--- title: "Open-circuited double stub for 60−j80 Ω" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-open-60-j80" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-open-60-j80 site: BCT Engineering --- # Open-circuited double stub for 60−j80 Ω RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2080 Baisakh, 8+2 marks - Video: [Double shunt stub matching network problem solving using smith chart by Dr. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=AuO61UXmMlw) — 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-open-60-j80 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a double stub shunt tuner to match a load impedance of ZL = 60 − j80 Ω to a 50 Ω line. The stubs are to be open circuited stubs and are spaced 3λ/8 apart. Assume the first is 0.4λ from the load. Formulate the S-matrix for your design. ## Short answer Normalize ZL=60−j80 to yL=0.300+j0.400; move 0.4λ to y1=0.268−j0.208. With open stubs and 3λ/8 spacing, solution A gives stub1 at 0.4λ with l1=0.3449λ and stub2 at 0.775λ with l2=0.2938λ. Matched S-matrix: S11=S22=0, |S21|=1; tuner S11=0.593∠143.6°, S21=S12=0.805∠−174.8°, S22=0.593∠46.8°. ## Answer **Given:** ZL = 60 − j80 Ω, Z0 = 50 Ω, **open-circuited** shunt stubs, spacing d = 3λ/8, first stub 0.4λ from the load. #### Step 1: Normalise and convert to admittance ```text zL = 1.200 − j1.600 yL = 1/zL = 0.300 + j0.400 ΓL = 0.593∠-46.8°, VSWR = 3.911 ``` On the chart, zL is at 0.3151λ WTG and yL is diametrically opposite, at 0.0651λ WTG. #### Step 2: Move to the first stub (0.4λ from the load) Rotate yL clockwise (toward the generator) on its VSWR circle: ```text y1 = 0.268 − j0.208 (0.4651λ WTG) ``` #### Step 3: Forbidden-region check and rotated g = 1 circle ```text βd = 360° × 0.375 = 135° g must be ≤ 1/sin²βd = 2.000 g1 = 0.2675 → 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 = 0.268 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 = 0.4635, √ = 0.6808 Sol A: b1 = −1.4730 Sol B: b1 = −0.1114 ``` #### Step 5: Move 0.375λ to stub 2 and find b2 ```text Sol A: y after stub 1 = 0.268 − j1.681 (0.3341λ) → after 0.375λ: y2 = 1.000 + j3.545 b2 = −3.5447 Sol B: y after stub 1 = 0.268 − j0.319 (0.4477λ) → after 0.375λ: y2 = 1.000 − j1.545 b2 = +1.5447 ``` Both y2 values lie on the g = 1 circle, as required. #### Step 6: Stub lengths - 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 A open : l1 = 0.3449λ, l2 = 0.2938λ Sol B open : l1 = 0.4824λ, l2 = 0.1586λ ``` | Quantity | Solution A | Solution B | |---|---|---| | Stub 1 from load | 0.4λ | 0.4λ | | Stub spacing | 0.375λ | 0.375λ | | b1, b2 | -1.473, -3.545 | -0.111, 1.545 | | Open l1, l2 | 0.3449λ, 0.2938λ | 0.4824λ, 0.1586λ | **Answer (design):** solution A, with stub 1 at 0.4λ from the load (open, l1 = 0.3449λ) and stub 2 at 0.775λ from the load (open, l2 = 0.2938λ). Solution B (l1 = 0.4824λ, l2 = 0.1586λ) is equally valid, with almost the same total length. ```text Z0=50Ω 3λ/8 0.4λ o------+---------------+--------[60−j80Ω] | | stub 2 stub 1 l2=0.2938λ l1=0.3449λ OC OC ``` #### S-matrix of the design Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.4λ) 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] ·[line d1], 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 = 0.593∠-46.8°, VSWR = 3.91 Tuner, solution A: S11 = 0.593∠143.6° S12 = 0.805∠-174.8° S21 = 0.805∠-174.8° S22 = 0.593∠46.8° 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 open-circuited stub length found from required susceptance b? A. y = -j cot βl, so l = (1/2π) cot^-1(-b) from the short point B. y = j tan βl, so l = (1/2π) tan^-1(b), measured from the open-circuit point C. y = j tan βl, measured from the short-circuit point D. y = j cot βl, so l = (1/2π) tan^-1(1/b) Answer: B. An open stub has input admittance j tan βl, so its length follows from the arctan of b. (Key point: Open stub formula) 2. How does the stub 1 location differ from earlier designs? A. It is at the load, y1 = yL B. It is 3λ/8 from the load C. It is 0.4λ from the load, y1 = 1 + j0 D. It is 0.4λ from the load, so yL is first rotated to y1 = 0.268 - j0.208 Answer: D. The problem fixes the first stub at 0.4λ from the load, so we rotate yL toward the generator by 0.4λ. (Key point: First stub location) 3. Where is stub 2 from the load? A. 0.775λ (0.4λ + 0.375λ) B. 0.375λ C. 0.4λ D. 0.025λ Answer: A. Stub 2 is 3λ/8 beyond stub 1, i.e. 0.4 + 0.375 = 0.775λ from the load. (Key point: Stub 2 position) 4. What are the open-stub lengths of solution A? A. l1 = 0.4824λ and l2 = 0.1586λ B. l1 = 0.0981λ and l2 = 0.0477λ C. l1 = 0.3449λ and l2 = 0.2938λ D. l1 = 0.4λ and l2 = 0.375λ Answer: C. Solution A has b1 = -1.473 and b2 = -3.545, giving open stubs of 0.3449λ and 0.2938λ. 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