--- title: "Double-stub for 10 GHz antenna, Γ = 0.45∠60°" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-10ghz-antenna" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-10ghz-antenna site: BCT Engineering --- # Double-stub for 10 GHz antenna, Γ = 0.45∠60° RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2082 Baisakh, 10+2+2 marks - 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/double-stub-10ghz-antenna - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a double-stub matching network for an antenna operating at 10 GHz, having Γ = |0.45|∠60° and connected to a 100 Ohm transmission line. Sketch its physical diagram using micro strips. Prepare its S-Matrices before and after the design. ## Short answer Use two short-circuited stubs separated by 0.375λg (0.816 cm), first stub at the antenna. With λg=2.176 cm, solution A lengths are l1=0.217 cm and l2=0.120 cm. Before matching S11=0.45∠60°; the tuner S-matrix is S11=0.45∠145.9°, S21=S12=0.893∠-47.1°, S22=0.45∠-60°, giving Γin=0. ## Answer **Given:** f = 10 GHz, ΓL = 0.45∠60°, Z0 = 100 Ω. **Assumptions:** short-circuited stubs, the first stub at the antenna, spacing 3λ/8. Microstrip with effective permittivity εeff ≈ 1.9 (for example a 100 Ω line on RT/Duroid 5880), so λg = λ0/√εeff = 3 cm/√1.9 = 2.176 cm. #### Load impedance ```text zL = (1 + Γ)/(1 − Γ) = 1.060 + j1.036 ZL = 105.98 + j103.58 Ω ``` #### Step 1: Normalise and convert to admittance ```text zL = 1.060 + j1.036 yL = 1/zL = 0.483 − j0.472 ΓL = 0.450∠60.0°, VSWR = 2.636 ``` On the chart, zL is at 0.1667λ WTG and yL is diametrically opposite, at 0.4167λ WTG. #### Step 2: First stub at the load The first stub is placed at the load, so y1 = yL = 0.483 − j0.472. #### Step 3: Forbidden-region check and rotated g = 1 circle ```text βd = 360° × 0.375 = 135° g must be ≤ 1/sin²βd = 2.000 g1 = 0.4826 → matchable ``` Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.483 circle until it meets the rotated circle. #### Step 4: Susceptance of stub 1 Analytically (Pozar), with t = tan βd: ```text t = tan 135° = −1.0000 b1 = −B + [1 ± √((1+t²)g − g²t²)]/t (1+t²)g − g²t² = 0.7323, √ = 0.8557 Sol A: b1 = −1.3841 Sol B: b1 = +0.3274 ``` #### Step 5: Move 0.375λ to stub 2 and find b2 ```text Sol A: y after stub 1 = 0.483 − j1.856 (0.3253λ) → after 0.375λ: y2 = 1.000 + j2.773 b2 = −2.7732 Sol B: y after stub 1 = 0.483 − j0.144 (0.4706λ) → after 0.375λ: y2 = 1.000 − j0.773 b2 = +0.7732 ``` Both y2 values lie on the g = 1 circle, as required. #### Step 6: Stub lengths - 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. ```text Sol A short: l1 = 0.0996λ, l2 = 0.0551λ Sol B short: l1 = 0.3004λ, l2 = 0.3548λ ``` | Quantity | Solution A | Solution B | |---|---|---| | Stub 1 from load | at load | at load | | Stub spacing | 0.375λ | 0.375λ | | b1, b2 | -1.384, -2.773 | 0.327, 0.773 | | Short l1, l2 | 0.0996λ, 0.0551λ | 0.3004λ, 0.3548λ | Physical lengths (λ = 2.176 cm): ```text d1 = 0.000 cm, spacing = 0.816 cm Sol A short: l1 = 0.217 cm, l2 = 0.120 cm Sol B short: l1 = 0.654 cm, l2 = 0.772 cm ``` **Answer (design):** solution A, with l1 = 0.0996λg = 0.217 cm, l2 = 0.0551λg = 0.120 cm, and spacing 0.375λg = 0.816 cm. #### Physical diagram (microstrip) ```text stub 2 stub 1 l2=1.20 mm l1=2.17 mm || || ● via ● via || || =====##=== 8.16 mm ======##[antenna] 100 Ω microstrip feed ``` #### S-matrices before and after the design Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D. ```text Before matching: S11 = ΓL = 0.450∠60.0°, VSWR = 2.64 Tuner, solution A: S11 = 0.450∠145.9° S12 = 0.893∠-47.1° S21 = 0.893∠-47.1° S22 = 0.450∠-60.0° S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless) S22 = ΓL* (conjugate match) After matching (tuner + load): Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1 ``` Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the microstrip wavelength at 10 GHz found? A. λg = λ0 x sqrt(εeff) = 4.13 cm B. λg = λ0/sqrt(εeff) = 3 cm/sqrt(1.9) = 2.176 cm C. λg = λ0 = 3 cm D. λg = λ0/εeff = 1.58 cm Answer: B. The effective permittivity slows the wave, so λg = λ0/sqrt(εeff). (Key point: Guided wavelength) 2. With ΓL = 0.45∠60° and Z0 = 100 Ω, what is the load? A. zL = 0.45 + j0.78, ZL = 45 + j78 Ω B. zL = 0.483 - j0.472, ZL = 48 - j47 Ω C. zL = 1.06 - j1.036, ZL = 106 - j104 Ω D. zL = (1 + Γ)/(1 - Γ) = 1.060 + j1.036, ZL = 105.98 + j103.58 Ω Answer: D. Applying zL = (1 + Γ)/(1 - Γ) with Γ = 0.225 + j0.390 gives 1.060 + j1.036. (Key point: Load impedance) 3. Why is the load matchable with 3λ/8 spacing? A. g1 = 0.4826 is below the limit 1/sin^2(135°) = 2 B. g1 = 0.4826 is above the limit of 2 C. Because Γ is less than 0.5 D. Because 3λ/8 has no forbidden region Answer: A. The forbidden region is g > 2; the load conductance 0.483 is outside it. (Key point: Matchability) 4. Before and after matching, what are the S-parameter results? A. Before: S11 = 0; after: S11 = 0.45∠60° B. Before: S11 = 1; after: S21 = 0 C. Before: S11 = ΓL = 0.45∠60°; after tuner + load, Γin = 0 and VSWR = 1 D. Before: S12 = 0.45; after: S11 = 1 Answer: C. Without the tuner the line sees ΓL; the lossless tuner has S22 = ΓL*, so Γin = 0 after matching. 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