# Transient Response of Thin Wire above a Layered Half-Space Using TDIE/FDTD Hybrid Method

Authors: Bing Wei; Qiong He; Jie Li; Ren-xian Li; Li-xin Guo
Journal: International Journal of Antennas and Propagation
 (2012)

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## Abstract

The TDIE/FDTD hybrid method is applied to calculate the transient responses of thin wire above a lossy layered half-space. The time-domain reflection of the layered half space is computed by one-dimensional modified FDTD method. Then, transient response of thin wire induced by two excitation sources (the incident wave and reflected wave) is calculated by TDIE method. Finally numerical results are given to illustrate the feasibility and high efficiency of the presented scheme.


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## Body

## 1. Introduction

Among the available literatures when analyzing transient response of thin-wire above a half-space, the scattering and radiation of thin-wire structure are generally analyzed by employing time-domain integral equation (TDIE) method [1–8] or finite element method (FEM) [9], and the influence of half-space is often considered by introducing the imaging principle, as well as reflection coefficient [3]. In 1980, Parviz Parhami et al. [10] derived the general integral equation for an arbitrarily shaped thin-wire antenna over a lossy half-space, and method of moment (MoM) in frequency domain is employed to solve the equation numerically. In his paper, far-field radiation patterns of center-fed horizontal dipole, center-fed vertical dipole, and center-fed inverted dipole are given.  In 1998, Poljak established space-time integral equation of Hallen’s type to deal with a straight thin wire horizontally placed above a dissipative half-space. The influence of a lossy half-space is taken into account by the Fresnel space-time reflection coefficient which appears inside the IE kernel [4]. In 2004, he calculated the transient responses of nonlinear loaded wire antenna in half-space with spatial-time Hallen equation combined with reflection coefficient method [5]. Recently, Haddad et al. used complex-time Green’s functions to obtain the transient response of thin-wire structures located above half-space [6].The above literature deals only with the interactions between thin wire and homogeneous or one-layer half-space. In many practical cases, the half-space is composed of complex dielectrics, which is usually in the form of layers. In this paper, a hybrid method that combines TDIE with FDTD is employed to study the transient responses of thin conducting wire above a lossy and layered half-space. It is well known that TDIE method is suited to simulate the scattering or radiation from thin-wire structures, whereas FDTD is a powerful tool that could model the interactions of EM waves with inhomogeneous media [11–29]. Huang et al. used the FDTD/MoM hybrid technique for modeling the radiation field of complex antennas above the heterogeneous grounds [26]. Monorchio et al. used the hybrid time-domain technique that combines the finite element, finite difference, and method of moment techniques to compute the radiation field of a thin-wire antenna near inhomogeneous dielectric bodies [27]. Hybrid method in [26, 27] uses iteration-based technique to couple 3-D FDTD(/FETD) and TDIE. The hybrid method in this paper is connected with one-reflection field (neglecting higher order reflections when the distance of the wire and the interface of the layered half-space are far enough for the problems we interested) without any iteration-based procedure to consider the multiple interactions between wire structure and underlying layered half-space. To calculate first-order reflected field, only 1-D FDTD is needed. In this case the original problem is decomposed into two subregions. The first sub-region is single-wire structure in free space, while the second sub-region is the layered dielectric without any wire structure above it. Then transient response on the wire is analyzed using TDIE [1, 2], which has two exciting terms in the formulation, both the original incidence and the reflected wave. This hybrid algorithm utilizes advantages of both TDIE and FDTD, respectively, and furthermore no iteration is needed. Numerical examples show that this hybrid method is a very efficient way to study transient responses of wire structures above a layered half-space.


