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The IBVP is
u_t+2u_x=0 , 0<x<1, t>0
u(x,0)=sin(pi*x)
u(0,t)=g(t)=sin(-2*pi*t)
I have to first implement the scheme using 4th-order central difference SBP operator in space and RK4 in time with spacing xi=i*h, i=0,1,...,N and update g(t) at every RK stage. Also, using the code compute the convergence rate on a sequence of grids.
Below I have shown my working which is not helping me find the convergence rate so can anyone help me with finding my mistakes.
%Parameters
nstp = 16; %number of grid points in space
t0 = 0; %initial time
tend = 1; %end time
x0 = 0; %left boundary
xN = 1; %right boundary
x = linspace(0,1,nstp); %points in space
h = xN-x0/nstp-1; %grid size
cfl = 4;
k = cfl*h; %length of time steps
N = ceil(tend/k); %number of steps in time
k = tend/N; %length of time steps
u0 = sin(pi*x); %initial data
e = zeros(nstp);
e(1) = 1;
e0 = e(:,1);
m = zeros(nstp);
m(1) = sin(-2*pi*tend);
g = m(:,1);
%4th order central SBP operator in space
m=10; %points
H=diag(ones(m,1),0);
H(1:4,1:4)=diag([17/48 59/48 43/48 49/48]);
H(m-3:m,m-3:m)=fliplr(flipud(diag([17/48 59/48 43/48 49/48])));
H=H*h;
HI=inv(H);
D1=(-1/12*diag(ones(m-2,1),2)+8/12*diag(ones(m-1,1),1)- ...
8/12*diag(ones(m-1,1),-1)+1/12*diag(ones(m-2,1),-2));
D1(1:4,1:6)=[-24/17,59/34,-4/17,-3/34,0,0; -1/2,0,1/2,0,0,0;
4/43,-59/86,0,59/86,-4/43,0; 3/98,0,-59/98,0,32/49,-4/49];
D1(m-3:m,m-5:m)=flipud( fliplr(-D1(1:4,1:6)));
D1=D1/h;
%SBP-SAT scheme
u = -2*D1*x(1:N-1)'-(2*HI*(u0-g)*e);
%Runge Kutta for ODE
for i=1:nstp %calculation loop
t=(i-1)*k;
k1=D*u;
k2=D*(u+k*k1/2);
k3=D*(u+k*k2/2);
k4=D*(u+k*k3);
u=u+(h*(k1+k2+k3+k4))/6; %main equation
figure(1)
plot(x(1:N-1),u); %plot
drawnow
end
%error calculcation
ucorrect = sin(pi*(x-2*tend)); %correct solution
ucomp = u(:,end); %computed solution
errornorm = sqrt((ucomp-ucorrect)'*H*(ucomp-ucorrect)); %norm of error**
One egregious error is
h = xN-x0/nstp-1; %grid size
You need parentheses here, else you compute h as xN-(x0/nstp)-1=1-(0/nstp)-1=0.
Or avoid this all and use the actual grid step. Also, if nstp is the number of steps or segments in the subdivision, the number of nodes is one larger. And if you define x0,xN then you should also actually use them.
x = linspace(x0,xN,nstp+1); %points in space
h = x(2)-x(1); %grid size
As to the SBP-SAT approach, the space discretization gives
u_t(t) + 2*D_1*u(t) = - H_inv * (dot(e_0,u(t)) - g(t))*e_0
which should give the MoL ODE function as
F = #(u,t) -2*D1*u - HI*e0*(u(1)-g(t))
The matrix construction has to be adapted to have the correct extends for these changed operations.
The implement RK4 as
k1 = F(u,t)
k2 = F(u+0.5*h*k1, t+0.5*h)
...
In total this gives a modified script below. Stable solutions only occur for cfl=1 or below. With that the errors seemingly behave like 4th order. The factor for the initial condition also works best in the range 0.5 to 1, larger factors tend to increase the error at x=0, at least in the initial time steps. With the smaller factors the error is uniform over the interval.
