---
title: "Double-stub on 75 Ω coax, load 0.4+j0.85"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-coax-normalized-load"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-coax-normalized-load
site: BCT Engineering
---

# Double-stub on 75 Ω coax, load 0.4+j0.85

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

- Asked in: 2075 Bhadra, 10+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-normalized-load
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

A 75-ohm coaxial line is terminated with a normalized complex load of 0.4 + j0.85 ohms. Design a double-stub matching system using short-circuited coaxial line of 75-ohm characteristic impedance. Sketch the network using micro strip.

## Short answer

For Z0=75 Ω and zL=0.4+j0.85, use stubs at the load and 3λ/8 apart. Solution A: b1=-0.874, b2=-2.847, short stub lengths l1=0.1357λ and l2=0.0538λ, giving Γ=0 and VSWR=1. In microstrip, replace the coax with a 75 Ω trace and short stubs by strips ending in via holes to ground.

## Answer

**Given:** Z0 = 75 Ω coaxial line, normalised load zL = 0.4 + j0.85 (ZL = 30 + j63.75 Ω). The stubs are short-circuited 75 Ω coaxial lines.
**Assumptions** (not given): the first stub is at the load, and the stub spacing is d = 3λ/8.

#### Step 1: Normalise and convert to admittance
```text
zL = 0.400 + j0.850
yL = 1/zL = 0.453 − j0.963
ΓL = 0.635∠94.0°,  VSWR = 4.483
```
On the chart, zL is at 0.1195λ WTG and yL is diametrically opposite, at 0.3695λ WTG.

#### Step 2: First stub at the load
The first stub is placed at the load, so y1 = yL = 0.453 − j0.963.

#### Step 3: Forbidden-region check and rotated g = 1 circle
```text
βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4533  → 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 = 0.453 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 = 0.7011, √ = 0.8373
Sol A: b1 = −0.8741
Sol B: b1 = +0.8005
```

#### Step 5: Move 0.375λ to stub 2 and find b2
```text
Sol A: y after stub 1 = 0.453 − j1.837
   (0.3263λ) → after 0.375λ: y2 = 1.000 + j2.847
   b2 = −2.8473
Sol B: y after stub 1 = 0.453 − j0.163
   (0.4681λ) → after 0.375λ: y2 = 1.000 − j0.847
   b2 = +0.8473
```
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 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).
```text
Sol A short: l1 = 0.1357λ, l2 = 0.0538λ
Sol B short: l1 = 0.3574λ, l2 = 0.3619λ
```

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | at load | at load |
| Stub spacing | 0.375λ | 0.375λ |
| b1, b2 | -0.874, -2.847 | 0.800, 0.847 |
| Short l1, l2 | 0.1357λ, 0.0538λ | 0.3574λ, 0.3619λ |

**Answer (choice):** solution A (l1 = 0.1357λ, l2 = 0.0538λ) has the shorter stubs, so it is preferred.

**Check:** the input admittance at stub 2 is 1.000 + j0.000, so Γ = 0 and VSWR = 1.

#### Sketch using microstrip
In microstrip, each "short-circuited" stub is a strip that ends in a via hole to the ground plane. The coax line is replaced by a 75 Ω microstrip trace.
```text
Top view (ground plane underneath)

     stub 2           stub 1
   l2=0.0538λ       l1=0.1357λ
       ||               ||
       ● via            ● via
       ||               ||
=======##====== 3λ/8 ===##[ZL]
in (75 Ω microstrip)       load
```

## Check yourself

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

1. For zL = 0.4 + j0.85, why is the first stub at the load handled with yL?
   A. Shunt stubs add susceptance, so yL = 1/zL = 0.453 - j0.963 is used
   B. Because zL is capacitive
   C. Because Z0 = 75 Ω
   D. Because the stub is in series
   Answer: A. Shunt stubs are in parallel, so the design uses admittance: yL = 0.453 - j0.963. (Key point: Load to admittance)

2. With spacing 3λ/8, what is the maximum g1 that can be matched, and does g1 = 0.453 pass?
   A. 0.5; no
   B. 2.0; yes
   C. 1.0; yes
   D. 2.0; no
   Answer: B. 1/sin^2(135°) = 2, and g1 = 0.4533 < 2, so matching is possible. (Key point: Forbidden region)

3. After stub 1 and moving 0.375λ, why does the admittance have g = 1?
   A. Because stub 2 adds conductance
   B. Because the load is lossless
   C. Stub 1 puts the point on the rotated g = 1 circle, which maps onto g = 1 after 3λ/8
   D. Because it is moved 90°
   Answer: C. Stub 1 changes only b, to a point on the rotated circle; moving 3λ/8 gives y2 = 1 + jb2, and stub 2 cancels jb2. (Key point: Stub 2 condition)

4. Which solution is preferred and why?
   A. B (0.3574λ, 0.3619λ), longer stubs
   B. B, because b is positive
   C. Neither works
   D. A (l1 = 0.1357λ, l2 = 0.0538λ), shorter stubs
   Answer: D. Solution A uses the shorter stubs so it has lower loss and wider bandwidth. (Key point: Preferred solution)

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