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The two-equation interface admittance system: solution formulas and the normal-flux identity

Statement

Let $Y_{1}, Y_{2} \in C$ with $Y_{1} + Y_{2} \neq 0$ , and let $E_{i}, E_{r}, E_{t} \in C$ satisfy the two linear equations

$E_{i} + E_{r} = E_{t}, Y_{1} (E_{i} - E_{r}) = Y_{2} E_{t} .$

Then: (a) [Solution] If $E_{i} \neq 0$ , then $r : = \frac{E_{r}}{E_{i}} = \frac{Y_{1} - Y_{2}}{Y_{1} + Y_{2}}, t : = \frac{E_{t}}{E_{i}} = \frac{2 Y_{1}}{Y_{1} + Y_{2}};$ if $E_{i} = 0$ then $E_{r} = E_{t} = 0$ and the field ratios are undefined. Throughout (b)-(d), $r, t$ denote the displayed algebraic transfer coefficients $(Y_{1} - Y_{2}) / (Y_{1} + Y_{2})$ and $2 Y_{1} / (Y_{1} + Y_{2})$ , which are defined independently of $E_{i}$ ; when $E_{i} \neq 0$ they equal the field ratios of (a). (b) [Normal-flux identity] If $Y_{1} > 0$ is real (the incident side admittance), then $∣ r ∣^{2} + \frac{Re Y_{2}}{Y_{1}} ∣ t ∣^{2} = 1,$ and explicitly $∣ t ∣^{2} = \frac{4 Y_{1}^{2}}{∣ Y_{1} + Y_{2} ∣^{2}}$ and $∣ r ∣^{2} = 1 - \frac{4 Y_{1} Re Y_{2}}{∣ Y_{1} + Y_{2} ∣^{2}}$ . The second summand $(Re Y_{2} / Y_{1}) ∣ t ∣^{2}$ is nonnegative if and only if $Re Y_{2} \geq 0$ ; it is the normal-flux coefficient of a passive transmitted medium and may be negative for an active one, in which case $∣ r ∣ > 1$ and the identity carries no energy-conservation reading. (c) [Pure-imaginary admittance] If $Y_{1} > 0$ is real and $Y_{2}$ is purely imaginary, then $∣ r ∣ = 1$ , $Re Y_{2} ∣ t ∣^{2} = 0$ and $∣ t ∣^{2} = \frac{4 Y_{1}^{2}}{Y_{1}^{2} + ∣ Y_{2} ∣^{2}} > 0$ ; in particular $r \neq 0$ and $t \neq 0$ . (d) [Real-positive specialisation] If $Y_{1}, Y_{2} > 0$ are real, then $r = (Y_{1} - Y_{2}) / (Y_{1} + Y_{2})$ and $t = 2 Y_{1} / (Y_{1} + Y_{2})$ are real and $∣ r ∣^{2} + (Y_{2} / Y_{1}) ∣ t ∣^{2} = 1$ ; with $Y_{j} = 1 / Z_{j}$ , $Z_{j} > 0$ , this is $r = (Z_{2} - Z_{1}) / (Z_{1} + Z_{2})$ , $t = 2 Z_{2} / (Z_{1} + Z_{2})$ and $R + T = 1$ for $R = ∣ r ∣^{2}$ , $T = (Z_{1} / Z_{2}) ∣ t ∣^{2}$ , in agreement with The two-equation interface system at normal incidence. (e) [Scope] The lemma is pure algebra of $C$ ; the labels 'incident', 'reflected' and 'transmitted' and the interpretation of the second summand as a flux fraction belong to the consuming physical theorem, not to this lemma.

Facts & Assumptions

Given: Complex numbers $Y_{1}, Y_{2}$ with $Y_{1} + Y_{2} \neq 0$ and complex numbers $E_{i}, E_{r}, E_{t}$ satisfying $E_{i} + E_{r} = E_{t}$ and $Y_{1} (E_{i} - E_{r}) = Y_{2} E_{t}$ ; the transfer coefficients $r = (Y_{1} - Y_{2}) / (Y_{1} + Y_{2})$ and $t = 2 Y_{1} / (Y_{1} + Y_{2})$ .

