Stub matching for antenna with Γ = 0.33∠66°

Given: ΓL = 0.33∠66°, Z0 = 75 Ω (microstrip). A single shunt stub is used, and both short (via) and open designs are given.

Load impedance

zL = (1 + ΓL)/(1 − ΓL) = 1.060 + j0.717
ZL = 75 × zL = 79.52 + j53.80 Ω

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.060 + j0.717
ΓL = (zL − 1)/(zL + 1) = 0.330∠66.0°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.985

Draw the constant-VSWR circle through zL (zL is at 0.1583λ 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 = 0.647 − j0.438   (0.4083λ 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 + j0.699  at 0.1518λ WTG
y2 = 1.000 − j0.699  at 0.3482λ WTG
d1 = 0.2434λ from the load
d2 = 0.4399λ 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 = −0.6992
   short stub l = 0.1529λ
   open stub  l = 0.4029λ
Sol 2: bstub = +0.6992
   short stub l = 0.3471λ
   open stub  l = 0.0971λ

Answer (design): for a microstrip, an open stub avoids a via hole. Solution 2: d = 0.4399λg, l = 0.0971λg. With a shorted stub (via to ground), solution 1: d = 0.2434λg, l = 0.1529λg. Here λg is the guided wavelength on the microstrip.

       open stub l = 0.0971λg
            ||
            ○
            ||
 ===========##====== d = 0.4399λg ====[patch]
  75 Ω feed

Scattering matrices

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 = 0.330∠66.0°,  VSWR = 1.99
Matching network (solution 2):
  S11 = 0.330∠-109.3°   S12 = 0.944∠-177.6°
  S21 = 0.944∠-177.6°   S22 = 0.330∠-66.0°
  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.