Appendix 1: Proof that lossy transmission lines are realistic

In this appendix, we prove that lossy transmission lines with frequency independent parameters are realistic.

Lemma A.1: The impedance matrix $Z\!\left(s\right)$ and admittance matrix $Y\!\left(s\right)$ of a transmission line with positive real $Z_{l}\left(s\right)$ and $Y_{l}\left(s\right)$ are positive real.

Proof: Let $I_1$ and $I_2$ be the currents entering each terminal of the transmission line and $V_1$ and $V_2$ be the potentials at each terminal. Let us write

Replacing $\partial V/\partial x$ and $\partial I/\partial x$ by their values in terms of $I$ and $V$ deduced from Equation (4.1), we obtain:

therefore the real part of $P$, $\Re(P)$, is given by

Clearly the integrand is positive for $\Re(s)\ge0$, hence $\Re(V_{1}\,\bar{I_{1}}+V_{2}\,\bar{I_{2}})$ is positive when $\Re(s)\ge0$. This in turn implies that both $Y\!\left(s\right)$ and $Z\!\left(s\right)$ are positive real.

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Proof: Let $Z(s)$ be the impedance of the line shown in (4.4) and set $\lambda=\max\{-R/L,-G/C\}<0$. As $\gamma(s)$ and $z_{0}(s)$ are analytic and non-vanishing in the half-plane $\Pi_{\lambda}=\{s:\,\Re(s) >\lambda\}$, and since moreover $\gamma$ is never pure imaginary in $\Pi_{\lambda}$, the matrix $Z$ is analytic there. As $\det(Z(s))=z_{0}^{2}$, we get that $Y=Z^{-1}$ is likewise analytic. From (A.1) and the symmetry of $Z$, $Y$, we observe that $Z(s)+Z^{*}(s)=2\Re(Z(s))$ and $Y(s)+Y^{*}(s)=2\Re(Y(s))$ are positive symmetric matrices in $\Pi_{\lambda}$ whose entries, being real parts of analytic functions, are harmonic functions of $s$. They are in fact positive definite at each $s\in\Pi_{\lambda}$, as (A.1) implies if $I_{1}=x_{1}+iy_{1}$ and $I_{2}=x_{2}+iy_{2}$ are not both zero that

(for in this case $\Re(Z_l(s))>0$ and $I(\xi)$ is not identically zero). We claim that

and this will imply that $(i)$ of Definition 3.1 is met. Indeed, if (A.2) holds and $u\in\R^{2}$ is a unit vector, the function $-u^{t}\Re(Y)u$ is a nonpositive harmonic function in $\Pi_{0}$ whose limit at every point of the imaginary axis exists and is at most $-\alpha$, therefore $-u^{t}\Re(Y)u$ is at most $-\alpha$ everywhere in $\Pi_{0}$ by the extended maximum principle [37]. To prove (A.2), we may restrict to large $|\omega|$ because on any compact interval of the imaginary axis it certainly holds by strict positivity of $\Re(Y(j\omega))$ and continuity of the latter with respect to $\omega$. Without loss of generality, we choose the branch of the square root which is positive for positive arguments. In view of (4.4), we can write:

Since $z_{0}(j\omega)\to\sqrt{L/C}>0$ when $\omega\to\pm\infty$ and the matrix $M$ in (A.3) is symmetric, it is enough to show that $M(j\omega)$ is bounded and that the eigenvalues of $\Re(M(j\omega))$ are greater than some $\delta>0$ for $|\omega|$ large enough. Now, using the Taylor expansion of $(1+x)^{1/2}$ at $x\sim0$, we get

so that $\Re(\gamma(j\omega))\geq\kappa=\frac{RC+GL}{3\sqrt{LC}}>0$, say, for $|\omega|$ large enough. Then $|e^{-2\gamma(j\omega)}|\leq e^{-\kappa}<1$, hence $M(j\omega)$ is bounded. To prove that the eigenvalues of $\Re(M(j\omega))$ are greater than $\delta>0$, we check that its trace and determinant are positive with $\det\big(\Re(M(j\omega))\big)\geq\eta>0$ for $|\omega|$ large. Since the trace is bounded (recall $M(j\omega)$ is bounded), we will be done. First, it is readily seen that

is indeed positive. Next, a short computation yields that

So, we are left to prove that

is positive and bounded away from $0$ for $|\omega|$ large enough. As $0<\!e^{-\Re(\gamma(j\omega))}\!\!<\!e^{-\kappa}\!\!<\!1$, this comes from the fact that

is strictly positive for $x\in(0,1)$. The same argument with $z_{0}^{-1}$ replaced by $z_{0}$ in (A.3) gives us $(ii)$ of Definition 3.1.

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