I tried to code the jexact of Mie Series, but it says "Matrix dimensions must agree.
how should it be fixed?
%wave_length
lamda = 2*pi/3.5;
%speed of light
c = 299792458;
%frequency
freq = c/lamda;
%Raduis
R = 1;
%Permeability
mu = 4.0e-7*pi;
%permittivity
epsilon = 1.0/c/mu/c;
% radian frequency (w)
w = 2.0*pi*freq;
% wavenumber
K = w*sqrt(mu*epsilon);
omega=2*pi*(0:N-1)/N;
x=cos(omega)*R;
y=sin(omega)*R;
phi = atan2(y,x);
Jexact = zeros(size(x));
for nu = -10:10;
Jexact= Jexact + (1j.^(-nu).*exp(1j*nu*phi)./besselh(nu,2,K*R))*(-2/(w.*mu.*pi.*R));
end
Related
I am in the process of writing a 2D non-linear Poisson's solver. One intermediate test I performed is using my non-linear solver to solve for a "linear" Poisson equation. Unfortunately, my non-linear solver is giving me incorrect results, unlike if I try to solve it directly in MATLAB using the backslash ""
non-linear solver iteratively code: "incorrect results"
clearvars; clc; close all;
Nx = 20;
Ny = 20;
Lx = 2*pi;
x = (0:Nx-1)/Nx*2*pi; % x coordinate in Fourier, equally spaced grid
kx = fftshift(-Nx/2:Nx/2-1); % wave number vector
kx(kx==0) = 1; %helps with error: matrix ill scaled because of 0s
ygl = -cos(pi*(0:Ny)/Ny)'; %Gauss-Lobatto chebyshev points
%make mesh
[X,Y] = meshgrid(x,ygl);
%Chebyshev matrix:
VGL = cos(acos(ygl(:))*(0:Ny));
dVGL = diag(1./sqrt(1-ygl.^2))*sin(acos(ygl)*(0:Ny))*diag(0:Ny);
dVGL(1,:) = (-1).^(1:Ny+1).*(0:Ny).^2;
dVGL(Ny+1,:) = (0:Ny).^2;
%Diferentiation matrix for Gauss-Lobatto points
Dgl = dVGL/VGL;
D = Dgl; %first-order derivative matrix
D2 = Dgl*Dgl;
%linear Poisson solved iteratively
Igl = speye(Ny+1);
Ig = speye(Ny);
ZNy = diag([0 ones(1,Ny-1) 0]);
div_x_act_on_grad_x = -Igl; % must be multiplied with kx(m)^2 for each Fourier mode
div_y_act_on_grad_y = D * ZNy *D;
u = Y.^2 .* sin( (2*pi / Lx) * X);
uh = fft(u,[],2);
uold = ones(size(u));
uoldk = fft(uold,[],2);
max_iter = 500;
err_max = 1e-5; %change to 1e-8;
for iterations = 1:max_iter
for m = 1:length(kx)
L = div_x_act_on_grad_x * (kx(m)^2) + div_y_act_on_grad_y;
d2xk = div_x_act_on_grad_x * (kx(m)^2) * uoldk;
d2x = real(ifft(d2xk,[],2));
d2yk = div_y_act_on_grad_y *uoldk;
d2y = real(ifft(d2yk,[],2));
ffh = d2xk + d2yk;
phikmax_old = max(max(abs(uoldk)));
unewh(:,m) = L\(ffh(:,m));
end
phikmax = max(max(abs(unewh)));
if phikmax == 0 %norm(unewh,inf) == 0
it_error = err_max /2;
else
it_error = abs( phikmax - phikmax_old) / phikmax;
end
if it_error < err_max
break;
end
end
unew = real(ifft(unewh,[],2));
DEsol = unew - u;
figure
surf(X, Y, unew);
colorbar;
title('Numerical solution of \nabla^2 u = f');
figure
surf(X, Y, u);
colorbar;
title('Exact solution of \nabla^2 u = f');
Direct solver
ubar = Y.^2 .* sin( (2*pi / Lx) * X);
uh = fft(ubar,[],2);
for m = 1:length(kx)
L = div_x_act_on_grad_x * (kx(m)^2) + div_y_act_on_grad_y;
d2xk = div_x_act_on_grad_x * (kx(m)^2) * uh;
d2x = real(ifft(d2xk,[],2));
d2yk = div_y_act_on_grad_y *uh;
d2y = real(ifft(d2yk,[],2));
ffh = d2xk + d2yk;
%----------------
unewh(:,m) = L\(ffh(:,m));
end
How can I fix code 1 to get the same results as code 2?
