METHOD OF MOMENTS SOLUTION FOR CHARGED LINE AND PLATE IMMERSED IN A DIELECTRIC

Transcription

1 ECE Project 3 METHOD OF MOMENTS SOLUTION FOR CHARGED LINE AND PLATE IMMERSED IN A DIELECTRIC May 7, 2014 Andrew H. Velzen Purdue University Professor Dan Jiao

2 1 Problem Statement The two problems which I seek to solve are as follows: Simulate the capacitance per unit length of a metallic strip of width W immersed in a dielectric region of relative permittivity ε r, and Simulate the capacitance of a metallic square plate of side length L immersed in a dielectric region of relative permittivity ε r, with various given values for W, L, and ε r. I employed the Method of Moments (MoM) to produce the solutions. I also went further, by generalizing the solution to any values of W, L, and ε r. I used MATLAB to implement the solution. 2 Approach How I went about solving the problem/presenting the solution in a user-frily way, is as follows. I first wrote the MoM solution for the metallic strip case, and plotted the results of the charge distribution for the requested conditions given in the problem statement and assured that they looked reasonable. Following that, I wrote the MoM solution for the metallic square plate case. Again, I plotted and verified the requested results. I also calculated the capacitance in each case, as requested in the problem. Lastly, I developed a Graphical User Interface (GUI), using the Graphical User Interface Design Environment (GUIDE) for MATLAB, to display any numeric and graphical solution elegantly. 3 Solution Implementation 3.1 MoM Solution for Metallic Strip I will begin by discussing the MoM solution for the metallic strip. I wrote a modular function to solve this problem, the code for which can be seen in Appix B, Subsection 1. The function takes in the width of the strip, as well as the relative permittivity of the dielectric, and returns an array containing the charge density, in C m, as well as the capacitance of the strip. I first define all the variables that will be used during the execution of the function. A small note on the step size in X. I decided to discretize it into 100 steps because it seemed to provide a reasonable balance between computation time, and solution resolution. I then instantiate the Z matrix and the potential array that will be used to generate the charge density array. The potential array, {V 0 }, is all 1 s because we are going to assume the voltage on the strip is 1V everywhere to solve the problem. The Z matrix and potential array correspond to the solution of [Z]{α} = {V 0 }, as discussed in class. Next, I populate the Z matrix with the appropriate values, as derived in class. The value of Z mn when m = n is 2πε [ln( ln xm xn 2 ) 1] and 2πε when m n, where is the step size, in X, and ε is the dielectric permittivity. From there, I use MATLAB s built in matrix inversion to calculate {α} = [Z] 1 {V 0 }. After that, I found the total charge on the strip by integrating the {α} array (taking the charge density value at 2

3 each index, and multiplying by the step size, in X). Lastly, before returning the requisite values from the function, I calculate the capacitance by dividing the total charge by the test voltage applied to the strip. 3.2 MoM Solution for Metallic Square Plate Now I will discuss the MoM solution for the metallic square plate. Again, I wrote a modular function to solve this problem, the code for which can be seen in Appix B, Subsection 2. The function takes in the side length of the square, as well as the relative permittivity of the dielectric, and returns a matrix containing the charge density, in C m, as well as the capacitance of the square. 2 As with the strip, I first define all the variables that will be used during the execution of the function. Different from the strip, though, is the step size I chose to use. I discretize the domain into 25 steps in X, and 25 in Y. When I tried 100 for each side, there was a noticeable lag with the computation, so I opted for 25 instead. I also saw no reason to choose different step sizes for X and Y, given that the problem was symmetric, but, in theory, one could easily do it. I next create the Z matrix and potential array, and populated them with zeros and ones, respectively. I then iterate through the Z matrix, finding the corresponding X and Y coordinates at the center of the grid square at each location. After finding those coordinates, I then populate the Z matrix with the ln(1+ 2) following values for Z mn : πε when m = n and 1 S n 4πε (xm x n) 2 +(y m y n) 2 when m n, where is the discretization in X or Y, x m and x n are the X- locations of the m and n indices, respectively, y m and y n are the Y-locations of the m and n indices, respectively, ε is the permittivity of the dielectric, and S n is 2. Now, as with the strip, I calculate the charge density array by taking [Z] 1 {V 0 }. I then calculate the total charge, again by integrating the {α} array (taking the charge density value at each index, and multiplying by the area step size, which is S n ). Again, as with before, I calculate the capacitance of the structure by dividing the total charge by the test voltage applied to the square. The last thing I do, before returning everything from the function, is to create a charge density matrix, which makes the result much easier to plot than an array does. All that must be done is correlate the indices in the [α] matrix to those in the {α} array, and then set the values of the [α] matrix to match those of the {α} array at equivalent indices. 3.3 Main MATLAB Function Code Next I will present the short section of code I wrote to actually execute the functions with the appropriate values and to plot the results, which can be seen in Appix B, Subsection 3. I first create two variables that describe how many solutions there will be of each type (metallic strip and metallic square). These will be used for indexing purposes later. I then create two arrays, one for the widths of the strips/squares, and one for the ε r of the dielectrics surrounding the strips/squares. I put into these arrays all of the combinations of W and ε r 3

