- 2073 Magh · 12 marks
Stability and gains for S21 = 10.11∠91°
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.
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.
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
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:
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:
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)
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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