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Exact plane-wave reflection coefficient of the local impedance condition

Statement

For $0 < q \leq k$ and tangential wavevector $η \in R^{2}$ satisfying $q^{2} + ∣ η ∣^{2} = k^{2}$ , consider $u (x, y) = e^{i η \cdot y} (e^{i q (x - R)} + R e^{- i q (x - R)}) .$ The impedance condition First-order local impedance condition at an artificial plane gives $R = \frac{q - k}{q + k}, ∣ R ∣ = \frac{k - q}{k + q} .$ Thus reflection is zero at normal incidence $q = k$ and its magnitude tends to one as $q ↓ 0$ (grazing incidence).

Facts & Assumptions

Given: A homogeneous propagating mode with $0 < q \leq k$ and the boundary convention in [F1].

[F1]

First-order local impedance condition at an artificial plane: on $x = R$ impose $\partial_{x} u - i k u = 0$ with outward normal $+ e_{x}$ .

Proof

Given: The mode and condition in [F1].

1.1givenF1algebra

At $x = R$ , the bracket in the definition of $u$ is $1 + R$ and its $x$ derivative is $i q (1 - R)$ . Substitution into [F1], after cancelling the nonzero tangential exponential, gives $q (1 - R) - k (1 + R) = 0$ .

2.1step 1.1algebra∎

Since $q + k > 0$ , solving the equation in step 1.1 gives $R = (q - k) / (q + k)$ . The numerator is nonpositive for $q \leq k$ , so $∣ R ∣ = (k - q) / (k + q)$ ; this is zero at $q = k$ and tends to $1$ as $q ↓ 0$ .

Depends on

First-order local impedance condition at an artificial plane

Used by

Normal, oblique, and grazing impedance reflection Example

Dependency tree · two levels

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Sources

Galkowski, Lafontaine, and Spence, Local absorbing boundary conditions on fixed domains give order-one errors for high-frequency waves (standard reference, not scraped)
Steven G. Johnson, Notes on Perfectly Matched Layers (PMLs), §2 (standard reference, not scraped)