4 we want to solve, from the given problem statement. I then create two cells which will hold the charge distribution solutions of each type that are returned from the two previously described functions. After that, I run two loops that s all of the combinations of W and ε r to the appropriate functions, get the solutions back, and put those solutions into their appropriate places in the charge distribution solution cells as well as a capacitance array for storing the capacitance values. All that remains is to plot the charge distributions. This is the job of the two remaining for loops. These for loops display the charge distributions on separate figures, one for the strip solutions and one for the square solutions, as well as make them more legible by adding appropriate axis labels, titles, capacitance values, and changing things such as color and font size. 3.4 GUI Creation The last coding part of the project was the addition of a GUI to make the solution more amenable to any requisite values of W and ε r for both the strip and square. The figure file was created using GUIDE, and the code that was associated with the figure file can be seen in Appix B, Subsection 4. All of the code basicallyjust sets up the GUI or is a blank function that currentlydoes not do anything, up until the Simulate Button Callback function. What this function does first is poll the editable boxes for the user requested values of W andε r. Itthen ssthesevaluestotheappropriatefunction, eitherthemetallic strip or metallic square, deping on which radio button the user selected. After that function ss back the charge distribution and capacitance, the program plots the charge distribution on the axes that are in the GUI, including in the title such information as the requested permittivity and width (rounded to look nicer on the title), as well as capacitance. Any time the simulate button is pressed by the user, this whole sequence will be executed again, allowing the user to see as many results as they wish before closing the program. 4 Results The results from the execution of the main MATLAB function can be seen in Appix A, Figures 1 and 2, as well as a few images of the GUI, Figures 3 and 4. The charge distributions in Figures 1 and 2 looks as one would expect. Given that a constant voltage is applied, one would expect the majority of the charge to be on the edges of strip and around the perimeter of the square, because of electrostatic repulsion. We can also see that, in both cases, the capacitance decreases with decreasing geometric size. This is again expected, given that the smaller a strip or square is, the less charge per unit voltage it should be able to sustain. Also, we can see that with increased dielectric permittivity, we get increased capacitance. We can see that this should also be the expected outcome, by simply comparing this result to the well known analytical result for a parallel plate capacitor, which is C = εa d, where C is capacitance, ε is the 4

5 permittivity of the dielectric medium in between the plates, A is the area of the plates, and d is the separation between the plates. As with our example, the capacitance increases with increasing dielectric permittivity. 5 Conclusions I have thus provided a satisfactory solution to the stated problems, as well as enabledusersofthe GUI tosolveanynumberofothermetallicstripandmetallic square problems with various dimensions and dielectric permittivities. Overall, I was quite pleased with how this project turned out. I arrived at the solutions I was hoping to, and provide, what I believe to be, a very intuitive, fast, and user-frily GUI associated with my project. If I were to do it again, one thing I may change is to allow a user of my program to calculate the capacitance of anygiven2dgeometry, so that it is moreflexible. This, however, wouldnot be a simple extension of my project, because I would have had to rewrite the metallic strip/metallic square functions to be more generalized. I might also add more plot options to the GUI so that the user can display more of what they desire to see, such as the electric field distribution in the dielectric. Furthermore, I d probably allow the user to set the discretization step, so if they wanted a higher resolution within the plot at the cost of computation time, they could select that. I learned a lot about the MoM during the course of this project and about the creation of GUIs in MATLAB. This project certainly served its purpose and helped develop my understanding of numerical electromagnetics. 5

