--- title: "Double-stub 3λ/8 for 300+j300 Ω antenna" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:double-stub-300-j300" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/double-stub-300-j300 site: BCT Engineering --- # Double-stub 3λ/8 for 300+j300 Ω antenna RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2071 Bhadra, 10 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-300-j300 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a double stub matching network using three-eighths wavelength (3λ/8) separation that match an antenna having load of 300+j300 Ohm connected to a 300 Ohm transmission line. Justify your design. ## Short answer Load 300+j300 Ω on 300 Ω line, 3λ/8 spacing, short stubs at load. Design: l1=0.1006λ, l2=0.0558λ (Solution A). Load conductance g=0.5 lies outside forbidden region g>2. Short stubs minimize loss and size; VSWR reduces from 2.62 to 1. ## Answer **Given:** antenna load ZL = 300 + j300 Ω, Z0 = 300 Ω, stub spacing 3λ/8. **Assumptions:** short-circuited 300 Ω stubs, with the first stub at the antenna terminals. #### Step 1: Normalise and convert to admittance ```text zL = 1.000 + j1.000 yL = 1/zL = 0.500 − j0.500 ΓL = 0.447∠63.4°, VSWR = 2.618 ``` On the chart, zL is at 0.1619λ WTG and yL is diametrically opposite, at 0.4119λ WTG. #### Step 2: First stub at the load The first stub is placed at the load, so y1 = yL = 0.500 − j0.500. #### Step 3: Forbidden-region check and rotated g = 1 circle ```text βd = 360° × 0.375 = 135° g must be ≤ 1/sin²βd = 2.000 g1 = 0.5000 → 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 = 0.500 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 = 0.7500, √ = 0.8660 Sol A: b1 = −1.3660 Sol B: b1 = +0.3660 ``` #### Step 5: Move 0.375λ to stub 2 and find b2 ```text Sol A: y after stub 1 = 0.500 − j1.866 (0.3247λ) → after 0.375λ: y2 = 1.000 + j2.732 b2 = −2.7321 Sol B: y after stub 1 = 0.500 − j0.134 (0.4721λ) → after 0.375λ: y2 = 1.000 − j0.732 b2 = +0.7321 ``` 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 = 0.1006λ, l2 = 0.0558λ Sol B short: l1 = 0.3058λ, l2 = 0.3506λ ``` | Quantity | Solution A | Solution B | |---|---|---| | Stub 1 from load | at load | at load | | Stub spacing | 0.375λ | 0.375λ | | b1, b2 | -1.366, -2.732 | 0.366, 0.732 | | Short l1, l2 | 0.1006λ, 0.0558λ | 0.3058λ, 0.3506λ | **Answer (design):** solution A, with the first stub at the load (l1 = 0.1006λ), the second stub 0.375λ away (l2 = 0.0558λ). ```text Z0=300Ω 3λ/8 o------+---------------+--------[300+j300Ω] | | stub 2 stub 1 l2=0.0558λ l1=0.1006λ SC SC ``` #### Justification - **Matchable:** the forbidden region for 3λ/8 spacing is g > 2. The load conductance g = 0.5 lies outside it, so the design is possible. - **Perfect match:** the admittance at stub 2 becomes 1 + j0. VSWR drops from 2.62 to 1, and no power is reflected (Γ goes from 0.447 to 0). - **Choice of solution A:** both stubs are short (about 0.1λ and 0.06λ). This gives the least stub loss, the widest bandwidth and the smallest size. - **Short-circuited stubs** are preferred: they do not radiate from their ends and are easy to tune with sliding shorts. - **3λ/8 spacing** gives the same forbidden region as λ/8 but more physical room between the junctions. - Solution B (l1 = 0.3058λ, l2 = 0.3506λ) is also valid but longer. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 3λ/8 spacing, loads with what conductance cannot be matched, and is g = 0.5 matchable? A. g > 1 cannot be matched, so g = 0.5 is matchable only just B. g > 1/sin^2(135°) = 2 cannot be matched; g = 0.5 is matchable C. g < 2 cannot be matched, so g = 0.5 fails D. 3λ/8 has no forbidden region Answer: B. sin^2(135°) = 0.5, so the limit is g = 2; the load has g = 0.5, outside the forbidden region. (Key point: Forbidden region) 2. What are zL and yL for ZL = 300 + j300 Ω on a 300 Ω line? A. zL = 1 + j1, yL = 1 - j1 B. zL = 300 + j300, yL = 0.5 + j0.5 C. zL = 0.5 - j0.5, yL = 1 + j1 D. zL = 1 + j1, yL = 0.5 - j0.5 Answer: D. zL = ZL/Z0 = 1 + j1 and yL = 1/zL = 0.5 - j0.5, the diametrically opposite point on the chart. (Key point: Normalised values) 3. Why is solution A (l1 = 0.1006λ, l2 = 0.0558λ) chosen over B? A. Both stubs are short, giving less loss, wider bandwidth and smaller size B. It is the only solution that gives a perfect match C. It needs open-circuited stubs D. It uses the larger stub susceptances to improve bandwidth Answer: A. Both solutions match, but A has shorter stubs, which store less energy, so lower loss and wider bandwidth. (Key point: Chosen solution) 4. Why are short-circuited stubs preferred in the justification? A. They have infinite input admittance at all lengths B. They give a larger forbidden region C. They do not radiate from open ends and are easy to tune with sliding shorts D. They halve the required spacing Answer: C. An open end radiates and has fringing; a shorted stub avoids this and can be adjusted by moving the short. 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