Derive SWR at a boundary

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:

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

|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:

z_max = −(θ + 2nπ)/(2β₁),   n = 0, 1, 2, ...

Minima occur where 2β₁z + θ = −(2n + 1)π:

z_min = −(θ + (2n + 1)π)/(2β₁)

Special cases (real Γ):

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.