## 2. Basic Theory of Hybrid Method

The geometry of a straight thin conducting wire above multilayer dielectric media is shown in Figure1. The thin wire is parallel to the y axis, the length of wire is l, and the distance between the wire and interface is h. The angle between the incident plane and the surface of the half-space is θ. The angle between the incident wave and the thin wire is φ. In this paper, the plane of incidence is determined by the incident vector ki⇀ and the direction of the thin wire.Figure 1
The geometry of a thin wire above layered half-space.When one analyzes the transient response of the wire in Figure1, the excitation sources include three parts: (1) the original incident wave (zero-order), (2) the reflected wave reflected by layered half-space from the original incident wave (first-reflection), and (3) the radiated field from the induced current which is reflected by the half-space and becomes incident upon the wire again (second- and higher-order reflections).The radiation field of thin wire can be regarded as superposition of the radiation field of many electric dipoles which are parallel to the interface. Using the spherical coordinate system, the radiation electric field of an electric dipole can be written as [30](1)E⇀=-iωIlexp⁡⁡(ikr)4πr{r̂(ikr+(ikr)2)2cos⁡⁡θ′+θ′̂(1+ikr+(ikr)2)sin⁡⁡θ′},
where i, ω, I, l, k, r, r̂, θ′, and θ′̂ are units of the imaginary number, circular frequency of the incident wave, electric current of the wire, length of the dipole, spatial frequency, the distance of the origin point to observation point, the orientation vector, the angle between the r̂ and the z′̂ direction, and the unit vector of θ′ direction, respectively, as illustrated in Figure 2. Equation (1) gives the radiation field of an electric dipole, which is suitable for both near field and far field. The radiation field of the dipole toward all directions in which the field is perpendicularly incident towards the interface of layered half-space can be reflected and becomes incident upon the dipole again (θ′=π/2). In this case, (1) can be written as(2)Eθ=ωIlexp⁡⁡(ikr)4πr[1kr+i(1(kr)2-1)].Figure 2
The electric dipole.To make sure the existence of higher-order reflection does not ruin the hybridization of TDIE and FDTD, the electric field that is radiated by the dipole and reflected by half-space and arriving at the dipole again is estimated. Suppose we have a one-meter-long wire composed of four hundred electric dipoles and the direction of the wire is parallel to the interface. LetE0, E1, and E2 represent the radiation electric field of a dipole at r0=0.002 m (the radius of the wire), r1=0.25 m and r2=2.0 m, respectively. According to (2) we know that the ratios of E1 to E0 and E2 to E0 basically do not change when the frequency varies from 1 MHz to 20 GHz. Suppose the observation point is j and the source point is i (i=1~400). The radiation electric fields of dipole jreflected by the ground half-space and reaching dipole j again can be estimated by (2). The radiation electric fields of dipole i (i≠j) reflected by the ground half-space and reaching dipole j again are less than the reflection field which is radiated by the dipole j. So the maximum reflected fields of point j on the surface of the thin wire are about(3)400×E1|0.25m≈1.39×10-3R×E0,400×E1|2m≈2.5×10-4R×E0,
where R<1 is the reflection coefficient of the layered half-space and the subscripts 0.25 m and 2 m indicated the wire is located 0.25 meter or 2.0 meters above the interface of the half-space, respectively. It is obvious that in these two cases the contribution of the radiation field reflected by the current is much smaller than original incident field. So high-order interactions between thin wire and half-space could be neglected in this paper.When one applies the hybrid scheme to analyze the transient responses of wire structure above layered half-space, TDIE is employed to study the above wire with both the original incident and reflected wave as the exciting source. In the hybrid formulations, considering thethinness nature of wire structure, the incident wave across the section of the conducting wire approximately has the same value, and the induced current on the wire is considered as line current. In forming the integral equation, the source point will be on the axis of the cylinder, whereas the observation point is positioned at the conductor surface. Therefore, the distance between the source point and observation point is always larger than or equal to the radius of the wire.Compared with the TDIE method in free space, the TDIE method in the hybrid approach deals with two kinds of incident waves. The first is the original incident wave, which is introduced in analytical form, while the other exciting source is from the reflected wave of the underlying layered half-space, which can be calculated using one-dimensional modified FDTD method. The configuration of the half-space can be arbitrary, and the dielectric parameters can be varied layer by layer.The FDTD model is given in Figure3. In order to obtain the reflected wave where the conducting wire is positioned, the location of output point in the scattering field (SF) region of FDTD domain is as high as the wire, whereas the half-space is modeled in the total field region. In the implementation of FDTD, UPML is used to truncate the infinite domain and reduce numerical errors.Figure 3
The FDTD model.The general formulations of FDTD and TDIE methods will be given in the following.