I am using a quiver plot in MATLAB to simulate a velocity field. Now I would like the vectors produced by the quiver plot to be all the same length, so that they just indicate the vectors direction. The value of the velocity in each point should be illustrated by different colors then.
Is there a possibility to have quiver plotting vectors of same length?
That's my current code:
%defining parameters:
age = 900;
vis= 15;
turbulences = zeros(9,3);
a = 0.01;
spacing = 1000;
[x,y] = meshgrid(-100000:spacing:100000);%, 0:spacing:10000);
u = a;
v = 0;
n = 0;
for i = 1:4
turbulences(i,1) = -80000 + n;
turbulences(i,2) = 15000;
n = 15000 * i;
end
n = 0;
for i = 5:9
turbulences(i,1) = -15000 + n*5000;
turbulences(i,2) = 4000;
n = n+1;
end
for i = 1:4
turbulences(i,3) = -1000;
end
for i = 5:9
turbulences(i,3) = 800;
end
%compute velocities in x and y direction
for k = 1:9
xc = turbulences(k,1);
yc = turbulences(k,2);
r1 = ((x-xc).^2 + (y-yc).^2);
r2 = ((x-xc).^2 + (y+yc).^2);
u = u + turbulences(k,3)/2*pi * (((y-yc)./r1).*(1-exp(-(r1./(4*vis*age)))) - ((y+yc)./r2).*(1-exp(-(r2./(4*vis*age)))));
v = v - turbulences(k,3)/2*pi* (((x-xc)./r1).*(1-exp(-(r1./(4*vis*age)))) - ((x-xc)./r2).*(1-exp(-(r2./(4*vis*age)))));
end
quiver(x,y,u,v);
grid on;
Thank you for your help!
One way to do this would be to normalize each component of your vectors to +- 1 just to keep their direction.
un = u./abs(u); % normalized u
vn = v./abs(v); % normalized v
quiver(x, y, un, vn)
I have been working on a FastICA algorithm implementation using MatLab. Currently the code does not separate the signals as good as id like. I was wondering if anyone here could give me some advice on what I could do to fix this problem?
disp('*****Importing Signals*****');
s = [1,30000];
[m1,Fs1] = audioread('OSR_us_000_0034_8k.wav', s);
[f1,Fs2] = audioread('OSR_us_000_0017_8k.wav', s);
ss = size(f1,1);
n = 2;
disp('*****Mixing Signals*****');
A = randn(n,n); %developing mixing matrix
x = A*[m1';f1']; %A*x
m_x = sum(x, n)/ss; %mean of x
xx = x - repmat(m_x, 1, ss); %centering the matrix
c = cov(x');
sq = inv(sqrtm(c)); %whitening the data
x = c*xx;
D = diff(tanh(x)); %setting up newtons method
SD = diff(D);
disp('*****Generating Weighted Matrix*****');
w = randn(n,1); %Random weight vector
w = w/norm(w,2); %unit vector
w0 = randn(n,1);
w0 = w0/norm(w0,2); %unit vector
disp('*****Unmixing Signals*****');
while abs(abs(w0'*w)-1) > size(w,1)
w0 = w;
w = x*D(w'*x) - sum(SD'*(w'*x))*w; %perform ICA
w = w/norm(w, 2);
end
disp('*****Output After ICA*****');
sound(w'*x); % Supposed to be one of the original signals
subplot(4,1,1);plot(m1); title('Original Male Voice');
subplot(4,1,2);plot(f1); title('Original Female Voice');
subplot(4,1,4);plot(w'*x); title('Post ICA: Estimated Signal');
%figure;
%plot(z); title('Random Mixed Signal');
%figure;
%plot(100*(w'*x)); title('Post ICA: Estimated Signal');
Your covariance matrix c is 2 by 2, you cannot work with that. You have to mix your signal multiple times with random numbers to get anywhere, because you must have some signal (m1) common to different channels. I was unable to follow through your code for fast-ICA but here is a PCA example:
url = {'https://www.voiptroubleshooter.com/open_speech/american/OSR_us_000_0034_8k.wav';...
