I wish to create a limit cycle in Matlab. A limit cycle looks something like this:
I have no idea how to do it though, I've never done anything like this in Matlab.
The equations to describe the limit cycle are the following:
x_1d=x_2
x_2d=-x_1+x_2-2*(x_1+2*x_2)x_2^2
It is to be centered around the equilibrium which is (0,0)
Can any of you help me?
If you use the partial derivatives of your function to make a vector field, you can then use streamlines to visualize the cycle that you are describing.
For example, the function f = x^2+y^2
Gives me partial derivatives dx = 2x, dy=2y For the visualization, I sample from the partial derivatives over a grid.
[x,y] = meshgrid(0:0.1:1,0:0.1:1);
dx = 2*x;
dy = 2*y;
To visualize the vector field, I use quiver;
figure;
quiver(x, y, dx, dy);
Using streamline, I can visualize the path a particle injected into the vector field would take. In my example, I inject the particle at (0.1, 0.1)
streamline(x,y, dx, dy, 0.1, 0.1);
This produces the following visualization
In your case, you can omit the quiver step to remove the hedgehog arrows at every grid point.
Here's another example that shows the particle converging to an orbit.
Edit: Your function specifically.
So as knedlsepp points out, the function you are interested in is a bit ambiguously stated. In Matlab, * represents the matrix product while .* represents the element-wise multiplication between matrices. Similarly, '^2' represents MM for a matrix M, while .^2 represents taking the element-wise power.
So,
[x_1,x_2] = meshgrid(-4:0.1:4,-4:0.1:4);
dx_1 = x_2;
dx_2 = -x_1+x_2-2*(x_1+2*x_2)*(x_2)^2;
figure; streamline(x_1,x_2, dx_1, dx_2, 0:0.1:4, 0:0.1:4);
Looks like
This function will not show convergence because it doesn't converge.
knedlsepp suggests that the function you are actually interested in is
dx_1 = -1 * x_2;
dx_2 = -1 * -x_1+x_2-2*(x_1+2*x_2).*(x_2).^2;
His post has a nice description of the rest.
This post shows the code to produce the integral lines of your vector field defined by:
dx/dt = y
dy/dt = -x+y-2*(x+2*y)*y^2.
It is important to properly vectorize this function. (i.e. Introducing dots at all the important places)
dxdt = #(x,y) y;
dydt = #(x,y) -x+y-2*(x+2*y).*y.^2;
[X,Y] = meshgrid(linspace(-4,4,100));
[sx,sy] = meshgrid(linspace(-3,3,20));
streamline(stream2(X, Y, ... % Points
dxdt(X,Y), dydt(X,Y),... % Derivatives
sx, sy)); % Starting points
axis equal tight
To get a picture more similar to yours, change the grid size and starting points:
[X,Y] = meshgrid(linspace(-1,1,100));
[sx,sy] = meshgrid(linspace(0,0.75,20),0.2);
Related
I want to solve my differential equation and plot velocity vectors but I am having some trouble with that. I tried this:
syms y(x);
ode = (1+exp(x))*y*diff(y,x)-2*exp(x) == 0;
ySol = dsolve(ode)
[X,Y] = meshgrid(-2:.2:2);
Z = 2*exp(X)/((1+exp(X)).*Y);
[DX,DY] = gradient(Z,.2,.2);
figure
contour(X,Y,Z)
hold on
quiver(X,Y,DX,DY)
hold off
and I get this error:
Warning: Matrix is singular to working precision.
Warning: Contour not rendered for non-finite ZData
It is probably something simple that I do not see but I am just starting using Matlab and I cold not find a right way to do my task. Please help me...
