, total y = 0
   C. -1.334, total y = 1
   D. 0, total y = 1 + j1.334
   Answer: C. The stub supplies -jb = -j1.334, so the total y = 1 and Γ = 0. (Key point: Stub susceptance)

4. For a short-circuited stub, which relation gives its length and where does measurement start?
   A. y = j tan βl, from the open-circuit point
   B. y = -j tan βl, from the matched point
   C. y = j cot βl, from the centre
   D. y = -j cot βl, from the short-circuit point (y = ∞) toward the generator
   Answer: D. A shorted stub has y = -j cot βl; on the chart you start at the short-circuit point y = ∞ and move toward the generator to the required susceptance. (Key point: Stub length)

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