I am trying to execute gabor filter on images.
%% Read
clear all;
close all;
clc;
I=imread('test.png');
imshow(I);
%% Crop
I2 = imcrop(I);
figure, imshow(I2)
m=size(I2,1);
n=size(I2,2);
%% Gabor
phi = 7*pi/8;
theta = 2;
sigma = 0.65*theta;
for i=1:3
for j=1:3
xprime= j*cos(phi);
yprime= i*sin(phi);
K = exp(2*pi*theta*i(xprime+ yprime));
G= exp(-(i.^2+j.^2)/(sigma^2)).*abs(K);
end
end
%% Convolve
for i=1:m
for j=1:n
J(i,j)=conv2(I2,G);
end
end
imshow(uint8(J))
I am getting this error always.
??? Subscript indices must either be real positive integers or logicals.
Not sure how to solve this...
You are missing a * in K = exp(2*pi*theta*i(xprime+ yprime)); between i and the parentheses. You like should be K = exp(2*pi*theta*i*(xprime+ yprime));. It is because of such cases Mathworks recommends using sqrt(-1) for the imaginary number.
Update:
You don't need a loop to do convolution in Matlab. You simply say J=conv2(I2,G);
Update 2:
Here's the working code
%% Gabor
phi = 7*pi/8;
theta = 2;
sigma = 0.65*theta;
filterSize = 6;
G = zeros(filterSize);
for i=(0:filterSize-1)/filterSize
for j=(0:filterSize-/filterSize
xprime= j*cos(phi);
yprime= i*sin(phi);
K = exp(2*pi*theta*sqrt(-1)*(xprime+ yprime));
G(round((i+1)*filterSize),round((j+1)*filterSize)) = exp(-(i^2+j^2)/(sigma^2))*K;
end
end
%% Convolve
J = conv2(I2,G);
imshow(imag(J));
According to the answers above, the final code being :
clear all;
close all;
clc;
I=imread('test.png');
imshow(I);
%% Crop
I2 = imcrop(I);
figure, imshow(I2)
phi = 7*pi/8;
theta = 2;
sigma = 0.65*theta;
filterSize = 6;
G = zeros(filterSize);
for i=(0:filterSize-1)/filterSize
for j=(0:filterSize-1)/filterSize
xprime= j*cos(phi);
yprime= i*sin(phi);
K = exp(2*pi*theta*sqrt(-1)*(xprime+ yprime));
G(round((i+1)*filterSize),round((j+1)*filterSize)) = exp(-(i^2+j^2)/(sigma^2))*K;
end
end
J = conv2(I,G);
figure(2);
imagesc(imag(J))
Related
I am trying to solve the 1D ADE
This is my code so far:
clc; clear; close all
%Input parameters
Ao = 1; %Initial value
L = 0.08; %Column length [m]
nx = 40; %spatial gridpoints
dx = L/nx; %Length step size [m]
T = 20/24; %End time [days]
nt = 100; %temporal gridpoints
dt = T/nt; %Time step size [days]
Vel = dx/dt; %Velocity in each cell [m/day]
alpha = 0.002; %Dispersivity [m]
De = alpha*Vel; % Dispersion coeff. [m2/day]
%Gridblocks
x = 0:dx:L;
t = 0:dt:T;
%Initial and boundary conditions
f = #(x) x; % initial cond.
% boundary conditions
g1 = #(t) Ao;
g2 = #(t) 0;
%Initialization
A = zeros(nx+1, nt+1);
A(:,1) = f(x);
A(1,:) = g1(t);
gamma = dt/(dx^2);
beta = dt/dx;
% Implementation of the explicit method
for j= 1:nt-1 % Time Loop
for i= 2:nx-1 % Space Loop
A(i,j+1) = (A(i-1,j))*(Vel*beta + De*gamma)...