6 6 Appix A Figure 1 Figure 2 6

7 Figure 3 Figure 4 7

8 7 Appix B 7.1 Metallic Strip Function Code % Project 3 - Metallic Strip Function % ECE Numerical Electromagnetics % Professor Dan Jiao % Purdue University % % Andrew Velzen % function [Alpha_Vector Capacitance] = metallic_strip(width, epsilon_r) %Define Variables W = width; %Width of Metallic Strip (in m) eps_r = epsilon_r; %Relative Permittivity of Dielectric Region del_x = W/100; %Discretization Step Size in X (in m) arr_length = W/del_x; %Length of the 2 arrays eps_0 = 8.854*(10^(-12)); eps = eps_r*eps_0; x_m = 0; %x-location of index m, for use later x_n = 0; %x-location of index n, for use later Capacitance = 0; Total_Charge = 0; Voltage = 1; %Define Necessary Matrices Z_Matrix = zeros(int64(arr_length),int64(arr_length)); Potential_Vector = ones(int64(arr_length),1); %If you use a voltage other than 1, this needs to be changed %For convenience, we will assume the left of the wire to be at x=0; for m = 1:arr_length for n =1:arr_length x_m = (m-1+.5)*del_x; x_n = (n-1+.5)*del_x; if m==n Z_Matrix(m,n) = (-del_x/(2*pi*eps))*(log(del_x/2)-1); else Z_Matrix(m,n) = (-del_x/(2*pi*eps))*log(abs(x_m-x_n)); Alpha_Vector = Z_Matrix\Potential_Vector; %Integrating up total charge on strip for n = 1:int64(arr_length) Total_Charge = Total_Charge + Alpha_Vector(n)*del_x; Capacitance = Total_Charge/Voltage; disp( The Capacitance of this metallic strip, in F/m, is: ); cap = num2str(capacitance); disp(cap); 8

9 7.2 Metallic Square Function Code % Project 3 - Metallic Square Function % ECE Numerical Electromagnetics % Professor Dan Jiao % Purdue University % % Andrew Velzen % function [Alpha_Matrix Capacitance] = metallic_square(side_length, epsilon_r) %Define Variables W = side_length; %Width of Metallic Square (in m) eps_r = epsilon_r; %Relative Permittivity of Dielectric Region del_x = W/25; %Discretization Step Size (in m) arr_length = (W/del_x)^2; %Length of the 2 arrays eps_0 = 8.854*(10^(-12)); eps = eps_r*eps_0; x_m = 0; %x-location of index m, for use later x_n = 0; %x-location of index n, for use later Capacitance = 0; Total_Charge = 0; Voltage = 1; %Define Necessary Matrices Z_Matrix = zeros(int64(arr_length),int64(arr_length)); Potential_Vector = ones(int64(arr_length),1); %If you use a voltage other than 1, this needs to be changed %For convenience, we will assume the left of the wire to be at x=0; for m = 1:arr_length for n = 1:arr_length x_m = (mod((m-1),sqrt(arr_length))+.5)*del_x; y_m = (floor((m-1)/sqrt(arr_length))+.5)*del_x; x_n = (mod((n-1),sqrt(arr_length))+.5)*del_x; y_n = (floor((n-1)/sqrt(arr_length))+.5)*del_x; if m==n Z_Matrix(m,n) = log(1+sqrt(2))*del_x/(pi*eps); else Z_Matrix(m,n) = (1/(4*pi*eps))*(del_x*del_x)/sqrt((x_m-x_n)^2+(y_m-y_n)^2); Alpha_Vector = Z_Matrix\Potential_Vector; %Integrating up total charge on strip for n = 1:int64(arr_length) Total_Charge = Total_Charge + Alpha_Vector(n)*del_x*del_x; Capacitance = Total_Charge/Voltage; disp( The Capacitance of this metallic square, in F, is: ); cap = num2str(capacitance); disp(cap); 9

10 % Make Alpha Vector into a Matrix, rather than a vector Alpha_Matrix = zeros(sqrt(arr_length),sqrt(arr_length)); for m = 1:sqrt(arr_length) for n = 1:sqrt(arr_length) Alpha_Matrix(m,n) = Alpha_Vector((m-1)*sqrt(arr_length)+n); 10

11 7.3 Main MATLAB Code % Project 3 - Main Code % ECE Numerical Electromagnetics % Professor Dan Jiao % Purdue University % % Andrew Velzen % 4/23/2014 % Creating the solutions requested in project description Num_Solutions_Strip = 6; % Number of strip solutions requested Num_Solutions_Square = 4; % Number of square solutions requested Num_Solutions = Num_Solutions_Strip + Num_Solutions_Square; width = zeros(num_solutions); % Widths for strip solutions width(1) =.01; width(2) =.001; width(3) = ; width(4) =.01; width(5) =.001; width(6) = ; % Widths for square solutions width(7) =.01; width(8) =.001; width(9) =.01; width(10) =.001; eps_r = zeros(num_solutions); % Epsilon_r s for strip solutions eps_r(1) = 1; eps_r(2) = 1; eps_r(3) = 1; eps_r(4) = 4; eps_r(5) = 4; eps_r(6) = 4; % Epsilon_r s for square solutions eps_r(7) = 1; eps_r(8) = 1; eps_r(9) = 4; eps_r(10) = 4; Strip_Charge_Distribution = cell(num_solutions_strip,100); Square_Charge_Distribution = cell(num_solutions_square,10,10); % Solutions for the strip for i = 1:Num_Solutions_Strip [Strip_Charge_Distribution{i}, Cap(i)] = metallic_strip(width(i),eps_r(i)); % Solutions for the square for i = 1:Num_Solutions_Square [Square_Charge_Distribution{i}, Cap(i+Num_Solutions_Strip)] = metallic_square(width(i+num_solutions_strip),ep 11

