Experiment 2 Bus Impedance Matrix
j*Topo_Structure_And_Branch_Para(CircleNumber,4));
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para( CircleNumber,2),Topo_Structure_And_Branch_Para(CircleNumber,1))=...
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And_Branch_Para( CircleNumber,2)); end end
format short
Nodal_impedance_Matrix*inv(Nodal_impedance_Matrix)
运行结果：
Nodal_impedance_Matrix =
1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i 0
-5.8824e-001 +2.3529e+000i 5.8824e-001 -2.3377e+000i 0
0 +3.6667e+000i 0 0
0 0 4.5386e-001 -1.8719e+000i
Nodal_impedance_Matrix =
1.0421e+000 -8.2429e+000i -5.8824e-001 +2.3529e+000i -4.5386e-001 +1.8911e+000i
-5.8824e-001 +2.3529e+000i 1.0690e+000 -4.7274e+000i 0
0 +3.6667e+000i 0 0
-4.5386e-001 +1.8911e+000i 0 9.3463e-001 -4.2616e+000i
ans =
1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 - 0.0000i
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0 +3.6667e+000i 0 0 -3.3333e+000i 0 0 +3.6667e+000i 0 0 -3.3333e+000i 0
-0.0000 - 0.0000i 1.0000 - 0.0000i -0.0000 + 0.0000i -0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 - 0.0000i 1.0000 - 0.0000i -0.0000 0 - 0.0000i 0 + 0.0000i 0.0000 - 0.0000i 1.0000 + 0.0000i
以上就是对阻抗矩阵的验证，其和其逆相乘为单位对角矩阵
Experiment 3
Gauss-Seidel Method 1. Objective
? To write a simple program in MATLAB? for the algorithm to solution of nonlinear algebraic equations;
? Known as the method of successive displacements. 2. Discussion
The most common methods for solving nonlinear algebraic equations are Gauss-Seidel,
Newtow-Rahpson, and quasi-Newton-Raphson methods. We start with one dimensional equations and then generalize to n-dimensional equations. 3. Mathmatics model
Consider the nonlinear equation f(x)?0.The equation is broken into two parts thus:x?g(x). We assume x(0) is an initial \
x(1)?g(x(0))
This process is repeated thus
x(2)?g(x(1))
(n)(n?1)and on the n iteration we have: x?g(x)
th
. If this process is convergent, then the successive solutions approach a value which is declared as the solution. Thus if at some step k?1 we have:
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Experiment 2 Bus Impedance Matrix
x(k?1)?x(k)??
where e ? is the desired \then we claim the solution has been found to the accuracy specified.
4. System Requirement
Computer with MATLAB? 6 or above installed. 5. Procedure
1.0 Launch the MATLAB program.
2.0 Go to FILE NEW M-file.
3.0 Write a function program of Gauss Seidel Method. 6. Exercises
Example: Using the Gauss-Seidel method to obtain the roots of the equation:
f(x)?x3?6x2?9x?4?0
First the equation is expressed in a different form thus
x??13x?6x2?4?g(x) 9?? 14
And the iteration can proceed. Take a good look at the shape of the iterations! Below is the program showing the process graphically (later showing how to do it iteratively). 7.The flow chart of Gauss Seidel method (Omitted) 8.Reference Program and result.
程序是： clear all clc x0=0.5; n=0;
while (abs(x0^3-6*x0^2+9*x0-4)>0.00001) x0=-(x0^3-6*x0^2-4)/9; y=x0; n=n+1; end
结果是：n=1627 y=x0=0.99818
clear all clc x0=2.5; n=0;
while (abs(x0^3-6*x0^2+9*x0-4)>0.00001) x0=-(x0^3-6*x0^2-4)/9; y=x0; n=n+1; end
结果是：n=7 y=x0=4
仿照高斯--赛德尔法，我们可以写出简单的牛顿法的程序,如下：
x0=0.5; n=0;
while (abs((x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9))>0.00001) dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9); x0=x0+dx0; n=n+1; end
结果是：dx0=1.7684e-005 n=15 x0=0.99998 y= -0.875
x0=0.5;
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牛顿法解方程 Experiment 2 Bus Impedance Matrix
n=0;
while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)
dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9); x0=x0+dx0; n=n+1; end
结果是: dx0=0.0011305 n=9 x0=0.99887 y= -0.875
x0=3.5; n=0;
while (abs(x0^3-6*x0^2+9*x0-4)>0.00001)
dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9); x0=x0+dx0; n=n+1; end
结果是：dx0= -2.5283e-006 n=5 x0=4 y= -0.875
Personal Summary：
The experiment of bilingual class is over,here is my personal summary.
In my opinion,first and foremost,I had to acknowledge that I have elementary know the base using of MATLAB,during approximately ten hours’ hard working.Although I have spent ten hours or less on learning this software,I merely grasp the knowledge which is just like the tip of the iceberg.In terms with the application of this software,we just do some simple steps.For instance,the node admittance matrix of node impedance matrix in date input software,the experimental program input last run results,the corresponding node admittance matrix of node impedance matrix is obtained.
Besides,I would say I haven’t master the method of how to edit a program.This is a pity.Nevertheless,I think it is just a program of time .I’ll pay more time on fulfill a deeper study.
Last but not least,I am really appreciate for teacher’s patient teaching and conducting.Thank you very much!
Because the experimental time coincides with exam review time, so I didn't understand a lot of knowledges in the experimental thoroughly, but I still hope the teacher can forgive me.
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