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How can I revise my 1D FDTD code

3 visualizaciones (últimos 30 días)
Yuanzhi
Yuanzhi el 4 de Mzo. de 2019
Editada: Sandeep Yadav Golla el 8 de Nov. de 2019
This is my code to simulate electromagnetic wave transmitting in a slab, and this is an example from CEM lecture at youtube, but code is not provided.
clc
close all
clear all
Nz = 94;
c0 = 299792458;
dz = 4.293e-3;
dt = 7.1599e-12;
L = Nz*dz;
x = linspace(0,L,Nz);
%Source parameters
STEPS = 4095;
f2 = 1.0e9;
NFREQ = 100;
FREQ = linspace(0,f2,NFREQ);
tau = 5e-10;
t0 = 3e-9;
nz_src = 3;
%Initial fourier transforms
K = exp (-1i*2*pi*dt*FREQ);
REF = zeros(1,NFREQ);
TRN = zeros(1,NFREQ);
SRC = zeros(1,NFREQ);
%Initialize materials to free space
ER = ones(1, Nz);
UR = ones(1, Nz);
ER(13:83) = 6.0;
UR(13:83) = 2.0;
%Compute update coefficients
mEy = (c0*dt)./ER;
mHx = (c0*dt)./UR;
%Compute Gaussian source functions
t = [0:STEPS-1]*dt; %time axis
delt = 1.074e-11; %total delay between E and H
A = -sqrt(ER(1)/UR(1)); %amplitude of H field
Esrc = exp(-((t-t0)/tau).^2); %E field source
Hsrc = A*exp(-((t-t0+delt)/tau).^2);%H field source
% initialize fields
Ey=zeros(1,Nz);
Hx=zeros(1,Nz);
%Initialize boundary terms
H3=0; H2=0; H1=0;
E3=0; E2=0; E1=0;
%Main FDTD loop
for T = 1 : STEPS
%Perform FDTD function
% H update
%Hx(1) = Hx(2);
for i = 1:(Nz-1)
Hx(i)=Hx(i)+mHx(i).*(Ey(i+1) - Ey(i))/dz;
end
Hx(Nz) = Hx(Nz) + mHx(Nz).*(E3 - Ey(Nz))/dz;
Hx(nz_src-1)=Hx(nz_src-1)-mHx(nz_src-1).*Esrc(T)/dz; % TF/SF source
H3=H2; H2=H1; H1=Hx(1); % ABC
% E update
%Ey(Nz) = Ey(Nz-1);
for i = 2:Nz
Ey(i) = Ey(i) + mEy(i).*(Hx(i) -Hx(i-1))/dz;
end
Ey(1) = Ey(1) + mEy(1)*(Hx(1) - H3)/dz;
Ey(nz_src-1)=Ey(nz_src-1)-mEy(nz_src-1).*Hsrc(T)/dz; % TF/SF source
E3=E2; E2=E1; E1=Ey(Nz); % ABC
%Update Fourier Transforms
for nf = 1 : NFREQ
REF(nf) = REF(nf) + (K(nf)^T)*Ey(1);
TRN(nf) = TRN(nf) + (K(nf)^T)*Ey(Nz);
SRC(nf) = SRC(nf) + (K(nf)^T)*Esrc(T);
end
%Visualize
figure(2)
hold off
plot(x,Ey,'b');
axis([0 L -1 1]);
grid on;
hold on
plot(x,Hx,'r');
pause(0);
T
end
%Finish fourier transforms
REF = REF*dt;
TRN = TRN*dt;
SRC = SRC*dt;
%Compute reflectance and transmittance
REF = abs(REF./SRC).^2;
TRN = abs(TRN./SRC).^2;
CON = REF + TRN;
figure(1)
plot(FREQ,REF)
hold on
plot(FREQ,TRN,'r-')
plot(FREQ,TRN+REF,'k-')
xlabel('Frequency(Hz)')
ylabel('Amplitude')
title('Impulse response of Etalon')
legend on
legend('Rx','Tx','Tx+Rx')
hold off
fh = figure(1);
set(fh, 'color', 'white');
There are some promblems with the boundary condition that I cannot fix. And when I change the value of nz_src, the simulation result will be totally different, the calue of E and H will reach to about 40! How could this even happen?
I referenced some code from matlab central.

