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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

QuantityUnilateralBilateral
ΓS0.55∠150°0.6307∠148.0°
ΓL0.45∠30°0.5551∠26.7°
Max gain (ratio)14.3016.21
Max gain (dB)11.55 dB12.10 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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