Regularization techniques for the high-frequency electromagnetic field coupling problem with terminated lines

Abstract

The problem of the high-frequency electromagnetic field coupling with a terminated transmission line is formulated and discussed in detail. Existence and uniqueness of the solution of the integro-differential equation describing the model are shown, pointing out, on the contrary, the not continuous dependence on the data. The problem is solved with three regularization methods (Landweber, Tikhonov and Truncated Singular Value Decomposition algorithms), and a comparative analysis is presented taking as reference the numerical results obtained using the Numerical Electromagnetic Code (NEC-2).

Appendix 1

1.1 The vector potential under thin wire approximation

The integral equation that governs the problem of radiation in the free space can be derived in the frequency domain from Maxwell’s Equations (1)–(4) (from now on, the subscript $s$ will be omitted for sake of simplicity, as here all the fields are scattered by definition).

We start observing that the magnetic field is always solenoidal; hence, it can be written as the $\mathrm {curl \,}$ of another arbitrary vector field $\mathbf{A}$, known as magnetic vector potential [16,  p. 256]:

$$\begin{aligned} \mu _0 \mathbf{H} = \mathrm {curl \,}\mathbf{A}. \end{aligned}$$

(83)

Substituting (83) into (1), we get

$$\begin{aligned} \mathrm {curl \,}(\mathbf{E} - \mathrm{i}\omega \mathbf{A}) = \mathbf{0}. \end{aligned}$$

(84)

Exploiting the identity $\mathrm {curl \,}(- \mathrm {grad \,}\Phi ) = \mathbf{0}$, we rewrite (84) as

$$\begin{aligned} \mathbf{E} = \mathrm{i}\omega \mathbf{A} - \mathrm {grad \,}\Phi , \end{aligned}$$

(85)

with $\Phi$ an arbitrary function, which is known as scalar potential. Since $\mathrm {curl \,}\mathrm {curl \,}\mathbf{A} = \mathrm {grad \,}\mathrm {div \,}\mathbf{A} - \mathrm {grad \,}^2 \mathbf{A}$, it can be easily deduced from (83) and (2) that

$$\begin{aligned} \mu _0\mathbf{J} - \mathrm{i}\omega \mu _0 \varepsilon _0 \mathbf{E}= \mathrm {grad \,}\mathrm {div \,}\mathbf{A} - \mathrm {grad \,}^2 \mathbf{A}. \end{aligned}$$

(86)

Using (85) into the above relation, we get

$$\begin{aligned} \mathrm {grad \,}^2 \mathbf{A} + k^2 \mathbf{A}= -\mu _0\mathbf{J} + \mathrm {grad \,}(\mathrm {div \,}\mathbf{A} - \mathrm{i}\omega \mu _0 \varepsilon _0 \Phi ). \end{aligned}$$

(87)

Recalling the Lorentz Gauge condition [16,  p. 257] that states that

$$\begin{aligned} \mathrm {div \,}\mathbf{A} = \mathrm{i}\omega \mu _0 \varepsilon _0 \Phi , \end{aligned}$$

(88)

relation (89) beomes

$$\begin{aligned} \mathrm {grad \,}^2 \mathbf{A} + k^2 \mathbf{A}= -\mu _0\mathbf{J}, \end{aligned}$$

(89)

which is an inhomogeneous vector Helmoltz equation for A. Equation (89), together with the radiation condition (see [28, Chap. 28, p. 188]) at infinity, admits a unique solution, which is given by

$$\begin{aligned} \mathbf{A}(\mathbf{r}) = \mu _0 \int _{\mathrm {supp \,}\mathbf{J}} \mathbf{J}(\mathbf{r}') \frac{\text {e}^{\mathrm{i}k R(\mathbf{r},\mathbf{r}')}}{4 \pi R(\mathbf{r},\mathbf{r}')} \, \mathrm{d}\mathbf{r}', \quad \mathbf{r}\in {\mathbb {R}}^3, \end{aligned}$$

(90)

where $\mathrm {supp \,}\mathbf{J} := \{ \mathbf{r}\in {\mathbb {R}}^3 \, : \, \mathbf{J}(\mathbf{r}) \ne \mathbf{0} \}$ denotes as usual the support of the currents $\mathbf{J}$, and where

$$\begin{aligned} R(\mathbf{r},\mathbf{r}')= \left| \mathbf{r}-\mathbf{r}' \right| \end{aligned}$$

(91)

with $\left| \, \cdot \, \right|$ represents the Euclidean distance in ${\mathbb {R}}^3$. The electric field can be deduced by using (85) and (88) as

$$\begin{aligned} \mathbf{E} = -\frac{1}{\mathrm{i}\omega \mu _0 \varepsilon _0} \left( \mathrm {grad \,}\mathrm {div \,}\mathbf{A} + k^2 \mathbf{A} \right) . \end{aligned}$$