%Parameters
xstp = 16; %number of grid points in space
x0 = 0; %left boundary
xM = 1; %right boundary
x = linspace(x0,xM,xstp+1)'; %points in space
h = x(2)-x(1); %grid size
cfl = 1;
t0 = 0; %initial time
tend = 1; %end time
k = cfl*h; %length of time steps
N = ceil(tend/k); %number of steps in time
k = tend/N; %length of time steps
u = sin(pi*x); %initial data
e0 = zeros(xstp+1,1);
e0(1) = 1;
g = #(t) sin(-2*pi*t);
%4th order central SBP operator in space
M=xstp+1; %points
H=diag(ones(M,1),0); % or eye(M)
H(1:4,1:4)=diag([17/48 59/48 43/48 49/48]);
H(M-3:M,M-3:M)=fliplr(flipud(diag([17/48 59/48 43/48 49/48])));
H=H*h;
HI=inv(H);
D1=(-1/12*diag(ones(M-2,1),2)+8/12*diag(ones(M-1,1),1)- ...
8/12*diag(ones(M-1,1),-1)+1/12*diag(ones(M-2,1),-2));
D1(1:4,1:6) = [ -24/17, 59/34, -4/17, -3/34, 0, 0;
-1/2, 0, 1/2, 0, 0, 0;
4/43, -59/86, 0, 59/86, -4/43, 0;
3/98, 0, -59/98, 0, 32/49, -4/49 ];
D1(M-3:M,M-5:M) = flipud( fliplr(-D1(1:4,1:6)));
D1=D1/h;
%SBP-SAT scheme
F = #(u,t) -2*D1*u - 0.5*HI*(u(1)-g(t))*e0;
clf;
%Runge Kutta for ODE
for i=1:N %calculation loop
t=(i-1)*k;
k1=F(u,t);
k2=F(u+0.5*k*k1, t+0.5*k);
k3=F(u+0.5*k*k2, t+0.5*k);
k4=F(u+k*k3, t+k);
u=u+(k/6)*(k1+2*k2+2*k3+k4); %main equation
figure(1)
subplot(211);
plot(x,u,x,sin(pi*(x-2*(t+k)))); %plot
ylim([-1.1,1.1]);
grid on;
legend(["computed";"exact"])
drawnow
subplot(212);
plot(x,u-sin(pi*(x-2*(t+k)))); %plot
grid on;
hold on;
drawnow
end
%error calculcation
ucorrect = sin(pi*(x-2*tend)); %correct solution
ucomp = u; %computed solution
errornorm = sqrt((ucomp-ucorrect)'*H*(ucomp-ucorrect)); %norm of error**
Consider the following problem:
I am now in the third part of this question. I wrote the vectorial loop equations (q=teta2, x=teta3 and y=teta4):
fval(1,1) = r2*cos(q)+r3*cos(x)-r4*cos(y)-r1;
fval(2,1) = r2*sin(q)+r3*sin(x)-r4*sin(y);
I have these 2 functions, and all variables except x and y are given. I found the roots with help of this video.
Now I need to plot graphs of q versus x and q versus y when q is at [0,2pi] with delta q of 2.5 degree. What should I do to plot the graphs?
Below is my attempt so far:
function [fval,jac] = lorenzSystem(X)
%Define variables
x = X(1);
y = X(2);
q = pi/2;
r2 = 15
r3 = 50
r4 = 45
r1 = 40
%Define f(x)
fval(1,1)=r2*cos(q)+r3*cos(x)-r4*cos(y)-r1;
fval(2,1)=r2*sin(q)+r3*sin(x)-r4*sin(y);
%Define Jacobian
jac = [-r3*sin(X(1)), r4*sin(X(2));
r3*cos(X(1)), -r4*cos(X(2))];
%% Multivariate NR
%Initial conditions:
X0 = [0.5;1];
maxIter = 50;
tolX = 1e-6;
X = X0;
Xold = X0;
for i = 1:maxIter
[f,j] = lorenzSystem(X);
X = X - inv(j)*f;
err(:,i) = abs(X-Xold);
Xold = X;
if (err(:,i)<tolX)
break;
end
end
Please take a look at my solution below, and study how it differs from your own.