+ A(i,j)*(1-2*De*gamma-Vel*beta) + A(i+1,j)*(De*gamma);
end
% Insert boundary conditions for i = 1 and i = N
A(2,j+1) = A(1,j)*(Vel*beta + De*gamma) + A(2,j)*(1-2*De*gamma-Vel*beta) + A(3,j)*(De*gamma);
A(nx,j+1) = A(nx-1,j)*(Vel*beta + 2*De*gamma) + A(nx,j)*(1-2*De*gamma-Vel*beta)+ (2*De*gamma*dx*g2(t));
end
figure
plot(t, A(end,:), 'r*', 'MarkerSize', 2)
title('A Vs time profile (Using FDM)')
xlabel('t'),ylabel('A')
Now, I have been able to solve the problem using MATLAB’s pdepe function (see plot), but I am trying to compare the result with the finite difference method (implemented in the code above). I am however getting negative values of the dependent variable, but I am not sure what exactly I could be doing wrong. I therefore will really appreciate if anyone can help me out here. Many thanks in anticipation.
PS: I can post the code I used for the pdepe if anyone would like to see it.
I am trying to find the spectrum of an image F' such that:
F' = F(u,v) T(u,v)
where
T(u,v) = (sqrt(u^2+v^2))^p
I = imread('cameraman.tif');
f = fft2(I);
F = fftshift(f);
My questions are:
How can I implement T(u,v)? What would be u and v where p=2?
How to get F', what is the suitable command to do this convolution?
Check last line, maybe you have to flip your kernel, because Matlab does a cross-correlation and call it convolution, also I'm not sure if you have to shift T before convolution or not T = fftshift(T);
clc
clear all
close all
%-----------------------------
I=imread('cameraman.tif');
f=fft2(I);
F=fftshift(f);
%-----------------------------
p = 2;
[r,c] = size(I);
% #xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
% T = zeros(r,c);
% for u = 1:r
% for v = 1:c
% T(u,v) = (u^2+v^2)^(p/2);
% end
% end
% xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
u = 1:r;
v = 1:c;
[U,V] = meshgrid(u,v);
T = (U.^2 + V.^2)^(p/2);
F2 = conv2(F,T);
I am trying to using fminunc to obtain the optimal theta in logistic regression, however I keep getting that:
fminunc stopped because it cannot decrease the objective function
along the current search direction.
Searching online, I found that this is usually the result of a gradient error which I am implementing in logistic_costFunction.m. I re-checked my work but I cannot spot the root cause.
I am not sure how to solve this issue, any help would be appreciated.
Here is my code:
clear all; close all; clc;
%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);
x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));
x = [x1 x2]'; % X
subplot(2,2,1);
dat = [dat1 dat2]'; % Y
scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);
%% Compute Cost and Gradient
% Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);
% Add intercept term to x and X_test
x = [ones(m, 1) x];
% Initialize fitting parameters
initial_theta = zeros(n + 1, 1);
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(initial_theta, x, dat);
fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Gradient at initial theta (zeros): \n');
fprintf(' %f \n', grad);
%% ============= Part 3: Optimizing using fminunc =============
% In this exercise, you will use a built-in function (fminunc) to find the
% optimal parameters theta.
% Set options for fminunc
options = optimset('GradObj', 'on', 'MaxIter', 400);
% Run fminunc to obtain the optimal theta
% This function will return theta and the cost
[theta, cost] = ...
fminunc(#(t)(logistic_costFunction(t, x, dat)), initial_theta, options);
logistic_costFunction.m
function [J, grad] = logistic_costFunction(theta, X, y)
% Initialize some useful values
m = length(y); % number of training examples
grad = zeros(size(theta));
H = sigmoid(X*theta);
T = y.*log(H) + (1 - y).*log(1 - H);
J = -1/m*sum(T);
% ====================Compute grad==================
for i = 1 : m
grad = grad + (H(i) - y(i)) * X(i,:)';
end
grad = 1/m*grad;
end
sigmoid.m
function g = sigmoid(z)
% Computes thes sigmoid of z
g = zeros(size(z));
g = 1 ./ (1 + (1 ./ exp(z)));
end
I want to use streamline to show a vector field. The vector field is singular in a point. I want to remove regions near the singularity (fo example regions which their distance to singularity is less than 1). I wrote below code but it doesn't show anything. Could anyone help me?
