---
title: "Stability and gains for S21 = 10.11∠91°"
subject: "RF and Microwave Engineering"
chapter: 6
key: "semester_7/rf_and_microwave_engineering:amplifier-stability-gains-0-64"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/amplifier-stability-gains-0-64
site: BCT Engineering
---

# Stability and gains for S21 = 10.11∠91°

RF and Microwave Engineering · Chapter 6: RF Design Practices

- Asked in: 2073 Magh, 12 marks
- Video: [Microwave Engineering (Stability)](https://www.youtube.com/watch?v=xmpxNnYmxRg) — Taghreed Al -Erani
- 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/amplifier-stability-gains-0-64
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6

## Question

Check the stability and find the maximum gain of a transistor amplifier having S11 = 0.64∠−169°, S12 = 0.03∠50°, S21 = 10.11∠91°, S22 = 0.22∠−82°. Consider both bilateral and unilateral model. Modify the S-parameters if necessary.

## Short answer

K=0.958, |Δ|=0.198, so conditionally stable; no simultaneous conjugate match, maximum stable gain MSG=25.28 dB. Unilateral model (S12=0) gives GTU,max=22.60 dB; use resistive loading/feedback to stabilise for bilateral design.

## Answer

```text
Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)
```

#### Stability

```text
S11·S22 = 0.64×0.22∠(−169°−82°) = 0.1408∠109°
        = −0.0458 + j0.1331
S12·S21 = 0.03×10.11∠(50°+91°) = 0.3033∠141°
        = −0.2357 + j0.1909
Δ = 0.1899 − j0.0577 = 0.1985∠−16.92°
K = (1 − 0.4096 − 0.0484 + 0.0394)/(2×0.3033)
  = 0.5814/0.6066 = 0.958
μ = 0.960
```

**K = 0.958 < 1 (|Δ| = 0.198 < 1) → potentially unstable (conditionally stable).**

Stability circles:

```text
CS = 1.800∠165.96°,  RS = 0.819
CL = 34.60∠62.75°,   RL = 33.64
```

Both circles lie mostly outside the chart; since |S11|, |S22| < 1 the chart centre is stable and lies outside both circles, so stable regions are outside the circles.

#### Bilateral model

Since K < 1, simultaneous conjugate matching is not possible (B1² < 4|C1|², so the formulas give |ΓMS| = |ΓML| = 1, on the edge of the chart, where no passive match can be realised), so GT,max does not exist. The upper limit is the **maximum stable gain**:

```text
MSG = |S21|/|S12| = 10.11/0.03 = 337 → 25.28 dB
```

To use the bilateral formula, the device must be modified: e.g. add a small series/shunt resistor (resistive loading) or feedback so that the modified S-parameters give K > 1.

#### Unilateral model (S12 modified to 0)

```text
GS,max = 1/(1 − 0.64²) = 1.694 → 2.29 dB
G0     = 10.11² = 102.21       → 20.10 dB
GL,max = 1/(1 − 0.22²) = 1.051 → 0.22 dB
GTU,max = 181.93 → 22.60 dB
U = 0.076 → error −0.64 dB to +0.69 dB
```

Check that the unilateral terminations are stable:

- Γs = S11* = 0.64∠169°: distance from CS = 1.162 > RS = 0.819 → stable side.
- ΓL = S22* = 0.22∠82°: distance from CL = 34.39 > RL = 33.64 → stable side.
- With both: |Γin| = 0.700, |Γout| = 0.528, both < 1.

So the unilateral design is usable.

**Answer: K = 0.958 → potentially unstable; bilateral: MSG = 25.28 dB (GT,max undefined); unilateral: GTU,max = 181.9 = 22.60 dB.**

## Check yourself

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

1. For S11 = 0.64∠−169°, S12 = 0.03∠50°, S21 = 10.11∠91°, S22 = 0.22∠−82°, what is K and what does it show?
   A. K = 0.958 < 1, potentially unstable
   B. K = 1.5, unconditionally stable
   C. K = 0.2, |Δ| > 1
   D. K = 0.958 < 1, unconditionally stable
   Answer: A. K = 0.958 < 1 even though |Δ| = 0.198 < 1, so the device is only conditionally stable. (Key point: Stability result)

2. Why does GT,max not exist here, and what is the upper limit used?
   A. K < 1, so simultaneous conjugate match is impossible; the limit is MSG = |S21|/|S12|
   B. S21 is too large; the limit is G0
   C. |Δ| < 1; the limit is GTU
   D. S12 = 0; the limit is infinite
   Answer: A. For K < 1, ΓMS and ΓML fall outside the chart; the maximum stable gain |S21|/|S12| = 337 (25.28 dB) is the limit. (Key point: Bilateral limit when K < 1)

3. How is the unilateral maximum gain obtained here?
   A. Set S21 = 0
   B. Set S12 = 0 and use GS,max·G0·GL,max = 22.60 dB
   C. Use K − sqrt(K^2 − 1)
   D. Add 25.28 dB to G0
   Answer: B. With S12 modified to 0, GTU,max = 1.694 × 102.21 × 1.051 = 181.9 (22.60 dB). (Key point: Unilateral model)

4. How is it confirmed that Γs = S11* and ΓL = S22* are usable?
   A. Their magnitudes are below 0.5
   B. The distance of each from its circle centre exceeds the radius (stable side) and |Γin|, |Γout| < 1
   C. K equals 1
   D. They coincide with the circle centres
   Answer: B. S11* is 1.162 from CS (> 0.819) and S22* is 34.39 from CL (> 33.64), so both are on the stable side; |Γin| = 0.700, |Γout| = 0.528. (Key point: Checking unilateral terminations)

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