EDIT
As bconrad suggested, I changed my Z function like this:
Z = 2*exp(X)/((1+exp(X)).*Y);
and the previous errors are fixed. However, my prime goal, to plot velocity vectors is not accomplished yet because I get a graph like this:
Don’t have the ability to check at the moment, but I reckon you want an element by element division in that line. You’re missing a dot on the division, try
Z = 2*exp(X)./((1+exp(X)).*Y);
I took a closer look once at my station. The zero-division mentioned by Pablo forces inf's in Z, so quiver get's confused when scaling the vectors (understandably) and just doesn't show them. Try this (with the ode part removed):
[X,Y] = meshgrid(-2 : .2 : 2);
Z = 2 * exp(X) ./ ((1 + exp(X)) .* Y);
Z(isinf(Z)) = nan; % To avoid 0-division problems
[DX, DY] = gradient(Z, .2, .2);
figure
contour(X, Y, Z, 30, 'k')
hold on
quiver(X, Y, DX, DY, 6)
hold off
I've done 3 things here:
Added the line Z(isinf(Z)) = nan; forcing infinite values to be essentially ignored by quiver
Added the arguments 30, 'k' to the contour function to show 30 lines, and make them black (a bit more visible)
Added the argument 6 to the quiver function. This overrides the automatic length-scaling of the vectors.
You'll want to play with the arguments in the contour and quiver functions to make your figure appear as you'd like.
PS: There is a handy arrow function on the file exchange that I find gives better control when creating vector field plots. See https://www.mathworks.com/matlabcentral/fileexchange/278-arrow - the ratings do it justice.
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 need a help. I have to generate a curve using MATLAB. The plot is defined by the formula (an analytic expression) :-
where, the meaning of the variables are as follows: R is the distributed resistive function, S is the distributive conductive function, k is the sheet resistance and r(x,y) is the distance between *
(x,y)*, and the perimeter dl with the integration made around all the perimeter of the chip.
A squared foil as shown in the figure with sides (a) 10 arbitrary units long and an unitary unit sheet resistance (k=1 ohm) is used for our consideration. The plot of the function R(x,y) is supposed to come out like this...
I literally have no clue how to plot this function. I could not even get the idea how to define the distance function r(x,y) with respect to dl. On top of that it is complicated further by the closed integral. Please help me. Any help in even simplifying the expression is also welcome. Is there any possible closed form expression for such a square structure ?
Thanks in advance. The link to the paper is here. paper here
Reconstructing the math
The definition of the function R is not particularly clear, but I guess what they mean is:
With dOmega being the boundary of the foil and p a point p = [px,py] on the foil.
Imagine that for a point p on the sheet you are computing R(p) by going around the boundary of the foil (what they call the perimeter), your position being q, and integrating one divided by the distance from you (q) to the point p multiplied by k.
I guess you could analytically compute the integral for this rectangular sheet, but if you just want to plot the function, you could simply approximate the integral by defining a finite number of points on the boundary, evaluating the integrand in those points, then taking the mean and multiplying by the perimeter. [The same way you could approximate integral(f(x), x=0...pi) by pi*(f(0)+f(pi/2)+f(pi))/3]
Alternative representation using coordinates:
If you are only familiar with integrals along the real line in coordinate representation you could write this in the following way, which is frankly quite UGGGLY:
Plotting an approximation
%% Config:
xlen = 10;
ylen = 10;
k = 1;
%% Setting up points on the boundary of the square
x = linspace(0,xlen,50);
y = linspace(0,ylen,50);
perimeter = 2*(xlen+ylen);
boundary = [x(1)*ones(length(y),1), y'
x', y(1)*ones(length(x),1); ...
x(end)*ones(length(y),1), y'; ...
x', y(end)*ones(length(x),1)];
%% Function definition
norm2 = #(X) sqrt(sum(X.^2,2));
R = #(p) 1/(perimeter*mean(1./(k*norm2(bsxfun(#minus,boundary,p)))));
%% Plotting
[X_grid,Y_grid] = ndgrid(x,y);
R_grid = zeros(size(X_grid));
for ii = 1:length(x)
for jj = 1:length(y)
R_grid(ii,jj) = R([x(ii),y(jj)]);
end
end
surf(X_grid, Y_grid, R_grid);
axis vis3d;
This will give you the following plot:
I have a problem with numerical derivative of a vector that is x: Nx1 with respect to another vector t (time) that is the same size of x.
