--- title: "Derive input impedance of transmission line" subject: "Electromagnetism" chapter: 5 key: "semester_3/electromagnetism:derive-input-impedance-line" source: https://www.bctengineer.com/q/semester_3/electromagnetism/derive-input-impedance-line site: BCT Engineering --- # Derive input impedance of transmission line Electromagnetism · Chapter 5: Transmission lines - Asked in: 2067 Mangsir, 6 marks - Video: [Input Impedance of Transmission Line in Microwave Engineering by Engineering Funda](https://www.youtube.com/watch?v=9efn7DQ72u8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/derive-input-impedance-line - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5 ## Question Derive the expression for input impedance of a transmission line with characteristic impedance Z0 excited by a source with source impedance Zs and terminated in load Z_L. ## Short answer From line voltage and current wave equations with load boundary condition, input impedance at distance l is Z_in = Z₀ (Z_L + jZ₀ tan βl)/(Z₀ + jZ_L tan βl). The source impedance Z_s does not change Z_in but forms a voltage divider V_in = V_s Z_in/(Z_s + Z_in). ## Answer Consider a lossless line of length l, characteristic impedance Z₀ and propagation constant jβ, with a source (V_s, Z_s) at the sending end and a load Z_L at the receiving end. Let z be the distance from the load toward the source. ```text Z_s I_in I_L o---/\/\--->--------- l ------------>---o V_s ~ V_in Z0, β V_L Z_L o----------------------------------------o ``` #### Voltage and current on the line ```text V(z) = V⁺ e^{jβz} + V⁻ e^{−jβz} I(z) = (V⁺ e^{jβz} − V⁻ e^{−jβz}) / Z₀ ``` At the load (z = 0): V_L = V⁺ + V⁻, I_L = (V⁺ − V⁻)/Z₀, and Z_L = V_L/I_L. Solving: ```text V⁺ = (I_L/2)(Z_L + Z₀), V⁻ = (I_L/2)(Z_L − Z₀) ``` #### Input impedance at z = l ```text V(l) = (I_L/2)[(Z_L + Z₀)e^{jβl} + (Z_L − Z₀)e^{−jβl}] = I_L [Z_L cos βl + jZ₀ sin βl] I(l) = (I_L/2Z₀)[(Z_L + Z₀)e^{jβl} − (Z_L − Z₀)e^{−jβl}] = I_L [cos βl + j(Z_L/Z₀) sin βl] ``` Dividing, and then dividing the numerator and denominator by cos βl: ```text Z_in = V(l)/I(l) = Z₀ (Z_L + jZ₀ tan βl) / (Z₀ + jZ_L tan βl) ``` For a lossy line, replace jβ by γ and tan βl by tanh γl: Z_in = Z₀ (Z_L + Z₀ tanh γl)/(Z₀ + Z_L tanh γl). In terms of the reflection coefficient: Z_in = Z₀(1 + Γ_in)/(1 − Γ_in), with Γ_in = Γ_L e^{−j2βl}. #### Effect of source impedance Z_s The source impedance and the input impedance form a voltage divider: ```text V_in = V_s Z_in / (Z_s + Z_in) I_in = V_s / (Z_s + Z_in) ``` The power delivered to the line is P_in = ½ Re(V_in I_in*). Special cases: Z_L = Z₀ gives Z_in = Z₀ for any l (matched); l = λ/4 gives Z_in = Z₀²/Z_L; l = λ/2 gives Z_in = Z_L; a short circuit gives Z_in = jZ₀ tan βl; an open circuit gives Z_in = −jZ₀ cot βl. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does I(z) contain a minus sign between the two waves and a factor 1/Z0? A. Because V- is always negative B. Because current is the sum of V+ and V- C. Because Z0 is imaginary D. The reflected current flows in the opposite direction, and I = V/Z0 for each wave Answer: D. I(z) = (V+ e^(jbz) - V- e^(-jbz))/Z0. (Key point: Voltage and current) 2. What do the load conditions give? A. V+ = (I_L/2)(Z_L - Z0) and V- = (I_L/2)(Z_L + Z0) B. V+ = I_L Z_L and V- = 0 C. V+ = (I_L/2)(Z_L + Z0) and V- = (I_L/2)(Z_L - Z0) D. V+ = V- = I_L Z0 Answer: C. Solving V_L = V+ + V- and I_L Z0 = V+ - V- gives them. (Key point: Constants from the load) 3. What is Z_in = V(l)/I(l)? A. Z0 (Z0 + j Z_L tan bl)/(Z_L + j Z0 tan bl) B. Z0 (Z_L + j Z0 tan bl)/(Z0 + j Z_L tan bl) C. Z_L (Z0 + j Z_L tan bl)/(Z_L + j Z0 tan bl) D. Z0 (Z_L tan bl)/(Z0) Answer: B. V(l)/I(l) = [Z_L cos + j Z0 sin]/[cos + j(Z_L/Z0) sin]. (Key point: Final result) 4. How does Z_s enter the problem? A. Z_in is independent of Z_s; Z_s only enters the divider V_in = V_s Z_in/(Z_s + Z_in) B. It changes Z_in by tan(beta Z_s) C. It replaces Z0 D. It makes Gamma_L zero Answer: A. Z_in depends on Z0, Z_L and l only. The source and Z_in form a voltage divider. 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