### 2.1. FDTD Method

Suppose a straight thin wire structure is located above the layered half-space as Figure1 shows. The electric fields have components along y and z axis when the model is impinged by oblique incident wave. The y-component can stimulate induced current whereas the z-component cannot.In FDTD method, a set of finite-difference equations for the time-dependent Maxwell’s curl equations system is originated by Yee [13]. These equations can be represented in a discrete form, both in space and time, employing the second-order accurate central difference formula.Supposing the incident angle isβ (the angle of the interface of the layered half-space and the ki⇀  as Figure 3 shows), the parameters of media and the field quantity are independent and denoted byy and z, and the modified Maxwell curl equations of one-dimensional case are(4)-∂Ex1D∂z=μ∂Hy1D∂t,(5)-1sin⁡2⁡β∂Hy1D∂z=ε∂Ex1D∂t,
where μ=μ0μr and ε=ε0εr are permeability and permittivity of the media, respectively. Equations (4) and (5) can be employed to compute the reflection and transmission electricmagnetic wave in case of an oblique plane wave incident to the surface of the layered half-space.Consideringky=kcosβ and kz=ksinβ, (5) can be written as(6)-k2kz2∂Hy1D∂z=εjωEx1D.
According to the phase match theory, ky=kcosβ is a constant in each layer. Equation (6) can be written as(7)-k2∂Hy1D∂z=(k2-ky2)jωε0εrEx1D,
in which k=ωμε, kz2=k2-ky2. Then we have(8)-∂Hy1D∂z=[εr-ε1cos⁡2⁡β]ε0jωEx1D.
Let ε′=εr-ε1sin2β, and (8) can be written as(9)-∂Hy1D∂z=ε0ε′∂Ex1D∂t.
The derivatives in (4) and (9) can be approximated by using the central difference formula with the position Ex(m) being the center point for the central difference formula in space and time instant (n+1/2)Δt being the center point in time. We can get FDTD updating equation as follows:(10)Ex1D|jn+1=CA(m)⋅Ex1D|jn-CB(m)⋅Hy1D|j+1/2n+1/2-Hy1D|j-1/2n+1/2Δz,Hy1D|j+1/2n+1/2=CP(m)⋅Hy1D|j+1/2n-1/2-CQ(m)⋅[Ex1D|j+1n-Ex1D|jnΔz],
where CA(m), CB(m), CP(m), and CQ(m) are coefficient of the updating equation.In this paper, in order to obtain the reflected wave of the layered half-space, the layered half-space is modeled in the total field (TF) region, and the output point of the reflection waves is in the scattering field (SF) region of FDTD domain. UPML absorbing conduction is used to truncate the infinite domain and reduce numerical errors.


### 2.2. TDIE Method

The scattering electric field of the thin wire can be expressed by the vector potential as follows:(11)∂E⃗∂t=-∂2A⃗∂t2+c2∇(∇⋅A⃗).The directed incident fieldExi and reflected field Exr can be regarded as a constant across the section of wire. Applying the boundary condition for the total electric field, (11) can be written as(12)∂2Ax∂x2-1c2∂2Ax∂t2|r=a=-1c2∂Exi∂t|r=a-1c2∂Exr∂t|r=a,x∈(0,L).
The vector potential is given by(13)Ax(x,t)=μ04π∫x′=0LI(x′,t-|x-x′|/c)|x-x′|2+a2dx′,
where a and c represent the radius of thin wire, and velocity of light, respectively.The thin wire is divided intoN+1 equal segments; the length of the segment is Δx (as Figure 4 shows). The basis function is defined as follows:(14)fm(x)={1,xm-Δx2≤x≤xm+Δx20,otherwise.Figure 4
The thin wire is divided intoN+1 subfield.Using these expansion functions, we approximate the currentI as follows:(15)I(x,t)≈∑k=1nIk(t)fk(x).
If central difference approximation is employed, (12) can be written as(16)Am+1,n-2Am,n+Am-1,n(Δx)2-Am,n+1-2Am,n+Am,n-1(cΔt)2=-Fm,n,
where the excitation term is(17)Fm,n=∂Exi(xm,tn)∂t+∂Exr(xm,tn)∂t.
Substituting (17) into (16), we have(18)Im,nκm,m=-Am,n+2Am,n-1-Am,n-2+(Δt)2Fm,n-1+(cΔtΔx)2[Am+1,n-1-2Am,n-1+Am-1,n-1],
where(19)Am,n=Im(tn)κm,m+∑k=1k≠mNIk(tn-|xm-xk|c)κm,k,Am,n=∑k=1k≠mNIk(tn-|xm-xk|c)κm,k,Im,n≡Im(tn),
where κm,k is impedance coefficient matrix. The algorithm may be started by assuming Im,0=Im,1=0 and calculating Im,2 using (18). Once we obtain Im,2, coupled with the knowledge of Im,0 and Im,1, we proceed to calculate Im,3 again using (18). This procedure can be continued to calculate currents at successive time instants t4,t5,…until the transient currents die down.


