I have adapted the code in Comparing FFT of Function to Analytical FT Solution in Matlab for this question. I am trying to do FFTs and comparing the result with analytical expressions in the Wikipedia tables.
My code is:
a = 1.223;
fs = 1e5; %sampling frequency
dt = 1/fs;
t = 0:dt:30-dt; %time vector
L = length(t); % no. sample points
t = t - 0.5*max(t); %center around t=0
y = ; % original function in time
Y = dt*fftshift(abs(fft(y))); %numerical soln
freq = (-L/2:L/2-1)*fs/L; %freq vector
w = 2*pi*freq; % angular freq
F = ; %analytical solution
figure; subplot(1,2,1); hold on
plot(w,real(Y),'.')
plot(w,real(F),'-')
xlabel('Frequency, w')
title('real')
legend('numerical','analytic')
xlim([-5,5])
subplot(1,2,2); hold on;
plot(w,imag(Y),'.')
plot(w,imag(F),'-')
xlabel('Frequency, w')
title('imag')
legend('numerical','analytic')
xlim([-5,5])
If I study the Gaussian function and let
y = exp(-a*t.^2); % original function in time
F = exp(-w.^2/(4*a))*sqrt(pi/a); %analytical solution
in the above code, looks like there is good agreement when the real and imaginary parts of the function are plotted:
But if I study a decaying exponential multiplied with a Heaviside function:
H = #(x)1*(x>0); % Heaviside function
y = exp(-a*t).*H(t);
F = 1./(a+1j*w); %analytical solution
then
Why is there a discrepancy? I suspect it's related to the line Y = but I'm not sure why or how.
Edit: I changed the ifftshift to fftshift in Y = dt*fftshift(abs(fft(y)));. Then I also removed the abs. The second graph now looks like:
What is the mathematical reason behind the 'mirrored' graph and how can I remove it?
The plots at the bottom of the question are not mirrored. If you plot those using lines instead of dots you'll see the numeric results have very high frequencies. The absolute component matches, but the phase doesn't. When this happens, it's almost certainly a case of a shift in the time domain.
And indeed, you define the time domain function with the origin in the middle. The FFT expects the origin to be at the first (leftmost) sample. This is what ifftshift is for:
Y = dt*fftshift(fft(ifftshift(y)));
ifftshift moves the origin to the first sample, in preparation for the fft call, and fftshift moves the origin of the result to the middle, for display.
Edit
Your t does not have a 0:
>> t(L/2+(-1:2))
ans =
-1.5000e-05 -5.0000e-06 5.0000e-06 1.5000e-05
The sample at t(floor(L/2)+1) needs to be 0. That is the sample that ifftshift moves to the leftmost sample. (I use floor there in case L is odd in size, not the case here.)
To generate a correct t do as follows:
fs = 1e5; % sampling frequency
L = 30 * fs;
t = -floor(L/2):floor((L-1)/2);
t = t / fs;
I first generate an integer t axis of the right length, with 0 at the correct location (t(floor(L/2)+1)==0). Then I convert that to seconds by dividing by the sampling frequency.
With this t, the Y as I suggest above, and the rest of your code as-is, I see this for the Gaussian example:
>> max(abs(F-Y))
ans = 4.5254e-16
For the other function I see larger differences, in the order of 6e-6. This is due to the inability to sample the Heaviside function. You need t=0 in your sampled function, but H doesn't have a value at 0. I noticed that the real component has an offset of similar magnitude, which is caused by the sample at t=0.
Typically, the sampled Heaviside function is set to 0.5 for t=0. If I do that, the offset is removed completely, and max difference for the real component is reduced by 3 orders of magnitude (largest errors happen for values very close to 0, where I see a zig-zag pattern). For the imaginary component, the max error is reduced to 3e-6, still quite large, and is maximal at high frequencies. I attribute these errors to the difference between the ideal and sampled Heaviside functions.
You should probably limit yourself to band-limited functions (or nearly-band-limited ones such as the Gaussian). You might want to try to replace the Heaviside function with an error function (integral of Gaussian) with a small sigma (sigma = 0.8 * fs is the smallest sigma I would consider for proper sampling). Its Fourier transform is known.
I am trying to simulate a network of mobile robots that uses artificial potential fields for movement planning to a shared destination xd. This is done by generating a series of m-files (one for each robot) from a symbolic expression, as this seems to be the best way in terms of computational time and accuracy. However, I can't figure out what is going wrong with my gradient computation: the analytical gradient that is being computed seems to be faulty, while the numerical gradient is calculated correctly (see the image posted below). I have written a MWE listed below, which also exhibits this problem. I have checked the file generating part of the code, and it does return a correct function file with a correct gradient. But I can't figure out why the analytic and numerical gradient are so different.
