## 4 Simulation Examples
### Simulation of High-Voltage Transmission Line without Load Being Switched-In Suddenly  
This example is an electromagnetic transient simulation of a single-phase middle-long transmission line with distributed parameters. The initial phase of the sinusoidal voltage source was \(\varphi_{0}=90^{\circ}\). The other parameters of the transmission line were given in Figure 6.  
The mathematical model of the electromagnetic transient process for the transmission line was described by the telegraph equation as follows:  
\[\left\{ \begin{array}{l}-\frac{\partial v(y,t)}{\partial y}=l_{0}\frac{ \partial i(y,t)}{\partial t}+r_{0}\;\;i(y,t)\\ -\frac{\partial i(y,t)}{\partial y}=c_{0}\frac{\partial v(y,t)}{\partial t}+g_ {0}v(y,t)\end{array} \right. \tag{14}\]  
where \(y\) and \(t\) are the spatial variables of the length and the time variable, respectively; \(v(y,t)\) and \(i(y,t)\) denote the changes of the voltage and the current with respect to the space and the time, respectively; \(l_{0}\), \(c_{0}\), \(r_{0}\) and \(g_{0}\) are the unit parameters of inductance, capacitance, resistance, and susceptance, respectively. In this paper, we adopt the time-domain method to settle the telegraph Equation (14). First, in order to obtain the voltage and current variables along the transmission line, we decompose the whole line into \(N\) sub-intervals uniformly, as is shown in Figure 7. Supposing the total length of the transmission line is \(l\), we define: the length of each small interval is \(\Delta y=l/N\). Let \(v_{k}\) denote the voltage at the point, where \(y=y_{k}=k\cdot\Delta y,k\in(0,N)\); \(i_{k},k\in(1,N)\) is the current from the point \(y_{k-1}\) to the point \(y_{k}\); and \(i_{0}\) and \(i_{N+1}\) are the currents of two terminals in the line, respectively. Furthermore, we apply the fourth-order accuracy interpolation formula to do spatial discretization of the telegraph equation [20]:  
\[\theta_{1}\frac{\partial\lambda_{j}(y)}{\partial y}\bigg{|}_{j=k-1}+\left. \theta_{2}\frac{\partial\lambda_{j}(y)}{\partial y}\right|_{j=k}+\left.\theta _{1}\frac{\partial\lambda_{j}(y)}{\partial y}\right|_{j=k+1}=\frac{\lambda_{ k}-\lambda_{k+1}}{\Delta y} \tag{15}\]  
where \(\lambda\) denotes \(v\) or \(i\); \(\theta_{1}\) and \(\theta_{2}\) are the constant coefficients decided by the interpolation formula; and \(\theta_{1}=1/24\), \(\theta_{2}=1-2\theta_{2}\). Substituting Equation (15) into the telegraph equation (Equation (14)), we obtained the spatial discretization form of the telegraph equation as follows:  
\[\left\{ \begin{array}{l}\frac{v_{k}-v_{k+1}}{\Delta y}=\theta_{1}\Big{(}l_{0 }\dot{i}_{k-1}+r_{0}\dot{i}_{k-1}\Big{)}+\theta_{2}\Big{(}l_{0}\dot{i}_{k}+r _{0}\dot{i}_{k}\Big{)}+\theta_{1}\Big{(}l_{0}\dot{i}_{k+1}+r_{0}\dot{i}_{k+1} \Big{)},k\in(1,N-2)\\ \frac{i_{k}-i_{k+1}}{\Delta y}=\theta_{1}\big{(}c_{0}\dot{v}_{k-1}+g_{0}v_{k- 1}\big{)}+\theta_{2}\big{(}c_{0}\dot{v}_{k}+g_{0}v_{k}\big{)}+\theta_{1}\big{(} c_{0}\dot{v}_{k+1}+g_{0}v_{k+1}\big{)},k\in(1,N-1)\end{array} \right. \tag{16}\]  
Figure 6: Schematic diagram of a single-phase high-voltage overhead transmission line without load being switched-in suddenly. For the head-end and the tail-end of the line, we used the second-order accuracy interpolation formula [20] to do spatial discretization as follows:  
\[\left\{ \begin{array}{l}\frac{v_{0}-v_{1}}{\Delta y}=\theta_{3}\Big{(}l_{0}\dot{ i}_{1}+r_{0}\dot{i}_{1}\Big{)}+\theta_{1}\Big{(}l_{0}\dot{i}_{2}+r_{0}i_{2} \Big{)}\\ \frac{i_{0}-\dot{i}_{1}}{\Delta y}=\theta_{4}\big{(}c_{0}\dot{v}_{0}+g_{0}v_{0 }\big{)}+\theta_{1}\big{(}c_{0}\dot{v}_{1}+g_{0}v_{1}\big{)}\end{array} \right., \tag{17}\]  
and  
\[\left\{ \begin{array}{l}\frac{v_{N-1}-v_{N}}{\Delta y}=\theta_{3}\Big{(}l_{0} \dot{i}_{N}+r_{0}i_{N}\Big{)}+\theta_{1}\Big{(}l_{0}\dot{i}_{N-1}+r_{0}i_{N-1 }\Big{)}\\ \frac{i_{N}-i_{N+1}}{\Delta y}=\theta_{4}\big{(}c_{0}\dot{v}_{N}+g_{0}v_{N} \big{)}+\theta_{1}\big{(}c_{0}\dot{v}_{N-1}+g_{0}v_{N-1}\big{)}\end{array} \right. \tag{18}\]  
where \(\theta_{3}=1-\theta_{1}\) and \(\theta_{4}=1/2-\theta_{1}\).