--- title: "Single stub for 78.27+j60.93 Ω on 75 Ω line" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:single-stub-75-ohm-microstrip" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/single-stub-75-ohm-microstrip site: BCT Engineering --- # Single stub for 78.27+j60.93 Ω on 75 Ω line RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2072 Magh, 4+4+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-75-ohm-microstrip - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question Design a single shunt and open matching networks using Smith Chart for a transmission line having surge impedance of 75 Ohm and load impedance of 78.27 + j60.93 Ohm. Sketch the physical diagram considering microstrips. ## Short answer For 75 Ω line with ZL=78.27+j60.93 Ω, shunt short stub: d=0.2458λ, l=0.1430λ. Shunt open stub: d=0.4355λ, l=0.1070λ. Microstrip layout: 75 Ω stub joined at a T-junction, short stub uses via to ground, open stub open-circuited end. ## Answer **Given:** Z0 = 75 Ω, ZL = 78.27 + j60.93 Ω. Design (a) a shunt short-circuited stub and (b) a shunt open-circuited stub. #### Step 1: Normalise and plot the load ```text zL = ZL/Z0 = 1.044 + j0.812 ΓL = (zL − 1)/(zL + 1) = 0.370∠65.2° VSWR = (1 + |Γ|)/(1 − |Γ|) = 2.174 ``` Draw the constant-VSWR circle through zL (zL is at 0.1594λ 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 = 0.597 − j0.464 (0.4094λ 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 + j0.796 at 0.1552λ WTG y2 = 1.000 − j0.796 at 0.3448λ WTG d1 = 0.2458λ from the load d2 = 0.4355λ 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 = −0.7964 short stub l = 0.1430λ open stub l = 0.3930λ Sol 2: bstub = +0.7964 short stub l = 0.3570λ open stub l = 0.1070λ ``` | Solution | Stub position d | Short stub l | Open stub l | |---|---|---|---| | 1 | 0.2458λ | 0.1430λ | 0.3930λ | | 2 | 0.4355λ | 0.3570λ | 0.1070λ | #### (a) Shunt short-circuited stub design Solution 1 is preferred: stub at d = 0.2458λ from the load, with a short-circuited stub of length l = 0.1430λ. #### (b) Open-circuited stub design Solution 2 is preferred: stub at d = 0.4355λ, with an open stub of length l = 0.1070λ. (Solution 1 with an open stub would need l = 0.3930λ.) **Check:** y(d) = 1 ± j0.796. The stub supplies ∓j0.796, so y = 1 and VSWR = 1 (it was 2.17 before matching). #### Physical diagram (microstrip) ```text (a) short stub (b) open stub || l=0.1430λ || l=0.1070λ ● via to ground ○ open end || || =====##== d=0.2458λ ==[ZL] =====##== d=0.4355λ ==[ZL] 75 Ω microstrip 75 Ω microstrip ``` The stub is a 75 Ω strip of the same width as the main line, joined at a T-junction. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the load converted to admittance before finding the position of a shunt stub? A. Shunt elements add in series, so they add reactance to the impedance B. Shunt elements add in parallel, so they add susceptance to the admittance C. Admittance values are always real, which simplifies the chart D. The stub length is read directly from the impedance chart Answer: B. A shunt stub is in parallel with the line, so its susceptance adds to the line admittance; the stub position is therefore found on the admittance chart. (Key point: Use of admittance) 2. Where on the line is the shunt stub placed? A. Where the VSWR circle cuts the real axis of the chart B. At the load itself, where y = 1 + jb C. At the point where the line admittance is purely imaginary D. Where the VSWR circle cuts the g = 1 circle, so the line admittance is 1 ± jb Answer: D. The stub can only cancel susceptance, so it is placed where the real part of the admittance is already 1 (g = 1). (Key point: Stub position) 3. If the line admittance at the stub position is y = 1 + j0.796, what must the stub supply? A. bstub = -0.796, to cancel the susceptance B. bstub = +0.796, to double the susceptance C. bstub = 0, since g is already 1 D. bstub = -1, to cancel the conductance Answer: A. The stub must cancel the imaginary part, giving bstub = -b so the total admittance becomes 1 + j0. (Key point: Stub susceptance) 4. For the same required susceptance, how do the short-circuited and open-circuited stub lengths compare (Sol 1: 0.1430λ short, 0.3930λ open)? A. They are always equal B. They differ by λ/2 (0.5λ) C. They differ by λ/4 (0.25λ) D. The open stub is exactly half the short stub Answer: C. Moving λ/4 turns a short into an open, so 0.1430λ + 0.25λ = 0.3930λ for the open stub. 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