--- title: "Stub matching for antenna with Γ = 0.33∠66°" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:microstrip-antenna-stub-matching" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/microstrip-antenna-stub-matching site: BCT Engineering --- # Stub matching for antenna with Γ = 0.33∠66° RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2082 Bhadra, 10 marks - Video: [Topic 10 Part 6 Shunt Stub Matching](https://www.youtube.com/watch?v=sBl_pXUBp5I) — ECET 364 - Fundamentals of Electromagnetics - 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/microstrip-antenna-stub-matching - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question A broadband microstrip antenna with a load exhibiting reflection coefficient of 0.33∠66° is connected to a transmission patch having impedance of 75 Ω. Design the appropriate matching stubs. Express the appropriate scattering matrix of your designed matched networks. ## Short answer Use a single shunt stub. For an open stub, place it d=0.4399λg from the load with length l=0.0971λg; a shorted stub alternative is d=0.2434λg, l=0.1529λg. Before matching S11=0.33∠66°; after matching, the tuner has S11=0.33∠-109.3°, S21=S12=0.944∠-177.6°, S22=0.33∠-66°, giving Γin=0. ## Answer **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 ```text zL = (1 + ΓL)/(1 − ΓL) = 1.060 + j0.717 ZL = 75 × zL = 79.52 + j53.80 Ω ``` #### Step 1: Normalise and plot the load ```text 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: ```text 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: ```text 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. ```text 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 ``` | Solution | Stub position d | Short stub l | Open stub l | |---|---|---|---| | 1 | 0.2434λ | 0.1529λ | 0.4029λ | | 2 | 0.4399λ | 0.3471λ | 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. ```text 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is zL found from ΓL = 0.33∠66°? A. zL = (1 - ΓL)/(1 + ΓL), giving ZL = 75/zL B. zL = (1 + ΓL)/(1 - ΓL) = 1.060 + j0.717, so ZL = 75 zL C. zL = ΓL x 75 = 24.75∠66° D. zL = (ΓL - 1)/(ΓL + 1) Answer: B. The standard inversion zL = (1 + Γ)/(1 - Γ) gives 1.060 + j0.717, so ZL = 79.52 + j53.80 Ω. (Key point: Load impedance from Γ) 2. Why is an open stub preferred in microstrip? A. An open stub has lower loss than any short stub B. An open stub gives a better VSWR C. A short stub cannot provide negative susceptance D. It avoids a via hole to ground that a short stub needs Answer: D. A short-circuited microstrip stub needs a via to the ground plane; an open stub is just a strip end. (Key point: Open stub in microstrip) 3. In what wavelength are the microstrip stub length and position expressed? A. The guided wavelength λg on the microstrip B. The free-space wavelength λ0 C. The cutoff wavelength D. The wavelength in air at DC Answer: A. Waves on microstrip travel in the dielectric, so lengths scale with the guided wavelength λg. (Key point: Lengths in microstrip) 4. What is true of the matching network S-matrix and the matched result? A. S11 = 0 for the network alone and S12 = 0 B. |S21| = 0 so no power reaches the load C. Lossless and reciprocal with S12 = S21, |S11|^2 + |S21|^2 = 1, and Γin = 0 after matching D. S22 = 0.33∠66° and Γin = 0.33 after matching Answer: C. The stub-plus-line network has S11 = 0.33∠-109.3°, |S21| = 0.944, S22 = ΓL*; with the load it gives Γin = 0, VSWR = 1. 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