(A larger version of the image below can be found here)
% create symbolic variables
xd = sym('xd',[1 2]);
x = sym('x',[2 2]);
% create a potential function and a gradient function for both (x,y) pairs
% in x
for i=1:size(x,1)
phi = norm(x(i,:)-xd)/norm(x(1,:)-x(2,:)); % potential field function
xvector = reshape(x.',1,size(x,1)*size(x,2)); % reshape x to allow for gradient computation
grad = gradient(phi,xvector(2*i-1:2*i)); % compute the gradient
gradx = grad(1);grady=grad(2); % split the gradient in two components
% create function file names
gradfun = strcat('GradTester',int2str(i),'.m');
phifun = strcat('PotTester',int2str(i),'.m');
% generate two output files
matlabFunction(gradx, grady,'file',gradfun,'outputs',{'gradx','grady'},'vars',{xvector, xd});
matlabFunction(phi,'file',phifun,'vars',{xvector, xd});
end
clear all % make sure the workspace is empty: the functions are in the files
pause(0.1) % ensure the function file has been generated before it is called
% these are later overwritten by a specific case, but they can be used for
% debugging
x = 0.5*rand(2);
xd = 0.5*rand(1,2);
% values for the Stackoverflow case
x = [0.0533 0.0023;
0.4809 0.3875];
xd = [0.4087 0.4343];
xp = x; % dummy variable to keep x intact
% compute potential field and gradient for both (x,y) pairs
for i=1:size(x,1)
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
% preallocate the gradient and potential matrices
gradx = zeros(length(xGrid),length(yGrid));
grady = zeros(length(xGrid),length(yGrid));
phi = zeros(length(xGrid),length(yGrid));
% generate appropriate function handles
fun = str2func(strcat('GradTester',int2str(i)));
fun2 = str2func(strcat('PotTester',int2str(i)));
% compute analytic gradient and potential for each position in the xGrid and
% yGrid vectors
for ii = 1:length(yGrid)
for jj = 1:length(xGrid)
xp(i,:) = [xGrid(ii) yGrid(jj)]; % select the position
Xvec = reshape(xp.',1,size(x,1)*size(x,2)); % turn the input into a vector
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd); % compute gradients
phi(jj,ii) = fun2(Xvec,xd); % compute potential value
end
end
[FX,FY] = gradient(phi); % compute the NUMERICAL gradient for comparison
%scale the numerical gradient
FX = FX/0.005;
FY = FY/0.005;
% plot analytic result
subplot(2,2,2*i-1)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-gradx,-grady)
contour(xGrid,yGrid,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-FX,-FY)
contour(xGrid,yGrid,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
end
The potential field I am trying to generate is defined by an (x,y) position, in my code called xd. x is the position matrix of dimension N x 2, where the first column represents x1, x2, and so on, and the second column represents y1, y2, and so on. Xvec is simply a reshaping of this vector to x1,y1,x2,y2,x3,y3 and so on, as the matlabfunction I am generating only accepts vector inputs.
The gradient for robot i is being calculated by taking the derivative w.r.t. x_i and y_i, these two components together yield a single derivative 'vector' shown in the quiver plots. The derivative should look like this, and I checked that the symbolic expression for [gradx,grady] indeed looks like that before an m-file is generated.
To fix the particular problem given in the question, you were actually calculating phi in such a way that meant you doing gradient(phi) was not giving the correct results compared to the symbolic gradient. I'll try and explain. Here is how you created xGrid and yGrid:
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
But then in the for loop, ii and jj were used like phi(jj,ii) or gradx(ii,jj), but corresponding to the same physical position. This is why your results were different. Another problem you had was you used gradient incorrectly. Matlab assumes that [FX,FY]=gradient(phi) means that phi is calculated from phi=f(x,y) where x and y are matrices created using meshgrid. You effectively had the elements of phi arranged differently to that, an so gradient(phi) gave the wrong answer. Between reversing the jj and ii, and the incorrect gradient, the errors cancelled out (I suspect you tried doing phi(jj,ii) after trying phi(ii,jj) first and finding it didn't work).
Anyway, to sort it all out, on the line after you create xGrid and yGrid, put this in:
[X,Y]=meshgrid(xGrid,yGrid);
Then change the code after you load fun and fun2 to:
for ii = 1:length(xGrid) %// x loop
for jj = 1:length(yGrid) %// y loop
xp(i,:) = [X(ii,jj);Y(ii,jj)]; %// using X and Y not xGrid and yGrid
Xvec = reshape(xp.',1,size(x,1)*size(x,2));
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd);
phi(ii,jj) = fun2(Xvec,xd);
end
end
[FX,FY] = gradient(phi,0.005); %// use the second argument of gradient to set spacing
subplot(2,2,2*i-1)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))]) %// use axis rather than xlim/ylim
quiver(X,Y,gradx,grady)
contour(X,Y,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))])
quiver(X,Y,FX,FY)
contour(X,Y,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
I have some other comments about your code. I think your potential function is ill-defined, which is causing all sorts of problems. You say in the question that x is an Nx2 matrix, but you potential function is defined as
norm(x(i,:)-xd)/norm(x(1,:)-x(2,:));
which means if N was three, you'd have the following three potentials:
norm(x(1,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(2,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(3,:)-xd)/norm(x(1,:)-x(2,:));
and I don't think the third one makes sense. I think this could be causing some confusion with the gradients.
