---
title: "Double-stub on 75 Ω coax, load 109+j120 Ω"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-coax-109-j120"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-coax-109-j120
site: BCT Engineering
---

# Double-stub on 75 Ω coax, load 109+j120 Ω

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

- Asked in: 2080 Bhadra, 8+2 marks
- Video: [Smith Chart - Double Stub Matching Problem & Solution (Microwave & Radar Engineering)](https://www.youtube.com/watch?v=YBTn74hFE0Q) — Engineers & Electronics
- 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-coax-109-j120
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

A 75 Ω coaxial line is terminated with a complex load of 109 + j120 Ω. Design a double-stub matching system using short-circuited coaxial lines. Prepare its S-matrix.

## Short answer

Normalize ZL=109+j120 to zL=1.453+j1.600, yL=0.311−j0.342; short stubs, first at the load, spacing 3λ/8. Solution A lengths are l1=0.0997λ and l2=0.0464λ. The matched two-port has S11=S22=0, |S21|=1; tuner S: S11=0.568∠145.3°, S21=S12=0.823∠−37.9°, S22=0.568∠−41.1°.

## Answer

**Given:** Z0 = 75 Ω coax, ZL = 109 + j120 Ω, short-circuited 75 Ω coaxial stubs. **Assumptions:** the first stub is at the load, and the stub spacing is 3λ/8.

#### Step 1: Normalise and convert to admittance
```text
zL = 1.453 + j1.600
yL = 1/zL = 0.311 − j0.342
ΓL = 0.568∠41.1°,  VSWR = 3.627
```
On the chart, zL is at 0.1930λ WTG and yL is diametrically opposite, at 0.4430λ WTG.

#### Step 2: First stub at the load
The first stub is placed at the load, so y1 = yL = 0.311 − j0.342.

#### Step 3: Forbidden-region check and rotated g = 1 circle
```text
βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3111  → 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 = 0.311 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 = 0.5254, √ = 0.7248
Sol A: b1 = −1.3824
Sol B: b1 = +0.0673
```

#### Step 5: Move 0.375λ to stub 2 and find b2
```text
Sol A: y after stub 1 = 0.311 − j1.725
   (0.3320λ) → after 0.375λ: y2 = 1.000 + j3.330
   b2 = −3.3302
Sol B: y after stub 1 = 0.311 − j0.275
   (0.4533λ) → after 0.375λ: y2 = 1.000 − j1.330
   b2 = +1.3302
```
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 = 0.0997λ, l2 = 0.0464λ
Sol B short: l1 = 0.2607λ, l2 = 0.3974λ
```

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | at load | at load |
| Stub spacing | 0.375λ | 0.375λ |
| b1, b2 | -1.382, -3.330 | 0.067, 1.330 |
| Short l1, l2 | 0.0997λ, 0.0464λ | 0.2607λ, 0.3974λ |

**Answer (design):** solution A, with l1 = 0.0997λ, l2 = 0.0464λ, and spacing 0.375λ.

```text
 Z0=75Ω          3λ/8          
 o------+---------------+--------[109+j120Ω]
        |               |
     stub 2          stub 1
     l2=0.0464λ       l1=0.0997λ
       SC              SC
```

#### S-matrix
Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd  j sin βd; j sin βd  cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.
```text
Before matching:
  S11 = ΓL = 0.568∠41.1°, VSWR = 3.63
Tuner, solution A:
  S11 = 0.568∠145.3°   S12 = 0.823∠-37.9°
  S21 = 0.823∠-37.9°   S22 = 0.568∠-41.1°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1
```
Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0  e^(−jφ); e^(−jφ)  0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

## Check yourself

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

1. How is zL found for ZL = 109 + j120 Ω on 75 Ω coax?
   A. zL = ZL x Z0 = 8175 + j9000
   B. zL = ZL/Z0 = 1.453 + j1.600
   C. zL = Z0/ZL = 0.311 - j0.342
   D. zL = ZL/100 = 1.09 + j1.2
   Answer: B. Divide by the line impedance 75 Ω to get 1.453 + j1.600. (Key point: Normalisation)

2. Why can stub 1 at the load handle yL = 0.311 - j0.342 with 3λ/8 spacing?
   A. g1 = 0.311 is above the limit of 2
   B. The susceptance is negative
   C. The coax is lossless
   D. g1 = 0.311 is below the forbidden limit of 2
   Answer: D. Only g > 1/sin^2(135°) = 2 is forbidden. (Key point: Matchability)

3. Which stub lengths does the answer give for the design?
   A. l1 = 0.0997λ and l2 = 0.0464λ, spacing 0.375λ
   B. l1 = 0.2607λ and l2 = 0.3974λ
   C. l1 = 0.375λ and l2 = 0.0464λ
   D. l1 = 0.0997λ and l2 = 0.375λ
   Answer: A. Solution A (b1 = -1.382, b2 = -3.330) gives the shorter stubs, 0.0997λ and 0.0464λ. (Key point: Why solution A)

4. What does S22 = ΓL* mean for the tuner S-matrix?
   A. The tuner reflects all of the incident power
   B. The tuner is lossy
   C. The tuner is conjugate-matched to the load, so the combined Γin = 0
   D. The load is a perfect short
   Answer: C. Conjugate matching at port 2 makes the cascade reflectionless: Γin = S11 + S12S21ΓL/(1 - S22ΓL) = 0. (Key point: S-matrix check)

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