## 2.1. FDTD Method

Suppose a straight thin wire structure is located above the layered half-space as Figure1 shows. The electric fields have components along y and z axis when the model is impinged by oblique incident wave. The y-component can stimulate induced current whereas the z-component cannot.In FDTD method, a set of finite-difference equations for the time-dependent Maxwell’s curl equations system is originated by Yee [13]. These equations can be represented in a discrete form, both in space and time, employing the second-order accurate central difference formula.Supposing the incident angle isβ (the angle of the interface of the layered half-space and the ki⇀  as Figure 3 shows), the parameters of media and the field quantity are independent and denoted byy and z, and the modified Maxwell curl equations of one-dimensional case are(4)-∂Ex1D∂z=μ∂Hy1D∂t,(5)-1sin⁡2⁡β∂Hy1D∂z=ε∂Ex1D∂t,
where μ=μ0μr and ε=ε0εr are permeability and permittivity of the media, respectively. Equations (4) and (5) can be employed to compute the reflection and transmission electricmagnetic wave in case of an oblique plane wave incident to the surface of the layered half-space.Consideringky=kcosβ and kz=ksinβ, (5) can be written as(6)-k2kz2∂Hy1D∂z=εjωEx1D.
According to the phase match theory, ky=kcosβ is a constant in each layer. Equation (6) can be written as(7)-k2∂Hy1D∂z=(k2-ky2)jωε0εrEx1D,
in which k=ωμε, kz2=k2-ky2. Then we have(8)-∂Hy1D∂z=[εr-ε1cos⁡2⁡β]ε0jωEx1D.
Let ε′=εr-ε1sin2β, and (8) can be written as(9)-∂Hy1D∂z=ε0ε′∂Ex1D∂t.
The derivatives in (4) and (9) can be approximated by using the central difference formula with the position Ex(m) being the center point for the central difference formula in space and time instant (n+1/2)Δt being the center point in time. We can get FDTD updating equation as follows:(10)Ex1D|jn+1=CA(m)⋅Ex1D|jn-CB(m)⋅Hy1D|j+1/2n+1/2-Hy1D|j-1/2n+1/2Δz,Hy1D|j+1/2n+1/2=CP(m)⋅Hy1D|j+1/2n-1/2-CQ(m)⋅[Ex1D|j+1n-Ex1D|jnΔz],
where CA(m), CB(m), CP(m), and CQ(m) are coefficient of the updating equation.In this paper, in order to obtain the reflected wave of the layered half-space, the layered half-space is modeled in the total field (TF) region, and the output point of the reflection waves is in the scattering field (SF) region of FDTD domain. UPML absorbing conduction is used to truncate the infinite domain and reduce numerical errors.


## 2.2. TDIE Method

The scattering electric field of the thin wire can be expressed by the vector potential as follows:(11)∂E⃗∂t=-∂2A⃗∂t2+c2∇(∇⋅A⃗).The directed incident fieldExi and reflected field Exr can be regarded as a constant across the section of wire. Applying the boundary condition for the total electric field, (11) can be written as(12)∂2Ax∂x2-1c2∂2Ax∂t2|r=a=-1c2∂Exi∂t|r=a-1c2∂Exr∂t|r=a,x∈(0,L).
The vector potential is given by(13)Ax(x,t)=μ04π∫x′=0LI(x′,t-|x-x′|/c)|x-x′|2+a2dx′,
where a and c represent the radius of thin wire, and velocity of light, respectively.The thin wire is divided intoN+1 equal segments; the length of the segment is Δx (as Figure 4 shows). The basis function is defined as follows:(14)fm(x)={1,xm-Δx2≤x≤xm+Δx20,otherwise.Figure 4
The thin wire is divided intoN+1 subfield.Using these expansion functions, we approximate the currentI as follows:(15)I(x,t)≈∑k=1nIk(t)fk(x).
If central difference approximation is employed, (12) can be written as(16)Am+1,n-2Am,n+Am-1,n(Δx)2-Am,n+1-2Am,n+Am,n-1(cΔt)2=-Fm,n,
where the excitation term is(17)Fm,n=∂Exi(xm,tn)∂t+∂Exr(xm,tn)∂t.
Substituting (17) into (16), we have(18)Im,nκm,m=-Am,n+2Am,n-1-Am,n-2+(Δt)2Fm,n-1+(cΔtΔx)2[Am+1,n-1-2Am,n-1+Am-1,n-1],
where(19)Am,n=Im(tn)κm,m+∑k=1k≠mNIk(tn-|xm-xk|c)κm,k,Am,n=∑k=1k≠mNIk(tn-|xm-xk|c)κm,k,Im,n≡Im(tn),
where κm,k is impedance coefficient matrix. The algorithm may be started by assuming Im,0=Im,1=0 and calculating Im,2 using (18). Once we obtain Im,2, coupled with the knowledge of Im,0 and Im,1, we proceed to calculate Im,3 again using (18). This procedure can be continued to calculate currents at successive time instants t4,t5,…until the transient currents die down.