function [th2,th3,th4] = q65270276()
[th2,th3,th4] = lorenzSystem();
hF = figure(); hAx = axes(hF);
plot(hAx, deg2rad(th2), deg2rad(th3), deg2rad(th2), deg2rad(th4));
xlabel(hAx, '\theta_2')
xticks(hAx, 0:pi/3:2*pi);
xticklabels(hAx, {'$0$','$\frac{\pi}{3}$','$\frac{2\pi}{3}$','$\pi$','$\frac{4\pi}{3}$','$\frac{5\pi}{3}$','$2\pi$'});
hAx.TickLabelInterpreter = 'latex';
yticks(hAx, 0:pi/6:pi);
yticklabels(hAx, {'$0$','$\frac{\pi}{6}$','$\frac{\pi}{3}$','$\frac{\pi}{2}$','$\frac{2\pi}{3}$','$\frac{5\pi}{6}$','$\pi$'});
set(hAx, 'XLim', [0 2*pi], 'YLim', [0 pi], 'FontSize', 16);
grid(hAx, 'on');
legend(hAx, '\theta_3', '\theta_4')
end
function [th2,th3,th4] = lorenzSystem()
th2 = (0:2.5:360).';
[th3,th4] = deal(zeros(size(th2)));
% Define geometry:
r1 = 40;
r2 = 15;
r3 = 50;
r4 = 45;
% Define the residual:
res = #(q,X)[r2*cosd(q)+r3*cosd(X(1))-r4*cosd(X(2))-r1; ... Δx=0
r2*sind(q)+r3*sind(X(1))-r4*sind(X(2))]; % Δy=0
% Define the Jacobian:
J = #(X)[-r3*sind(X(1)), r4*sind(X(2));
r3*cosd(X(1)), -r4*cosd(X(2))];
X0 = [acosd((45^2-25^2-50^2)/(-2*25*50)); 180-acosd((50^2-25^2-45^2)/(-2*25*45))]; % Accurate guess
maxIter = 500;
tolX = 1e-6;
for idx = 1:numel(th2)
X = X0;
Xold = X0;
err = zeros(maxIter, 1); % Preallocation
for it = 1:maxIter
% Update the guess
f = res( th2(idx), Xold );
X = Xold - J(Xold) \ f;
% X = X - pinv(J(X)) * res( q(idx), X ); % May help when J(X) is close to singular
% Determine convergence
err(it) = (X-Xold).' * (X-Xold);
if err(it) < tolX
break
end
% Update history
Xold = X;
end
% Unpack and store θ₃, θ₄
th3(idx) = X(1);
th4(idx) = X(2);
% Update X0 for faster convergence of the next case:
X0 = X;
end
end
Several notes:
All computations are performed in degrees.
The specific plotting code I used is less interesting, what matters is that I defined all θ₂ in advance, then looped over them to find θ₃ and θ₄ (without recursion, as was done in your own implementation).
The initial guess (actually, analytical solution) for the very first case (θ₂=0) can be found by solving the problem manually (i.e. "on paper") using the law of cosines. The solver also works for other guesses, but you might need to increase maxIter. Also, for certain guesses (e.g. X(1)==X(2)), the Jacobian is ill-conditioned, in which case you can use pinv.
If my computation is correct, this is the result:
i'm making image segmentation with self organizing map. the image segement by 3 cluster. Sample image is :
and i have type the matlab code like this bellow :
clear;
clc;
i=imread('DataSet/3.jpg');
I = imresize(i,0.5);
cform = makecform('srgb2lab');
lab_I = applycform(I,cform);
ab = double(lab_I(:,:,2:3));
nrows = size(ab,1);
ncols = size(ab,2);
ab = reshape(ab,nrows*ncols,2);
a = ab(:,1);
b = ab(:,2);
normA = (a-min(a(:))) ./ (max(a(:))-min(a(:)));
normB = (b-min(b(:))) ./ (max(b(:))-min(b(:)));
ab = [normA normB];
newnRows = size(ab,1);
newnCols = size(ab,2);
cluster = 3;
% Max number of iteration
N = 90;
% initial learning rate
eta = 0.3;
% exponential decay rate of the learning rate
etadecay = 0.2;
%random weight
w = rand(2,cluster);
%initial D
D = zeros(1,cluster);
% initial cluster index
clusterindex = zeros(newnRows,1);
% start
for t = 1:N
for data = 1 : newnRows
for c = 1 : cluster
D(c) = sqrt(((w(1,c)-ab(data,1))^2) + ((w(2,c)-ab(data,2))^2));
end
%find best macthing unit
[~, bmuindex] = min(D);
clusterindex(data)=bmuindex;
%update weight
oldW = w(:,bmuindex);
new = oldW + eta * (reshape(ab(data,:),2,1)-oldW);
w(:,bmuindex) = new;
end
% update learning rate
eta= etadecay * eta;
end
%Label Every Pixel in the Image Using the Results from KMEANS
pixel_labels = reshape(clusterindex,nrows,ncols);
%Create Images that Segment the I Image by Color.
segmented_images = cell(1,3);
rgb_label = repmat(pixel_labels,[1 1 3]);
for k = 1:cluster
color = I;
color(rgb_label ~= k) = 0;
segmented_images{k} = color;
end
figure,imshow(segmented_images{1}), title('objects in cluster 1');
figure,imshow(segmented_images{2}), title('objects in cluster 2');
figure,imshow(segmented_images{3}), title('objects in cluster 3');
and after runing the matlab code, there is no image segmentation result. Matlab show 3 figure, Figure 1 show the full image, figure 2 blank, figure 3 blank .
please anyone help me to revise my matlab code, is any wrong code or something?
new = oldW + eta * (reshape(ab(data,:),2,1)-oldW);
This line looks suspicious to me, why you are subtracting old weights here, i dont think this makes any sense there, just remove oldW from there and check your results again.
Thank You
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');
I'm trying to produce some computer generated holograms by using MATLAB. I used equally spaced mesh grid to initialize the spatial grid, and I got the following image
This pattern is sort of what I need except the center region. The fringe should be sharp but blurred. I think it might be the problem of the mesh grid. I tried generate a grid in polar coordinates and the map it into Cartesian coordinates by using MATLAB's pol2cart function. Unfortunately, it doesn't work as well. One may suggest that using fine grids. It doesn't work too. I think if I can generate a spiral mesh grid, perhaps the problem is solvable. In addition, the number of the spiral arms could, in general, be arbitrary, could anyone give me a hint on this?
I've attached the code (My final projects are not exactly the same, but it has a similar problem).
clc; clear all; close all;
%% initialization
tic
lambda = 1.55e-6;
k0 = 2*pi/lambda;
c0 = 3e8;
eta0 = 377;
scale = 0.25e-6;
NELEMENTS = 1600;
GoldenRatio = (1+sqrt(5))/2;
g = 2*pi*(1-1/GoldenRatio);
pntsrc = zeros(NELEMENTS, 3);
phisrc = zeros(NELEMENTS, 1);
for idxe = 1:NELEMENTS
pntsrc(idxe, :) = scale*sqrt(idxe)*[cos(idxe*g), sin(idxe*g), 0];
phisrc(idxe) = angle(-sin(idxe*g)+1i*cos(idxe*g));
end
phisrc = 3*phisrc/2; % 3 arms (topological charge ell=3)
%% post processing
sigma = 1;
polfilter = [0, 0, 1i*sigma; 0, 0, -1; -1i*sigma, 1, 0]; % cp filter
xboundl = -100e-6; xboundu = 100e-6;
yboundl = -100e-6; yboundu = 100e-6;
xf = linspace(xboundl, xboundu, 100);
yf = linspace(yboundl, yboundu, 100);
zf = -400e-6;
[pntobsx, pntobsy] = meshgrid(xf, yf);
% how to generate a right mesh grid such that we can generate a decent result?
pntobs = [pntobsx(:), pntobsy(:), zf*ones(size(pntobsx(:)))];
% arbitrary mesh may result in "wrong" results
NPNTOBS = size(pntobs, 1);
nxp = length(xf);
nyp = length(yf);
%% observation
Eobs = zeros(NPNTOBS, 3);
matlabpool open local 12
parfor nobs = 1:NPNTOBS
rp = pntobs(nobs, :);
Erad = [0; 0; 0];
for idx = 1:NELEMENTS
rs = pntsrc(idx, :);
p = exp(sigma*1i*2*phisrc(idx))*[1 -sigma*1i 0]/2; % simplified here
u = rp - rs;
r = sqrt(u(1)^2+u(2)^2+u(3)^2); %norm(u);
u = u/r; % unit vector
ut = [u(2)*p(3)-u(3)*p(2),...
u(3)*p(1)-u(1)*p(3), ...
u(1)*p(2)-u(2)*p(1)]; % cross product: u cross p
Erad = Erad + ... % u cross p cross u, do not use the built-in func
c0*k0^2/4/pi*exp(1i*k0*r)/r*eta0*...
[ut(2)*u(3)-ut(3)*u(2);...
ut(3)*u(1)-ut(1)*u(3); ...
ut(1)*u(2)-ut(2)*u(1)];
end
Eobs(nobs, :) = Erad; % filter neglected here
end
matlabpool close
Eobs = Eobs/max(max(sum(abs(Eobs), 2))); % normailized
%% source, gaussian beam
E0 = 1;
w0 = 80e-6;
theta = 0; % may be titled
RotateX = [1, 0, 0; ...
0, cosd(theta), -sind(theta); ...
0, sind(theta), cosd(theta)];
Esrc = zeros(NPNTOBS, 3);
for nobs = 1:NPNTOBS
rp = RotateX*[pntobs(nobs, 1:2).'; 0];
z = rp(3);
r = sqrt(sum(abs(rp(1:2)).^2));
zR = pi*w0^2/lambda;
wz = w0*sqrt(1+z^2/zR^2);
Rz = z^2+zR^2;
zetaz = atan(z/zR);
gaussian = E0*w0/wz*exp(-r^2/wz^2-1i*k0*z-1i*k0*0*r^2/Rz/2+1i*zetaz);% ...
Esrc(nobs, :) = (polfilter*gaussian*[1; -1i; 0]).'/sqrt(2)/2;
end
Esrc = [Esrc(:, 2), Esrc(:, 3), Esrc(:, 1)];
Esrc = Esrc/max(max(sum(abs(Esrc), 2))); % normailized
toc
%% visualization
fringe = Eobs + Esrc; % I'll have a different formula in my code
normEsrc = reshape(sum(abs(Esrc).^2, 2), [nyp nxp]);
normEobs = reshape(sum(abs(Eobs).^2, 2), [nyp nxp]);
normFringe = reshape(sum(abs(fringe).^2, 2), [nyp nxp]);
close all;
xf0 = linspace(xboundl, xboundu, 500);
yf0 = linspace(yboundl, yboundu, 500);
[xfi, yfi] = meshgrid(xf0, yf0);
data = interp2(xf, yf, normFringe, xfi, yfi);
figure; surf(xfi, yfi, data,'edgecolor','none');
% tri = delaunay(xfi, yfi); trisurf(tri, xfi, yfi, data, 'edgecolor','none');
xlim([xboundl, xboundu])
ylim([yboundl, yboundu])
% colorbar
view(0,90)
colormap(hot)
axis equal
axis off
title('fringe thereo. ', ...
