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Max gains for S11 = 0.55∠−150° transistor
Short answer
Stability: K=2.29>1, |Δ|=0.147<1 – unconditionally stable. Unilateral maximum gain (S12=0): GTU,max=1/(1−|S11|²)·|S21|²·1/(1−|S22|²)=14.3 (11.6 dB). Bilateral maximum gain: MAG=(|S21|/|S12|)(K−√(K²−1))=16.2 (12.1 dB).
Using the following S-parameters: S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180° and S22 = 0.45∠−30°, design a microwave amplifier for maximum power gain (calculate and compare maximum power gain) for both bilateral and unilateral cases.
The amplifier is designed by (1) checking stability, (2) finding the source and load reflection coefficients for maximum gain in the unilateral (S12 = 0) and bilateral cases, (3) computing the maximum gains, and (4) designing single-stub matching networks.
Given: S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180°, S22 = 0.45∠−30° (Z₀ = 50 Ω).
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)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))
Step 1: Stability
S11·S22 = 0.2475∠180.0° = −0.2475 + j0.0000
S12·S21 = 0.1128∠−160.0° = −0.1060 − j0.0386
Δ = S11·S22 − S12·S21 = −0.1415 + j0.0386
= 0.1467∠164.7°, |Δ|² = 0.0215
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
= (1 − 0.3025 − 0.2025 + 0.0215) / (2 × 0.1128)
= 0.5165 / 0.2256
= 2.290
Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.436 (> 1, unconditionally stable).
K = 2.290 > 1 and |Δ| = 0.147 < 1, so the transistor is unconditionally stable. Conjugate matching can be used at both ports.
Step 2: Unilateral design (S12 assumed 0)
Maximum gain needs ΓS = S11* = 0.55∠150° and ΓL = S22* = 0.45∠30°.
Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.3025)
= 1.4337 (1.56 dB)
G0 = |S21|² = 2.820² = 7.9524 (9.00 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2025)
= 1.2539 (0.98 dB)
GTU,max = GS·G0·GL = 14.296 (11.55 dB)
Unilateral figure of merit (to check the error of assuming S12 = 0):
U = |S11·S12·S21·S22| / [(1 − |S11|²)(1 − |S22|²)]
= 0.0502
1/(1+U)² < GT/GTU < 1/(1−U)²
−0.43 dB < error < +0.45 dB
Step 3: Bilateral design (S12 included)
B1 = 1 + |S11|² − |S22|² − |Δ|² = 1.0785
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.8785
C1 = S11 − Δ·S22* = 0.4866∠−148.0°
C2 = S22 − Δ·S11* = 0.3728∠−26.7°
ΓS = [B1 − √(B1² − 4|C1|²)] / (2C1) = 0.6307∠148.0°
ΓL = [B2 − √(B2² − 4|C2|²)] / (2C2) = 0.5551∠26.7°
ZS = 50(1 + ΓS)/(1 − ΓS) = 12.2 + j13.6 Ω
ZL = 50(1 + ΓL)/(1 − ΓL) = 109.3 + j78.9 Ω
Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
= (2.820/0.040) × (2.290 − √(2.290² − 1))
= 70.500 × 0.2299
= 16.210 (12.10 dB)
Check: with these ΓS and ΓL, Γin = ΓS* and Γout = ΓL* (simultaneous conjugate match), and the transducer gain G_T works out to 16.210.
Step 4: Matching networks (single open stub + series 50 Ω line)
50 Ω --+--[ line dS ]--[ FET ]--[ line dL ]--+-- 50 Ω
source | | load
open open
stub lS stub lL
Each network uses a shunt open stub at the 50 Ω port and a 50 Ω line toward the transistor, chosen so that the transistor sees ΓS (input) and ΓL (output). The stub gives normalised susceptance b with |Γ| = b/√(4 + b²).
Bilateral (ΓS = 0.6307∠148.0°, ΓL = 0.5551∠26.7°):
input : |ΓS| = 0.6307 -> b = 2|Γ|/√(1−|Γ|²) = 1.626
open stub lS = arctan(b)/2π = 0.162 λ
series line dS = 0.115 λ
output: |ΓL| = 0.5551 -> b = 1.335
open stub lL = 0.148 λ
series line dL = 0.291 λ
Unilateral (ΓS = 0.55∠150°, ΓL = 0.45∠30°):
input : |ΓS| = 0.5500 -> b = 2|Γ|/√(1−|Γ|²) = 1.317
open stub lS = arctan(b)/2π = 0.147 λ
series line dS = 0.120 λ
output: |ΓL| = 0.4500 -> b = 1.008
open stub lL = 0.126 λ
series line dL = 0.296 λ
(Lengths are in guide wavelengths at the design frequency. The other root, with negative b, gives a second valid solution.)
Step 5: Comparison
| Quantity | Unilateral | Bilateral |
|---|---|---|
| ΓS | 0.55∠150° | 0.6307∠148.0° |
| ΓL | 0.45∠30° | 0.5551∠26.7° |
| Max gain (ratio) | 14.30 | 16.21 |
| Max gain (dB) | 11.55 dB | 12.10 dB |
The bilateral gain is 0.55 dB higher, because the feedback through S12 adds to the gain when both ports are matched together. The unilateral error bounds (U = 0.0502) are −0.43 to +0.45 dB. The bilateral optimum is slightly above the upper bound because it uses different terminations, but the difference is small, so the unilateral design is a fair first approximation and the bilateral design is the exact one.
Answer: unconditionally stable (K = 2.290, |Δ| = 0.147); GTU,max = 11.55 dB (unilateral); Gmax = 12.10 dB (bilateral).
✓ Answer checked by Claude Opus 5.5 (high effort)
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