Also, I'm not sure if there is a reason to create the .m file functions in your real code, but they are not necessary for the code you posted.
I am implementing a code for image enhancement and to apply Fourier and inverse Fourier transform I am using the code below but in result it gives black image.
F = fft2(image); F = fftshift(F); % Center FFT
F = abs(F); % Get the magnitude
F = log(F+1); % Use log, for perceptual scaling, and +1 since log(0) is undefined
F = mat2gray(F); % Use mat2gray to scale the image between 0 and 1
Y = ifft2(F);
subplot(1,1,1);
imshow(Y,[]); % Display the result
You try to image the inverse FFT of a matrix which is pure real (and positive), abs(F). The inverse FT of that is a complex one, and since you lose the phase of the original FT, you will get strange result (almost black image, with eventually the first pixel white...).
Second error, you shift the fft to make some computations, but you don't inverse the shift before
For what you want, you have to keep the phase of the FFT:
F = fft2(image); F = fftshift(F); % Center FFT
Fp = angle(F); % Get the phase
F = abs(F); % Get the magnitude
F = log(F+1); % Use log, for perceptual scaling, and +1 since log(0) is undefined
F = mat2gray(F); % Use mat2gray to scale the image between 0 and 1
Y = real(ifft2(ifftshift(F.*exp(1i*Fp))));
subplot(1,1,1);
imshow(Y,[]); % Display the result
Note: you need to take the real part of the inverse FFT since Matlab creates automatically a complex array as output of a FT (direct or inverse), even if it is a real output. You can check this if you see the value of max(abs(imag(Y(:)))), 6e-11 on my computer.
My intention is to plot the original mathematical function values from the differential equation of the second order below:
I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0
where thetadbldot is the second derivative of theta with respect to t and m,d,I,g,a,o are given constants. Initial conditions are theta(0)=pi/2 and thetadot(0)=0.
My issue is that my knowledge and tutoring is limited to storing the values of the derivatives and returning them, not values from the original mathematical function in the equation. Below you can see a code that calculates the differential in Cauchy-form and gives me the derivatives. Does anyone have suggestions what to do? Thanks!
function xdot = penduluma(t,x)
% The function penduluma(t,x) calculates the differential
% I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0 where thetadbldot is the second
% derivative of theta with respect to t and m,d,I,g,a,o are given constants.
% For the state-variable form, x1=theta and x2=thetadot. x is a 2x1 vector on the form
% [theta,thetadot].
m=1;d=0.2;I=0.1;g=9.81;a=0.1;o=4;
xdot = [x(2);m*d*(o^2*a*sin(o*t)-g)*sin(x(1))/I];
end
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
% Then the desired vector from xa is plotted to t. As it looks now the desired
% values are not found in xa however.
Once you have the angle, you can calculate the angular velocity and acceleration using diff:
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
x_ddot = zeros(size(t));
x_ddot(2:end) = diff(xa(:,2))./diff(t);
plot(t,xa,t,x_ddot)
legend('angle','angular velocity','angular acceleration')
which gives the following plot in Octave (should be the same in MATLAB):
Alternatively, you can work it out using your original differential equation:
x_ddot = -m*d*(o^2*a*sin(o*t)-g).*sin(xa(:,1))/I;
which gives a similar result:
I'm totally new to MATLAB and I know only few basic commands. My task is to plot a function of this kind:
I(T) = ((2*J(k*r*sin(T))/(k*r*sin(T))))^2
where
T = angle
k = (2*pi*f)/c (f= frequency in Hz and c is speed of sound)
r = radius
J = bessel function first kind
I explain a bit: the function represent the power of a soundwave in the space. I've tried many times to plot this but i get always a single point in the plot.
I assume you've defined your Bessel function in J. If not, the MATLAB command for a Bessel function of the first kind is besselj. You'll also have to specify the order of the Bessel function.
You can define your anonymous function as
f=#(t,k,r)(2*besselj(0,k*r*sin(t))./(k*r*sin(t))).^2
and plot it as
T=linspace(0,pi,100);%# a sample angle vector
plot(T,f(T,k,r)) %# where k and r are values you'll have to provide
what about
I = ((2*J(k*r*sin(T))./(k*r*sin(T)))).^2
I finally found how to manage the problem described above, here's the solution that i found, in case that other people may need it.
% MATLAB Istruction for graph generation of bessel like function %
% Variables (fixed values)%
k = 912.91
r = 0.0215
% Set the range for the angle theta %
theta = (-(2/3)*pi:pi/180:(2/3)*pi)
% Calculation of bessel function of first kind %
J = besselj(1,k*r*sin(theta))
% Calculation I function %
% notice the ./ and .^ operators
I = ((2*J)./(k*r*sin(theta))).^2
% now plot the results with the plot command
plot(theta,I)