英语翻译The 11 pi-networks are connected in series as shown inFigure 2 to form the transmission line model.Note that theshunt elements of adjacent pi-networks are combined inparallel.The capacitor voltages and inductor currents aredesignated as t

英语翻译The 11 pi-networks are connected in series as shown inFigure 2 to form the transmission line model.Note that theshunt elements of adjacent pi-networks are combined inparallel.The capacitor voltages and inductor currents aredesignated as t
英语翻译
The 11 pi-networks are connected in series as shown in
Figure 2 to form the transmission line model.Note that the
shunt elements of adjacent pi-networks are combined in
parallel.The capacitor voltages and inductor currents are
designated as the state variables.Examination of Figure 2
indicates that the state equations for the transmission line
have a special structure.The state for each of the shunt
capacitors is affected only by the two inductors connected to it
and the capacitor itself.This will be demonstrated more
clearly with an example system in a later section.
SOLUTION PROCEDURE
The model of the transmission line presented in the
previous section is combined with other component models.
State equations are formulated for the system and can be
written in the form of (1).This set of linear ordinary
differential equations is transformed into a set of linear
difference equations using trapezoidal integration [8].If T is
the time step,equation (1) becomes
All matrices in (5) are constant for a fixed time step;
they are evaluated before simulation begins.An LU
factorization is also performed on A' before simulation begins.
The vector y[k] must be evaluated at each discrete point in
time.
Because of the transmission line model employed here,A
is a tridiagonal matrix.Therefore,A' and A" are also
tridiagonal.Only the nonzero elements of A" and B' are
stored to decrease memory requirements and execution time.
In order to retain the sparsity of A,Crout reduction [9] is
utilized to obtain an LU factorization of A' considering its
tridiagonal nature.Crout reduction may be applied to a
system of linear equations if the diagonal entries of L are
nonzero [9].For a set of n equations in n unknowns,Crout
reduction requires only (5n - 4) multiplications/divisions and
(3n - 3) additions/subtractions to solve the set of equations.
This can result in considerable savings in execution time for
large n.
The simulation procedure will now be summarized:
1.Calculate A',A",and B' from (5).
2.Compute LU factorization of A'.
3.Set k = 0.
4.Calculate b'[k] from (6).
5.Solve for x[k+l] using (4).
7.Repeat 4,5,and 6 until end of sirnulation.
6.k = k + 1.
EXAMPLE SYSTEM
The state variable technique presented in this paper is
illustrated using an example from [5].In this example,a
lightning strike occurs at the center of a 20 kV,10 km line
which has R' = 0.05 Q/km,G' = 0.556 pS/km,L' = 1 mH/km,
and C' = 11.11 nF/km.A transformer is connected to one end
of the line which is modeled by a capacitance to ground of
CT = 6 nF.The other end of the line is terminated in an open
circuit.The lightning strike is modeled by an ideal square
wave current source connected to a node in the middle of the
line; the source has a magnitude of 20 kA for 20 ps.The
transformer is protected by a surge arrester whose v-i curve is
approximated as:RA = 2 MQ for VA < 55 kV and RA = 4.5 Q
for VA > 55 kV.

英语翻译The 11 pi-networks are connected in series as shown inFigure 2 to form the transmission line model.Note that theshunt elements of adjacent pi-networks are combined inparallel.The capacitor voltages and inductor currents aredesignated as t
11pi-networks串联连接如图所示
图2的形式传输线模型.请注意,
分流元素相邻pi-networks组合
平行的.电容电压和电感电流
指定为状态变量.检验图2
指出状态方程的传输线
有一个特殊的结构.国家为每个分流
电容器是唯一受影响的电感连接到它
和电容本身.这将是展示更多
明显的例子系统在后来的一段.
该模型中提出的传输线
前一部分是与其他组件模型.
状态方程制定系统和可
书面的形式（1）.这套线性
微分方程转化为一组线性
差分方程使用梯形积分[ 8].如果是
时间步长,方程（1）成为
所有矩阵（5）是不变的,一个固定的时间步；
他们评价模拟开始之前.一路
分解也进行“模拟前开始.
向量的[钾]必须评估在每一个离散点
时间.
由于传输线模型受雇在这里,一个
是一个对角矩阵.因此,“和”是也
三对角矩阵.只有非零元素的“乙”
存储减少内存要求和执行的时间.
为了保持稀疏的一组减少,[ 9 ]是
利用获得的一个考虑其分解
三对角性质.能减少可能被应用到
线性方程系统,如果对角线条目升
非零[ 9].一集的方程中未知数,克劳特
减少只需要（—4）乘法/部门
（П- 3）加法或减法求解方程组.
这可能导致大量节省执行时间
大的
仿真程序现在将总结：
1.计算”,“,”（5）.
2.计算分解的'.
3.集金=0.
4.计算乙'[]从（6）.
5.解决×[+]使用（4）.
7.重复4,5,和6个结束薄膜.
6.钾,钾1.
示例系统
状态变量技术介绍
用一个例子说明从[ 5].在这个例子中,一个
雷击发生在一个20伏,10公里的线路
哪有' =0.05/公里,克' =0.556/公里,我=1氢/公里,
和' =11.11核/公里.一个变压器连接到一个结束
本线是模仿的对地电容
克拉=6因子.线的另一端终止在一个开放的
电路.雷击模型的一个理想的方
电流源连接到一个节点的中间的
线；源有一个大小为20的20家
变压器保护用避雷器的伏安特性曲线
近似为：甲酸=2t弗吉尼亚55.