--- title: "Single stub for 120−j160 Ω on 100 Ω line" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:single-stub-120-j160" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/single-stub-120-j160 site: BCT Engineering --- # Single stub for 120−j160 Ω on 100 Ω line RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2081 Bhadra, 5+3+2 marks - Video: [Single shunt stub matching network using smith chart by Dr. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=r78seP5upKA) — 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/single-stub-120-j160 - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question A 100Ω lossless transmission line is terminated with a complex load of 120 − j160 Ω. Design a single matching stub. Mention all the design steps with proper reasoning. Provide the S-Matrix for both before and after the design. ## Short answer Place a short-circuited shunt stub at d=0.1104λ from the load with length l=0.0950λ; an open-stub alternative is d=0.2594λ, l=0.1550λ. Before matching S11=0.593∠-46.8°; after matching, the tuner has S11=0.593∠126.4°, S21=S12=0.805∠-3.4°, S22=0.593∠46.8°, giving Γin=0. ## Answer **Given:** Z0 = 100 Ω, ZL = 120 − j160 Ω (capacitive). A single shunt stub is used. **Reasoning:** a shunt stub can only add susceptance. So, first move along the line to a point where the real part of the admittance equals Y0 (g = 1). Then cancel the imaginary part with the stub. #### Step 1: Normalise and plot the load ```text zL = ZL/Z0 = 1.200 − j1.600 ΓL = (zL − 1)/(zL + 1) = 0.593∠-46.8° VSWR = (1 + |Γ|)/(1 − |Γ|) = 3.911 ``` Draw the constant-VSWR circle through zL (zL is at 0.3151λ 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 = 0.300 + j0.400 (0.0651λ 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 + j1.472 at 0.1755λ WTG y2 = 1.000 − j1.472 at 0.3245λ WTG d1 = 0.1104λ from the load d2 = 0.2594λ 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 from the short-circuit point (y = ∞, right end of the chart) toward the generator. - 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 1: bstub = −1.4720 short stub l = 0.0950λ open stub l = 0.3450λ Sol 2: bstub = +1.4720 short stub l = 0.4050λ open stub l = 0.1550λ ``` | Solution | Stub position d | Short stub l | Open stub l | |---|---|---|---| | 1 | 0.1104λ | 0.0950λ | 0.3450λ | | 2 | 0.2594λ | 0.4050λ | 0.1550λ | **Answer (design):** solution 1, with the stub at d = 0.1104λ from the load and a short-circuited stub of l = 0.0950λ. These are the shortest lengths, giving the widest bandwidth. (Open-stub alternative: solution 2, d = 0.2594λ, l = 0.1550λ.) ```text Z0=100Ω d = 0.1104λ o---------+--------------[120 − j160 Ω] | l = 0.0950λ | short ``` #### S-matrix before and after the design Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D. ```text Before matching (line sees the load directly): S11 = ΓL = 0.593∠-46.8°, VSWR = 3.91 Matching network (solution 1): S11 = 0.593∠126.4° S12 = 0.805∠-3.4° S21 = 0.805∠-3.4° S22 = 0.593∠46.8° S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1 After matching (network + load): Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1 ``` With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must the line be moved to a point with g = 1 before adding the stub? A. A shunt stub adds conductance as well, so g must be zero B. A shunt stub only adds susceptance, so the real part must already equal Y0 C. A stub can only change the VSWR, not the phase D. The stub works only on the real axis Answer: B. A stub adds only imaginary admittance, so g must be 1 at its location and the stub cancels the imaginary part. (Key point: Reasoning) 2. For zL = 1.2 - j1.6, what is yL and how is it found on the chart? A. yL = 1.2 + j1.6, by reflecting about the real axis B. yL = 0.300 - j0.400, by taking the complex conjugate C. yL = 0.6 - j0.8, by dividing by two D. yL = 0.300 + j0.400, by rotating 180° (λ/4) on the VSWR circle Answer: D. yL = 1/zL = 0.3 + j0.4, the point diametrically opposite zL. (Key point: Admittance conversion) 3. Why is solution 1 (d = 0.1104λ, short stub 0.0950λ) preferred? A. Shortest line and stub lengths give the widest bandwidth B. It gives VSWR below 1 C. The stub is open-circuited D. It needs the longest line for lowest loss Answer: A. Both match perfectly, but shorter lengths reduce the frequency sensitivity, giving wider bandwidth. (Key point: Chosen solution) 4. What is the property of the matching network S-matrix (solution 1)? A. S11 = 0 and S22 = 0.593∠-46.8° always B. S12 is not equal to S21 C. S22 = ΓL*, S12 = S21 and |S11|^2 + |S21|^2 = 1 D. |S11|^2 + |S21|^2 = 2 Answer: C. The lossless reciprocal network satisfies |S11|^2 + |S21|^2 = 1 and conjugate-matches the load so Γin = 0. 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