--- title: "Single stub matching for ΓL = 0.5∠51°" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:single-stub-reflection-0-5-51" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/single-stub-reflection-0-5-51 site: BCT Engineering --- # Single stub matching for ΓL = 0.5∠51° RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2074 Bhadra, 8 marks - Video: [Single Stub Tuning - Lesson 7](https://www.youtube.com/watch?v=FPa2pozZ-AU) — EMViso - 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/single-stub-reflection-0-5-51 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a single short and open-circuited shunt matching network for a transmission line using Smith Chart by considering an output reflection coefficient ΓL = 0.5∠51° Ohm and surge impedance Z0 = 50 Ohm. ## Short answer Plot ΓL=0.5∠51° to obtain normalized load zL=1.208+j1.252. Convert to admittance yL=0.399−j0.414, then move toward generator to the g=1 circle. The distances are d=0.2375λ for short stub (l=0.1136λ) and d=0.4042λ for open stub (l=0.1364λ). Z0=50 Ω. ## Answer **Given:** ΓL = 0.5∠51° (the "Ohm" in the question is a slip; Γ has no unit), Z0 = 50 Ω. Design a shunt single-stub match with (a) a short-circuited and (b) an open-circuited stub. #### Step 0: Find the load from Γ On the Smith chart, plot the point at radius 0.5 and angle 51°. It reads: ```text zL = (1 + ΓL)/(1 − ΓL) = 1.208 + j1.252 ZL = 50 × zL = 60.42 + j62.60 Ω ``` #### Step 1: Normalise and plot the load ```text zL = ZL/Z0 = 1.208 + j1.252 ΓL = (zL − 1)/(zL + 1) = 0.500∠51.0° VSWR = (1 + |Γ|)/(1 − |Γ|) = 3.000 ``` Draw the constant-VSWR circle through zL (zL is at 0.1792λ WTG). #### Step 2: Convert to admittance (shunt stub) Move zL by 180° on the VSWR circle, which is the same as moving λ/4: ```text yL = 1/zL = 0.399 − j0.414 (0.4292λ WTG) ``` #### Step 3: Move toward the generator to the g = 1 circle The VSWR circle cuts the g = 1 circle at two points: ```text y1 = 1.000 + j1.155 at 0.1667λ WTG y2 = 1.000 − j1.155 at 0.3333λ WTG d1 = 0.2375λ from the load d2 = 0.4042λ from the load ``` #### Step 4: Stub susceptance and length The stub must cancel the susceptance: bstub = −b. - 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). - For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b). Measure toward the generator from the open-circuit point (y = 0; the left end, 0λ WTG). ```text Sol 1: bstub = −1.1547 short stub l = 0.1136λ open stub l = 0.3636λ Sol 2: bstub = +1.1547 short stub l = 0.3864λ open stub l = 0.1364λ ``` | Solution | Stub position d | Short stub l | Open stub l | |---|---|---|---| | 1 | 0.2375λ | 0.1136λ | 0.3636λ | | 2 | 0.4042λ | 0.3864λ | 0.1364λ | **Answer (choice):** solution 1 with the short stub (d = 0.2375λ, l = 0.1136λ). With an open stub, solution 2 (d = 0.4042λ, l = 0.1364λ) is the shorter choice. ```text (a) Short stub (b) Open stub Z0 d=0.2375λ Z0 d=0.4042λ o---+--------[ZL] o---+--------[ZL] | | l=0.1136λ l=0.1364λ | | short open ``` **Check:** y(d) = 1 ± j1.155. The stub supplies ∓j1.155, so y = 1, Γ = 0 and VSWR = 1 (it was 3.0 before matching). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load found from ΓL = 0.5∠51°? A. zL = (1 + ΓL)/(1 - ΓL) B. zL = (1 - ΓL)/(1 + ΓL) C. zL = ΓL/(1 - ΓL) D. zL = 1/ΓL Answer: A. zL = (1 + Γ)/(1 - Γ) = 1.208 + j1.252, so ZL = 60.4 + j62.6 Ω. (Key point: Load from Γ) 2. With |ΓL| = 0.5, what is the VSWR? A. 2.0 B. 3.0 C. 1.5 D. 0.33 Answer: B. VSWR = (1 + 0.5)/(1 - 0.5) = 3.0. (Key point: VSWR) 3. For the same bstub, how do the short-stub and open-stub lengths relate (e.g. 0.1136λ and 0.3636λ)? A. Differ by λ/2 B. Equal C. Differ by λ/4 D. Differ by λ/8 Answer: C. Short and open stubs giving the same susceptance differ by λ/4. (Key point: Short vs open stub lengths) 4. Which choices does the answer give as the shortest? A. Short stub: solution 2; open stub: solution 1 B. Both solution 1 C. Both solution 2 D. Short stub: solution 1 (d = 0.2375λ, l = 0.1136λ); open stub: solution 2 (d = 0.4042λ, l = 0.1364λ) Answer: D. Solution 1 gives the shorter short stub (0.1136λ); solution 2 gives the shorter open stub (0.1364λ). 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