[F1]

Field arithmetic in $C$ : $C$ is a field, every element has the unique form $a + b i$ with $a, b \in R$ , addition and multiplication follow the coordinate formulas, $(a + b i)^{- 1} = (a - b i) / (a^{2} + b^{2})$ for $a + b i \neq 0$ , and a product of nonzero field elements is nonzero (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F2]

Conjugation and modulus: $z \overline{z} = ∣ z ∣^{2}$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , $∣ z ∣ \geq 0$ with $∣ z ∣ = 0$ exactly for $z = 0$ , conjugation is an involutive real-field automorphism fixing every real number, and $∣ \overline{z} ∣ = ∣ z ∣$ (Real and imaginary parts, complex conjugation, and modulus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F3]

Real order: the real numbers form an ordered field with positive cone $P$ (Ordered field): by trichotomy a positive real is nonzero, and by closure sums and products of positive reals are positive; the same axioms give that the product of two nonzero reals of equal sign is positive and of opposite signs is negative, and that division by the positive real $Y_{1}$ preserves the sign of a quotient, since $a = (a / Y_{1}) Y_{1}$ has the sign of $a / Y_{1}$ when $Y_{1} > 0$ .

[F4]

Squares in an ordered field: the square of a nonzero element is positive (Squares of nonzero elements are positive), and for $a, b \geq 0$ one has $a < b$ if and only if $a^{2} < b^{2}$ (Squaring is monotone on the nonnegatives).

[F5]

The normal-incidence interface lemma The two-equation interface system at normal incidence: for real $ε_{j}, μ_{j} > 0$ , $Z_{j} = \sqrt{μ_{j} / ε_{j}} > 0$ , and $E_{i}, E_{r}, E_{t} \in C$ with $E_{i} \neq 0$ satisfying $E_{i} + E_{r} = E_{t}$ and $(E_{i} - E_{r}) / Z_{1} = E_{t} / Z_{2}$ , one has $E_{r} / E_{i} = (Z_{2} - Z_{1}) / (Z_{1} + Z_{2})$ , $E_{t} / E_{i} = 2 Z_{2} / (Z_{1} + Z_{2})$ , and $R + T = 1$ for $R = ∣ E_{r} / E_{i} ∣^{2}$ , $T = (Z_{1} / Z_{2}) ∣ E_{t} / E_{i} ∣^{2}$ .

Proof

technique · direct; solve the two linear equations by substitution, expand the moduli with $|z|^2=z\overline z$, and read the sign statements off the ordered-field structure of $\mathbb R$

1.1givenF1algebra

[Solution of the linear system] Substituting the first equation $E_{r} = E_{t} - E_{i}$ into the second gives $Y_{1} (2 E_{i} - E_{t}) = Y_{2} E_{t}$ , that is $2 Y_{1} E_{i} = (Y_{1} + Y_{2}) E_{t}$ after expanding and collecting terms by the field arithmetic of [F1]. Since $Y_{1} + Y_{2} \neq 0$ is invertible in the field $C$ by [F1], $E_{t} = 2 Y_{1} E_{i} / (Y_{1} + Y_{2})$ , and then $E_{r} = E_{t} - E_{i} = (2 Y_{1} - (Y_{1} + Y_{2})) E_{i} / (Y_{1} + Y_{2}) = (Y_{1} - Y_{2}) E_{i} / (Y_{1} + Y_{2})$ , again by [F1].