'fontsize', 18)
I didn't read your code because it is too long to do such a simple thing. I wrote mine and here is the result:
the code is
%spiral.m
function val = spiral(x,y)
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r;
x = r*cos(a);
y = r*sin(a);
val = exp(-x*x*y*y);
val = 1/(1+exp(-1000*(val)));
endfunction
%show.m
n=300;
l = 7;
A = zeros(n);
for i=1:n
for j=1:n
A(i,j) = spiral( 2*(i/n-0.5)*l,2*(j/n-0.5)*l);
end
end
imshow(A) %don't know if imshow is in matlab. I used octave.
the key for the sharpnes is line
val = 1/(1+exp(-1000*(val)));
It is logistic function. The number 1000 defines how sharp your image will be. So lower it for more blurry image or higher it for sharper.
I hope this answers your question ;)
Edit: It is real fun to play with. Here is another spiral:
function val = spiral(x,y)
s= 0.5;
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r*r*r;
x = r*cos(a);
y = r*sin(a);
val = 0;
if (abs(x)<s )
val = s-abs(x);
endif
if(abs(y)<s)
val =max(s-abs(y),val);
endif
%val = 1/(1+exp(-1*(val)));
endfunction
Edit2: Fun, fun, fun! Here the arms do not get thinner.
function val = spiral(x,y)
s= 0.1;
r = sqrt( x*x + y*y);
a = atan2(y,x)*2+r*r; % h
x = r*cos(a);
y = r*sin(a);
val = 0;
s = s*exp(r);
if (abs(x)<s )
val = s-abs(x);
endif
if(abs(y)<s)
val =max(s-abs(y),val);
endif
val = val/s;
val = 1/(1+exp(-10*(val)));
endfunction
Damn your question I really need to study for my exam, arghhh!
Edit3:
I vectorised the code and it runs much faster.
%spiral.m
function val = spiral(x,y)
s= 2;
r = sqrt( x.*x + y.*y);
a = atan2(y,x)*8+exp(r);
x = r.*cos(a);
y = r.*sin(a);
val = 0;
s = s.*exp(-0.1*r);
val = r;
val = (abs(x)<s ).*(s-abs(x));
val = val./s;
% val = 1./(1.+exp(-1*(val)));
endfunction
%show.m
n=1000;
l = 3;
A = zeros(n);
[X,Y] = meshgrid(-l:2*l/n:l);
A = spiral(X,Y);
imshow(A)
Sorry, can't post figures. But this might help. I wrote it for experiments with amplitude spatial modulators...
R=70; % radius of curvature of fresnel lens (in pixel units)
A=0; % oblique incidence by linear grating (1=oblique 0=collinear)
B=1; % expanding by fresnel lens (1=yes 0=no)
L=7; % topological charge
Lambda=30; % linear grating fringe spacing (in pixels)
aspect=1/2; % fraction of fringe period that is white/clear
xsize=1024; % resolution (xres x yres number data pts calculated)
ysize=768; %
% define the X and Y ranges (defined to skip zero)
xvec = linspace(-xsize/2, xsize/2, xsize); % list of x values
yvec = linspace(-ysize/2, ysize/2, ysize); % list of y values
% define the meshes - matrices linear in one dimension
[xmesh, ymesh] = meshgrid(xvec, yvec);
% calculate the individual phase components
vortexPh = atan2(ymesh,xmesh); % the vortex phase
linPh = -2*pi*ymesh; % a phase of linear grating
radialPh = (xmesh.^2+ymesh.^2); % a phase of defocus
% combine the phases with appropriate scales (phases are additive)
% the 'pi' at the end causes inversion of the pattern
Ph = L*vortexPh + A*linPh/Lambda + B*radialPh/R^2;
% transmittance function (the real part of exp(I*Ph))
T = cos(Ph);
% the binary version
binT = T > cos(pi*aspect);
% plot the pattern
% imagesc(binT)
imagesc(T)
colormap(gray)