--- title: "Double-stub for Γ = 0.64∠58°" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-reflection-0-64-58" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-reflection-0-64-58 site: BCT Engineering --- # Double-stub for Γ = 0.64∠58° RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2081 Baisakh, 8+2 marks - Video: [Double shunt stub matching network 2.0 problem solving using smith chart by Dr. Niraj Kumar VITCC](https://www.youtube.com/watch?v=q8w6O9WPA6w) — RF Design Basics - 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-reflection-0-64-58 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a double-stub impedance matching network for a load of Γ = 0.64∠58° connected to a 100 [?] Ohm transmission line. Prepare the S-Matrix of its matched network. ## Short answer Normalize the load to zL=0.807+j1.484, yL=0.283−j0.520; use short stubs, first at the load, spacing 0.375λ. Solution A gives stub lengths l1=0.1121λ and l2=0.0447λ. The matched network has S11=S22=0, |S21|=1; tuner S-parameters are S11=0.640∠149.7°, S21=S12=0.768∠−44.1°, S22=0.640∠−58°. ## Answer **Given:** ΓL = 0.64∠58°, Z0 = 100 Ω (taking the unclear value as 100 Ω). **Assumptions:** short-circuited stubs, the first stub at the load, spacing 3λ/8. #### Load impedance ```text zL = (1 + Γ)/(1 − Γ) = 0.807 + j1.484 ZL = 80.73 + j148.43 Ω ``` #### Step 1: Normalise and convert to admittance ```text zL = 0.807 + j1.484 yL = 1/zL = 0.283 − j0.520 ΓL = 0.640∠58.0°, VSWR = 4.556 ``` On the chart, zL is at 0.1694λ WTG and yL is diametrically opposite, at 0.4194λ WTG. #### Step 2: First stub at the load The first stub is placed at the load, so y1 = yL = 0.283 − j0.520. #### Step 3: Forbidden-region check and rotated g = 1 circle ```text βd = 360° × 0.375 = 135° g must be ≤ 1/sin²βd = 2.000 g1 = 0.2828 → 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 = 0.283 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 = 0.4856, √ = 0.6968 Sol A: b1 = −1.1769 Sol B: b1 = +0.2167 ``` #### Step 5: Move 0.375λ to stub 2 and find b2 ```text Sol A: y after stub 1 = 0.283 − j1.697 (0.3333λ) → after 0.375λ: y2 = 1.000 + j3.464 b2 = −3.4643 Sol B: y after stub 1 = 0.283 − j0.303 (0.4497λ) → after 0.375λ: y2 = 1.000 − j1.464 b2 = +1.4643 ``` 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 = 0.1121λ, l2 = 0.0447λ Sol B short: l1 = 0.2840λ, l2 = 0.4046λ ``` | Quantity | Solution A | Solution B | |---|---|---| | Stub 1 from load | at load | at load | | Stub spacing | 0.375λ | 0.375λ | | b1, b2 | -1.177, -3.464 | 0.217, 1.464 | | Short l1, l2 | 0.1121λ, 0.0447λ | 0.2840λ, 0.4046λ | **Answer (design):** solution A, with l1 = 0.1121λ, l2 = 0.0447λ, and spacing 0.375λ. ```text Z0=100Ω 3λ/8 o------+---------------+--------[ZL] | | stub 2 stub 1 l2=0.0447λ l1=0.1121λ SC SC ``` #### S-matrix of the matched network 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 = 0.640∠58.0°, VSWR = 4.56 Tuner, solution A: S11 = 0.640∠149.7° S12 = 0.768∠-44.1° S21 = 0.768∠-44.1° S22 = 0.640∠-58.0° 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 load obtained from ΓL = 0.64∠58° with Z0 = 100 Ω? A. zL = 0.64 + j58, ZL = 64 + j58 Ω B. zL = (1 + Γ)/(1 - Γ) = 0.807 + j1.484, ZL = 80.73 + j148.43 Ω C. zL = 0.283 - j0.520, ZL = 28 - j52 Ω D. zL = (Γ - 1)/(Γ + 1), ZL = 100/zL Answer: B. Inverting the reflection coefficient formula gives zL = 0.807 + j1.484, then ZL = 100 zL. (Key point: Load impedance) 2. Why is g1 = 0.283 acceptable for 3λ/8 spacing? A. It is above 2, which is required B. The VSWR is below 2 C. Because the first stub is at the load D. It is below the limit g <= 1/sin^2(135°) = 2 Answer: D. The forbidden region for 3λ/8 is g > 2; 0.283 is well within the matchable range. (Key point: Forbidden region check) 3. Which design values are given for solution A? A. l1 = 0.1121λ, l2 = 0.0447λ, spacing 0.375λ, both short-circuited B. l1 = 0.2840λ, l2 = 0.4046λ, spacing 0.125λ C. l1 = 0.1121λ, l2 = 0.0447λ, spacing 0.125λ D. l1 = 0.0447λ, l2 = 0.1121λ, open stubs Answer: A. Solution A uses b1 = -1.177 and b2 = -3.464, giving short stubs of 0.1121λ and 0.0447λ. (Key point: Design) 4. What is the S-matrix of the matched network referred to the load? A. S = [1 0; 0 1] B. S = [0.64∠58° 0; 0 0.64∠58°] C. S = [0 e^(-jφ); e^(-jφ) 0], with S11 = S22 = 0 and |S21| = 1 D. S = [0 0; 0 0] Answer: C. A matched lossless network has no reflection at either port and transmits all power, so |S21| = 1. 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