# Multiresolution Time-Domain Scheme for Terminal Response of Two-Conductor Transmission Lines

Authors: Zongliang Tong; Lei Sun; Ying Li; Jianshu Luo
Journal: Mathematical Problems in Engineering
 (2016)

---

## Abstract

This paper derives a multiresolution time-domain (MRTD) scheme for the two-conductor lossless transmission line equations based on Daubechies’ scaling functions. And a method is proposed to generate the scheme at the terminal and near the terminal of the lines. The stability and numerical dispersion of this scheme are studied, and the proposed scheme shows a better dispersion property than the conventional FDTD method. Then the MRTD scheme is extended to the two-conductor lossy transmission line equations. The MRTD scheme is implemented with different basis functions for both lossless and lossy transmission lines. Numerical results show that the MRTD schemes which use the scaling functions with high vanishing moment obtain more accurate results.


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## Body

## 1. Introduction

The multiresolution time-domain (MRTD) scheme proposed by [1] provides an efficient algorithm for electromagnetic field computation and shows excellent capability to approximate exact solution with low sampling rates. However, the Battle-Lemarie wavelet function used in [1] is not compact supported, which means the iterative equations contain infinite terms. We must cut off the iterative equations in the actual computation and this may introduce truncation errors. So different wavelet bases, which are compact supported with some numbers of vanishing moments, have been used to improve this method [2–5]. This makes a great development for MRTD schemes. As a kind of numerical method, the MRTD schemes show great advantages in numerical dispersion properties [6–9]; meanwhile, these schemes need a more rigorous stable condition than the conventional FDTD method [10]. For containing more terms in the iterative equations, the terminal conditions or absorbing boundary conditions are more complicated to process in MRTD schemes; this disadvantage has limited the application of the MRTD scheme. To overcome this limitation, some works on the perfect match layer have been made [11–13]; however, other terminal conditions also need to be analyzed specifically. For the transmission lines equations, the resistive terminal conditions could be equivalent as a general Thevenin circuit; this paper will solve this kind of terminal condition in the MRTD scheme.Since the appearance of the telegraph equations, studies on transmission lines have had a considerable development. Several equivalent forms of transmission line theory have been proposed to describe the influence of the incident electromagnetic field to the transmission lines [14–16]. In [17], the classical theory of the transmission line has been summarized and the theory on the high frequency radiation effects to the transmission lines is introduced. In the monograph [18], the multiconductor transmission lines (MTL) theory has been comprehensively studied in detail. For the two-conductor lossless transmission lines, there are several methods, which contain the series solution, the SPICE solution, the time-domain to frequency-domain (TDFD) transformation method, and the FDTD method [18]. However, the MRTD scheme has not been used to calculate the terminal response of transmission lines. In this paper we will derive a MRTD scheme for this problem.In this paper, we focus on the calculation of the terminal response of two-conductor transmission lines equations by using MRTD scheme. In Section2, the MRTD scheme is derived based on Daubechies’ scaling functions for the two-conductor lossless transmission line equations, and, for the resistive terminations, the iterative equations for the terminal voltages are derived, a method is proposed to update the iterative equations which contain some terms whose indices exceed the index range in the MRTD scheme, and then the stability and the numerical dispersion are studied. In Section 3, the MRTD scheme is extended to the two-conductor lossy transmission line. In Section 4, the numerical results are presented on the terminal response of both lossless and lossy transmission lines using the MRTD scheme and compared to the FDTD method at different space discretization numbers and different Courant numbers.


## 2. MRTD Scheme for Two-Conductor Lossless Transmission Lines

### 2.1. MRTD Formulation

In this section, the MRTD scheme is applied to the following scalar transmission lines equations for two-conductor lossless lines [18]: (1a)∂Vz,t∂z+l∂Iz,t∂t=0(1b)∂Iz,t∂z+c∂Vz,t∂t=0, where l and c are the per-unit-length inductance and capacitance, respectively.Based on the method outlined in [1], the voltage and current can be expanded as follows: (2a)Vz,t=∑k,n=-∞+∞Vknϕkzhnt(2b)Iz,t=∑k,n=-∞+∞Ik+1/2n+1/2ϕk+1/2zhn+1/2t, where Vkn and Ik+1/2n+1/2 are the coefficients for the voltages and currents in terms of scaling functions, respectively. The indices n and k are the discrete spatial and temporal indices related to space and time coordinates via z=kΔz and t=nΔt, where Δz and Δt represent the spatial and temporal discretization intervals in z and t direction. The function hn(t) is defined as(3)hnt=htΔt-n with the rectangular pulse function(4)ht=1fort<1212fort=120fort>12.The functionϕk(z) is defined as(5)ϕkz=ϕzΔz-k, where ϕ(z) represents Daubechies’ scaling function. Figure 1 shows Daubechies’ scaling function with two vanishing moments.Figure 1
Daubechies’ scaling function with two vanishing moments.For deriving the MRTD scheme for (1a) and (1b), we need the following integrals:(6)∫-∞+∞hnthn′tdt=δn,n′Δt∫-∞+∞hnt∂hn′+1/2t∂tdt=δn,n′-δn,n′+1, where δn,n′ represents the Kronecker symbol. Consider(7)∫-∞+∞ϕkzϕk′zdz=δk,k′Δz(8)∫-∞+∞ϕnz∂ϕn′+1/2z∂zdz=∑i=-LSLS-1aiδn+i,n′,where LS denotes the effective support size of the basis functions. The coefficients a(i) are called connection coefficients and can be calculated by (9). Taking Daubechies’ scaling functions as the basis functions, Table 1 shows a(i) for 0≤i≤LS-1, which are zeros for i>LS-1, and for i<0 it can be obtained by the symmetry relation a(-1-i)=-a(i),(9)ai=1π∫0∞ϕ^λ2λsin⁡λi+12dλ, where ϕ^(λ) represents the Fourier transform of ϕ(z).Table 1
Connection coefficientsa(i) and the first-order moments M1 of Daubechies’ scaling functions.
i
D
2
D
3
D
4
0
1.2291666667
1.2918129281
1.3110340773
1
−0.0937500000
−0.1371343465
−0.1560100110
2
0.0104166667
0.0287617728
0.0419957460
3
−0.0034701413
−0.0086543236
4
0.0000080265
0.0008308695
5
0.0000108999
6
0.0000000041
M
1
0.6339743121
0.8174005815
1.0053923835
q
m
a
x
0.7500
0.6844
0.6585Daubechies’ scaling functions satisfy the shifted interpolation property [19](10)ϕi+M1=δi,0 for i integer, where M1=∫-∞+∞zϕ(z)dz is the first moment of the scaling functions and the values of M1 are listed in Table 1.Following the theory in [3] and making use of (10), (5) is modified to(11)ϕkz=ϕzΔz-k+M1. In spite of the support of the scaling functions [20], single-point sampling of the total voltages and currents can be taken at integer points with negligible error. Taking voltage at spatial point kΔz and at time nΔt, we obtain(12)VkΔz,nΔt=∬-∞+∞Vz,tδzΔz-kδtΔt-ndzdt=Vkn, where δ is the Dirac delta function. Equation (12) means the voltage value at each integer point is equal to the coefficient. The current values have the same character at each half integer point. Therefore, we will use Vkn and Ik+1/2n+1/2 directly to represent the voltage at the point (kΔz,nΔt) and the current at the point ((k+1/2)Δz,(n+1/2)Δt) in this paper.The modifiedϕk(z) in (11) also satisfy integrals (7) and (8). Applying the Galerkin technique to (1a) and (1b), we can obtain the following iterative equations for the voltages and currents: (13a)Vkn+1=Vkn-ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(13b)Ik+1/2n+1/2=Ik+1/2n-1/2-ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in.