Respuestas (2)

Sandeep Yadav Golla
Sandeep Yadav Golla el 8 de Nov. de 2019
Editada: Sandeep Yadav Golla el 8 de Nov. de 2019
clc;
close all;
clear all;
Nz = 94;
c0 = 299792458;
dz = 4.293e-3;
dt = 7.1599e-12;
L = Nz*dz;
x = linspace(0,L,Nz);
%Source parameters
STEPS = 4095;
f2 = 1.0e9;
NFREQ = 100;
FREQ = linspace(0,f2,NFREQ);
tau = 5e-10;
t0 = 3e-9;
nz_src = 2; %%% corrected here %%%
%Initial fourier transforms
K = exp (-1i*2*pi*dt*FREQ);
REF = zeros(1,NFREQ);
TRN = zeros(1,NFREQ);
SRC = zeros(1,NFREQ);
%Initialize materials to free space
ER = ones(1, Nz);
UR = ones(1, Nz);
ER(13:83) = 6.0;
UR(13:83) = 2.0;
%Compute update coefficients
mEy = (c0*dt)./ER;
mHx = (c0*dt)./UR;
%Compute Gaussian source functions
t = [0:STEPS-1]*dt; %time axis
delt = 1.074e-11; %total delay between E and H
A = -sqrt(ER(1)/UR(1)); %amplitude of H field
Esrc = exp(-((t-t0)/tau).^2); %E field source
Hsrc = A*exp(-((t-t0+delt)/tau).^2);%H field source
% initialize fields
Ey=zeros(1,Nz);
Hx=zeros(1,Nz);
%Initialize boundary terms
H2=0; H1=0; %%%% corrected here %%%
E2=0; E1=0;
%Main FDTD loop
for T = 1 : STEPS
%Perform FDTD function
% H update
%Hx(1) = Hx(2);
H2=H1; H1=Hx(1); % ABC
for i = 1:(Nz-1)
Hx(i)=Hx(i)+mHx(i)*(Ey(i+1) - Ey(i))/dz;
end
Hx(Nz) = Hx(Nz) + mHx(Nz)*(E2 - Ey(Nz))/dz;
Hx(nz_src-1)=Hx(nz_src-1)-mHx(nz_src-1)*Esrc(T)/dz; % TF/SF source
% E update
%Ey(Nz) = Ey(Nz-1);
E2=E1; E1=Ey(Nz); % ABC
Ey(1) = Ey(1) + mEy(1)*(Hx(1) - H2)/dz;
for i = 2:Nz
Ey(i) = Ey(i) + mEy(i)*(Hx(i) -Hx(i-1))/dz;
end
Ey(nz_src)=Ey(nz_src)-mEy(nz_src)*Hsrc(T)/dz; % TF/SF source %%% corrected here %%%
%Update Fourier Transforms
for nf = 1 : NFREQ
REF(nf) = REF(nf) + (K(nf)^T)*Ey(1);
TRN(nf) = TRN(nf) + (K(nf)^T)*Ey(Nz);
SRC(nf) = SRC(nf) + (K(nf)^T)*Esrc(T);
end
%Visualize %%% wait some time to see the simulations
figure(2)
hold off
plot(x,Ey,'b');
axis([0 L -1 1]);
grid on;
hold on
plot(x,Hx,'r');
pause(0);
end
%Finish fourier transforms
REF = REF*dt;
TRN = TRN*dt;
SRC = SRC*dt;
%Compute reflectance and transmittance
REF = abs(REF./SRC).^2;
TRN = abs(TRN./SRC).^2;
CON = REF + TRN;
figure(1)
plot(FREQ,REF)
hold on
plot(FREQ,TRN,'r-')
plot(FREQ,TRN+REF,'k-')
xlabel('Frequency(Hz)')
ylabel('Amplitude')
title('Impulse response of Etalon')
legend on
legend('Rx','Tx','Tx+Rx')
hold off
fh = figure(1);
set(fh, 'color', 'white');

alican kara
alican kara el 14 de Jul. de 2020

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