--- title: "Smith chart analysis of 75+j100 Ω load" subject: "RF and Microwave Engineering" chapter: 2 key: "semester_7/rf_and_microwave_engineering:smith-chart-75-j100-load" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/smith-chart-75-j100-load site: BCT Engineering --- # Smith chart analysis of 75+j100 Ω load RF and Microwave Engineering · Chapter 2: RF and Microwave Transmission Lines - Asked in: 2079 Chaitra, 10 marks - Video: [Smith Chart Example for Transmission line Parameters (VSWR, Reflection Coefficient, Input Impedance)](https://www.youtube.com/watch?v=X26nHbBRMjk) — Engineering Funda - 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/smith-chart-75-j100-load - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-2 ## Question A lossless 50 Ohm transmission line is terminated by an impedance of 75 + j100 Ohm. Using Smith Chart, find (a) ΓL, (b) VSWR, (c) Zin at a distance of 0.375λ from the load, (d) the shortest length of line for which impedance is purely resistive, and (e) the value of this resistance. ## Short answer Normalize zL=(75+j100)/50=1.5+j2.0. Smith chart gives GammaL=0.644 at 37.3 degrees, VSWR=4.62; at 0.375lambda from load Zin=13.33+j23.33 ohm; shortest length for purely resistive input is 0.0518lambda, giving R=230.8 ohm. ## Answer **Given:** Z0 = 50 Ω, ZL = 75 + j100 Ω. The values below were computed exactly; Smith chart readings agree to about ±0.005λ. ```text zL = ZL/Z0 = (75 + j100)/50 = 1.5 + j2.0 ``` Plot zL at the intersection of the r = 1.5 circle and the x = 2.0 arc. It reads **0.198λ** on the WTG scale. #### (a) Reflection coefficient ΓL ```text ΓL = (zL − 1)/(zL + 1) = (0.5 + j2)/(2.5 + j2) = 0.644 ∠ 37.3° ``` On the chart: |Γ| = OP/OR (distance from the centre to the point, divided by the chart radius) = 0.644. The angle is read on the "angle of reflection coefficient" scale: 37.3°. #### (b) VSWR ```text VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.644/0.356 = 4.62 ``` On the chart: draw the constant-|Γ| circle through zL. It cuts the right-hand real axis at r = 4.62. #### (c) Zin at 0.375λ from the load Move 0.375λ toward the generator (clockwise) on the |Γ| circle: ```text 0.198λ + 0.375λ = 0.573λ → 0.073λ WTG zin = (zL + j tan βl)/(1 + j zL tan βl), βl = 135°, tan βl = −1 = 0.267 + j0.467 Zin = 50 × zin = 13.33 + j23.33 Ω ``` #### (d) Shortest length for a purely resistive impedance Moving clockwise from 0.198λ, the |Γ| circle first meets the real axis at the **voltage maximum** (right side, 0.25λ): ```text dmax = 0.25λ − 0.198λ = 0.0518λ (= θΓ/(4π) = 37.3°/720° × λ) ``` The next resistive point is the voltage minimum at dmin = 0.0518λ + 0.25λ = 0.3018λ. #### (e) Value of this resistance ```text At dmax: R = Z0 × VSWR = 50 × 4.617 = 230.8 Ω (at dmin: R = Z0/VSWR = 10.83 Ω) ``` **Answer:** ΓL = 0.644∠37.3°, VSWR = 4.62, Zin(0.375λ) = 13.33 + j23.33 Ω, shortest length = 0.0518λ, R = 230.8 Ω. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For zL = 1.5 + j2.0, which expression gives ΓL? A. (zL - 1)/(zL + 1) B. (zL + 1)/(zL - 1) C. (1 - zL)/(1 + zL) D. zL/(1 + zL) Answer: A. ΓL = (zL - 1)/(zL + 1) = (0.5 + j2)/(2.5 + j2) = 0.644∠37.3°. (Key point: Reflection coefficient) 2. With |Γ| = 0.644, what is the VSWR and where is it read on the chart? A. 1.62, where the circle crosses the left-hand real axis B. 4.62, where the constant-|Γ| circle crosses the right-hand real axis C. 0.216, at the centre D. 4.62, at the short-circuit point Answer: B. VSWR = (1 + 0.644)/(1 - 0.644) = 4.62, read as r where the |Γ| circle meets the right-hand real axis. (Key point: VSWR from |Γ|) 3. Why is the shortest length giving a purely resistive impedance 0.0518λ? A. It is the distance to the voltage minimum B. It equals λ/8 minus 0.198λ C. Moving toward the generator, the |Γ| circle first meets the real axis at the voltage maximum (0.25λ on the WTG scale), 0.25λ - 0.198λ away D. It is the distance to the short-circuit point Answer: C. Starting at 0.198λ WTG, the first real-axis crossing is the voltage maximum at 0.25λ, so d = 0.0518λ (= θΓ/(4π)). (Key point: Shortest resistive length) 4. What is the resistance at the voltage maximum, and at the next resistive point (voltage minimum)? A. Z0/VSWR = 10.8 Ω at max; Z0*VSWR = 230.8 Ω at min B. Both equal 50 Ω C. 75 Ω at max; 100 Ω at min D. Z0*VSWR = 230.8 Ω at max; Z0/VSWR = 10.83 Ω at min Answer: D. At the voltage maximum Z = Z0*VSWR = 50 x 4.617 = 230.8 Ω; at the voltage minimum Z = Z0/VSWR = 10.83 Ω. 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