### 2.2. Terminal Iterative Equations for Resistive Load in MRTD Scheme

We will consider the terminal conditions for the two-conductor lossless transmission lines equations in this section. Equations (1a) and (1b) are homogeneous linear equations; we need to add the terminal conditions to obtain the unique solution.Considering the two-conductor lines shown in Figure2, we assume the length of the total line is L and the resistive loads are RS and RL. The line is divided uniformly into NDZ segments with the space interval Δz and the total solution time is divided into NDT steps with the uniform time interval Δt. Similar to the conventional FDTD, we will calculate the interlace voltages, V0n,V1n,…,VNDZn, and currents, I1/2n+1/2,I3/2n+1/2,…,INDZ-1/2n+1/2, in both space domain and time-domain as shown in (13a) and (13b), for n=1,2,…,NDT.Figure 2
A two-conductor line in time-domain.For the resistive terminations, we note the voltage at the source(z=0) as VS(t) and the current at the source as IS(t), the external voltage at the load (z=L) as VL(t), and the current at the load as IL(t). The discrete voltages and currents at the source are denoted as VSn≡VS(nΔt) and ISn≡IS(nΔt),  and the discrete voltages and currents at the load are denoted as VLn≡VL(nΔt), and ILn≡IL(nΔt), then the terminal characterizations could be written in terms of a generalized Thevenin equivalent as  (14a)V0n=VSn-RSISn(14b)VNDZn=VLn+RLILn. Equations (14a) and (14b) denote the discretization terminal conditions for the case of resistive terminations, so we need to introduce these conditions to the iterative equations (13a) and (13b) to obtain the numerical solution.Notice that, in the iterative equations (13a) and (13b), not only the iterative equations of the terminal voltages V0n+1 and VNDZn+1 should be derived and the iterative equations of voltages and currents “near" the terminals also need to be updated. The voltages and currents “near" the terminals we mean are the voltages Vin+1 and VNDZ-in for i=1,2,…,LS-1, and the currents Ii+1/2n+1/2 and INDZ-i+1/2n+1/2 for i=0,1,…,LS-2. All of these voltages and currents contain some terms that exceed the index range in iterative equations (13a) and (13b). Figure 3 shows the discretization of the terminal voltages and the voltages and currents near the terminal.Figure 3
Discretizing the terminal voltages and currents.We will derive the MRTD scheme at the terminal firstly. For updating the iterative equations for the terminal voltages, we need to decompose iterative equations ((13a), (13b)). Since the coefficients a(i) satisfy the following relation [4](15)∑i=0LS-12i+1ai=1 substituting (15) into (13a), we can obtain(16)∑i=0LS-1ai2i+1Vkn+1=∑i=0LS-1ai2i+1Vkn-∑i=0LS-1Δt2i+1Δzc-1ai2i+1Ik+i+1/2n+1/2-Ik-i-1/2n+1/2.Considering the corresponding terms withi, we can decompose (13a) as [21](17)ai2i+1Vkn+1=ai2i+1Vkn-ai2i+1Δt2i+1Δzc-1Ik+i+1/2n+1/2-Ik-i-1/2n+1/2for i=0,1,…,LS-1. Equation (13b) could make the analogous decomposition. We could view the MRTD scheme for two-conductor transmission lines as the weighted mean of the conventional FDTD method with spatial discretization step (2i+1)Δz for i=0,1,…,LS-1, and the weighting coefficient for each term is (2i+1)a(i). Besides, for the MRTD scheme whose coefficients a(i) satisfy relationship (15), the analogous decomposition could be made. This relationship between the MRTD scheme and the conventional FDTD method is useful for us to update the iterative equations.TakingV0n+1 as an example to derive the iterative equations at the terminal,(18)V0n+1=V0n-ΔtΔzc-1∑i=0LS-1aiIi+1/2n+1/2-I-i-1/2n+1/2.Following steps of (16) and (17), we can decompose (18) as(19a)a0V0n+1=a0V0n-a0ΔtΔzc-1I1/2n+1/2-I-1/2n+1/2(19b)3a1V0n+1=3a1V0n-3a1Δt3Δzc-1I3/2n+1/2-I-3/2n+1/2(19c)⋮2LS-1aLS-1V0n+1=2LS-1aLS-1V0n=-2LS-1aLS-1Δt2LS-1Δz=·c-1ILS-1/2n+1/2-I-LS+1/2n+1/2.Here, we could view each equation in (19a), (19b), and (19c) as a central difference scheme, (19a) is the central difference scheme related to points z=-Δz/2 and z=Δz/2, (19b) is the central difference scheme related to points z=-3Δz/2 and z=3Δz/2, and so on, but the terms I-1/2n+1/2,I-3/2n+1/2,…,I-LS+1/2n+1/2, whose subscripts exceed the index range, make (19a), (19b), and (19c) out of work. So we need to make some update for the iterative equations. Using the forward difference scheme to replace the central difference scheme, we change the difference points in (19a) to become z=0 and z=Δz/2 and change the difference points in (19b) to become z=0 and z=3Δz/2; the others follow the same step. Keeping the weighting coefficient unchanged in each equation, we can obtain (20a)a0V0n+1=a0V0n-a0ΔtΔz/2c-1I1/2n+1/2-ISn+1/2(20b)3a1V0n+1=3a1V0n-3a1Δt3Δz/2c-1I3/2n+1/2-ISn+1/2(20c)⋮2LS-1aLS-1V0n+1=2LS-1aLS-1V0n=-2LS-1aLS-1Δt2LS-1Δz/2=·c-1ILS-1/2n+1/2-ISn+1/2, where the terminal current ISn+1/2=(ISn+ISn+1)/2 and ISn can be derived from (14a)(21)ISn=VSn-V0nRS.Summing up all the equations in (20a), (20b), and (20c),(22)V0n+1=V0n-ΔtΔzc-1∑i=0LS-12aiIi+1/2n+1/2-ISn+1/2.Substituting (21) into (22), we can obtain the iterative equation at the source(23)V0n+1=∑i=0LS-1ai+ΔzΔtcRS-1ΔzΔtcRS-∑i=0LS-1aiV0n-2RS∑i=0LS-1aiIi+1/2n+1/2+∑i=0LS-1aiVSn+1+VSn.With the same steps, we can obtain the iterative equation at the load(24)VNDZn+1=∑i=0LS-1ai+ΔzΔtcRL-1ΔzΔtcRL-∑i=0LS-1aiVNDZn+2RL∑i=0LS-1aiINDZ-i+1/2n+1/2+∑i=0LS-1aiVLn+1+VLn.After deriving the iterative equations at the terminal, we will put forward a truncation method to update iterative equations which contain some terms whose indices exceed the index range in the MRTD scheme.TakingVkn+1 as an example, for k=1,2,…,LS-1, decomposing (13a), (25a)a0Vkn+1=a0Vkn-a0ΔtΔzc-1Ik+1/2n+1/2-Ik-1/2n+1/2(25b)3a1Vkn+1=3a1Vkn-3a1Δt3Δzc-1Ik+3/2n+1/2-Ik-3/2n+1/2(25c)⋮2k-1ak-1Vk-1n+1=2k-1ak-1Vk-1n=-2k-1ak-1Δt2k-1Δzc-1I2k-1/2n+1/2-I1/2n+1/2(25d)2k+1akVkn+1=2k+1akVkn-2k+1ak=·Δt2k+1Δzc-1I2k+1/2n+1/2-I-1/2n+1/2(25e)⋮2LS-1aLS-1VLS-1n+1=2LS-1aLS-1VLS-1n=-2LS-1aLS-1Δt2LS-1Δz=·c-1Ik+LS-3/2n+1/2-Ik-LS+1/2n+1/2.Noticing the first k terms in (25a), (25b), (25c), (25d), and (25e), there is no term exceeding the index range in each equation. Meanwhile, the equations which contain the exceeding indices terms are all appearing in the rest of LS-k terms. As mentioned before, we can view the MRTD scheme as the weighted mean of the conventional FDTD method, but (25a), (25b), (25c), (25d), and (25e) show that the last LS-k equations are unavailable for forming the iterative equations in MRTD scheme. To solve this problem, we make truncation here. We update the iterative equation of Vkn+1 by using the summation of the first k terms in (25a), (25b), (25c), (25d), and (25e) that means we use the weighted mean of the first k to approximate the summation of all the LS terms.Summing up the firstk terms in (25a), (25b), (25c), (25d), and (25e), we can obtain the modified iterative equations(26)Vkn+1=Vkn-∑i=0k-12i+1ai-1ΔtΔzc-1∑i=0k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=1,2,…,LS-1.Using the same method, we can obtain the modified iterative equations near the load(27)Vkn+1=Vkn-∑i=0NDZ-k-12i+1ai-1ΔtΔzc-1∑i=0NDZ-k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The voltages at the interior points are determined from (13a)(28)Vkn+1=Vkn-ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=LS,LS+1,…,NDZ-LS.For the iterative equation of the current, there is a little difference from the voltage’s. As shown in Figure3, the interlace currents appear at the half integer points which means all the currents are located at the interior points. So we only need to modify the currents near the terminals. Following the same steps of the derivation of voltages iterative equations near the terminal, we could obtain the current iterative equations near the terminals.For the current iterative equations near the source(29)Ik+1/2n+1/2=Ik+1/2n-1/2-∑i=0k2i+1ai-1ΔtΔzl-1∑i=0kaiVk+i+1n-Vk-in for k=0,1,…,LS-2.For the current iterative equations near the load(30)Ik+1/2n+1/2=Ik+1/2n-1/2-∑i=0NDZ-k-12i+1ai-1ΔtΔzl-1∑i=0NDZ-k-1aiVk+i+1n-Vk-in for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The currents at the interior points are determined from (13b)(31)Ik+1/2n+1/2=Ik+1/2n-1/2-ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in for k=LS-1,LS,…,NDZ-LS.


### 2.3. Terminal Iterative Equations for Inductive Resistance in MRTD Scheme

Since we have discussed the resistive load in Section2.2, we will consider a more complicated terminal load which consisted of a resistance and inductance shown in Figure 4. The resistance and the inductance at the load are noted as RL and LL, respectively.Figure 4
A two-conductor line with an inductive resistance.Keeping the source as a resistive terminal and changing the load including inductance, we note the voltage at the source(z=0) as VS(t) and the current at the source as IS(t), the external voltage at the load (z=L) as VL(t), and the current at the load as IL(t). The terminal conditions could be written as follows: (32a)V0t=VSt-RSISt(32b)VNDZt=VLt+RLILt+LLdILtdt.ExpandingV0(t), I0(t), VL(t), and IL(t) as (2a) and (2b) and sampling them at the time discreting point (n+1/2)Δt, the terminal conditions become (33a)V0n+1+V0n2=VSn+1+VSn2-RSISn+1/2(33b)VNDZn+1+VNDZn2=VLn+1+VLn2+RLILn+1/2+LL2ΔtILn+3/2-ILn-1/2.It could be seen from Section2.2 that the change of the terminal condition only affects the iterative equations at the terminals, so we just need to consider two iterative equations. Since we keep the source as a resistive termination, the iterative equation at the source should be the same as (23). Actually, if we substitute (33a) into (22), we could obtain (23). So the only iterative equation we should derive is located at the load. If we set LL=0 in (33b), the terminal condition at the load will have the analogous form with the source, which means the load with a resistance and inductance degenerates to be a resistance. For those LL≠0, transforming (33b) as(34)ILn+3/2=ILn-1/2+ΔtLLVNDZn+1+VNDZn-VLn+1+VLn-2RLILn+1/2.Following the steps we get the iterative equation ofV0n+1 in Section 2.2; we can obtain the iterative equation at the load(35)VNDZn+1=VNDZn-ΔtΔzc-1∑i=0LS-12aiINDZ-i+1/2n+1/2-ILn+1/2, where(36)ILn+1/2=ILn-3/2+ΔtLLVNDZn+VNDZn-1-VLn+VLn-1-2RLILn-1/2 for n=2,3,…,NDT.


