---
title: "Double-stub shunt tuner for admittance load"
subject: "RF and Microwave Engineering"
chapter: 2
key: "semester_7/rf_and_microwave_engineering:double-stub-admittance-load"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-admittance-load
site: BCT Engineering
---

# Double-stub shunt tuner for admittance load

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

- Asked in: 2079 Bhadra, 8+2 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-admittance-load
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2

## Question

A 50 Ω lossless transmission line is required to be matched with the load admittance 0.00813 + j0.0065 ℧, by a double-stub shunt tuner with separation of 3λ/8 and the distance of the first stub from the load is 0.01λ. Calculate the length of each stub by using the smith chart. Write the s-parameter for the matched network.

## Short answer

Normalize the load admittance to yL=0.4065+j0.325; use short stubs, first 0.01λ from the load, spacing 3λ/8. Solution A gives l1=0.0678λ and l2=0.0524λ. The matched network has S11=S22=0, |S21|=1; tuner S-parameters: S11=0.469∠106.4°, S21=S12=0.883∠−15.9°, S22=0.469∠41.7°.

## Answer

**Given:** Z0 = 50 Ω (Y0 = 0.02 ℧), YL = 0.00813 + j0.0065 ℧, spacing 3λ/8, first stub 0.01λ from the load. Short-circuited stubs are assumed (the usual choice; open-stub lengths are λ/4 different).

#### Step 1: Normalise and convert to admittance
```text
yL = YL·Z0 = (0.00813 + j0.0065)×50 = 0.4065 + j0.325
zL = 1.501 − j1.200
yL = 1/zL = 0.406 + j0.325
ΓL = 0.469∠-41.7°,  VSWR = 2.765
```
On the chart, zL is at 0.3079λ WTG and yL is diametrically opposite, at 0.0579λ WTG.

#### Step 2: Move to the first stub (0.01λ from the load)
Rotate yL clockwise (toward the generator) on its VSWR circle:
```text
y1 = 0.425 + j0.385  (0.0679λ WTG)
```

#### Step 3: Forbidden-region check and rotated g = 1 circle
```text
βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4250  → 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 = 0.425 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 = 0.6694, √ = 0.8182
Sol A: b1 = −2.2031
Sol B: b1 = −0.5667
```

#### Step 5: Move 0.375λ to stub 2 and find b2
```text
Sol A: y after stub 1 = 0.425 − j1.818
   (0.3273λ) → after 0.375λ: y2 = 1.000 + j2.925
   b2 = −2.9250
Sol B: y after stub 1 = 0.425 − j0.182
   (0.4655λ) → after 0.375λ: y2 = 1.000 − j0.925
   b2 = +0.9250
```
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 = 0.0678λ, l2 = 0.0524λ
Sol B short: l1 = 0.1679λ, l2 = 0.3688λ
```

| Quantity | Solution A | Solution B |
|---|---|---|
| Stub 1 from load | 0.01λ | 0.01λ |
| Stub spacing | 0.375λ | 0.375λ |
| b1, b2 | -2.203, -2.925 | -0.567, 0.925 |
| Short l1, l2 | 0.0678λ, 0.0524λ | 0.1679λ, 0.3688λ |

**Answer (Smith chart values agree within reading accuracy):** solution A, with l1 = 0.0678λ and l2 = 0.0524λ (shortest). Solution B: l1 = 0.1679λ, l2 = 0.3688λ.

```text
 Z0=50Ω          3λ/8          0.01λ
 o------+---------------+--------[YL]
        |               |
     stub 2          stub 1
     l2=0.0524λ       l1=0.0678λ
       SC              SC
```

#### S-parameters of the matched network
Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.01λ) 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] ·[line d1], 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 = 0.469∠-41.7°, VSWR = 2.76
Tuner, solution A:
  S11 = 0.469∠106.4°   S12 = 0.883∠-15.9°
  S21 = 0.883∠-15.9°   S22 = 0.469∠41.7°
  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 yL found from YL = 0.00813 + j0.0065 ℧ on a 50 Ω line?
   A. yL = YL x Y0 = 0.00016 + j0.00013
   B. yL = YL/Y0 = YL x Z0 = 0.4065 + j0.325
   C. yL = 1/YL = 61 - j49
   D. yL = YL/Z0
   Answer: B. Normalised admittance is YL/Y0 with Y0 = 1/50, i.e. YL x 50. (Key point: Normalising admittance)

2. The first stub is 0.01λ from the load. What does this mean for y1?
   A. y1 = yL, since stub 1 is at the load
   B. y1 is found by rotating 0.375λ
   C. y1 = 1 + j0
   D. yL is rotated 0.01λ toward the generator to y1 = 0.425 + j0.385
   Answer: D. A short rotation of 0.01λ toward the generator gives y1, the admittance seen at the first stub. (Key point: First stub distance)

3. Is g1 = 0.425 matchable with 3λ/8 spacing?
   A. Yes, because 0.425 < 2
   B. No, because 0.425 < 1
   C. No, because g1 must be above 2
   D. Yes, because spacing does not matter
   Answer: A. Only g > 1/sin^2(135°) = 2 lies in the forbidden region. (Key point: Forbidden region)

4. Which stub lengths does solution A give and what is the S-matrix property?
   A. l1 = 0.1679λ, l2 = 0.3688λ; S11 = 1
   B. l1 = 0.0524λ, l2 = 0.0678λ; |S21| = 0
   C. l1 = 0.0678λ, l2 = 0.0524λ; S22 = ΓL* and Γin = 0 after matching
   D. l1 = 0.0678λ, l2 = 0.0524λ; Γin = ΓL
   Answer: C. Solution A has the shorter stubs; the lossless network has |S11|^2 + |S21|^2 = 1 and cancels reflection. (Key point: Stub results)

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