--- title: "Derive SWR at a boundary" subject: "Electromagnetism" chapter: 4 key: "semester_3/electromagnetism:standing-wave-ratio-derive-boundary" source: https://www.bctengineer.com/q/semester_3/electromagnetism/standing-wave-ratio-derive-boundary site: BCT Engineering --- # Derive SWR at a boundary Electromagnetism · Chapter 4: Wave equation and Wave propagation - Asked in: 2075 Chaitra, 8 marks - Video: [RELATION BETWEEN VSWR AND REFLECTION COEFFICIENT](https://www.youtube.com/watch?v=Sr9qLfdELIc) — RF Microwave & Antenna - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/standing-wave-ratio-derive-boundary - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4 ## Question Derive an expression for Standing Wave Ratio (SWR) indicating where on the z-axis you will get the maximum and minimum value of electric field intensity E. Assume that the boundary is at z = 0, the region z < 0 is a perfect dielectric and the region z > 0 may be of any material. ## Short answer For a wave incident from a lossless medium onto a boundary at z=0, the standing wave ratio s = (1+|Γ|)/(1−|Γ|) with Γ = (η₂−η₁)/(η₂+η₁). Maxima of |E| occur where 2β₁z+θ = −2nπ, giving z_max = −(θ+2nπ)/(2β₁); minima where 2β₁z+θ = −(2n+1)π, giving z_min = −(θ+(2n+1)π)/(2β₁). ## Answer #### Set-up Normal incidence of a plane wave from region 1 (z < 0, perfect dielectric with ε₁, μ₁, η₁, β₁) on region 2 (z > 0, intrinsic impedance η₂, any material). The incident wave travels in +z, and part is reflected back into region 1. The standing wave ratio is **s = |E|max / |E|min** #### Fields in region 1 (phasors) Take E along a_x. At z = 0 the incident amplitude is E₀: ```text E_s1 = E₀ ( e^{−jβ₁z} + Γ e^{+jβ₁z} ) a_x ``` where Γ = E_r/E_i = (η₂ − η₁)/(η₂ + η₁) is the reflection coefficient at z = 0 (from the boundary condition of continuous tangential E and H). It may be complex: Γ = |Γ| e^{jθ} (θ = 0 if η₂ > η₁ both real, θ = π if η₂ < η₁). #### Magnitude along z ```text |E_s1| = E₀ |e^{−jβ₁z}| |1 + Γ e^{j2β₁z}| = E₀ |1 + |Γ| e^{j(θ + 2β₁z)}| = E₀ √(1 + |Γ|² + 2|Γ| cos(2β₁z + θ)) ``` - **Maximum:** cos(2β₁z + θ) = 1, so |E|max = E₀(1 + |Γ|). - **Minimum:** cos(2β₁z + θ) = −1, so |E|min = E₀(1 − |Γ|). #### SWR **s = |E|max / |E|min = (1 + |Γ|)/(1 − |Γ|)**, 1 ≤ s ≤ ∞, and inversely |Γ| = (s − 1)/(s + 1). s = 1 for a matched boundary (Γ = 0) and s = ∞ for total reflection (|Γ| = 1). #### Locations on the z-axis Maxima occur where 2β₁z + θ = −2nπ: ```text z_max = −(θ + 2nπ)/(2β₁), n = 0, 1, 2, ... ``` Minima occur where 2β₁z + θ = −(2n + 1)π: ```text z_min = −(θ + (2n + 1)π)/(2β₁) ``` Special cases (real Γ): | Case | Γ | E max at | E min at | |---|---|---|---| | η₂ > η₁ (θ = 0) | positive | z = −nλ₁/2 (including boundary z = 0) | z = −(2n+1)λ₁/4 | | η₂ < η₁ (θ = π) | negative | z = −(2n+1)λ₁/4 | z = −nλ₁/2 (including boundary z = 0) | Successive maxima (or minima) are λ₁/2 apart, and a maximum is λ₁/4 away from the nearest minimum. Here λ₁ = 2π/β₁. The fields of the two-wave sum form a standing wave pattern in region 1, while region 2 has only a travelling transmitted wave. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The standing wave ratio is defined as: A. |E|min / |E|max B. |E|max / |E|min C. |E|max - |E|min D. eta2 / eta1 always Answer: B. s = |E|max/|E|min, always between 1 and infinity. (Key point: Definition of SWR) 2. The field in region 1 is: A. E0 (e^(-j beta1 z) - Gamma e^(-j beta1 z)) B. E0 e^(-j beta1 z) only C. E0 (e^(-j beta1 z) + Gamma e^(+j beta1 z)), incident plus reflected waves D. E0 Gamma e^(-j beta2 z) Answer: C. Region 1 carries the incident wave plus a reflected wave with amplitude Gamma, travelling in -z. (Key point: Field in region 1) 3. Max and min of |E| are E0(1 + |Gamma|) and E0(1 - |Gamma|), so s is: A. (1 - |Gamma|)/(1 + |Gamma|) B. (1 + |Gamma|)/(1 - |Gamma|) C. 1 + |Gamma| D. |Gamma|/(1 - |Gamma|) Answer: B. s = (1 + |Gamma|)/(1 - |Gamma|), with s = 1 when matched and s = infinity for total reflection. (Key point: SWR formula) 4. If eta2 > eta1 (Gamma positive), where is the first E maximum in region 1? A. At z = -lambda1/4 B. At z = +lambda1/4 C. At the boundary z = 0 D. At z = -lambda1/8 Answer: C. With theta = 0 the maxima are at z = -n lambda1/2, including z = 0, and minima are lambda1/4 away at z = -(2n+1) lambda1/4. 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