### 2.4. Stability Analysis

For the purpose of stability analysis, (13a) and (13b) can be rewritten as (37a)Vkn+1-VknΔt=-1cΔz∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(37b)Ik+1/2n+1/2-Ik+1/2n-1/2Δt=-1lΔz∑i=0LS-1aiVk+i+1n-Vk-in.Following the procedures in [22], the finite-difference approximations of the time derivations on the left side of (37a) and (37b) can be written as an eigenvalue problem (38a)Vkn+1-VknΔt=λVkn+1/2(38b)Ik+1/2n+1/2-Ik+1/2n-1/2Δt=λIk+1/2n.In order to avoid instability during normal time stepping, the imaginary part ofλ must satisfy(39)Im⁡λ≤2Δt.As we consider the lossless two-conductor transmission lines in (1a) and (1b), the transient values of voltages and currents distributed in space can be Fourier transformed with respect to k-coordinates to provide a spectrum of sinusoidal modes. Assuming an eigenmode of the spectral domain with kz, the voltages and currents can be written as (40a)VK=Vz0exp⁡-jkzKΔz(40b)IK+1/2=Iz0exp⁡-jkzK+12Δz, where VK represents the voltage at point KΔz, IK+1/2 represents the current at point (K+1/2)Δz, and Vz0 and Iz0 are the amplitudes of the voltages and currents.Substituting (40a) and (40b) into (38a) and (38b) and (37a) and (37b), we obtain(41)λ2=-4lcΔz2∑i=0LS-1aisin⁡kzi+12Δz2. In (41), λ is pure imaginary and(42)Im⁡λ≤2vΔz∑i=0LS-1ai,where v=1/lc is the velocity of the wave along with the lines.Numerical stability is maintained for every spatial mode only when the range of eigenvalues given by (42) is contained entirely within the stable range of time-difference eigenvalues given by (39). Since both ranges are symmetrical around zero, it is adequate to set the upper bound of (42) to be smaller or equal to (39); we can obtain the stability condition(43)vΔtΔz≤1∑i=0LS-1ai.Noting the Courant number(44)q=vΔtΔz the maximum values of q required by a stable algorithm, which are listed in Table 1 as qmax, can be calculated from the connection coefficients.


### 2.5. Dispersion Analysis

To calculate the numerical dispersion, substituting a time harmonic trial solution into (37a) and (37b), we can obtain(45)1vΔtsin⁡ωΔt2=1Δz∑i=0LS-1aisin⁡kzi+12Δz,where ω is the wave angular frequency and kz is the numerical wave number.Using the number of cells per wavelengthnl=λREAL/Δz and the wave number kz=(2π)/λNUM, we obtain the dispersion relationship(46)1qsin⁡πqnl=∑i=0LS-1aisin⁡2i+1πunl,where u=λREAL/λNUM is the ratio between the theoretical and numerical wavelength.The value ofu could be computed by Newton iterative method and using the formula |(u-1)×360| calculates the phase error in degrees. Figure 5 shows the phase errors of different MRTD schemes versus the samples per wavelength. Compared with the conventional FDTD method at the same sampling numbers, the MRTD schemes show better numerical dispersion properties than FDTD method, which means the discretization error of the MRTD schemes is smaller than the FDTD method.Figure 5
Phase error (in degrees) for different MRTD schemes. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)


## 2.1. MRTD Formulation

In this section, the MRTD scheme is applied to the following scalar transmission lines equations for two-conductor lossless lines [18]: (1a)∂Vz,t∂z+l∂Iz,t∂t=0(1b)∂Iz,t∂z+c∂Vz,t∂t=0, where l and c are the per-unit-length inductance and capacitance, respectively.Based on the method outlined in [1], the voltage and current can be expanded as follows: (2a)Vz,t=∑k,n=-∞+∞Vknϕkzhnt(2b)Iz,t=∑k,n=-∞+∞Ik+1/2n+1/2ϕk+1/2zhn+1/2t, where Vkn and Ik+1/2n+1/2 are the coefficients for the voltages and currents in terms of scaling functions, respectively. The indices n and k are the discrete spatial and temporal indices related to space and time coordinates via z=kΔz and t=nΔt, where Δz and Δt represent the spatial and temporal discretization intervals in z and t direction. The function hn(t) is defined as(3)hnt=htΔt-n with the rectangular pulse function(4)ht=1fort<1212fort=120fort>12.The functionϕk(z) is defined as(5)ϕkz=ϕzΔz-k, where ϕ(z) represents Daubechies’ scaling function. Figure 1 shows Daubechies’ scaling function with two vanishing moments.Figure 1
Daubechies’ scaling function with two vanishing moments.For deriving the MRTD scheme for (1a) and (1b), we need the following integrals:(6)∫-∞+∞hnthn′tdt=δn,n′Δt∫-∞+∞hnt∂hn′+1/2t∂tdt=δn,n′-δn,n′+1, where δn,n′ represents the Kronecker symbol. Consider(7)∫-∞+∞ϕkzϕk′zdz=δk,k′Δz(8)∫-∞+∞ϕnz∂ϕn′+1/2z∂zdz=∑i=-LSLS-1aiδn+i,n′,where LS denotes the effective support size of the basis functions. The coefficients a(i) are called connection coefficients and can be calculated by (9). Taking Daubechies’ scaling functions as the basis functions, Table 1 shows a(i) for 0≤i≤LS-1, which are zeros for i>LS-1, and for i<0 it can be obtained by the symmetry relation a(-1-i)=-a(i),(9)ai=1π∫0∞ϕ^λ2λsin⁡λi+12dλ, where ϕ^(λ) represents the Fourier transform of ϕ(z).Table 1
Connection coefficientsa(i) and the first-order moments M1 of Daubechies’ scaling functions.
i
D
2
D
3
D
4
0
1.2291666667
1.2918129281
1.3110340773
1
−0.0937500000
−0.1371343465
−0.1560100110
2
0.0104166667
0.0287617728
0.0419957460
3
−0.0034701413
−0.0086543236
4
0.0000080265
0.0008308695
5
0.0000108999
6
0.0000000041
M
1
0.6339743121
0.8174005815
1.0053923835
q
m
a
x
0.7500
0.6844
0.6585Daubechies’ scaling functions satisfy the shifted interpolation property [19](10)ϕi+M1=δi,0 for i integer, where M1=∫-∞+∞zϕ(z)dz is the first moment of the scaling functions and the values of M1 are listed in Table 1.Following the theory in [3] and making use of (10), (5) is modified to(11)ϕkz=ϕzΔz-k+M1. In spite of the support of the scaling functions [20], single-point sampling of the total voltages and currents can be taken at integer points with negligible error. Taking voltage at spatial point kΔz and at time nΔt, we obtain(12)VkΔz,nΔt=∬-∞+∞Vz,tδzΔz-kδtΔt-ndzdt=Vkn, where δ is the Dirac delta function. Equation (12) means the voltage value at each integer point is equal to the coefficient. The current values have the same character at each half integer point. Therefore, we will use Vkn and Ik+1/2n+1/2 directly to represent the voltage at the point (kΔz,nΔt) and the current at the point ((k+1/2)Δz,(n+1/2)Δt) in this paper.The modifiedϕk(z) in (11) also satisfy integrals (7) and (8). Applying the Galerkin technique to (1a) and (1b), we can obtain the following iterative equations for the voltages and currents: (13a)Vkn+1=Vkn-ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(13b)Ik+1/2n+1/2=Ik+1/2n-1/2-ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in.


