Stability from sketched Smith chart (Fig. Q6)

Stability on the Smith chart is studied with stability circles: the locus of Γs (or ΓL) for which |Γout| = 1 (or |Γin| = 1). One side of each circle is stable, the other unstable.

Since the figure is missing, assume a transistor with S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°.

Steps

1. Compute Δ and K: Δ = 0.6964∠−83.07°, K = 0.607 < 1 → potentially (conditionally) unstable.
2. Compute circle centres and radii:

CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²)
RS = |S12S21| / ||S11|² − |Δ|²|
CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²)
RL = |S12S21| / ||S22|² − |Δ|²|

CS = 1.132∠68.46°,  RS = 0.199
CL = 1.363∠46.69°,  RL = 0.500

3. Plot the circles on the chart: each is centred outside the unit circle and cuts the chart edge near the upper-right.
4. Find the stable side by testing the chart centre (Γ = 0). For ΓL = 0, |Γin| = |S11| = 0.894 < 1, so the centre is stable. The centre lies outside the load circle (|CL| − RL = 0.863 > 0), so the stable region is outside the load circle. The same test with |S22| = 0.781 < 1 (and |CS| − RS = 0.933 > 0) shows that the stable source region is outside the source circle.

          unit Smith chart
        .-----------.   ( CL circle,
      /        (####)    shaded = unstable)
     |    o    (####)
     | centre   (##)  ( CS circle )
      \  stable      /
        '-----------'

5. Synthesis: choose Γs and ΓL in the unshaded region and away from the circle edge for margin. For example S11* = 0.894∠60.6° is 0.275 from CS (> RS) and S22* is 0.675 from CL (> RL), so conjugate unilateral matching is allowed here.
6. Verify: compute Γin and Γout with the chosen terminations; both must have magnitude < 1 (here |Γin| = 0.912, |Γout| = 0.848).