(92)

In radiation problems with straight cylindrical wires with radius $a$ and length $2\ell$, it is common to assume that the current is distributed along the lateral surface of the wire itself [29,  p. 64]. So, once a cylindrical coordinate system $\{O,{\hat{\varvec{\rho }}},{\hat{\varvec{\theta }}},{\hat{\varvec{\xi }}} \}$ is introduced, such that $O$ coincides with the center of the wire and ${\hat{\varvec{\xi }}}$ is oriented as the wire axis, then $\mathrm {supp \,}\mathbf{J} = \{ (a, \theta , \xi ) \, : \, \theta \in [0,2\pi ], \ \xi \in [-\ell ,\ell ]\}$.

Whenever the radius $a$ is very small with respect its length $2\ell$ and the wavelength $\lambda = 2 \pi /k$, i.e., $a \ll \min \{ 2\ell , \, \lambda \}$, the so-called thin wire approximation (TWA) can be introduced [29, 30] (p. 63 and p. 123, respectively), which states that for all $\mathbf{r}' \in \mathrm {supp \,}\mathbf{J}$

$$\begin{aligned} \mathbf{J}(\mathbf{r}') = \frac{I(\xi )}{2\pi a} {\hat{\varvec{\xi }}}. \end{aligned}$$

(93)

Moreover, since the direction of the magnetic vector potential A in (90) is the same as $\mathbf{J}$, then

$$\begin{aligned} \mathbf{A}(\mathbf{r}) = A(\mathbf{r}) {\hat{\varvec{\xi }}} \quad \forall \mathbf{r}\in {\mathbb {R}}^3 {\setminus } \Omega , \end{aligned}$$

(94)

where the expression of the scalar function $A$ can be easily obtained by (90) as

$$\begin{aligned} A(\mathbf{r}) = \frac{\mu _0}{4\pi } \int _{\mathrm {supp \,}\mathbf{J}} \frac{I(\xi )}{2\pi a} \frac{\text {e}^{\mathrm{i}k R(\mathbf{r},\mathbf{r}')}}{R(\mathbf{r},\mathbf{r}')} \, \mathrm{d}\mathbf{r}', \end{aligned}$$

(95)

where $\mathbf{r}= (\rho , \theta , \xi )$ the observation point, $\mathbf{r}' = (\rho ',\theta ',\xi ')$ the source point, and (91) becomes

$$\begin{aligned} R(\mathbf{r},\mathbf{r}') = \sqrt{(\xi -\xi ')^2 + \rho ^2 + \rho '^2 + 2 \rho \rho ' \cos (\theta - \theta ') }. \end{aligned}$$

(96)

So, if one wants to evaluate $\mathbf{A}$ in a point $\mathbf{r}\in {\mathbb {R}}^3 \setminus \mathrm {supp \,}\mathbf{J}$, the fact that $a \ll \min \{ 2\ell , \, \lambda \}$ suggests that $\left| \mathbf{r}-\mathbf{r}' \right|$ can be approximated as the distance between the wire axis and the poind $\mathbf{r}$, i.e one can set $\rho ' =0$; hence, $R(\mathbf{r},\mathbf{r}') = \sqrt{(\xi -\xi ')^2 + \rho ^2}$. Vice versa, if one is interested in evaluating $\mathbf{A}$ in the points of the same wire that produces it, i.e., $\mathbf{r}\in \mathrm {supp \,}\mathbf{J}$, the TWA allows to locate the source points on the wire surface and the observation points along the wire axis [29,  p. 64], i.e., $\rho =0$ and $\rho ' = a$; hence, $R(\mathbf{r},\mathbf{r}') = \sqrt{(\xi -\xi ')^2 + a^2 }$.

If the following function is introduced

$$\begin{aligned} G: [0,+\infty )\times {\mathbb {R}}\ni (\rho ,\xi ) \mapsto {\left\{ \begin{array}{ll} \frac{\exp ({\mathrm{i}k \sqrt{\xi ^2 + \rho ^2})}}{ \sqrt{\xi ^2 + \rho ^2}} &{} \rho > a \\ \frac{\exp ({\mathrm{i}k \sqrt{\xi ^2 + a^2})}}{ \sqrt{\xi ^2 + a^2}} &{} \rho \le a \end{array}\right. } \in {\mathbb {C}}, \end{aligned}$$

(97)

then the vector potential $\mathbf{A}$ in (95) is approximated by

$$\begin{aligned} A(\rho ,\theta ,\xi ) = \frac{\mu _0}{4\pi } \int _{\xi ' = -\ell }^{\ell } I(\xi ') G(\rho ,\xi -\xi ') \, \mathrm{d}\xi '. \end{aligned}$$

(98)