## 2.2. Terminal Iterative Equations for Resistive Load in MRTD Scheme

We will consider the terminal conditions for the two-conductor lossless transmission lines equations in this section. Equations (1a) and (1b) are homogeneous linear equations; we need to add the terminal conditions to obtain the unique solution.Considering the two-conductor lines shown in Figure2, we assume the length of the total line is L and the resistive loads are RS and RL. The line is divided uniformly into NDZ segments with the space interval Δz and the total solution time is divided into NDT steps with the uniform time interval Δt. Similar to the conventional FDTD, we will calculate the interlace voltages, V0n,V1n,…,VNDZn, and currents, I1/2n+1/2,I3/2n+1/2,…,INDZ-1/2n+1/2, in both space domain and time-domain as shown in (13a) and (13b), for n=1,2,…,NDT.Figure 2
A two-conductor line in time-domain.For the resistive terminations, we note the voltage at the source(z=0) as VS(t) and the current at the source as IS(t), the external voltage at the load (z=L) as VL(t), and the current at the load as IL(t). The discrete voltages and currents at the source are denoted as VSn≡VS(nΔt) and ISn≡IS(nΔt),  and the discrete voltages and currents at the load are denoted as VLn≡VL(nΔt), and ILn≡IL(nΔt), then the terminal characterizations could be written in terms of a generalized Thevenin equivalent as  (14a)V0n=VSn-RSISn(14b)VNDZn=VLn+RLILn. Equations (14a) and (14b) denote the discretization terminal conditions for the case of resistive terminations, so we need to introduce these conditions to the iterative equations (13a) and (13b) to obtain the numerical solution.Notice that, in the iterative equations (13a) and (13b), not only the iterative equations of the terminal voltages V0n+1 and VNDZn+1 should be derived and the iterative equations of voltages and currents “near" the terminals also need to be updated. The voltages and currents “near" the terminals we mean are the voltages Vin+1 and VNDZ-in for i=1,2,…,LS-1, and the currents Ii+1/2n+1/2 and INDZ-i+1/2n+1/2 for i=0,1,…,LS-2. All of these voltages and currents contain some terms that exceed the index range in iterative equations (13a) and (13b). Figure 3 shows the discretization of the terminal voltages and the voltages and currents near the terminal.Figure 3
Discretizing the terminal voltages and currents.We will derive the MRTD scheme at the terminal firstly. For updating the iterative equations for the terminal voltages, we need to decompose iterative equations ((13a), (13b)). Since the coefficients a(i) satisfy the following relation [4](15)∑i=0LS-12i+1ai=1 substituting (15) into (13a), we can obtain(16)∑i=0LS-1ai2i+1Vkn+1=∑i=0LS-1ai2i+1Vkn-∑i=0LS-1Δt2i+1Δzc-1ai2i+1Ik+i+1/2n+1/2-Ik-i-1/2n+1/2.Considering the corresponding terms withi, we can decompose (13a) as [21](17)ai2i+1Vkn+1=ai2i+1Vkn-ai2i+1Δt2i+1Δzc-1Ik+i+1/2n+1/2-Ik-i-1/2n+1/2for i=0,1,…,LS-1. Equation (13b) could make the analogous decomposition. We could view the MRTD scheme for two-conductor transmission lines as the weighted mean of the conventional FDTD method with spatial discretization step (2i+1)Δz for i=0,1,…,LS-1, and the weighting coefficient for each term is (2i+1)a(i). Besides, for the MRTD scheme whose coefficients a(i) satisfy relationship (15), the analogous decomposition could be made. This relationship between the MRTD scheme and the conventional FDTD method is useful for us to update the iterative equations.TakingV0n+1 as an example to derive the iterative equations at the terminal,(18)V0n+1=V0n-ΔtΔzc-1∑i=0LS-1aiIi+1/2n+1/2-I-i-1/2n+1/2.Following steps of (16) and (17), we can decompose (18) as(19a)a0V0n+1=a0V0n-a0ΔtΔzc-1I1/2n+1/2-I-1/2n+1/2(19b)3a1V0n+1=3a1V0n-3a1Δt3Δzc-1I3/2n+1/2-I-3/2n+1/2(19c)⋮2LS-1aLS-1V0n+1=2LS-1aLS-1V0n=-2LS-1aLS-1Δt2LS-1Δz=·c-1ILS-1/2n+1/2-I-LS+1/2n+1/2.Here, we could view each equation in (19a), (19b), and (19c) as a central difference scheme, (19a) is the central difference scheme related to points z=-Δz/2 and z=Δz/2, (19b) is the central difference scheme related to points z=-3Δz/2 and z=3Δz/2, and so on, but the terms I-1/2n+1/2,I-3/2n+1/2,…,I-LS+1/2n+1/2, whose subscripts exceed the index range, make (19a), (19b), and (19c) out of work. So we need to make some update for the iterative equations. Using the forward difference scheme to replace the central difference scheme, we change the difference points in (19a) to become z=0 and z=Δz/2 and change the difference points in (19b) to become z=0 and z=3Δz/2; the others follow the same step. Keeping the weighting coefficient unchanged in each equation, we can obtain (20a)a0V0n+1=a0V0n-a0ΔtΔz/2c-1I1/2n+1/2-ISn+1/2(20b)3a1V0n+1=3a1V0n-3a1Δt3Δz/2c-1I3/2n+1/2-ISn+1/2(20c)⋮2LS-1aLS-1V0n+1=2LS-1aLS-1V0n=-2LS-1aLS-1Δt2LS-1Δz/2=·c-1ILS-1/2n+1/2-ISn+1/2, where the terminal current ISn+1/2=(ISn+ISn+1)/2 and ISn can be derived from (14a)(21)ISn=VSn-V0nRS.Summing up all the equations in (20a), (20b), and (20c),(22)V0n+1=V0n-ΔtΔzc-1∑i=0LS-12aiIi+1/2n+1/2-ISn+1/2.Substituting (21) into (22), we can obtain the iterative equation at the source(23)V0n+1=∑i=0LS-1ai+ΔzΔtcRS-1ΔzΔtcRS-∑i=0LS-1aiV0n-2RS∑i=0LS-1aiIi+1/2n+1/2+∑i=0LS-1aiVSn+1+VSn.With the same steps, we can obtain the iterative equation at the load(24)VNDZn+1=∑i=0LS-1ai+ΔzΔtcRL-1ΔzΔtcRL-∑i=0LS-1aiVNDZn+2RL∑i=0LS-1aiINDZ-i+1/2n+1/2+∑i=0LS-1aiVLn+1+VLn.After deriving the iterative equations at the terminal, we will put forward a truncation method to update iterative equations which contain some terms whose indices exceed the index range in the MRTD scheme.TakingVkn+1 as an example, for k=1,2,…,LS-1, decomposing (13a), (25a)a0Vkn+1=a0Vkn-a0ΔtΔzc-1Ik+1/2n+1/2-Ik-1/2n+1/2(25b)3a1Vkn+1=3a1Vkn-3a1Δt3Δzc-1Ik+3/2n+1/2-Ik-3/2n+1/2(25c)⋮2k-1ak-1Vk-1n+1=2k-1ak-1Vk-1n=-2k-1ak-1Δt2k-1Δzc-1I2k-1/2n+1/2-I1/2n+1/2(25d)2k+1akVkn+1=2k+1akVkn-2k+1ak=·Δt2k+1Δzc-1I2k+1/2n+1/2-I-1/2n+1/2(25e)⋮2LS-1aLS-1VLS-1n+1=2LS-1aLS-1VLS-1n=-2LS-1aLS-1Δt2LS-1Δz=·c-1Ik+LS-3/2n+1/2-Ik-LS+1/2n+1/2.Noticing the first k terms in (25a), (25b), (25c), (25d), and (25e), there is no term exceeding the index range in each equation. Meanwhile, the equations which contain the exceeding indices terms are all appearing in the rest of LS-k terms. As mentioned before, we can view the MRTD scheme as the weighted mean of the conventional FDTD method, but (25a), (25b), (25c), (25d), and (25e) show that the last LS-k equations are unavailable for forming the iterative equations in MRTD scheme. To solve this problem, we make truncation here. We update the iterative equation of Vkn+1 by using the summation of the first k terms in (25a), (25b), (25c), (25d), and (25e) that means we use the weighted mean of the first k to approximate the summation of all the LS terms.Summing up the firstk terms in (25a), (25b), (25c), (25d), and (25e), we can obtain the modified iterative equations(26)Vkn+1=Vkn-∑i=0k-12i+1ai-1ΔtΔzc-1∑i=0k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=1,2,…,LS-1.Using the same method, we can obtain the modified iterative equations near the load(27)Vkn+1=Vkn-∑i=0NDZ-k-12i+1ai-1ΔtΔzc-1∑i=0NDZ-k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The voltages at the interior points are determined from (13a)(28)Vkn+1=Vkn-ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=LS,LS+1,…,NDZ-LS.For the iterative equation of the current, there is a little difference from the voltage’s. As shown in Figure3, the interlace currents appear at the half integer points which means all the currents are located at the interior points. So we only need to modify the currents near the terminals. Following the same steps of the derivation of voltages iterative equations near the terminal, we could obtain the current iterative equations near the terminals.For the current iterative equations near the source(29)Ik+1/2n+1/2=Ik+1/2n-1/2-∑i=0k2i+1ai-1ΔtΔzl-1∑i=0kaiVk+i+1n-Vk-in for k=0,1,…,LS-2.For the current iterative equations near the load(30)Ik+1/2n+1/2=Ik+1/2n-1/2-∑i=0NDZ-k-12i+1ai-1ΔtΔzl-1∑i=0NDZ-k-1aiVk+i+1n-Vk-in for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The currents at the interior points are determined from (13b)(31)Ik+1/2n+1/2=Ik+1/2n-1/2-ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in for k=LS-1,LS,…,NDZ-LS.