'https://www.voiptroubleshooter.com/open_speech/american/OSR_us_000_0017_8k.wav'};
%fs = 8000;
m1 = webread(url{1});
m1 = m1(1:30000);
f1 = webread(url{2});
f1 = f1(1:30000);
ss = size(f1,1);
n = 2;
disp('*****Mixing Signals*****');
A = randn(50,n); %developing mixing matrix
x = A*[m1';f1']; %A*x
[www,comp] = pca(x');
sound(comp(:,1)',8000)
I'm currently learning Matlab's ODE-functions to solve simple vibration-problems.
For instance mx''+cx'+kx=F*sin(wt) can be solved using
function dx = fun(t,x)
m=0.02; % Mass - kg
k=25.0; % Stiffness - N/m
c=0.0125; % System damping - Ns/m
f=10; % Frequency
F=5;
dx= [x(2); (F*sin(2*pi*f*t)-c*x(2)-k*x(1))/m]
And then calling the ode45 function to get displacement and velocity
[t,x]=ode45(#fun,[0 10],[0.0;0.0])
My question, which I have not fully understood searching the web, is if it is possible to use ODE-function for a multiple degree of freedom system? For instance, if we have two masses, springs and dampers, which we excite att mass 1, we get the following equations:
m1*x1''+c1*x1'-c2*x2'+(k1+k2)*x1-k2*x2 = f1(t)
m2*x2''-c2*x1'+(c1+c2)*x2'-k2*x1+k2*x2 = 0
Here, the displacements x1 & x2 depend on each other, my question is how one should go about to solve these ODE's in Matlab?
There is no restriction that the inputs to the function solved by ODE45 be scalar. Just pass in an input matrix and expect out an output matrix. For example here is a function that solves the position of a 6 bar mechanism.
function zdot = cp_solve(t,z)
%% Constants
g = -9.81;
L1 = .2;
m0 = 0;
I0 = 0;
m1 = 1;
I1 = (1/3) * m1 * L1^2;
%% Inputs
q = z(1:6);
qdot = z(7:12);
%% Mass Matrix
M = zeros(6,6);
M(1,1) = m0;
M(2,2) = m0;
M(3,3) = I0;
M(4,4) = m1;
M(5,5) = m1;
M(6,6) = I1;
%% Constraint Matrix
Phiq = zeros(5,6);
Phiq(1,1) = 1;
Phiq(2,2) = 1;
Phiq(3,3) = 1;
Phiq(4,1) = 1;
Phiq(4,4) = -1;
Phiq(4,6) = (-L1/2)*sin(q(6));
Phiq(5,2) = 1;
Phiq(5,5) = -1;
Phiq(5,6) = (L1/2)*cos(q(6));
%% Generalized Forces
Q = zeros(6,1);
Q(5) = m1 * g;
%% Right Side Vector
rs = zeros(5,1);
rs(4) = (L1/2) * cos(q(6)) * qdot(6)^2;
rs(5) = (L1/2) * sin(q(6)) * qdot(6)^2;
%% Coefficient Matrix
C = [M Phiq'; Phiq zeros(5,5)];
R = [Q; rs];
%% Solution
Sol = C \ R;
zdot = [qdot; Sol(1:6)];
end
The inputs are the positions and velocities of the members. The outputs are the new positions and velocities.
You use it the same way you would any ODE45 problem. Setup the initial conditions, define a time and solve the problem.
