clear
clc
% created by ujwol palanchoke
display('for the given problem, lx=120, ly=60,h=15, w=30')
display('the program will evaluate the impedance and phatse velocity in microstrip transmission line')
display('enter n=1 to simulate the above dimension, enter n= integer greater than 1 to simulate other dimensions')
display('you can simulate for different strip voltages, for that enter voltage')
V_m= input('Enter Voltage in strip='); %voltage in metallic strip
n=input('Enter the integer number to scale the dimensions (note only lx and ly could be changed)='); % change the dimensions
%define geometry
Lx = n*120; Ly = n*60; h=15; w=30; %all dimensions are in mm
% define permitivity
e_1=8.854e-12; %permitivity of air
e_2=3.5*e_1; %permitivity of dielectric. you can change premitivity accourding to dielectric used
%define differential lengths and number of nodes in each direction,
%discretize the domain
dx = 5; %differential x length of cells
Nx = (Lx/dx)+1; %number of x nodes
dy = 5; %differential y length of cells
Ny = (Ly/dy)+1; %Number of Y nodes
Ny_d=(h/dy)+1; %number of Y nodes in dielectics
Ny_a=Ny-Ny_d; %number of Y nodes in air
Nx_m=w/dx; %number of x node in metal strip
Nx_nm=(Lx-w)/10; % number of nodes in non-metal ,
% WE Assume the symmetricity, ie the distance between metallic wall and one
% end of strip is equal in both sides
%Initialize grid
V(:,:)=zeros(Ny,Nx); %create 13*25 matrix which carries the grid voltages
%initialize the voltage at metal strip. The v(:,:) forms the matrix
%which depicts the geometry for the given problem. here v(1,1) is the
%top-left corner of the transmission line with shielding. following this
%convention, the metal strip lies in (Ny_a+1,?) where? is the grid points which
%represents the actual position of metal strip
for j=(Nx_nm+1):(Nx_nm+Nx_m) % we write j for x-axis in problem, x axis 10-16
V(Nx_nm+1,j)=V_m; % the voltage at metal strip is the input value
end
% caluclate the voltage at the grid points with dielectric
for iteration=1:200 % for good approximatation we compute the value for 200 times
for i=2:Ny-1
for j=2:Nx-1
V_temp=V(i,j);
e_com=2*(e_1+e_2);
if (Ny==Ny_a+1) %at the air dielectric boundary, 10th row in the matrix. Nx_a=9,
V(i,j)=((e_1/e_com)*V(i+1,j))+((1/4)*V(i,j-1))+((e_2/e_com)*V(i-1,j))+((1/4)*V(i,j+1));
else
V(i,j)=(1/4)*(V(i+1,j)+V(i,j-1)+V(i-1,j)+V(i,j+1));
end
for j=(Nx_nm+1):(Nx_nm+Nx_m) % we write j for x-axis in problem
V(Nx_nm+1,j)=V_m; % the voltage at metal strip is the input value
end
end
end
end
V;
imagesc(V) % image for the voltage inside the shielded microstrip.
xlabel('x')
ylabel('y')
title('voltage distribution with dielectric')
% now we calculate the total charge inclosed with dielectric in botttom
% layer
% for this condition e_2=3.5*e_1, and e_1=e_0(permitivity of free space)
% evaluating in top face (L1 in lecture) and bottom face (L2 in lecture)
L_1=0; %initilize the sum
L_2=0; %initilize the sum
for j=2:Nx-1
L_2=L_2+(e_2*(V(Ny,j)-V(Ny-1,j))); %bottom face
L_1=L_1+(e_1*(V(1,j)-V(2,j))); %top face
end
% evaluating in right face L3 in lecture. To do this we have to evaluate in
% dielectric, boundary and air.
% initialize the values
L_3_1=0; %first summation in lecture, in dielectric
L_3_2=0; %second term in lecture, in boundary
L_3_3=0; %third summation in lecture, in air
for i=2:Ny_a %Nx_a=9, and row 1-9 of the matrix corresponds to air
L_3_3=L_3_3+ e_1*(V(i,Nx)-V(i,Nx-1)); %voltage at 25- at 24
end
for i=Ny_a+2:Ny % row 11-13 corresponds to dielectric
L_3_1=L_3_1+e_2*(V(i,Nx)-V(i,Nx-1));
end
L_3_2=L_3_2+((e_1+e_2)/2)*(V(Ny_a+1,Nx)-V(Ny_a+1,Nx-1));% row 10 corresponds to boundary
% evaluate the total sum in right face
L_3=L_3_1+L_3_2+L_3_3;
% evaluating in left face L4 in lecture. To do this we have to evaluate in
% dielectric, boundary and air.