## 2.3. Terminal Iterative Equations for Inductive Resistance in MRTD Scheme

Since we have discussed the resistive load in Section2.2, we will consider a more complicated terminal load which consisted of a resistance and inductance shown in Figure 4. The resistance and the inductance at the load are noted as RL and LL, respectively.Figure 4
A two-conductor line with an inductive resistance.Keeping the source as a resistive terminal and changing the load including inductance, we note the voltage at the source(z=0) as VS(t) and the current at the source as IS(t), the external voltage at the load (z=L) as VL(t), and the current at the load as IL(t). The terminal conditions could be written as follows: (32a)V0t=VSt-RSISt(32b)VNDZt=VLt+RLILt+LLdILtdt.ExpandingV0(t), I0(t), VL(t), and IL(t) as (2a) and (2b) and sampling them at the time discreting point (n+1/2)Δt, the terminal conditions become (33a)V0n+1+V0n2=VSn+1+VSn2-RSISn+1/2(33b)VNDZn+1+VNDZn2=VLn+1+VLn2+RLILn+1/2+LL2ΔtILn+3/2-ILn-1/2.It could be seen from Section2.2 that the change of the terminal condition only affects the iterative equations at the terminals, so we just need to consider two iterative equations. Since we keep the source as a resistive termination, the iterative equation at the source should be the same as (23). Actually, if we substitute (33a) into (22), we could obtain (23). So the only iterative equation we should derive is located at the load. If we set LL=0 in (33b), the terminal condition at the load will have the analogous form with the source, which means the load with a resistance and inductance degenerates to be a resistance. For those LL≠0, transforming (33b) as(34)ILn+3/2=ILn-1/2+ΔtLLVNDZn+1+VNDZn-VLn+1+VLn-2RLILn+1/2.Following the steps we get the iterative equation ofV0n+1 in Section 2.2; we can obtain the iterative equation at the load(35)VNDZn+1=VNDZn-ΔtΔzc-1∑i=0LS-12aiINDZ-i+1/2n+1/2-ILn+1/2, where(36)ILn+1/2=ILn-3/2+ΔtLLVNDZn+VNDZn-1-VLn+VLn-1-2RLILn-1/2 for n=2,3,…,NDT.


## 2.4. Stability Analysis

For the purpose of stability analysis, (13a) and (13b) can be rewritten as (37a)Vkn+1-VknΔt=-1cΔz∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(37b)Ik+1/2n+1/2-Ik+1/2n-1/2Δt=-1lΔz∑i=0LS-1aiVk+i+1n-Vk-in.Following the procedures in [22], the finite-difference approximations of the time derivations on the left side of (37a) and (37b) can be written as an eigenvalue problem (38a)Vkn+1-VknΔt=λVkn+1/2(38b)Ik+1/2n+1/2-Ik+1/2n-1/2Δt=λIk+1/2n.In order to avoid instability during normal time stepping, the imaginary part ofλ must satisfy(39)Im⁡λ≤2Δt.As we consider the lossless two-conductor transmission lines in (1a) and (1b), the transient values of voltages and currents distributed in space can be Fourier transformed with respect to k-coordinates to provide a spectrum of sinusoidal modes. Assuming an eigenmode of the spectral domain with kz, the voltages and currents can be written as (40a)VK=Vz0exp⁡-jkzKΔz(40b)IK+1/2=Iz0exp⁡-jkzK+12Δz, where VK represents the voltage at point KΔz, IK+1/2 represents the current at point (K+1/2)Δz, and Vz0 and Iz0 are the amplitudes of the voltages and currents.Substituting (40a) and (40b) into (38a) and (38b) and (37a) and (37b), we obtain(41)λ2=-4lcΔz2∑i=0LS-1aisin⁡kzi+12Δz2. In (41), λ is pure imaginary and(42)Im⁡λ≤2vΔz∑i=0LS-1ai,where v=1/lc is the velocity of the wave along with the lines.Numerical stability is maintained for every spatial mode only when the range of eigenvalues given by (42) is contained entirely within the stable range of time-difference eigenvalues given by (39). Since both ranges are symmetrical around zero, it is adequate to set the upper bound of (42) to be smaller or equal to (39); we can obtain the stability condition(43)vΔtΔz≤1∑i=0LS-1ai.Noting the Courant number(44)q=vΔtΔz the maximum values of q required by a stable algorithm, which are listed in Table 1 as qmax, can be calculated from the connection coefficients.


## 2.5. Dispersion Analysis

To calculate the numerical dispersion, substituting a time harmonic trial solution into (37a) and (37b), we can obtain(45)1vΔtsin⁡ωΔt2=1Δz∑i=0LS-1aisin⁡kzi+12Δz,where ω is the wave angular frequency and kz is the numerical wave number.Using the number of cells per wavelengthnl=λREAL/Δz and the wave number kz=(2π)/λNUM, we obtain the dispersion relationship(46)1qsin⁡πqnl=∑i=0LS-1aisin⁡2i+1πunl,where u=λREAL/λNUM is the ratio between the theoretical and numerical wavelength.The value ofu could be computed by Newton iterative method and using the formula |(u-1)×360| calculates the phase error in degrees. Figure 5 shows the phase errors of different MRTD schemes versus the samples per wavelength. Compared with the conventional FDTD method at the same sampling numbers, the MRTD schemes show better numerical dispersion properties than FDTD method, which means the discretization error of the MRTD schemes is smaller than the FDTD method.Figure 5
Phase error (in degrees) for different MRTD schemes. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)


## 3. MRTD Scheme for Two-Conductor Lossy Transmission Lines

In this section, we will extend the MRTD scheme to the two-conductor lossy transmission lines.


### 3.1. MRTD Formulation

For the lossy case, the two-conductor transmission line equations become(47a)∂Vz,t∂z+rIz,t+l∂Iz,t∂t=0(47b)∂Iz,t∂z+gVz,t+c∂Vz,t∂t=0, where r, l, g, and c are the per-unit-length resistance, inductance, conductance, and capacitance, respectively.Compared to the lossless transmission lines, the lossy case must consider the resistance losses along the lines and the losses in the medium. However, the steps to obtain the MRTD scheme are the same. Similar to the lossless case, we firstly extend the voltage and current with Daubechies’ scaling functions and the rectangular function and useVkn and Ik+1/2n+1/2 representing the voltage at the point (kΔz,nΔt) and the current at the point ((k+1/2)Δz,(n+1/2)Δt), respectively, and Δz and Δt represent the spatial and temporal discretization intervals. By applying the Galerkin technique, we can obtain the iterative equations for the lossy transmission lines equations (48a)Vkn+1=P1Vkn-P2ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(48b)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in,where a(i) is the connection coefficient and P1, P2, Q1,  and  Q2 are constants (49a)P1=1+Δt2gc-1-11-Δt2gc-1(49b)P2=1+Δt2gc-1-1(50a)Q1=1+Δt2rl-1-11-Δt2rl-1(50b)Q2=1+Δt2rl-1-1.The differences between (48a) and (48b) and (13a) and (13b) are the coefficients of terms in the iterative equations that are caused by the unit-per-length resistance r and inductance g. If we set r=0 and g=0, the coefficients P1, P2 and Q1, Q2 are all equal to 1, and the iterative equations (48a) and (48b) will degenerate to iterative equations (13a) and (13b).For the lossy case, we should also modify the iterative equations at the terminal and near the terminal of the lines. Considering the two-conductor lines shown in Figure2, we assume the length of the total line is L and the resistive loads are RS and RL. The line is divided uniformly into NDZ segments with the space interval Δz and the total solution time is divided into NDT steps with the uniform time interval Δt. Following the same steps in Section 2.2, we could obtain the modified iterative equations.For the voltage iterative equation at the source(51)V0n+1=P2∑i=0LS-1ai+ΔzΔtcRS-1P1ΔzΔtcRS-P2∑i=0LS-1aiV0n-2P2RS∑i=0LS-1aiIi+1/2n+1/2+P2∑i=0LS-1aiVSn+1+VSn.For the voltage iterative equation at the load(52)VNDZn+1=P2∑i=0LS-1ai+ΔzΔtcRL-1P1ΔzΔtcRL-P2∑i=0LS-1aiVNDZn+2P2RL∑i=0LS-1aiINDZ-i+1/2n+1/2+P2∑i=0LS-1aiVLn+1+VLn.For the voltage iterative equations near the source(53)Vkn+1=P1Vkn-P2∑i=0k-12i+1ai-1ΔtΔzc-1∑i=0k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=1,2,…,LS-1.For the voltage iterative equations near the load(54)Vkn+1=P1Vkn-P2∑i=0NDZ-k-12i+1ai-1ΔtΔzc-1∑i=0NDZ-k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The voltages at the interior points are determined from (48a)(55)Vkn+1=P1Vkn-P2ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=LS,LS+1,…,NDZ-LS.For the current iterative equations near the source(56)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2∑i=0k2i+1ai-1ΔtΔzl-1∑i=0kaiVk+i+1n-Vk-infor k=0,1,…,LS-2.For the current iterative equations near the load(57)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2∑i=0NDZ-k-12i+1ai-1ΔtΔzl-1∑i=0NDZ-k-1aiVk+i+1n-Vk-in for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The currents at the interior points are determined from (48b)(58)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in for k=LS-1,LS,…,NDZ-LS.Besides, for the two-conductor line shown in Figure4, we can obtain the iterative equation at the load(59)VNDZn+1=P1VNDZn-P2ΔtΔzc-1∑i=0LS-12aiINDZ-i+1/2n+1/2-ILn+1/2, where(60)ILn+1/2=ILn-3/2+ΔtLLVNDZn+VNDZn-1-VLn+VLn-1-2RLILn-1/2 for n=2,3,…,NDT.