2.1step 1.1F1algebra

[Consequences of the solution] If $E_{i} \neq 0$ , dividing the two identities of step 1.1 by $E_{i}$ gives $r = E_{r} / E_{i} = (Y_{1} - Y_{2}) / (Y_{1} + Y_{2})$ and $t = E_{t} / E_{i} = 2 Y_{1} / (Y_{1} + Y_{2})$ . If $E_{i} = 0$ , the identities of step 1.1 give $E_{t} = 0$ and $E_{r} = 0$ , so no field ratio exists; the algebraic expressions $r = (Y_{1} - Y_{2}) / (Y_{1} + Y_{2})$ and $t = 2 Y_{1} / (Y_{1} + Y_{2})$ remain defined because $Y_{1} + Y_{2} \neq 0$ , and they agree with the field ratios whenever the latter exist. This proves (a) and the convention stated before (b).

3.1step 2.1F2algebra

[Modular expansions] For any $z, w \in C$ , $∣ z \pm w ∣^{2} = (z \pm w) (\overline{z} \pm \overline{w}) = ∣ z ∣^{2} \pm (z \overline{w} + \overline{z} w) + ∣ w ∣^{2}$ , and $z \overline{w} + \overline{z} w = 2 Re (z \overline{w})$ by [F2]; hence for real $Y_{1}$ and arbitrary $Y_{2}$ , $∣ Y_{1} \pm Y_{2} ∣^{2} = Y_{1}^{2} \pm 2 Y_{1} Re Y_{2} + ∣ Y_{2} ∣^{2}$ , and subtraction gives $∣ Y_{1} + Y_{2} ∣^{2} - ∣ Y_{1} - Y_{2} ∣^{2} = 4 Y_{1} Re Y_{2}$ . Moreover $∣ r ∣^{2} = ∣ Y_{1} - Y_{2} ∣^{2} / ∣ Y_{1} + Y_{2} ∣^{2}$ and $∣ t ∣^{2} = 4 Y_{1}^{2} / ∣ Y_{1} + Y_{2} ∣^{2}$ : indeed $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z ∣^{2} = z \overline{z}$ by [F2] applied to the quotients of step 2.1 give $∣ r ∣^{2} = ∣ Y_{1} - Y_{2} ∣^{2} ∣ Y_{1} + Y_{2} ∣^{- 2}$ and $∣ t ∣^{2} = ∣ 2 Y_{1} ∣^{2} ∣ Y_{1} + Y_{2} ∣^{- 2} = 4 Y_{1}^{2} ∣ Y_{1} + Y_{2} ∣^{- 2}$ because $(2 Y_{1}) \overline{(2 Y_{1})} = 4 Y_{1}^{2}$ for real $Y_{1}$ , and the denominator is nonzero since $Y_{1} + Y_{2} \neq 0$ .

4.1step 3.1F1F2F3algebra

[The flux identity] Assume $Y_{1} > 0$ real. By [F2] and [F3], $∣ Y_{1} + Y_{2} ∣^{2} > 0$ , so $∣ t ∣^{2} = 4 Y_{1}^{2} / ∣ Y_{1} + Y_{2} ∣^{2} > 0$ , and the identity of step 3.1 gives $∣ r ∣^{2} = ∣ Y_{1} - Y_{2} ∣^{2} / ∣ Y_{1} + Y_{2} ∣^{2} = 1 - 4 Y_{1} Re Y_{2} / ∣ Y_{1} + Y_{2} ∣^{2}$ ; adding $(Re Y_{2} / Y_{1}) ∣ t ∣^{2} = 4 Y_{1} Re Y_{2} / ∣ Y_{1} + Y_{2} ∣^{2}$ cancels the correction and yields $∣ r ∣^{2} + (Re Y_{2} / Y_{1}) ∣ t ∣^{2} = 1$ , the displayed identity and explicit formulas of (b).