% initialize the values
L_4_1=0; %first summation in lecture, in dielectric
L_4_2=0; %second term in lecture, in boundary
L_4_3=0; %third summation in lecture, in air
for i=2:Ny_a %Nx_a=9, and row 1-9 of the matrix corresponds to air
L_4_3=L_4_3+ e_1*(V(i,1)-V(i,2));
end
for i=Ny_a+2:Ny % row 11-13 corresponds to dielectric
L_4_1=L_4_1+e_2*(V(i,1)-V(i,2));
end
L_4_2=L_4_2+((e_1+e_2)/2)*(V((Ny_a+1),1)-V((Ny_a+1),2));% row 10 corresponds to boundary
% evaluate the total sum at left face
L_4=L_4_1+L_4_2+L_4_3;
%evaluate the total charge inclosed
Q_ed=-1*(L_1+L_2+L_3+L_4)
%evaluate the capacitance with dielectric
C_ed=Q_ed/V_m %V_m is the potential of the strip
%**************************************************************************
%**************************************************************************
% now evaluating the voltage and charge enclosed without dielectric
% for this we repalce the dielectric with air, ie e_2=e_1
e_2=e_1; % repalce dielectric with air
% same code can be used to evaluate the charge withou dielectric
for j=(Nx_nm+1):(Nx_nm+Nx_m) % we write j for x-axis in problem, x axis 10-16
V(Ny_a+1,j)=V_m; % the voltage at metal strip is the input value
end
% caluclate the voltage at the grid points without dielectric
for iteration=1:200 % for good approximatation we compute the value for 200 times
for i=2:Ny-1
for j=2:Nx-1
V_temp=V(i,j);
e_com=2*(e_1+e_2);
if (Ny==Ny_a+1) %at the air dielectric boundary, 10th row in the matrix. Nx_nm=9,
V(i,j)=((e_1/e_com)*V(i+1,j))+((1/4)*V(i,j-1))+((e_2/e_com)*V(i-1,j))+((1/4)*V(i,j+1));
else
V(i,j)=(1/4)*(V(i+1,j)+V(i,j-1)+V(i-1,j)+V(i,j+1));
end
for j=(Nx_nm+1):(Nx_nm+Nx_m) % we write j for x-axis in problem
V(Nx_nm+1,j)=V_m; % the voltage at metal strip is the input value
end
end
end
end
V;
figure
imagesc(V) % image for the voltage inside the shielded microstrip.
xlabel('x')
ylabel('y')
title('voltage distribution without dielectric')
% evaluating in top face (L1 in lecture) and bottom face (L2 in lecture)
L_1=0; %initilize the sum
L_2=0; %initilize the sum
for j=2:Nx-1
L_2=L_2+(e_2*(V(Ny,j)-V(Ny-1,j)));
L_1=L_1+(e_1*(V(1,j)-V(2,j)));
end
% evaluating in right face L3 in lecture. To do this we have to evaluate in
% dielectric, boundary and air.
% initialize the values
L_3_1=0; %first summation in lecture, in dielectric
L_3_2=0; %second term in lecture, in boundary
L_3_3=0; %third summation in lecture, in air
for i=2:Ny_a %Ny_a=9, and row 1-9 of the matrix corresponds to air
L_3_3=L_3_3+ e_1*(V(i,Nx)-V(i,Nx-1));
end
for i=Ny_a+2:Ny % row 11-13 corresponds to dielectric
L_3_1=L_3_1+e_2*(V(i,Nx)-V(i,Nx-1));
end
L_3_2=L_3_2+((e_1+e_2)/2)*(V(Ny_a+1,Nx)-V(Ny_a+1,Nx-1));% row 10 corresponds to boundary
% evaluate the total sum in right face
L_3=L_3_1+L_3_2+L_3_3;
% evaluating in left face L4 in lecture. To do this we have to evaluate in
% dielectric, boundary and air.
% initialize the values
L_4_1=0; %first summation in lecture, in dielectric
L_4_2=0; %second term in lecture, in boundary
L_4_3=0; %third summation in lecture, in air
for i=2:Ny_a %Ny_a=9, and row 1-9 of the matrix corresponds to air
L_4_3=L_4_3+ e_1*(V(i,1)-V(i,2));
end
for i=Ny_a+2:Ny % row 11-13 corresponds to dielectric
L_4_1=L_4_1+e_2*(V(i,1)-V(i,2));
end
L_4_2=L_4_2+((e_1+e_2)/2)*(V(Ny_a+1,1)-V(Ny_a+1,2));% row 10 corresponds to boundary
% evaluate the total sum at left face
L_4=L_4_1+L_4_2+L_4_3;
%evaluate the total charge enclosed
Q_eo=-1*(L_1+L_2+L_3+L_4)
C_eo=Q_eo/V_m %V_m is the potential of the strip
% evaluating the impedance Z=1/(velocity of light*(sqrt(C_ed*C_eo))
V_l=3e8; % speed of light
den=sqrt(C_ed*C_eo);
Z_imp=1/(V_l*den)
%evaluating phase velocity
m_t=sqrt(C_eo/C_ed);
U_phase=V_l*m_t