%% Constants
L1 = .2;
C1 = L1/2;
theta1 = 30*pi/180;
theta_dot1 = 0;
tspan = 0:.001:2;
%% Initial Conditions
q = zeros(6,1);
q(6) = theta1;
q(4) = C1 * cos(q(6));
q(5) = C1 * sin(q(6));
qdot = zeros(6,1);
qdot(6) = theta_dot1;
z0 = [q; qdot];
%% Solve the problem
options = odeset('RelTol', 1.0e-9, 'AbsTol', 1.0e-6);
[Tout, Zout] = ode45(#cp_solve, tspan, z0, options);
In your case you have 2 equations and 2 unknowns. Set the problem up as a matrix problem and solve it simultaneously in your function. I would recommend the modal approach for your case.
I have written a Matlab code which is a simulation of NOT gate implementation using ring lasers. Si variable is input in my code. But my code works only for Si = 0. for any other non zero value it doesn't show output.
%-----------Rate Equation MATLAB code ---------
function dydt = requations(t,y)
dydt = zeros(size(y));
%prompt = 'Sinput???';
%Si = input(prompt);
Si = 0; %MY INPUT
q = 1.6e-19; % charge of electron
tau_e = 1e-9; % carrier lifetime
No = 3.3e18; % # No of carriers at transparency
a = 1e-15; % Linear gain coefficient
Vg = 7.5e9; % group velocity
Vp = 3e-11; %Photon reservoir volume
V = 1e-11; %Carrier reservoir Volume
tau_p = 1.7e-12; % photon lifetime
beta = 1e-5; % spontateous emission coefficient
eps = 7.5e-17; % Nonlinear gain suppression coefficient
Ni = 0.8; %Internal quantum efficiency
w = 2*pi*10e5;
Io = 10e-3;
%I = Io*sin(w*t);
I = 2.5*Io; %for test purposes
tp = 1/tau_p;
te = 1/tau_e;
Aint = 6; %Internal losses inside cavity waveguides
%G = ((a/Vp)* (N-(V*No)))/(1-eps*(Sc + Scc));
alpha = -2; %alpha factor
L = 76e-4 ;%size of the ring
wcb = 2*pi*70;
%R = 0.25;
Wcb = wcb*1000000;
%r = 1/R;
tpcw = Vg*(Aint + ((1/L)*log(4)));
Tpcw = 1/tpcw;
%------Rate equations-------
N = y(1);
Sc = y(2); %Clock wise photon number
yc = y(3);
Scc = y(4); %anti clockwise photon number
ycc = y(5);
G = ((a/Vp)* (N-(V*No)))/(1-eps*(Sc + Scc));
dydt(1) = (Ni*I)/q - y(1)*te - Vg*G*(y(2) + y(4)); %dN/dt
dydt(2) = (Vg*G-Tpcw)*y(2) + beta*y(1)*te; %dSc/dt
dydt(3) = -(alpha/2)*(Vg*G-Tpcw); %dyc/dt
dydt(4) = (Vg*G-Tpcw)*y(4) + beta*y(1)*te + ((2*Vg)/L)*cos(y(5))*(sqrt(Si*y(4))); %dScc/dt
dydt(5) = -Wcb - ((alpha/2)*(Vg*G-Tpcw)) - ((Vg/L)*sin(y(5))*(sqrt(Si/y(4)))); %dycc/dt
Below is the Ode file
%------Rate equations for requation file------
format bank;
close all;
clear all;
clc;
%time interval
ti=0;
tf=200;
tspan=[ti tf];
x0 = [3.75e7, 2.25e6, 0, 2.25e6, 0]; %initial vectors
%options= odeset('RelTol',100, 'AbsTol',[3.75e7, 2.25e6]);
[t,y]= ode23t(#requations,tspan,x0);
%Plotting the graphs:
figure
subplot(5,1,1), plot(t,y(:,1),'r'),grid on;
title('Laserrate equations'),ylabel('N');
subplot(5,1,2), plot(t,y(:,2),'b'),grid on;
ylabel('Scw'); xlabel('t');
subplot(5,1,3), plot(t,y(:,3),'g'),grid on;
ylabel('ycw');xlabel('t');
subplot(5,1,4), plot(t,y(:,3),'g'),grid on;
ylabel('Sccw');xlabel('t');
subplot(5,1,5), plot(t,y(:,3),'g'),grid on;
ylabel('yccw');xlabel('t');