5.1step 4.1F3F4algebra

[Sign analysis] Still with $Y_{1} > 0$ real: since $∣ t ∣^{2} > 0$ by step 4.1, the second summand $(Re Y_{2} / Y_{1}) ∣ t ∣^{2}$ has the sign of $Re Y_{2}$ by [F3], so it is nonnegative if and only if $Re Y_{2} \geq 0$ . If $Re Y_{2} < 0$ , then $∣ r ∣^{2} = 1 - 4 Y_{1} Re Y_{2} / ∣ Y_{1} + Y_{2} ∣^{2} > 1$ , and since $∣ r ∣ \geq 0$ and $1 \geq 0$ , [F4] gives $∣ r ∣ > 1$ ; in that case the identity of step 4.1 has a negative transmitted term and carries no energy-conservation reading, while $∣ r ∣ > 1$ alone is consistent with an active transmitted model.

5.2step 4.1step 3.1F2F3algebra

[Purely imaginary admittance] Assume $Y_{1} > 0$ real and $Y_{2} = i β$ with $β \in R$ . Then $\overline{Y_{1} + Y_{2}} = Y_{1} - i β = Y_{1} - Y_{2}$ by [F2], so $∣ Y_{1} - Y_{2} ∣ = ∣ \overline{Y_{1} + Y_{2}} ∣ = ∣ Y_{1} + Y_{2} ∣$ by [F2] and step 3.1 gives $∣ r ∣ = 1$ ; the identity of step 4.1 then gives $Re Y_{2} ∣ t ∣^{2} = 0$ (multiplying the identity by $Y_{1} > 0$ and using $∣ r ∣^{2} = 1$ ), while $∣ t ∣^{2} = 4 Y_{1}^{2} / ∣ Y_{1} + Y_{2} ∣^{2} = 4 Y_{1}^{2} / (Y_{1}^{2} + β^{2}) = 4 Y_{1}^{2} / (Y_{1}^{2} + ∣ Y_{2} ∣^{2}) > 0$ by [F2] and [F3]; hence $r \neq 0$ because $∣ r ∣ = 1$ and $t \neq 0$ because $∣ t ∣ > 0$ by the definiteness of the modulus in [F2]. This proves (c); no division by $Re Y_{2} = 0$ is performed, and the case $Y_{2} = 0$ is included with $β = 0$ .

5.3step 4.1step 2.1F1F5algebra

[Real-positive specialisation] Assume $Y_{1}, Y_{2} > 0$ real. Then $Y_{1} - Y_{2}$ is real and $Y_{1} + Y_{2} > 0$ is real and nonzero, so $r, t$ of step 2.1 are real by [F1], and step 4.1 gives $∣ r ∣^{2} + (Y_{2} / Y_{1}) ∣ t ∣^{2} = 1$ because $Re Y_{2} = Y_{2}$ . Substituting $Y_{j} = 1 / Z_{j}$ with $Z_{j} > 0$ real, $Y_{j}$ is invertible and $1 / Z_{1} \pm 1 / Z_{2} = (Z_{2} \pm Z_{1}) / (Z_{1} Z_{2})$ by [F1], so $r = (Z_{2} - Z_{1}) / (Z_{1} + Z_{2})$ and $t = 2 Z_{2} / (Z_{1} + Z_{2})$ by cancellation of the nonzero factor $Z_{1} Z_{2}$ ; moreover $Y_{2} / Y_{1} = Z_{1} / Z_{2}$ by [F1], so the balance is $R + T = 1$ for $R = ∣ r ∣^{2}$ and $T = (Z_{1} / Z_{2}) ∣ t ∣^{2}$ . These are exactly the solution formulas and the balance of [F5], so clause (d) is in agreement with The two-equation interface system at normal incidence.

6.1givenF1F5algebra∎

[Scope] Clause (e) holds: the entire argument is algebra in the field $C$ together with the ordered-field arithmetic of the real scalars $Y_{1}, Re Y_{2}$ ; no wave model, no labelling of the amplitudes and no interpretation of the second summand is used, and no choice principle occurs, all divisions being by explicitly nonzero field elements.

The two-equation interface admittance system: solution formulas and the normal-flux identity

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources