Single stub for 120−j160 Ω on 100 Ω line

Given: Z0 = 100 Ω, ZL = 120 − j160 Ω (capacitive). A single shunt stub is used.

Reasoning: a shunt stub can only add susceptance. So, first move along the line to a point where the real part of the admittance equals Y0 (g = 1). Then cancel the imaginary part with the stub.

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.200 − j1.600
ΓL = (zL − 1)/(zL + 1) = 0.593∠-46.8°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 3.911

Draw the constant-VSWR circle through zL (zL is at 0.3151λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.300 + j0.400   (0.0651λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j1.472  at 0.1755λ WTG
y2 = 1.000 − j1.472  at 0.3245λ WTG
d1 = 0.1104λ from the load
d2 = 0.2594λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −1.4720
   short stub l = 0.0950λ
   open stub  l = 0.3450λ
Sol 2: bstub = +1.4720
   short stub l = 0.4050λ
   open stub  l = 0.1550λ

Answer (design): solution 1, with the stub at d = 0.1104λ from the load and a short-circuited stub of l = 0.0950λ. These are the shortest lengths, giving the widest bandwidth. (Open-stub alternative: solution 2, d = 0.2594λ, l = 0.1550λ.)

 Z0=100Ω     d = 0.1104λ
 o---------+--------------[120 − j160 Ω]
           |
         l = 0.0950λ
           |
         short

S-matrix before and after the design

Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching (line sees the load directly):
  S11 = ΓL = 0.593∠-46.8°,  VSWR = 3.91
Matching network (solution 1):
  S11 = 0.593∠126.4°   S12 = 0.805∠-3.4°
  S21 = 0.805∠-3.4°   S22 = 0.593∠46.8°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred.