--- title: "Double-stub for inductive antenna load on 78 Ω line" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-inductive-load-78-ohm" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-inductive-load-78-ohm site: BCT Engineering --- # Double-stub for inductive antenna load on 78 Ω line RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2073 Bhadra, 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-inductive-load-78-ohm - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question By assuming a complex inductive load of an antenna which is mismatched with the line impedance of 78.0 ohm, design a double-stub short-circuited matching network. Show both electrical and physical connections. ## Short answer Normalize ZL=117+j78 Ω to zL=1.5+j1.0, then yL=0.462−j0.308. With first stub at load and 3λ/8 spacing, the required short-circuited stub lengths are l1=0.0919λ (2.76 cm) and l2=0.0541λ (1.62 cm) at 1 GHz on a 78 Ω line. ## Answer **Assumptions:** an inductive antenna load ZL = 117 + j78 Ω on a Z0 = 78 Ω line at f = 1 GHz (air-filled coax, λ = 30 cm). Short-circuited 78 Ω stubs, the first stub at the load, and spacing d = 3λ/8. #### Step 1: Normalise and convert to admittance ```text zL = 1.500 + j1.000 yL = 1/zL = 0.462 − j0.308 ΓL = 0.415∠41.6°, VSWR = 2.420 ``` On the chart, zL is at 0.1922λ WTG and yL is diametrically opposite, at 0.4422λ WTG. #### Step 2: First stub at the load The first stub is placed at the load, so y1 = yL = 0.462 − j0.308. #### Step 3: Forbidden-region check and rotated g = 1 circle ```text βd = 360° × 0.375 = 135° g must be ≤ 1/sin²βd = 2.000 g1 = 0.4615 → 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 = 0.462 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 = 0.7101, √ = 0.8427 Sol A: b1 = −1.5350 Sol B: b1 = +0.1503 ``` #### Step 5: Move 0.375λ to stub 2 and find b2 ```text Sol A: y after stub 1 = 0.462 − j1.843 (0.3260λ) → after 0.375λ: y2 = 1.000 + j2.826 b2 = −2.8257 Sol B: y after stub 1 = 0.462 − j0.157 (0.4688λ) → after 0.375λ: y2 = 1.000 − j0.826 b2 = +0.8257 ``` 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 toward the generator from the short-circuit point (y = ∞; with y plotted on an ordinary Smith chart, this is the right end, 0.25λ WTG). ```text Sol A short: l1 = 0.0919λ, l2 = 0.0541λ Sol B short: l1 = 0.2737λ, l2 = 0.3599λ ``` | Quantity | Solution A | Solution B | |---|---|---| | Stub 1 from load | at load | at load | | Stub spacing | 0.375λ | 0.375λ | | b1, b2 | -1.535, -2.826 | 0.150, 0.826 | | Short l1, l2 | 0.0919λ, 0.0541λ | 0.2737λ, 0.3599λ | Physical lengths (λ = 30.000 cm): ```text d1 = 0.000 cm, spacing = 11.250 cm Sol A short: l1 = 2.757 cm, l2 = 1.624 cm Sol B short: l1 = 8.212 cm, l2 = 10.796 cm ``` **Answer (choice):** solution A (l1 = 0.0919λ = 2.76 cm, l2 = 0.0541λ = 1.62 cm). These are the shortest stubs, so the design has the lowest loss and the widest bandwidth. **Check:** y at stub 2 = 1.000 + j0.000, so the line is matched. #### Electrical connection ```text Z0=78Ω 3λ/8 = 11.25 cm o------+---------------+--------[117+j78Ω] | | stub 2 stub 1 l2=1.62 cm l1=2.76 cm SC SC ``` #### Physical connection Two coaxial T-junctions are placed 11.25 cm apart on the main 78 Ω coax. The first is at the antenna terminals. Each side arm is a 78 Ω coaxial stub closed by an adjustable sliding short, set to l1 = 2.76 cm and l2 = 1.62 cm. ```text Physical connection (coaxial tuner) sliding short sliding short ┃ ┃ stub 2 ═══╋═══ stub 1 ══╋══ ║ ║ gen ════════╩═══════════════╩══════ antenna main coax d (load) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For ZL = 117 + j78 Ω on a 78 Ω line, what is zL? A. 1.5 + j1.0 B. 0.67 - j0.5 C. 117 + j78 D. 1.0 + j1.5 Answer: A. zL = ZL/Z0 = 117/78 + j78/78 = 1.5 + j1.0. (Key point: Normalisation) 2. For d = 3λ/8, g1 = 0.4615 passes because: A. It is above 1 B. It is below 1/sin^2(135°) = 2 C. It equals 0.5 D. It is inductive Answer: B. The limit is 2 and 0.4615 < 2. (Key point: Forbidden check) 3. Why is solution A (l1 = 0.0919λ, l2 = 0.0541λ) chosen? A. Largest susceptance B. Matches both ways C. Shortest stubs, lowest loss and widest bandwidth D. Simplest physical form Answer: C. Short stubs reduce loss and give wider bandwidth. (Key point: Solution choice) 4. How is the physical coaxial tuner built? A. One T-junction with a resistor B. Two series stubs in the centre conductor C. A slotted line with a probe only D. Two coaxial T-junctions 3λ/8 = 11.25 cm apart, each with a stub closed by an adjustable sliding short Answer: D. Stub lengths are set by sliding shorts at the fixed junctions; spacing is 0.375 x 30 cm = 11.25 cm. 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