---
title: "Stability of input matching network from Smith chart"
subject: "RF and Microwave Engineering"
chapter: 6
key: "semester_7/rf_and_microwave_engineering:input-matching-stability-smith"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/input-matching-stability-smith
site: BCT Engineering
---

# Stability of input matching network from Smith chart

RF and Microwave Engineering · Chapter 6: RF Design Practices

- Asked in: 2074 Bhadra, 8 marks
- Video: [Stability- Smith Chart- Problem 1](https://www.youtube.com/watch?v=gKdaa1QBhJk) — Ms. S. Chithra
- 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/input-matching-stability-smith
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6

## Question

Synthesize stability parameters of input matching network for the attached sketched smith chart. [Figure: sketched Smith chart attached to the paper; not included in the scan]

## Short answer

For assumed transistor, K=0.607<1, conditionally stable. Input stability circle CS=1.132∠68.46°, RS=0.199; stable region is outside the circle because centre Γs=0 gives |Γout|=0.781<1. Choose Γs=S11*=0.894∠60.6° (stable, margin 0.076), giving Zs=10.89+j84.52 Ω; verify stability.

## Answer

The input matching network sets Γs. Its stability is judged from the **source (input) stability circle**: the set of Γs that make |Γout| = 1. Since the sketched chart is not available, the steps are shown with an assumed transistor: S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6° (Z0 = 50 Ω).

#### Step 1: Stability parameters

```text
Δ = S11S22 − S12S21 = 0.6964∠−83.07°,  |Δ| < 1
K = (1 − |S11|² − |S22|² + |Δ|²)/(2|S12S21|) = 0.607
```

K < 1 → conditionally stable; the input stability circle must be drawn.

#### Step 2: Input (source) stability circle

```text
CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²)
RS = |S12S21| / ||S11|² − |Δ|²|
CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²)
RL = |S12S21| / ||S22|² − |Δ|²|
```

```text
CS = 1.132∠68.46°,  RS = 0.199
```

#### Step 3: Locate the stable region

- |CS| − RS = 0.933 > 0, so the chart centre (Γs = 0) is outside the circle.
- With Γs = 0, |Γout| = |S22| = 0.781 < 1, so the centre is stable.
- Hence the **stable region is outside the circle**; the small shaded zone inside the circle, near the top-right edge of the chart (inductive, high-|Γ| region), must be avoided.

```text
           .----------.
         /       (##)  \   ## = unstable Γs
        |   o           |  o  = centre (stable)
        |  stable       |
         \             /
           '----------'
```

#### Step 4: Synthesis of the input match

1. Target Γs = S11* = 0.894∠60.6° for maximum unilateral gain. Distance from CS = 0.275 > RS, so it is in the stable region, but close (margin 0.076). A practical design may back off to a constant-gain circle (smaller |Γs|) for a safer margin.
2. Zs = 50(1+Γs)/(1−Γs) = 10.89 + j84.52 Ω.
3. Network: from the 50 Ω centre, add a shunt stub to reach the |Γ| = 0.894 circle, then a series line to rotate to 60.6°. Only the end point Γs matters for stability, so check that it lies in the stable region.
4. Verify: with the chosen Γs and ΓL, |Γout| = 0.848 < 1 and |Γin| = 0.912 < 1, so the amplifier is stable.

GS,max = 1/(1 − 0.894²) = 4.98 (6.97 dB) is the gain contributed by this input network.

## Check yourself

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

1. The stability of the input matching network is judged from which circle?
   A. The constant-gain circle only
   B. The load stability circle, set by |Γin| = 1
   C. The unit circle of the Smith chart
   D. The source stability circle, set by |Γout| = 1
   Answer: D. The input network sets Γs, and the Γs values giving |Γout| = 1 form the source stability circle. (Key point: Source stability circle)

2. Why is the stable region outside the circle (CS = 1.132∠68.46°, RS = 0.199)?
   A. At Γs = 0, |Γout| = |S22| = 0.781 < 1, and the centre lies outside the circle
   B. Because RS is small
   C. Because CS has a positive angle
   D. Because |CS| > 1 always means the inside is stable
   Answer: A. The centre is stable (|S22| < 1) and is outside the circle (|CS| − RS > 0), so the outside is stable. (Key point: Locating the stable region)

3. For maximum unilateral gain, what Γs is targeted, and is it stable?
   A. Γs = S11, inside the circle
   B. Γs = S11* = 0.894∠60.6°, 0.275 from CS (> RS), so stable but with small margin
   C. Γs = 0, unstable
   D. Γs = S22*, stable with large margin
   Answer: B. Γs = S11* lies 0.275 from CS which exceeds RS = 0.199, so it is stable, though close to the circle. (Key point: Target source termination)

4. If the margin of the chosen Γs is too small, what can a practical design do?
   A. Increase |Γs| beyond 0.894
   B. Remove the input network
   C. Back off to a constant-gain circle of smaller |Γs|
   D. Make K smaller
   Answer: C. A smaller |Γs| on a constant-gain circle moves away from the unstable zone at the cost of some gain. (Key point: Improving the margin)

## Related in RF and Microwave Engineering

- [Stability from sketched Smith chart (Fig. Q6)](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Astability-sketched-smith-chart) — RF and Microwave Engineering, chapter 6

## Related in other subjects

- [Smith chart, load 80+j100 on 50 Ω line](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Asmith-chart-load-80-j100-50ohm) — Electromagnetism, chapter 5