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【正文】 +000ians = + + + Experiment 2 Bus Impedance Matrix12 0 0 + + 以上就是對(duì)阻抗矩陣的驗(yàn)證，其和其逆相乘為單位對(duì)角矩陣Experiment 3GaussSeidel Method 1. Objective ? To write a simple program in MATLAB174。 for the algorithm to solution of nonlinear algebraic equations。? Known as the method of successive displacements. 2. Discussion The most mon methods for solving nonlinear algebraic equations are GaussSeidel, NewtowRahpson, and quasiNewtonRaphson methods. We start with one dimensional equations and then generalize to ndimensional equations. 3. Mathmatics modelConsider the nonlinear equation .The equation is broken into two parts thus: . We 0)(?xf )(xg?assume is an initial guess of the solution, then refine the solution using: )0(x )(0)1(xgThis process is repeated thus )(1)2(x?and on the iteration we have: thn)()(?nngx. If this process is convergent, then the successive solutions approach a value which is declared as the solution. Thus if at some step we have: 1?k13????)()1(kkx where e is the desired accuracy, then we claim the solution has been found to the accuracy ?specified.4. System Requirement Computer with MATLAB174。 6 or above installed. 5. Procedure Launch the MATLAB program. Go to FILE NEW Mfile. Write a function program of Gauss Seidel Method. 6. Exercises Example: Using the GaussSeidel method to obtain the roots of the equation:0496)(23????xxfFirst the equation is expressed in a different form thus??)(123g?Experiment 2 Bus Impedance Matrix14And 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). flow chart of Gauss Seidel method (Omitted) Program and result.程序是：clear allclcx0=。n=0。while (abs(x0^36*x0^2+9*x04))x0=(x0^36*x0^24)/9。y=x0。n=n+1。end結(jié)果是：n=1627 y=x0=clear allclcx0=。n=0。while (abs(x0^36*x0^2+9*x04))x0=(x0^36*x0^24)/9。y=x0。n=n+1。end結(jié)果是：n=7 y=x0=4仿照高斯賽德?tīng)柗?，我們可以?xiě)出簡(jiǎn)單的牛頓法的程序,如下： 牛頓法解方程x0=。n=0。while (abs((x0^36*x0^2+9*x04)/(3*x0^212*x0+9)))dx0=(x0^36*x0^2+9*x04)/(3*x0^212*x0+9)。x0=x0+dx0。n=n+1。end結(jié)果是：dx0= n=15 x0= y= x0=。15n=0。while (abs(x0^36*x0^2+9*x04))dx0=(x0^36*x0^2+9*x04)/(3*x0^212*x0+9)。x0=x0+dx0。n=n+1。end結(jié)果是: dx0= n=9 x0= y= x0=。n=0。while (abs(x0^36*x0^2+9*x04))dx0=(x0^36*x0^2+9*x04)/(3*x0^212*x0+9)。x0=x0+dx0。n=n+1。end結(jié)果是：dx0= n=5 x0=4 y= 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 I have spent ten hours or less on learning this software,I merely grasp the knowledge which is just like the tip of the terms with the application of this software,we just do some simple 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 is a ,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 you very much!Because the experimental time coincides with exam review time, so I didn39。t understand a lot of knowledges in the experimental thoroughly, but I still hope the teacher can five me.
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