--- title: "Unilateral max gain with design flow chart" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:unilateral-gain-flowchart" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/unilateral-gain-flowchart site: BCT Engineering --- # Unilateral max gain with design flow chart RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2078 Chaitra, 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/unilateral-gain-flowchart - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Assuming suitable S-parameters calculate maximum gain of an amplifier working in an unilateral model. Also illustrate a flow chart showing the designing steps. ## Short answer For assumed S-parameters (S11=0.5∠-150°, S21=3∠60°, S22=0.4∠-40°, S12≈0), maximum unilateral gain GTU,max = 14.29 (11.55 dB). Design flow: obtain S-parameters, verify unconditional stability, assume unilateral (S12≈0), conjugate match (Γs=S11*, ΓL=S22*), compute gain, and realize matching networks. ## Answer In a unilateral design S12 is taken as zero, so the input and output can be conjugate-matched independently: Γs = S11* and ΓL = S22*. #### Assumed S-parameters (Z0 = 50 Ω) S11 = 0.5∠−150°, S12 ≈ 0, S21 = 3.0∠60°, S22 = 0.4∠−40°. With S12 = 0, Δ = S11S22 and K → ∞; since |S11| < 1 and |S22| < 1 the device is unconditionally stable. #### Maximum unilateral gain ```text GTU,max = GS,max · G0 · GL,max GS,max = 1/(1 − |S11|²) = 1/(1 − 0.25) = 1.3333 → 1.25 dB G0 = |S21|² = 9.0 → 9.54 dB GL,max = 1/(1 − |S22|²) = 1/(1 − 0.16) = 1.1905 → 0.76 dB GTU,max = 1.3333 × 9 × 1.1905 = 14.29 = 1.25 + 9.54 + 0.76 = 11.55 dB ``` Matching terminations needed: - Γs = S11* = 0.5∠150° → Zs = 50(1+Γs)/(1−Γs) = 17.72 + j11.81 Ω - ΓL = S22* = 0.4∠40° → ZL = 76.76 + j46.99 Ω The input matching network must transform the 50 Ω source to 17.72 + j11.81 Ω, and the output network must transform the 50 Ω load to 76.76 + j46.99 Ω. Each can be a shunt open stub plus a series line, or an L-section, designed on the Smith chart. **Answer: GTU,max = 14.29 ≈ 11.55 dB** (for the assumed S-parameters). #### Design flowchart ```text +------------------------------+ | Specs: f, gain, Z0, BW, NF | +------------------------------+ | +------------------------------+ | Choose transistor, bias; | | get [S] at design frequency | +------------------------------+ | +------------------------------+ | Compute Δ, K (and μ) | +------------------------------+ | K>1 and |Δ|<1 ? yes | | no v v Uncond. stable Draw stability circles; | pick Γs, ΓL in stable | region (or stabilise) +----+-----+ | Unilateral? (U small, S12≈0) yes: Γs=S11*, ΓL=S22* no : Γs=ΓMS, ΓL=ΓML (bilateral) | +------------------------------+ | Compute GS, G0, GL -> GT | +------------------------------+ | +------------------------------+ | Design input/output matching | | (stubs or L-C) on Smith chart| +------------------------------+ | | Bias network, layout, simulate | Specs met? -- no --> revise | yes Build & test ``` Key points of the flow: 1. The gain is split into three independent blocks: input match (GS), transistor (G0) and output match (GL). 2. The unilateral figure of merit U is checked. If the error bounds 1/(1±U)² are small (under about ±0.5 dB), the unilateral design is accepted; otherwise a bilateral (simultaneous conjugate match) design is used. 3. After matching, a bias network with RF chokes and DC blocks is added, and the circuit is simulated and tuned. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the unilateral design assume, and why does it simplify matching? A. S21 = 0, so output is isolated B. S12 = 0, so input and output are conjugate-matched independently C. S11 = S22 D. K < 1 Answer: B. With no reverse feedback, the input and output networks do not interact: Γs = S11*, ΓL = S22*. (Key point: Unilateral assumption) 2. Which expression gives the maximum unilateral gain? A. GS,max·G0·GL,max with G0 = |S21|² B. |S21|/|S12| C. |S11|²·|S22|² D. G0 − GS − GL Answer: A. GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)] = 14.29 (11.55 dB) here. (Key point: Gain blocks) 3. Why is the device unconditionally stable when S12 ≈ 0? A. Since |S11| < 1 and |S22| < 1 and K → ∞ B. Since S21 is large C. Since Δ = 0 D. Since the load is 50 Ω Answer: A. With S12 = 0, K → ∞ and Δ = S11S22 < 1, so Rollett's condition holds. (Key point: Stability) 4. In the design flow, when is the unilateral design accepted? A. When U is small (error under about ±0.5 dB) B. Always C. When K < 1 D. When bias is complete Answer: A. If the 1/(1±U)² bounds are small the unilateral design stands; otherwise a bilateral design with ΓMS, ΓML is used. 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