## 3. Numerical Results

In this section, numerical examples are given to verify the accuracy of the presented algorithm. Then, the transient responses of the straight thin wires above different layered half-spaces are analyzed. In all examples of this section, the length of thin wire isl=1 m, and the time step is the same in both TDIE and FDTD methods, which make it convenient for calculation in hybrid method.Figure5 gives the transient current (solid line) induced at the midpoint of this thin wire, which is at the height of h=0.25 m above the interface. The underlying half-space is homogeneous with relative permittivity εr=10, and the Gaussian pulse E=E0exp(-g2(t-t0)2) (where E0=1 V/m, g=4.0×109 s−1, t0=1.2×10-9 s) is incident perpendicularly to the interface of the half-space (φ=90°, θ=0°). The polarizing direction of the electric field is along y axis. The results of [5] (dot) are also given for comparison in Figure 5. It is obvious that the results obtained by hybrid method are in good agreement with the results of [5]. This illustrates the correctness of the presented scheme.Figure 5
Current induced at midpoint of a straight thin wire above half-space.The second example gives the comparison between directly incident wave and the reflected wave by the half-space. The radiusa=0.002 m, and the height h=0.5 m. It is illuminated by a Gaussian pulse (E0=1 V/m, g=2.5×109 s−1), t0=1.2×10-9 s). The half-space is a nonmagnetic medium with εr=4 and σ=10-5 s/m. Figure 6 presents the incident wave and reflected wave when φ=90° and the angle between the incident plane and the surface of the half-space is changed. The dash line, solid line, and dash dot line represent incident angle θ=30°, θ=45°, and θ=60°, respectively. It can be seen from Figure 6 that the amplitude of reflected wave is less than the original incident wave. There is also obvious time delay compared with the original incident wave.Figure 6
The incident and reflected wave on the height of thin wire with different incident angles.Figure7 plots the induced currents at the center of thin wire varying with time. The meaning of dashed line, solid line, and dash-dot line is same as above. As shown in the figure, the transient responses of different incident angles are the same at the early time and afterwards have obvious difference.Figure 7
The transient current induced at midpoint of thin wire above half-space with different incident angles.Figure8 gives the reflected wave of the height of thin wire located above the layered half space with different incident angle. The reflected electric fields have multiple peaks, which is due to the presence of three interfaces and the interactions between the layers.Figure 8
The reflected wave on the height of thin wire located above the half-space with different incident angle.Then let us consider the transient response of a thin wire above layered half-space. As the configuration shown in Figure1 exhibits, the thin wire is located at the height h=0.5 m above the interface. The multilayer medium is composed as follows: the first layer is dry soil (εr=4, σ=10-5 s/m, thickness d1=1.0 m); the second layer is wet soil (εr=10, σ=10-3 s/m, thickness d2=1.0m); the third layer is ground water (εr=81, σ=10-3 s/m). The length and the radius of the thin wire are the same as in the above example. The incident wave is the Guassain pulse (E0=1V/m, g=2.5×109 s−1, t0=0.86×10-9 s), which is oblique to the layered interface. Figure 7 gives the reflected wave of the height of thin wire located above the layered half-space with different incident angles. The reflected electric fields have multiple peaks, which is due to the presence of three interfaces and the interactions between the layers.Figure9 gives the transient currents induced at the center of thin wire with different incident angles. The meaning of dash line, solid line, and dash-dot line is the same as in the above example. As shown in this figure, the transient responses are thesame at the early time and show obvious difference at latter time. This is because the effect of the interfaces reaches the wire at the latter time.Figure 9
The current induced at midpoint of thin wire with different incident angle.The last example is about the transient responses of thin wire above layered half-space when the incident angleφ and θ are changed. The background and the location of wire are the same as in the above example. Figure 10(a) plots the induced current at the center of thin wire when the incident wave vector is at the plane which is perpendicular to the interface of the layered half-space (θ=90°). The dash line, solid line, and dash dot line represent incident angle φ=30°, φ=45°, and φ=60°, respectively. Figure 10(b) plots the induced current at the center of thin wire when θ=45°. Also, the dash line, solid line, and dash dot line represent incident angles φ=30°, φ=45°, and φ=60°, respectively. We can see from these two figures, the amplitudes of the current are different and the fluctuation time periods are obviously different when different incidence angles are used.The current induced at midpoint of thin wire with different incident angle.
(a)(b)


## 4. Conclusion

The TDIE/FDTD hybrid method is efficient for commutating the transient responses of thin wire above layered half-space in the case of the EM oblique incident wave. One-dimensional TDIE is applied to study the above wire structure whereas one-dimensional modified FDTD method is used to get reflected field of the layered half-space. This can save computing time and memory, so the presented algorithm consumes less memory, offers high speed of computing, and is a highly efficient numerical solution.



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