--- title: "Input intrinsic impedance via reflection" subject: "Electromagnetism" chapter: 4 key: "semester_3/electromagnetism:input-intrinsic-impedance-reflection" source: https://www.bctengineer.com/q/semester_3/electromagnetism/input-intrinsic-impedance-reflection site: BCT Engineering --- # Input intrinsic impedance via reflection Electromagnetism · Chapter 4: Wave equation and Wave propagation - Asked in: 2070 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/input-intrinsic-impedance-reflection - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4 ## Question Derive the expression for input intrinsic impedance using the concept of reflection of uniform plane waves. ## Short answer The input intrinsic impedance η_in at a distance l from the boundary is the ratio of total tangential E to H, derived by combining incident and reflected waves. It yields η_in = η₁ (η₂ + jη₁ tan β₁l)/(η₁ + jη₂ tan β₁l), analogous to transmission line input impedance. ## Answer **Input intrinsic impedance** η_in(−l) is the ratio of the total electric field to the total magnetic field (E_x/H_y) at a distance l in front of the boundary, when there are both an incident and a reflected wave. It is the wave impedance "seen" looking toward the boundary. #### Set-up Region 1 (z < 0): lossless medium with η₁, β₁. Region 2 (z > 0): impedance η₂. The boundary is at z = 0. The reflection coefficient at the boundary is Γ = (η₂ − η₁)/(η₂ + η₁) Total fields in region 1 (E along a_x, H along a_y): ```text E_s(z) = E₀ ( e^{−jβ₁z} + Γ e^{+jβ₁z} ) H_s(z) = (E₀/η₁) ( e^{−jβ₁z} − Γ e^{+jβ₁z} ) ``` #### Derivation The wave impedance at any z in region 1: ```text η(z) = E_s/H_s = η₁ (e^{−jβ₁z} + Γ e^{jβ₁z}) / (e^{−jβ₁z} − Γ e^{jβ₁z}) = η₁ (1 + Γ e^{j2β₁z}) / (1 − Γ e^{j2β₁z}) ``` At distance l from the boundary, z = −l: ```text η_in = η₁ (1 + Γ e^{−j2β₁l}) / (1 − Γ e^{−j2β₁l}) ``` Substitute Γ = (η₂ − η₁)/(η₂ + η₁) and multiply numerator and denominator by (η₂ + η₁): ```text numerator: (η₂ + η₁) + (η₂ − η₁) e^{−j2β₁l} denominator: (η₂ + η₁) − (η₂ − η₁) e^{−j2β₁l} ``` Multiply both by e^{jβ₁l} and use e^{±jβ₁l} = cos β₁l ± j sin β₁l: ```text numerator: 2η₂ cos β₁l + j2η₁ sin β₁l denominator: 2η₁ cos β₁l + j2η₂ sin β₁l ``` Dividing both by 2 cos β₁l: **η_in = η₁ · (η₂ + jη₁ tan β₁l) / (η₁ + jη₂ tan β₁l)** #### Checks - l = 0: η_in = η₂ (the load impedance). - l = λ₁/2 (β₁l = π, tan = 0): η_in = η₂, so a half-wave layer repeats the load impedance. - l = λ₁/4 (tan → ∞): η_in = η₁²/η₂, the quarter-wave transformer relation used for anti-reflection coatings (choose η₁ = √(η_a η₂) for matching). - Perfect conductor (η₂ = 0): η_in = jη₁ tan β₁l, purely reactive. This has exactly the same form as the input impedance of a transmission line, with η playing the role of Z₀. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the input intrinsic impedance eta_in(-l)? A. The ratio of incident E to incident H only B. The ratio of reflected E to incident E C. The ratio of total E to total H at distance l in front of the boundary, with incident and reflected waves present D. The impedance of region 2 alone Answer: C. It is the total E_x/H_y seen looking toward the boundary at distance l, including both incident and reflected waves. (Key point: Definition) 2. In the total H_s in region 1, the reflected term has a minus sign: (E0/eta1)(e^-jb1z - Gamma e^+jb1z). Why? A. Because Gamma is always negative B. The reflected wave travels in -z, so its H is reversed relative to its E C. Because H leads E by 180 degrees in every wave D. Because the reflected wave has lower amplitude Answer: B. H = (1/eta) a_k x E, so reversing the direction of travel reverses the sign of H for the same E. (Key point: Total fields) 3. Which is the derived expression for eta_in? A. eta1 (eta2 + j eta1 tan b1l)/(eta1 + j eta2 tan b1l) B. eta1 (eta1 + j eta2 tan b1l)/(eta2 + j eta1 tan b1l) C. eta2 (eta1 + j eta2 tan b1l)/(eta2 + j eta1 tan b1l) D. eta1 (eta2 + eta1 tan b1l)/(eta1 + eta2 tan b1l) Answer: A. Substituting Gamma and simplifying gives eta_in = eta1 (eta2 + j eta1 tan b1l)/(eta1 + j eta2 tan b1l), the same form as the transmission-line formula. (Key point: Final result) 4. What is eta_in when l = lambda1/4? A. eta2 B. eta1*eta2 C. 0 D. eta1^2/eta2 Answer: D. tan(b1l) tends to infinity, so eta_in = eta1^2/eta2, the quarter-wave transformer relation. 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