### 3.2. Stability Analysis

To study the stability of the MRTD scheme for lossy case, we need to make some changes. Rewrite (48a) and (48b) as follows: (61a)P2-1Vkn+1-P2-1P1VknΔt=-1cΔz∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(61b)Q2-1Ik+1/2n+1/2-Q2-1Q1Ik+1/2n-1/2Δt=-1lΔz∑i=0LS-1aiVk+i+1n-Vk-in.The finite-difference approximations of the time derivations on the left side of (61a) and (61b) are different from the left side of (37a) and (37b), but we can also write them as an eigenvalue problem (62a)P2-1Vkn+1-P2-1P1VknΔt=λVkn+1/2(62b)Q2-1Ik+1/2n+1/2-Q2-1Q1Ik+1/2n-1/2Δt=λIk+1/2n.In order to avoid instability during normal time stepping, the imaginary part ofλ also must satisfy (63)Im⁡λ≤2Δt.Then following the steps in Section2.4, using the Fourier transform, expand the transient values of voltages and currents distributed in space. We can also obtain the same stability condition as (43) (64)vΔtΔz≤1∑i=0LS-1ai.However, the dispersion analysis for the iterative equations of the lossy transmission lines is quite different from the lossless case. SinceP2-1≠P2-1P1 and Q2-1≠Q2-1Q1, that is, the coefficient of Vkn+1 is not equal to the coefficient of Vkn and the coefficient of Ik+1/2n+1/2 is not equal to the coefficient of Ik+1/2n-1/2 on the left side of (61a) and (61b), if we substitute a time harmonic into the iterative equations into (61a) and (61b), there are some other terms that could affect the ratio between the theoretical and numerical wavelength except samples per wavelength, so we could not obtain a brief dispersion relation as (45). However, the numerical results in Section 4.2 show that the MRTD method could obtain a more accurate result than the FDTD method at the same space interval and time interval.


## 3.1. MRTD Formulation

For the lossy case, the two-conductor transmission line equations become(47a)∂Vz,t∂z+rIz,t+l∂Iz,t∂t=0(47b)∂Iz,t∂z+gVz,t+c∂Vz,t∂t=0, where r, l, g, and c are the per-unit-length resistance, inductance, conductance, and capacitance, respectively.Compared to the lossless transmission lines, the lossy case must consider the resistance losses along the lines and the losses in the medium. However, the steps to obtain the MRTD scheme are the same. Similar to the lossless case, we firstly extend the voltage and current with Daubechies’ scaling functions and the rectangular function and useVkn and Ik+1/2n+1/2 representing the voltage at the point (kΔz,nΔt) and the current at the point ((k+1/2)Δz,(n+1/2)Δt), respectively, and Δz and Δt represent the spatial and temporal discretization intervals. By applying the Galerkin technique, we can obtain the iterative equations for the lossy transmission lines equations (48a)Vkn+1=P1Vkn-P2ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(48b)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in,where a(i) is the connection coefficient and P1, P2, Q1,  and  Q2 are constants (49a)P1=1+Δt2gc-1-11-Δt2gc-1(49b)P2=1+Δt2gc-1-1(50a)Q1=1+Δt2rl-1-11-Δt2rl-1(50b)Q2=1+Δt2rl-1-1.The differences between (48a) and (48b) and (13a) and (13b) are the coefficients of terms in the iterative equations that are caused by the unit-per-length resistance r and inductance g. If we set r=0 and g=0, the coefficients P1, P2 and Q1, Q2 are all equal to 1, and the iterative equations (48a) and (48b) will degenerate to iterative equations (13a) and (13b).For the lossy case, we should also modify the iterative equations at the terminal and near the terminal of the lines. Considering the two-conductor lines shown in Figure2, we assume the length of the total line is L and the resistive loads are RS and RL. The line is divided uniformly into NDZ segments with the space interval Δz and the total solution time is divided into NDT steps with the uniform time interval Δt. Following the same steps in Section 2.2, we could obtain the modified iterative equations.For the voltage iterative equation at the source(51)V0n+1=P2∑i=0LS-1ai+ΔzΔtcRS-1P1ΔzΔtcRS-P2∑i=0LS-1aiV0n-2P2RS∑i=0LS-1aiIi+1/2n+1/2+P2∑i=0LS-1aiVSn+1+VSn.For the voltage iterative equation at the load(52)VNDZn+1=P2∑i=0LS-1ai+ΔzΔtcRL-1P1ΔzΔtcRL-P2∑i=0LS-1aiVNDZn+2P2RL∑i=0LS-1aiINDZ-i+1/2n+1/2+P2∑i=0LS-1aiVLn+1+VLn.For the voltage iterative equations near the source(53)Vkn+1=P1Vkn-P2∑i=0k-12i+1ai-1ΔtΔzc-1∑i=0k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=1,2,…,LS-1.For the voltage iterative equations near the load(54)Vkn+1=P1Vkn-P2∑i=0NDZ-k-12i+1ai-1ΔtΔzc-1∑i=0NDZ-k-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The voltages at the interior points are determined from (48a)(55)Vkn+1=P1Vkn-P2ΔtΔzc-1∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2 for k=LS,LS+1,…,NDZ-LS.For the current iterative equations near the source(56)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2∑i=0k2i+1ai-1ΔtΔzl-1∑i=0kaiVk+i+1n-Vk-infor k=0,1,…,LS-2.For the current iterative equations near the load(57)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2∑i=0NDZ-k-12i+1ai-1ΔtΔzl-1∑i=0NDZ-k-1aiVk+i+1n-Vk-in for k=NDZ-LS+1,NDZ-LS+2,…,NDZ-1.The currents at the interior points are determined from (48b)(58)Ik+1/2n+1/2=Q1Ik+1/2n-1/2-Q2ΔtΔzl-1∑i=0LS-1aiVk+i+1n-Vk-in for k=LS-1,LS,…,NDZ-LS.Besides, for the two-conductor line shown in Figure4, we can obtain the iterative equation at the load(59)VNDZn+1=P1VNDZn-P2ΔtΔzc-1∑i=0LS-12aiINDZ-i+1/2n+1/2-ILn+1/2, where(60)ILn+1/2=ILn-3/2+ΔtLLVNDZn+VNDZn-1-VLn+VLn-1-2RLILn-1/2 for n=2,3,…,NDT.


## 3.2. Stability Analysis

To study the stability of the MRTD scheme for lossy case, we need to make some changes. Rewrite (48a) and (48b) as follows: (61a)P2-1Vkn+1-P2-1P1VknΔt=-1cΔz∑i=0LS-1aiIk+i+1/2n+1/2-Ik-i-1/2n+1/2(61b)Q2-1Ik+1/2n+1/2-Q2-1Q1Ik+1/2n-1/2Δt=-1lΔz∑i=0LS-1aiVk+i+1n-Vk-in.The finite-difference approximations of the time derivations on the left side of (61a) and (61b) are different from the left side of (37a) and (37b), but we can also write them as an eigenvalue problem (62a)P2-1Vkn+1-P2-1P1VknΔt=λVkn+1/2(62b)Q2-1Ik+1/2n+1/2-Q2-1Q1Ik+1/2n-1/2Δt=λIk+1/2n.In order to avoid instability during normal time stepping, the imaginary part ofλ also must satisfy (63)Im⁡λ≤2Δt.Then following the steps in Section2.4, using the Fourier transform, expand the transient values of voltages and currents distributed in space. We can also obtain the same stability condition as (43) (64)vΔtΔz≤1∑i=0LS-1ai.However, the dispersion analysis for the iterative equations of the lossy transmission lines is quite different from the lossless case. SinceP2-1≠P2-1P1 and Q2-1≠Q2-1Q1, that is, the coefficient of Vkn+1 is not equal to the coefficient of Vkn and the coefficient of Ik+1/2n+1/2 is not equal to the coefficient of Ik+1/2n-1/2 on the left side of (61a) and (61b), if we substitute a time harmonic into the iterative equations into (61a) and (61b), there are some other terms that could affect the ratio between the theoretical and numerical wavelength except samples per wavelength, so we could not obtain a brief dispersion relation as (45). However, the numerical results in Section 4.2 show that the MRTD method could obtain a more accurate result than the FDTD method at the same space interval and time interval.


