---
title: "Double-stub short and open for inductive load"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-inductive-load-50-ohm"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-inductive-load-50-ohm
site: BCT Engineering
---

# Double-stub short and open for inductive load

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

- Asked in: 2073 Magh, 8 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-inductive-load-50-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

Assume an inductive load impedance is connected to a mismatched 50Ω transmission line. Find the size and placement of the matching stub that will remove all the standing waves and match load to the line. Use double stub shunt tuning short and open circuited stub. Draw its electrical diagram and physical connection.

## Short answer

Normalize ZL=50+j100 Ω to yL=0.2−j0.4. Using double shunt stubs at the load with λ/8 spacing: short-circuited stubs lengths l1=0.3574λ, l2=0.0738λ; open-circuited stubs lengths l1=0.1762λ, l2=0.2110λ. Z0=50 Ω.

## Answer

**Assumptions:** inductive load ZL = 50 + j100 Ω on a Z0 = 50 Ω line. Double shunt stubs, the first at the load, spacing d = λ/8. Both short- and open-circuited stubs are given.

#### Step 1: Normalise and convert to admittance
```text
zL = 1.000 + j2.000
yL = 1/zL = 0.200 − j0.400
ΓL = 0.707∠45.0°,  VSWR = 5.828
```
On the chart, zL is at 0.1875λ WTG and yL is diametrically opposite, at 0.4375λ WTG.

#### Step 2: First stub at the load
The first stub is placed at the load, so y1 = yL = 0.200 − j0.400.

#### Step 3: Forbidden-region check and rotated g = 1 circle
```text
βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2000  → matchable
```
Rotate the g = 1 circle by 0.125λ (90° 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 = 0.200 circle until it meets the rotated circle.

#### Step 4: Susceptance of stub 1
Analytically (Pozar), with t = tan βd:
```text
t = tan 45° = +1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.3600, √ = 0.6000
Sol A: b1 = +2.0000
Sol B: b1 = +0.8000
```

#### Step 5: Move 0.125λ to stub 2 and find b2
```text
Sol A: y after stub 1 = 0.200 + j1.600
   (0.1619λ) → after 0.125λ: y2 = 1.000 − j4.000
   b2 = +4.0000
Sol B: y after stub 1 = 0.200 + j0.400
   (0.0625λ) → after 0.125λ: y2 = 1.000 + j2.000
   b2 = −2.0000
```
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).
- 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 A short: l1 = 0.4262λ, l2 = 0.4610λ
Sol A open : l1 = 0.1762λ, l2 = 0.2110λ
Sol B short: l1 = 0.3574λ, l2 = 0.0738λ
Sol B open : l1 = 0.1074λ, l2 = 0.3238λ
```

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | at load | at load |
| Stub spacing | 0.125λ | 0.125λ |
| b1, b2 | 2.000, 4.000 | 0.800, -2.000 |
| Short l1, l2 | 0.4262λ, 0.4610λ | 0.3574λ, 0.0738λ |
| Open l1, l2 | 0.1762λ, 0.2110λ | 0.1074λ, 0.3238λ |

**Answer (choice):** with short stubs, solution B (l1 = 0.3574λ, l2 = 0.0738λ). With open stubs, solution A (l1 = 0.1762λ, l2 = 0.2110λ, total 0.387λ) is slightly shorter than solution B (l1 = 0.1074λ, l2 = 0.3238λ, total 0.431λ).

**Check:** y at stub 2 = 1 + j0, so all standing waves on the main line are removed (VSWR goes from 5.83 to 1).

#### Electrical diagram
```text
 Z0=50Ω          λ/8          
 o------+---------------+--------[50+j100Ω]
        |               |
     stub 2          stub 1
     l2=0.0738λ       l1=0.3574λ
       SC              SC
```

#### Physical connection (microstrip)
```text
   stub 2 (l2)       stub 1 (l1)
      ||                ||
      ● via (short)     ○ open end
      ||                ||
======##===== λ/8 ======##====[ZL]
 in (50 Ω)                  load
```
A short stub ends in a via to ground. An open stub just ends (a small end-correction length is subtracted for fringing).

## Check yourself

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

1. With d = λ/8 (βd = 45°), what is the largest matchable g1?
   A. 2
   B. 1
   C. 0.5
   D. 4
   Answer: A. 1/sin^2(45°) = 2; g1 = 0.2 is fine. (Key point: Forbidden check)

2. For yL = 0.2 - j0.4, t = 1, which gives b1 solutions?
   A. -2.0 and -0.8
   B. +2.0 and +0.8
   C. 0 and 1
   D. +0.4 and -0.4
   Answer: B. b1 = -B + [1 ± 0.6]/1 = 0.4 + 1.6 = 2.0 or 0.4 + 0.4 = 0.8. (Key point: Stub 1 susceptance)

3. For an open stub with b = 0.8 (solution B) l1 = 0.1074λ while the short stub is 0.3574λ. Why?
   A. Open stubs are always shorter by λ/2
   B. Mistake
   C. Open: y = j tan βl; short: y = -j cot βl, with λ/4 difference
   D. Shorts are not used
   Answer: C. They differ by λ/4 for the same susceptance. (Key point: Open vs short)

4. What does the check confirm?
   A. VSWR becomes 5.83
   B. y = 0
   C. Γ = 0.707
   D. y at stub 2 = 1 + j0, so VSWR goes from 5.83 to 1
   Answer: D. After matching y = 1, Γ = 0 and all standing waves vanish. (Key point: Result)

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