---
title: "Single-stub tuner for antenna with S-matrix"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:single-stub-antenna-smatrix"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/single-stub-antenna-smatrix
site: BCT Engineering
---

# Single-stub tuner for antenna with S-matrix

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

- Asked in: 2076 Bhadra, 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-stub-antenna-smatrix
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

A single-stub tuner is to match a lossless line to a load of an antenna. Design the stub with any assumed placement and length and derive its S-matrix.

## Short answer

Assuming ZL=73+j42.5 Ω on 50 Ω line, choose shunt short stub at d=0.2143λ with l=0.1426λ. The network S-matrix is S11=0.371∠111.8°, S12=S21=0.928∠-55.4°, S22=0.371∠-42.5°=ΓL*; with the load connected Γin=0 and VSWR=1.

## Answer

**Assumptions:** the antenna is a half-wave dipole with ZL = 73 + j42.5 Ω. It is fed by a lossless Z0 = 50 Ω line at f = 1 GHz (air line, λ = 30 cm). A single **shunt stub** is used. Both short and open stubs are given.

A single-stub tuner places a shunt stub at a distance d from the load, where the line admittance is y = 1 + jb. The stub supplies −jb, so the total admittance becomes 1 (matched).

#### Step 1: Normalise and plot the load
```text
zL = ZL/Z0 = 1.460 + j0.850
ΓL = (zL − 1)/(zL + 1) = 0.371∠42.5°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 2.181
```
Draw the constant-VSWR circle through zL (zL is at 0.1909λ WTG).

#### Step 2: Convert to admittance (shunt stub)
Move zL by 180° on the VSWR circle, which is the same as moving λ/4:
```text
yL = 1/zL = 0.512 − j0.298   (0.4409λ 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 = 1.000 + j0.800  at 0.1553λ WTG
y2 = 1.000 − j0.800  at 0.3447λ WTG
d1 = 0.2143λ from the load
d2 = 0.4038λ from the load
```

#### Step 4: Stub susceptance and length
The stub must cancel the susceptance: bstub = −b.
- For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b). Measure toward the generator from the short-circuit point (y = ∞; with y plotted on an ordinary Smith chart, this is the right end, 0.25λ WTG).
- For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b). Measure toward the generator from the open-circuit point (y = 0; the left end, 0λ WTG).
```text
Sol 1: bstub = −0.7999
   short stub l = 0.1426λ
   open stub  l = 0.3926λ
Sol 2: bstub = +0.7999
   short stub l = 0.3574λ
   open stub  l = 0.1074λ
```

| Solution | Stub position d | Short stub l | Open stub l |
|---|---|---|---|
| 1 | 0.2143λ | 0.1426λ | 0.3926λ |
| 2 | 0.4038λ | 0.3574λ | 0.1074λ |

Physical lengths (λ = 30.000 cm):
```text
Sol 1: d = 6.430 cm, short l = 4.279 cm, open l = 11.779 cm
Sol 2: d = 12.113 cm, short l = 10.721 cm, open l = 3.221 cm
```

**Answer (choice):** solution 1 with a short-circuited stub (d = 0.2143λ, l = 0.1426λ) uses the shortest lengths. It has the lowest loss and the widest bandwidth.

```text
 Z0=50Ω        d = 0.2143λ (6.43 cm)
 o----------+------------------[Dipole 73+j42.5Ω]
            |
          stub l = 0.1426λ (4.28 cm)
            |
          short
```

**Check:** y(d) = 1 + j0.800. The stub adds −j0.800, so y = 1 and Γ = 0.

#### S-matrix
Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd  j sin βd; j sin βd  cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.
```text
Before matching (line sees the load directly):
  S11 = ΓL = 0.371∠42.5°,  VSWR = 2.18
Matching network (solution 1):
  S11 = 0.371∠111.8°   S12 = 0.928∠-55.4°
  S21 = 0.928∠-55.4°   S22 = 0.371∠-42.5°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1
```
With port 2 referenced to the load, the matched lossless network has the ideal form S = [0  e^(−jφ); e^(−jφ)  0]: no reflection at either port, and all power is transferred.

## Check yourself

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

1. At the stub position d, the line admittance is y = 1 + jb. What must the shunt stub supply?
   A. -jb
   B. +jb
   C. -1
   D. Zero susceptance
   Answer: A. The stub supplies -jb, so the total admittance is 1 and Γ = 0. (Key point: Stub principle)

2. For an open-circuited stub, which relation applies and where do you measure from?
   A. y = -j cot βl, from the short-circuit point
   B. y = j tan βl, from the open-circuit point (y = 0) toward the generator
   C. y = j cot βl, from the centre
   D. y = -j tan βl, from the load
   Answer: B. An open stub has y = j tan βl, so l = (1/2π) tan^-1(b), measured from the open-circuit point. (Key point: Open stub length)

3. For solution 1 (bstub = -0.8), the short stub is 0.1426λ. The open stub is 0.3926λ. How are they related?
   A. They differ by λ/2
   B. They are equal
   C. They differ by λ/4 (0.25λ)
   D. They differ by λ/8
   Answer: C. A shorted and an open stub differ by λ/4 for the same susceptance, 0.1426λ + 0.25λ = 0.3926λ. (Key point: Short vs open lengths)

4. For the matched lossless network, which statements hold?
   A. S22 = ΓL and S21 = 0
   B. S11 = 1 and S21 = 0
   C. S22 = 0 and |S21| > 1
   D. S22 = ΓL* and |S11|^2 + |S21|^2 = 1
   Answer: D. Conjugate match S22 = ΓL* gives Γin = 0, and lossless means |S11|^2 + |S21|^2 = 1 (0.371^2 + 0.928^2 = 1). (Key point: S-matrix of the matching network)

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