--- title: "Smith chart, load 80+j100 on 50 Ω line" subject: "Electromagnetism" chapter: 5 key: "semester_3/electromagnetism:smith-chart-load-80-j100-50ohm" source: https://www.bctengineer.com/q/semester_3/electromagnetism/smith-chart-load-80-j100-50ohm site: BCT Engineering --- # Smith chart, load 80+j100 on 50 Ω line Electromagnetism · Chapter 5: Transmission lines - Asked in: 2076 Asoj, 7 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_3/electromagnetism/smith-chart-load-80-j100-50ohm - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5 ## Question A load Z_L = 80 + j100 Ω is located at z = 0 on a lossless 50-Ω line. The operating frequency is 200 MHz and the wavelength on the line is 2 m. (a) If the line is 0.8 m in length, use the Smith chart to find the input impedance. (b) What is s (the standing wave ratio)? (c) What is the distance from the load to the nearest voltage maximum? ## Short answer Moving 0.4λ toward the generator on the Smith chart gives input impedance 16.6 + j33.8 Ω. SWR is 4.5. The nearest voltage maximum is about 0.1 m (0.05λ) from the load. ## Answer Given: Z_L = 80 + j100 Ω, Z₀ = 50 Ω, f = 200 MHz, λ = 2 m, l = 0.8 m = 0.4λ. #### Step 1: Normalise and plot z_L = Z_L/Z₀ = 1.6 + j2.0. Mark this point on the chart (intersection of the r = 1.6 circle and the x = +2 arc, upper half). Draw the constant-|Γ| circle through it with centre at the chart centre. Reading the chart: |Γ| = 0.637, angle θ_Γ = 35.7°. (Check: Γ = (z_L − 1)/(z_L + 1) = (0.6 + j2)/(2.6 + j2) = 0.637∠35.7°.) #### (a) Input impedance at 0.4λ Move 0.4λ from the load toward the generator (clockwise on the WTG scale). The Γ angle changes by −2βl = −2 × 360° × 0.4 = −288°: ```text Γ_in = 0.637 ∠(35.7° − 288°) = 0.637 ∠107.7° (≡ −252.3°) ``` Reading z_in at this point on the chart: z_in ≈ 0.33 + j0.68. Using the exact formula Z_in = Z₀(Z_L + jZ₀ tanβl)/(Z₀ + jZ_L tanβl), with βl = 144°, tan = −0.7265: ```text Numerator: 80 + j100 + j50(−0.7265) = 80 + j63.7 (102.2∠38.5°) Denominator: 50 + j(80 + j100)(−0.7265) = 122.7 − j58.1 (135.7∠−25.3°) Z_in = 50 × 0.753∠63.9° = 37.7∠63.9° = 16.6 + j33.8 Ω ``` Result: **Z_in = 16.6 + j33.8 Ω** (z_in = 0.33 + j0.68), which the chart reading confirms. #### (b) Standing wave ratio At the point where the |Γ| circle crosses the real axis on the right side (r = s): ```text s = (1 + |Γ|)/(1 − |Γ|) = 1.637/0.363 = 4.5 ``` **s ≈ 4.5** #### (c) Distance from load to nearest voltage maximum The voltage maximum is where Γ is real positive, i.e. on the right of the real axis. From the load angle 35.7°, rotate clockwise to 0°, a change of 35.7° = 2βd. ```text d = 35.7° / 720° × λ = 0.0496 λ = 0.0496 × 2 m ≈ 0.099 m ≈ 0.1 m ``` On the wavelengths-toward-generator scale this is about 0.0496λ. **Distance ≈ 0.1 m (0.05λ).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Z_L = 80 + j100 on a 50 ohm line has which normalised value and electrical length (lambda = 2 m, line 0.8 m)? A. z_L = 80 + j100, length 0.8 lambda B. z_L = 0.625 + j0.5, length 0.4 lambda C. z_L = 1.6 + j2.0, length 0.4 lambda D. z_L = 1.6 + j2.0, length 0.8 lambda Answer: C. Divide by Z0 = 50 ohm, and 0.8 m/2 m = 0.4 lambda. (Key point: Normalised load) 2. What is the SWR for |Gamma| = 0.637? A. 1/0.637 = 1.57 B. (1 + 0.637)/(1 - 0.637) = 4.5 C. (1 - 0.637)/(1 + 0.637) = 0.22 D. 0.637 Answer: B. s = (1 + |Gamma|)/(1 - |Gamma|). (Key point: SWR) 3. Moving 0.4 lambda toward the generator changes the angle of Gamma by: A. -2 beta l = -288 degrees B. +288 degrees C. -144 degrees D. -72 degrees Answer: A. Moving toward the generator is clockwise by 720 degrees per wavelength. Result: Z_in = 16.6 + j33.8 ohm. (Key point: Input impedance by rotation) 4. How is the distance to the nearest voltage maximum found? A. Rotate to Gamma real negative: d = 0.25 lambda B. Use d = lambda/4 always C. Use d = 35.7 degrees/360 degrees x lambda D. Rotate clockwise from the load angle (35.7 degrees) to Gamma real positive: d = 0.0496 lambda, about 0.1 m Answer: D. The maximum occurs where Gamma is real positive, 2 beta d = 35.7 degrees, so d = 0.0496 lambda. 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