I do the following (x is chosen to be sine function as an example):
t=t0:ts:tf;
x=sin(t);
xd=diff(x)/ts;
but the answer xd is (N-1)x1 and I figured out that it does not compute derivative corresponding to the first element of x.
is there any other way to compute this derivative?
You are looking for the numerical gradient I assume.
t0 = 0;
ts = pi/10;
tf = 2*pi;
t = t0:ts:tf;
x = sin(t);
dx = gradient(x)/ts
The purpose of this function is a different one (vector fields), but it offers what diff doesn't: input and output vector of equal length.
gradient calculates the central difference between data points. For an
array, matrix, or vector with N values in each row, the ith value is
defined by
The gradient at the end points, where i=1 and i=N, is calculated with
a single-sided difference between the endpoint value and the next
adjacent value within the row. If two or more outputs are specified,
gradient also calculates central differences along other dimensions.
Unlike the diff function, gradient returns an array with the same
number of elements as the input.
I know I'm a little late to the game here, but you can also get an approximation of the numerical derivative by taking the derivatives of the polynomial (cubic) splines that runs through your data:
function dy = splineDerivative(x,y)
% the spline has continuous first and second derivatives
pp = spline(x,y); % could also use pp = pchip(x,y);
[breaks,coefs,K,r,d] = unmkpp(pp);
% pre-allocate the coefficient vector
dCoeff = zeroes(K,r-1);
% Columns are ordered from highest to lowest power. Both spline and pchip
% return 4xn matrices, ordered from 3rd to zeroth power. (Thanks to the
% anonymous person who suggested this edit).
dCoeff(:, 1) = 3 * coefs(:, 1); % d(ax^3)/dx = 3ax^2;
dCoeff(:, 2) = 2 * coefs(:, 2); % d(ax^2)/dx = 2ax;
dCoeff(:, 3) = 1 * coefs(:, 3); % d(ax^1)/dx = a;
dpp = mkpp(breaks,dCoeff,d);
dy = ppval(dpp,x);
The spline polynomial is always guaranteed to have continuous first and second derivatives at each point. I haven not tested and compared this against using pchip instead of spline, but that might be another option as it too has continuous first derivatives (but not second derivatives) at every point.
The advantage of this is that there is no requirement that the step size be even.
There are some options to work-around your issue.
First: you can make your domain larger. Instead of N, use N+1 gridpoints.
Second: depending on the end-point of interest, you can use
Forward difference: F(x + dx) - F(x)
Backward difference: F(x) - F(x - dx)
I have discrete regular grid of a,b points and their corresponding c values and I interpolate it further to get a smooth curve. Now from interpolation data, I further want to create a polynomial equation for curve fitting. How to fit 3D plot in polynomial?
I try to do this in MATLAB. I used Surface fitting toolbox in MATLAB (r2010a) to curve fit 3-dimensional data. But, how does one find a formula that fits a set of data to the best advantage in MATLAB/MAPLE or any other software. Any advice? Also most useful would be some real code examples to look at, PDF files, on the web etc.
This is just a small portion of my data.
a = [ 0.001 .. 0.011];
b = [1, .. 10];
c = [ -.304860225, .. .379710865];
Thanks in advance.
To fit a curve onto a set of points, we can use ordinary least-squares regression. There is a solution page by MathWorks describing the process.
As an example, let's start with some random data:
% some 3d points
data = mvnrnd([0 0 0], [1 -0.5 0.8; -0.5 1.1 0; 0.8 0 1], 50);
As #BasSwinckels showed, by constructing the desired design matrix, you can use mldivide or pinv to solve the overdetermined system expressed as Ax=b:
% best-fit plane
C = [data(:,1) data(:,2) ones(size(data,1),1)] \ data(:,3); % coefficients
% evaluate it on a regular grid covering the domain of the data
[xx,yy] = meshgrid(-3:.5:3, -3:.5:3);
zz = C(1)*xx + C(2)*yy + C(3);
% or expressed using matrix/vector product
%zz = reshape([xx(:) yy(:) ones(numel(xx),1)] * C, size(xx));
Next we visualize the result:
% plot points and surface
figure('Renderer','opengl')
line(data(:,1), data(:,2), data(:,3), 'LineStyle','none', ...