12 % Plotting the results for i = 1:Num_Solutions_Strip subplot(2,3,i) plot(strip_charge_distribution{i}, Color, magenta, LineWidth,1) Title_Print = sprintf( Strip Width %0.4g m, Relative Permittivity %1.1f \n Capacitance = %.5g F/m,width(i),e title(title_print, fontsize,14) tick_spread = width(i)/100; x_label_print = sprintf( Location on Strip (in %.4g m), tick_spread); xlabel(x_label_print, fontsize,14); ylabel( Charge Density (in Coulombs/m), fontsize,14); figure; for i = 1:Num_Solutions_Square subplot(2,2,i) surf(square_charge_distribution{i}); Title_Print = sprintf( Square Width %0.4g m, Relative Permittivity %1.1f \n Capacitance = %.5g F,width(i+Num title(title_print, fontsize,14) tick_spread = width(i+num_solutions_strip)/25; x_label_print = sprintf( X-Location on Square (in %.4g m), tick_spread); xlabel(x_label_print, fontsize,14); y_label_print = sprintf( Y-Location on Square (in %.4g m), tick_spread); ylabel(y_label_print, fontsize,14); zlabel( Charge Density (in Coulombs/m^2), fontsize,14); 12

13 7.4 GUI Creation Code % Project 3 - GUI Execution Code % ECE Numerical Electromagnetics % Professor Dan Jiao % Purdue University % % Andrew Velzen % 4/23/2014 function varargout = Project_3_GUI(varargin) gui_singleton = 1; gui_state = struct( gui_name, mfilename,... gui_singleton, gui_singleton,... gui_layoutfcn, [],... gui_callback, []); if nargin && ischar(varargin{1}) gui_state.gui_callback = str2func(varargin{1}); if nargout [varargout{1:nargout}] = gui_mainfcn(gui_state, varargin{:}); else gui_mainfcn(gui_state, varargin{:}); function Project_3_GUI_OpeningFcn(hObject, eventdata, handles, varargin) handles.output = hobject; guidata(hobject, handles); function varargout = Project_3_GUI_OutputFcn(hObject, eventdata, handles) varargout{1} = handles.output; function eps_r_edit_box_callback(hobject, eventdata, handles) function eps_r_edit_box_createfcn(hobject, eventdata, handles) if ispc && isequal(get(hobject, BackgroundColor ), get(0, defaultuicontrolbackgroundcolor )) set(hobject, BackgroundColor, white ); function sidelength_edit_box_callback(hobject, eventdata, handles) function sidelength_edit_box_createfcn(hobject, eventdata, handles) if ispc && isequal(get(hobject, BackgroundColor ), get(0, defaultuicontrolbackgroundcolor )) set(hobject, BackgroundColor, white ); function Simulate_Button_Callback(hObject, eventdata, handles) width = str2num(get(handles.sidelength_edit_box, String )); eps_r = str2num(get(handles.eps_r_edit_box, String )); Cap = 0; axes(handles.axes1); 13

14 %%%%%%%%%%%%%%%%%%%% % Perform the Metallic Strip Calculation if (get(handles.strip_radio_button, Value )==1) Strip_Charge_Distribution = cell(1,100); [Strip_Charge_Distribution{1}, Cap] = metallic_strip(width,eps_r); % Plot Results plot(strip_charge_distribution{1}, Color, magenta, LineWidth,1) Title_Print = sprintf( Strip Width %0.4g m, Relative Permittivity %1.1f \n Capacitance = %.5g F/m,width,eps_r title(title_print, fontsize,14) tick_spread = width/100; x_label_print = sprintf( Location on Strip (in %.4g m), tick_spread); xlabel(x_label_print, fontsize,14); ylabel( Charge Density (in Coulombs/m), fontsize,14); %%%%%%%%%%%%%%%%%%%% % Otherwise, Perform the Metallic Square Calculation else Square_Charge_Distribution = cell(1,10,10); [Square_Charge_Distribution{1}, Cap] = metallic_square(width,eps_r); % Plot Results surf(square_charge_distribution{1}); Title_Print = sprintf( Square Width %0.4g m, Relative Permittivity %1.1f \n Capacitance = %.5g F,width,eps_r, title(title_print, fontsize,14) %tick_spread = width/25; %x_label_print = sprintf( X-Location on Square (in %.4g m), tick_spread); %xlabel(x_label_print, fontsize,14); %y_label_print = sprintf( Y-Location on Square (in %.4g m), tick_spread); %ylabel(y_label_print, fontsize,14); zlabel( Charge Density (in Coulombs/m^2), fontsize,14); 14