## 4. Numerical Result

### 4.1.  Lossless Transmission Lines

In this section, the two-conductor lossless transmission lines as shown in Figure2 are considered to calculate the terminal voltages. The length of the lines is L=400 m and the per-unit-length capacitance and inductance are c=100 pF/m and l=0.25 μH/m, respectively. This corresponds to RG58U coaxial cable [18]. The characteristic impedance of the line is ZC=l/c=50Ω and the velocity of propagation is v=1/lc=200m/μs. The terminal load is 100Ω and the source resistance is 50Ω.We will use a trapezoidal pulse as shown in Figure6 as the source voltage, whose initial value is 0 V and amplitude is A=30V, total time is 10μs, rise time is τr=1μs, and fall time is τf=1μs, respectively. The bandwidth of the pulse is approximate to BW=1/τr, so the segment length Δz should be electrically short at this frequency requiring Δz≤(1/10)(v/fmax)  (fmax=1/τr) [18]. As we take the rise time of the trapezoidal pulse to be 1μs, the approximate bandwidth of the pulse is 1MHz and the maximum section length of Δz should be less than 20m, noting Δzmax=20. Since we divide the line into NDZ segments uniformly, NDZ should be greater than L/Δzmax=20. The time step will be calculated by (44) with different Courant number q.Figure 6
Trapezoidal pulse with the rise timeτr and fall time τf.Under the conditions of the space discretization numberNDZ=20 and the Courant number q=0.5, we calculate the terminal voltages using the MRTD schemes with D2, D3, and D4 wavelets’ scaling functions. We also calculate the terminal voltages by the FDTD method under the same conditions. Figure 7 shows the numerical results by different methods and Di-MRTD represents the MRTD scheme using Daubechies’ scaling functions with i vanishing moment as basis functions, where i=2,3,4.Figure 7
Terminal voltage calculated by different numerical methods. The space discretization number and the Courant number are NDZ = 20 andq=0.5.For the time-dependent discrete terminal voltages, the relative error is defined as follows [23]:(65)ϵ=∑i=1NDTxi-x^i2∑i=1NDTx^2i. Here, x(i) represents the numerical results of FDTD and MRTD at each time discretization point and x^(i) is the numerical result of series solution at each time discretization point, which can be regarded as the exact solution [18]. Taking x^(i) as the exact result, we calculate the relative errors of FDTD method and MRTD scheme.Table2 shows the relative errors and runtime for different schemes. Since the iterative equations in MRTD schemes contain more terms than the conventional FDTD method, the MRTD schemes expend more runtime. When we use D2-MRTD, the numerical result shows a larger relative error. The reason is that the vanishing moment of the D2 wavelet’s scaling function is not high enough. For the wavelets’ scaling functions whose vanishing moment is high enough, like D3 wavelet and D4 wavelet, the numerical results show smaller relative errors.Table 2
Relative errors and runtime of different methods (NDZ = 20,q=0.5).
Analysis scheme
Relative error
Runtime (s)
FDTD
1.08%
0.0011
D
2-MRTD
1.93%
0.0050
D
3-MRTD
0.66%
0.0054
D
4-MRTD
0.62%
0.0058Figure8 shows the relative errors versus the space discretization numbers. The space interval will decrease with the increase of the space discretization number. We can see from the figures that the relative errors for D2-MRTD increase with the increase of NDZ. And, for D3-MRTD and D4-MRTD, the relative errors decrease with the increase of NDZ and show a little smaller relative error than the conventional FDTD. It means we can get a more accurate result at same space interval and time interval by D3-MRTD and D4-MRTD.Figure 8
Relative error for different discretization numbers. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)Figure9 describes the relative errors versus the Courant numbers. The time interval will decrease with the decrease of the Courant number. The results show that, with the decrease of the Courant numbers, the relative errors are almost unchanged for the conventional FDTD method, while the relative errors are quite different with different Courant numbers for all the three MRTD schemes. That means the choice of the Courant number significantly affects the relative errors of the MRTD schemes, and we could choose the proper Courant number to optimize the MRTD schemes.Figure 9
Relative errors for different Courant numbers. The space discretization numbers are NDZ = 20 for (a) and NDZ = 50 for (b).
(a)
(b)To validate the stability of the MRTD schemes, we increase the initial value of the pulse to 20 V and keep other parameters unchanged. Figure10 shows the relative errors versus the pace discretization numbers. It can be seen from the figures that the D2-MRTD scheme shows a larger relative error and the relative errors of D3-MRTD and D4-MRTD schemes are smaller than the conventional FDTD method.Figure 10
Relative error for different space discretization numbers for the initial value of the trapezoidal pulse to be 20 V. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)The numerical results for the lossless transmission lines also show that theD2-MRTD does not perform better than the FDTD method; meanwhile, D3-MRTD and D4-MRTD schemes show better quality in accuracy and stability. The reason for this phenomenon is that the scaling function of the D2 wavelet does not have enough high vanishing moment. When we use Daubechies’ scaling functions to expand the voltages and currents in the two-conductor transmission lines equations, the vanishing moment decides the accuracy of the approximation. The scaling functions with high vanishing moments could approximate voltages and currents more accurately; however, the scaling functions with low vanishing moments like D2 wavelet’s may introduce a larger error. So we can see from Figure 5 that D2-MRTD has a better numerical dispersion property than the FDTD method, but it gets larger relative errors in numerical computation. For the scaling functions which have high enough vanishing moments like D3 wavelet and D4 wavelet, the numerical results show smaller relative errors and are in agreement with the dispersion analysis. However, a high vanishing moment for scaling function may increase the computation complexity in MRTD schemes. So the vanishing moments of the wavelet’s scaling functions have a great effect on the accuracy of the MRTD scheme; it is necessary to choose a scaling function with proper vanishing moment when we use the MRTD scheme for the numerical computation.


### 4.2. Lossy Transmission Lines

In this section, we will consider the lossy two-conductor transmission lines shown in Figure11. Two conductors of rectangular cross section of width w=20 μm and thickness t=10 μm are separated by s=20 μm and placed on one side of a silicon substrate (εr=12) of thickness h=100 μm; the total line length is L=20 cm. The near end is a source with a RS=50 Ω resistance and the far end is a load with a RL=50 Ω resistance and LL=0.5 μH inductance in series. The per-unit-length inductance and capacitance were computed as l=0.805969 μH/m and c=88.2488 pF/m. This gives a velocity of v=1.18573×108 m/s and a one-way time delay of TD=1.68672 ns, which gives an effective dielectric constant of (εr′=6.4) and characteristic impedance of ZC=95.566 Ω. The per-unit-length dc resistance is computed as r=1/(σwt)=86.207 Ω/m [18]. Dielectric loss is not included in these calculations, which means the per-unit-length conductance is g=0.Figure 11
A lossy printed circuit board: (a) line dimensions and terminations and (b) cross-sectional dimensions.
(a)
(b)The source is a ramp function as shown in Figure12, the initial value of the ramp function is 0V, and the amplitude is A=1 V with a rise time of τr=0.5 ns. The total computing time is 20 ns. The bandwidth of the source is approximate to BW=1/τr=2 GHz. The space discretization step for the MRTD was chosen to be λ/10, so the space discretization number NDZ should be greater than 34. The time step will also be calculated by (44) with different Courant numbers.Figure 12
Representation of the source voltage waveform.We calculate the near end voltage and the far end voltage of the lossy PCB with space discretization number NDZ = 40 and the Courant numberq=0.5. Figure 13 shows the computing results. Since D2-MRTD may introduce a larger error in the computation as shown in Section 4.1, we use D3-MRTD and D4-MRTD to compute the terminal voltages.Figure 13
Terminal voltages for lossy transmission lines: (a) for the near end and (b) for the far end. The space discretization number isNDZ=40 and the Courant number is q=0.5.
(a)
(b)Here, we choose the time-domain to frequency-domain transformation method (TDFD), which is a straightforward adaptation of a common analysis technique for lumped, linear circuits and systems [18], to validate the computing results of MRTD schemes and FDTD method. Table 3 shows the relative errors and the runtime of the MRTD schemes and FDTD method. It can be seen that MRTD schemes spend much time to obtain more accurate results.Table 3
Relative errors and runtime of different methods.V0 represents the near end voltage and VL represents the far end voltage (NDZ = 40, q=0.5).
Analysis scheme
Relative error forV0
Relative error forVL
Runtime (s)
FDTD
0.30%
0.92%
0.0066
D
3-MRTD
0.18%
0.57%
0.0245
D
4-MRTD
0.17%
0.54%
0.0275Figure14 describes the relative errors versus the space discretization numbers. For both D3-MRTD and D4-MRTD, the MRTD schemes show a little smaller relative error than the conventional FDTD. That means the MRTD schemes could obtain a more accurate solution under the same time interval and space interval. This is because D3-MRTD and D4-MRTD have better dispersion property than the FDTD method and the scaling functions of D3 wavelet and D4 wavelet have high enough vanishing moments.Figure 14
Relative error for the terminal voltages with different space discretization numbers: (a) for the near end and (b) for the far end. The Courant number isq=0.5.
(a)
(b)Figure15 shows the relative errors versus the Courant numbers. With the decrease of the Courant number, the relative errors of the FDTD method increase, because the time-domain is oversampled for the FDTD method. However, the MRTD schemes perform decreasing relative errors. That means the MRTD schemes could obtain a more accurate result with a smaller time interval. And even when the FDTD method is oversampled in the time-domain, the MRTD schemes perform well.Figure 15
Relative errors for the terminal voltages with different Courant numbers: (a) for the near end and (b) for the far end. The space discretization number is NDZ = 40.
(a)
(b)