'Marker','.', 'MarkerSize',25, 'Color','r')
surface(xx, yy, zz, ...
'FaceColor','interp', 'EdgeColor','b', 'FaceAlpha',0.2)
grid on; axis tight equal;
view(9,9);
xlabel x; ylabel y; zlabel z;
colormap(cool(64))
As was mentioned, we can get higher-order polynomial fitting by adding more terms to the independent variables matrix (the A in Ax=b).
Say we want to fit a quadratic model with constant, linear, interaction, and squared terms (1, x, y, xy, x^2, y^2). We can do this manually:
% best-fit quadratic curve
C = [ones(50,1) data(:,1:2) prod(data(:,1:2),2) data(:,1:2).^2] \ data(:,3);
zz = [ones(numel(xx),1) xx(:) yy(:) xx(:).*yy(:) xx(:).^2 yy(:).^2] * C;
zz = reshape(zz, size(xx));
There is also a helper function x2fx in the Statistics Toolbox that helps in building the design matrix for a couple of model orders:
C = x2fx(data(:,1:2), 'quadratic') \ data(:,3);
zz = x2fx([xx(:) yy(:)], 'quadratic') * C;
zz = reshape(zz, size(xx));
Finally there is an excellent function polyfitn on the File Exchange by John D'Errico that allows you to specify all kinds of polynomial orders and terms involved:
model = polyfitn(data(:,1:2), data(:,3), 2);
zz = polyvaln(model, [xx(:) yy(:)]);
zz = reshape(zz, size(xx));
There might be some better functions on the file-exchange, but one way to do it by hand is this:
x = a(:); %make column vectors
y = b(:);
z = c(:);
%first order fit
M = [ones(size(x)), x, y];
k1 = M\z;
%least square solution of z = M * k1, so z = k1(1) + k1(2) * x + k1(3) * y
Similarly, you can do a second order fit:
%second order fit
M = [ones(size(x)), x, y, x.^2, x.*y, y.^2];
k2 = M\z;
which seems to have numerical problems for the limited dataset you gave. Type help mldivide for more details.
To make an interpolation over some regular grid, you can do like so:
ngrid = 20;
[A,B] = meshgrid(linspace(min(a), max(a), ngrid), ...
linspace(min(b), max(b), ngrid));
M = [ones(numel(A),1), A(:), B(:), A(:).^2, A(:).*B(:), B(:).^2];
C2_fit = reshape(M * k2, size(A)); % = k2(1) + k2(2)*A + k2(3)*B + k2(4)*A.^2 + ...
%plot to compare fit with original data
surfl(A,B,C2_fit);shading flat;colormap gray
hold on
plot3(a,b,c, '.r')
A 3rd-order fit can be done using the formula given by TryHard below, but the formulas quickly become tedious when the order increases. Better write a function that can construct M given x, y and order if you have to do that more than once.
This sounds like more of a philosophical question than specific implementation, specifically to bit - "how does one find a formula that fits a set of data to the best advantage?" In my experience that is a choice you have to make depending on what you're trying to achieve.
What defines "best" for you? For a data fitting problem you can keep adding more and more polynomial coefficients and making a better R^2 value... but will eventually "over fit" the data. A downside of high order polynomials is behavior outside the bounds of the sample data which you've used to fit your response surface - it can quickly go off in some wild direction which may not be appropriate for whatever it is you're trying to model.
Do you have insight into the physical behavior of the system / data you're fitting? That can be used as a basis for what set of equations to use to create a math model. My recommendation would be to use the most economical (simple) model you can get away with.