--- title: "Max gain for S11 = 0.656∠146.7°" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:max-gain-s11-0-656" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/max-gain-s11-0-656 site: BCT Engineering --- # Max gain for S11 = 0.656∠146.7° RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2070 Bhadra, 10 marks - Video: [RF Design Power Gain Expressions and Calculations](https://www.youtube.com/watch?v=OCJ-RE46kHc) — AmeyaKadam - 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/max-gain-s11-0-656 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Find the maximum gain for a microwave transistor amplifier with S11 = 0.656∠146.7°, S12 = 0.122∠46.1°, S21 = 2.3∠44.7°, S22 = 0.172∠−117.1°. ## Short answer The device is unconditionally stable, K = 1.071 and |Δ| = 0.247. Bilateral maximum gain is 12.96, or 11.13 dB, achieved with ΓMS = 0.850∠−150.4° and ΓML = 0.655∠76.3°. ## 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) ``` #### Step 1: Δ and K ```text S11·S22 = 0.656×0.172∠(146.7°−117.1°) = 0.1128∠29.6° = 0.0981 + j0.0557 S12·S21 = 0.122×2.3∠(46.1°+44.7°) = 0.2806∠90.8° = −0.0039 + j0.2806 Δ = 0.1020 − j0.2248 = 0.2469∠−65.59° K = (1 − 0.4303 − 0.0296 + 0.0610)/(2×0.2806) = 0.6010/0.5612 = 1.071 ``` K = 1.071 > 1 and |Δ| = 0.247 < 1 → **unconditionally stable**, so simultaneous conjugate matching is possible. #### Step 2: Maximum transducer gain (bilateral) ```text GT,max = (|S21|/|S12|)(K − √(K² − 1)) = (2.3/0.122)(1.071 − √(1.071² − 1)) = 18.852 × 0.6876 = 12.96 → 11.13 dB ``` #### Step 3: Matching reflection coefficients ```text B1 = 1 + 0.4303 − 0.0296 − 0.0610 = 1.3398 C1 = S11 − ΔS22* = 0.6612∠150.37° ΓMS = 0.8504∠−150.37° (Zs = 4.32 − j13.13 Ω) B2 = 1 + 0.0296 − 0.4303 − 0.0610 = 0.5383 C2 = S22 − ΔS11* = 0.2467∠−76.27° ΓML = 0.6548∠76.27° (ZL = 25.55 + j56.90 Ω) ``` (The other roots, 1.176 and 1.527 in magnitude, are outside the chart and are rejected.) #### Step 4: Gain blocks (check) ```text GS = 1/(1 − |ΓMS|²) = 3.613 → 5.58 dB G0 = |S21|² = 5.29 → 7.23 dB GL = (1 − |ΓML|²)/|1 − S22ΓML|² = 0.678 → −1.69 dB GT = 5.58 + 7.23 − 1.69 = 11.13 dB ✓ ``` #### Unilateral comparison ```text GTU,max = 1/(1−0.4303) × 5.29 × 1/(1−0.0296) = 1.755 × 5.29 × 1.030 = 9.57 → 9.81 dB ``` The unilateral formula underestimates by 1.32 dB because |S12| = 0.122 is large, so the bilateral value is the true maximum. **Answer: Gmax = GT,max = 12.96 ≈ 11.13 dB (with ΓMS = 0.850∠−150.4°, ΓML = 0.655∠76.3°).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this transistor K = 1.071 and |Δ| = 0.247. What does it allow? A. Only a unilateral design B. Unconditional stability, so simultaneous conjugate matching is possible C. Matching is impossible D. A design only inside stability circles Answer: B. K > 1 and |Δ| < 1 mean unconditional stability, so ΓMS and ΓML exist inside the chart. (Key point: Stability decides matching) 2. Why is the root with the minus sign chosen in ΓMS = [B1 − sqrt(B1^2 − 4|C1|^2)]/(2C1)? A. The other roots are outside the chart (magnitude above 1) and rejected B. It gives a larger gain C. It is easier to compute D. Both roots are always valid Answer: A. The other roots (1.176, 1.527 in magnitude) lie outside the Smith chart and cannot be passive terminations. (Key point: Choosing the root) 3. How is GT = 11.13 dB confirmed from the gain blocks? A. GS + G0 − GL = 5.58 + 7.23 + 1.69 B. GS × G0 only C. GS + G0 + GL = 5.58 + 7.23 − 1.69 dB D. G0 − GS − GL Answer: C. GS = 5.58 dB, G0 = 7.23 dB, GL = −1.69 dB add to 11.13 dB. (Key point: Gain check) 4. Why does the unilateral formula (9.81 dB) underestimate the true maximum? A. Because the device is unstable B. Because |S12| = 0.122 is large, so feedback is ignored and the bilateral value is the true maximum C. Because S22 is zero D. Because K < 1 Answer: B. Ignoring a large S12 gives 1.32 dB less than the bilateral maximum of 11.13 dB. 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