## 4.1.  Lossless Transmission Lines

In this section, the two-conductor lossless transmission lines as shown in Figure2 are considered to calculate the terminal voltages. The length of the lines is L=400 m and the per-unit-length capacitance and inductance are c=100 pF/m and l=0.25 μH/m, respectively. This corresponds to RG58U coaxial cable [18]. The characteristic impedance of the line is ZC=l/c=50Ω and the velocity of propagation is v=1/lc=200m/μs. The terminal load is 100Ω and the source resistance is 50Ω.We will use a trapezoidal pulse as shown in Figure6 as the source voltage, whose initial value is 0 V and amplitude is A=30V, total time is 10μs, rise time is τr=1μs, and fall time is τf=1μs, respectively. The bandwidth of the pulse is approximate to BW=1/τr, so the segment length Δz should be electrically short at this frequency requiring Δz≤(1/10)(v/fmax)  (fmax=1/τr) [18]. As we take the rise time of the trapezoidal pulse to be 1μs, the approximate bandwidth of the pulse is 1MHz and the maximum section length of Δz should be less than 20m, noting Δzmax=20. Since we divide the line into NDZ segments uniformly, NDZ should be greater than L/Δzmax=20. The time step will be calculated by (44) with different Courant number q.Figure 6
Trapezoidal pulse with the rise timeτr and fall time τf.Under the conditions of the space discretization numberNDZ=20 and the Courant number q=0.5, we calculate the terminal voltages using the MRTD schemes with D2, D3, and D4 wavelets’ scaling functions. We also calculate the terminal voltages by the FDTD method under the same conditions. Figure 7 shows the numerical results by different methods and Di-MRTD represents the MRTD scheme using Daubechies’ scaling functions with i vanishing moment as basis functions, where i=2,3,4.Figure 7
Terminal voltage calculated by different numerical methods. The space discretization number and the Courant number are NDZ = 20 andq=0.5.For the time-dependent discrete terminal voltages, the relative error is defined as follows [23]:(65)ϵ=∑i=1NDTxi-x^i2∑i=1NDTx^2i. Here, x(i) represents the numerical results of FDTD and MRTD at each time discretization point and x^(i) is the numerical result of series solution at each time discretization point, which can be regarded as the exact solution [18]. Taking x^(i) as the exact result, we calculate the relative errors of FDTD method and MRTD scheme.Table2 shows the relative errors and runtime for different schemes. Since the iterative equations in MRTD schemes contain more terms than the conventional FDTD method, the MRTD schemes expend more runtime. When we use D2-MRTD, the numerical result shows a larger relative error. The reason is that the vanishing moment of the D2 wavelet’s scaling function is not high enough. For the wavelets’ scaling functions whose vanishing moment is high enough, like D3 wavelet and D4 wavelet, the numerical results show smaller relative errors.Table 2
Relative errors and runtime of different methods (NDZ = 20,q=0.5).
Analysis scheme
Relative error
Runtime (s)
FDTD
1.08%
0.0011
D
2-MRTD
1.93%
0.0050
D
3-MRTD
0.66%
0.0054
D
4-MRTD
0.62%
0.0058Figure8 shows the relative errors versus the space discretization numbers. The space interval will decrease with the increase of the space discretization number. We can see from the figures that the relative errors for D2-MRTD increase with the increase of NDZ. And, for D3-MRTD and D4-MRTD, the relative errors decrease with the increase of NDZ and show a little smaller relative error than the conventional FDTD. It means we can get a more accurate result at same space interval and time interval by D3-MRTD and D4-MRTD.Figure 8
Relative error for different discretization numbers. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)Figure9 describes the relative errors versus the Courant numbers. The time interval will decrease with the decrease of the Courant number. The results show that, with the decrease of the Courant numbers, the relative errors are almost unchanged for the conventional FDTD method, while the relative errors are quite different with different Courant numbers for all the three MRTD schemes. That means the choice of the Courant number significantly affects the relative errors of the MRTD schemes, and we could choose the proper Courant number to optimize the MRTD schemes.Figure 9
Relative errors for different Courant numbers. The space discretization numbers are NDZ = 20 for (a) and NDZ = 50 for (b).
(a)
(b)To validate the stability of the MRTD schemes, we increase the initial value of the pulse to 20 V and keep other parameters unchanged. Figure10 shows the relative errors versus the pace discretization numbers. It can be seen from the figures that the D2-MRTD scheme shows a larger relative error and the relative errors of D3-MRTD and D4-MRTD schemes are smaller than the conventional FDTD method.Figure 10
Relative error for different space discretization numbers for the initial value of the trapezoidal pulse to be 20 V. The Courant numbers areq=0.5 for (a) and q=0.1 for (b).
(a)
(b)The numerical results for the lossless transmission lines also show that theD2-MRTD does not perform better than the FDTD method; meanwhile, D3-MRTD and D4-MRTD schemes show better quality in accuracy and stability. The reason for this phenomenon is that the scaling function of the D2 wavelet does not have enough high vanishing moment. When we use Daubechies’ scaling functions to expand the voltages and currents in the two-conductor transmission lines equations, the vanishing moment decides the accuracy of the approximation. The scaling functions with high vanishing moments could approximate voltages and currents more accurately; however, the scaling functions with low vanishing moments like D2 wavelet’s may introduce a larger error. So we can see from Figure 5 that D2-MRTD has a better numerical dispersion property than the FDTD method, but it gets larger relative errors in numerical computation. For the scaling functions which have high enough vanishing moments like D3 wavelet and D4 wavelet, the numerical results show smaller relative errors and are in agreement with the dispersion analysis. However, a high vanishing moment for scaling function may increase the computation complexity in MRTD schemes. So the vanishing moments of the wavelet’s scaling functions have a great effect on the accuracy of the MRTD scheme; it is necessary to choose a scaling function with proper vanishing moment when we use the MRTD scheme for the numerical computation.


## 4.2. Lossy Transmission Lines

In this section, we will consider the lossy two-conductor transmission lines shown in Figure11. Two conductors of rectangular cross section of width w=20 μm and thickness t=10 μm are separated by s=20 μm and placed on one side of a silicon substrate (εr=12) of thickness h=100 μm; the total line length is L=20 cm. The near end is a source with a RS=50 Ω resistance and the far end is a load with a RL=50 Ω resistance and LL=0.5 μH inductance in series. The per-unit-length inductance and capacitance were computed as l=0.805969 μH/m and c=88.2488 pF/m. This gives a velocity of v=1.18573×108 m/s and a one-way time delay of TD=1.68672 ns, which gives an effective dielectric constant of (εr′=6.4) and characteristic impedance of ZC=95.566 Ω. The per-unit-length dc resistance is computed as r=1/(σwt)=86.207 Ω/m [18]. Dielectric loss is not included in these calculations, which means the per-unit-length conductance is g=0.Figure 11
A lossy printed circuit board: (a) line dimensions and terminations and (b) cross-sectional dimensions.
(a)
(b)The source is a ramp function as shown in Figure12, the initial value of the ramp function is 0V, and the amplitude is A=1 V with a rise time of τr=0.5 ns. The total computing time is 20 ns. The bandwidth of the source is approximate to BW=1/τr=2 GHz. The space discretization step for the MRTD was chosen to be λ/10, so the space discretization number NDZ should be greater than 34. The time step will also be calculated by (44) with different Courant numbers.Figure 12
Representation of the source voltage waveform.We calculate the near end voltage and the far end voltage of the lossy PCB with space discretization number NDZ = 40 and the Courant numberq=0.5. Figure 13 shows the computing results. Since D2-MRTD may introduce a larger error in the computation as shown in Section 4.1, we use D3-MRTD and D4-MRTD to compute the terminal voltages.Figure 13
Terminal voltages for lossy transmission lines: (a) for the near end and (b) for the far end. The space discretization number isNDZ=40 and the Courant number is q=0.5.
(a)
(b)Here, we choose the time-domain to frequency-domain transformation method (TDFD), which is a straightforward adaptation of a common analysis technique for lumped, linear circuits and systems [18], to validate the computing results of MRTD schemes and FDTD method. Table 3 shows the relative errors and the runtime of the MRTD schemes and FDTD method. It can be seen that MRTD schemes spend much time to obtain more accurate results.Table 3
Relative errors and runtime of different methods.V0 represents the near end voltage and VL represents the far end voltage (NDZ = 40, q=0.5).
Analysis scheme
Relative error forV0
Relative error forVL
Runtime (s)
FDTD
0.30%
0.92%
0.0066
D
3-MRTD
0.18%
0.57%
0.0245
D
4-MRTD
0.17%
0.54%
0.0275Figure14 describes the relative errors versus the space discretization numbers. For both D3-MRTD and D4-MRTD, the MRTD schemes show a little smaller relative error than the conventional FDTD. That means the MRTD schemes could obtain a more accurate solution under the same time interval and space interval. This is because D3-MRTD and D4-MRTD have better dispersion property than the FDTD method and the scaling functions of D3 wavelet and D4 wavelet have high enough vanishing moments.Figure 14
Relative error for the terminal voltages with different space discretization numbers: (a) for the near end and (b) for the far end. The Courant number isq=0.5.
(a)
(b)Figure15 shows the relative errors versus the Courant numbers. With the decrease of the Courant number, the relative errors of the FDTD method increase, because the time-domain is oversampled for the FDTD method. However, the MRTD schemes perform decreasing relative errors. That means the MRTD schemes could obtain a more accurate result with a smaller time interval. And even when the FDTD method is oversampled in the time-domain, the MRTD schemes perform well.Figure 15
Relative errors for the terminal voltages with different Courant numbers: (a) for the near end and (b) for the far end. The space discretization number is NDZ = 40.
(a)
(b)


## 5. Conclusion

In this paper, we derived the MRTD scheme for the two-conductor transmission lines and studied the stability and the numerical dispersion of this scheme. By viewing the MRTD schemes as the weighted mean of the conventional FDTD method at different space interval, we derived the iterative equations for the terminal voltages when the terminals are pure resistive, and a method is proposed to update the iterative equations which contain some terms whose indices exceed the index range in the MRTD scheme. Using the same method, we derived the iterative equation for the inductive load. Then we extended the MRTD scheme to the lossy transmission lines. Using different wavelets’ scaling functions as basis functions, the MRTD schemes are implemented for both lossless case and lossy case and the numerical results are compared to the conventional FDTD method. The numerical results show the MRTD schemes need more runtime to obtain more accurate results. And the vanishing moment of the wavelet’s scaling functions will significantly affect the quality of the MRTD scheme; using a scaling function with a proper vanishing moment as basis function in MRTD scheme could obtain a more accurate result.



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*Source: 8045749-2016-05-18.xml*