clear all;
close all;
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[x,y] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
idx = ((x-x_0).^2 + (y-y_0).^2 > r1^2 & (x-x_0).^2 + (y-y_0).^2 < r2^2);
x = sort(x(idx));
[x, index] = unique(x);
y = sort(y(idx));
[y, index] = unique(y);
U=cos(x)/sqrt(x.^2+y.^2);
V=sin(x)/sqrt(x.^2+y.^2);
streamslice(x,y,U,V);
The problem with your code is that U and V are all zeros, so you get white space. The reason for that is that you don't use elementwise division with ./. So as a first step you should write:
U = cos(x)./sqrt(x.^2+y.^2);
V = sin(x)./sqrt(x.^2+y.^2);
Now U and V are not zeros but are also not matrices anymore, so they are not a valid input for streamslice. The reason for that is that x and y are converted to vectors when calling:
x = sort(x(idx));
y = sort(y(idx));
My guess is that you can remove all this indexing, and simply write:
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[x,y] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
U = cos(x)./sqrt(x.^2+y.^2);
V = sin(x)./sqrt(x.^2+y.^2);
streamslice(x,y,U,V);
so you get:
I think you misunderstood the concept of streamslice. Is this you expecting?
close all;
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[xx,yy] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
% idx = ((xx-x_0).^2 + (yy-y_0).^2 > r1^2 & (xx-x_0).^2 + (yy-y_0).^2 < r2^2);
% x = sort(xx(idx));
% [x, index] = unique(x);
% y = sort(yy(idx));
% [y, index] = unique(y);
U=cos(xx)./sqrt(xx.^2+yy.^2);
V=sin(xx)./sqrt(xx.^2+yy.^2);
streamslice(xx,yy,U,V);
I am working on a project that has to do with solving the wave equation in 2D (x, y, t) numericaly using the central difference approximation in MATLAB with the following boundary conditions:
The general assembly formula is:
I understand some of the boundary conditions (BC), like
du/dy=0 at j=m,
,
but I am not sure how to implement these boundary conditions in MATLAB.
A friend has given me these equations:
Here is my try with the MATLAB code,
but I am not able to progress any further:
% The wave function
% Explicit
% Universal boundary conditions for all 3 cases:
% u=0 at t=0
% du/dt=0 at t=0
% Case 1 boundary conditions
% At x=0, u=2sin(2*pi*t/5);
% At y=0, du/dy=0;
% At y=2, du/dy=0;
% At x=5, du/dx=0;
% u=0 and du/dt=0 at t=0;
%-------------------------------------------------------------------------%
% Setting up
clc; clear all; close all;
% length, time, height
L = 5; % [m]
h = 2; % [m]
T = 10; % [s]
% Constants
c_x = 1; % arbitrary
c_y = 1; % arbitrary
dx = 0.1; % <x> increment
dy = 0.1; % <y> increment
dt = 0.1; % time increment
nx = L/dx + 1; % number of <x> samples
ny = h/dy + 1; % number of <y> samples
nt = T/dt + 1; % number of time samples
t(:,1) = linspace(0, T, nt);
theta_x = c_x*(dt^2)/(dx^2);
theta_y = c_y*(dt^2)/(dy^2);
% theta_x = theta_y
theta = theta_x;
%-------------------------------------------------------------------------%
% The matrix
U = zeros(nt, nx, ny);
% Setting up the <U> matrix with the boundary conditions - case 1
U(1, :, :) = 0; % U=0 at t=0
for tt=1:nt % U=2sin(2pi/5*t) at x=0
for jj=1:ny
U(tt, 1, jj)=2*sin(2*pi/5.*t(tt));
end
end
for it=2:t
for ix=2:nx-1
for iy=2:ny-1
% Boundary conditions
% General case (internal):
U1 = -U(it-1, ix, iy);
U2 = 2*(1-2*theta)*u(it, ix, iy);
U3 = theta*U(it, ix-1, iy);
U4 = theta*U(it, ix+1, iy);
U5 = theta*U(it, ix, iy-1);
U6 = theta*U(it, ix, iy+1);
end
end
end
The general assembly formula you have kind of applies to the boundaries as well.
The complication is that when you apply the formula when j = 1 and j = m, you have j = 0 and j = m+1 term that are off of your grid.
To ameliorate this problem, boundary conditions give you a relationship between the points off the grid and on the grid.
As you have indicated, the dudy = 0 condition has given you the relation that u(i,m-1) == u(u,m+1) on the boundary. So you use the general assembly formula and replace all of the m+1 terms with m-1 on the boundary. You'll have a similar relation for the lower boundary as well.