The equation has the following form:
x'' + w.^2 x=n
w=1
and n is Gaussian noise with mean = 0 and standard deviation = 1.
Without the Gaussian noise I can solve the equation by using ODE45 from matlab.The problem is, how can I deal with this equation when the Gaussian noise is taken into consideration?
it really depends on how the noise is added to the system. If you want to arbitrarily add noise to the system, in which every time the function is called, you add it to the equation representing your data:
function dydt = solve(t,y)
dydt = [y(2); -y(1)+randn(1)];
then call
[t,y] = ode45(#solve, [0 10],[1 -1]);
the problem here is that if the noise is large compared to the signal size, more iterations will be needed thus more time.
On the other hand, if the noise is predetermined, you can either sample and hold, or incorporate a first-order hold and then add that to the system
Related
Having trouble in using a random white noise in ode45 function handle input.
I am using ode45 function to solve a 2 degree of freedom mechanical system (spring/mass/viscous damper). I have the differential equations, where s(1) s(3) are displacements of the two masses, s(2) and 2(4) are the velocity of the two masses. f(t) is the exciting force of the system. K1 k2 are the stiffness (constant), m1 m2 the masses (constant), c damping coefficient (constant)
sdot = #(t,s) ...
[s(2);
(-k1*s(1)-k2*(s(1)-s(3))-c*(s(2)-s(4))+f(t))/m1;
s(4);
(k2*(s(1)-s(3))+c*(s(2)-s(4)))/m2];
I want the force f(t) to be a random white noise that I can scale (scaling factor f0), for example:
rng(3);
f0=100;
f#(time) f0*randn;
The way I am calling the function ode45 is:
[time,result]=ode45(sdot,tspan,IC);
The problem is, when I increase f0 the amount of steps in the calculation increases dramatically (f0=1 gives 46,000 steps ; f0=100 gives 4,342,705 steps). Which I do not want, I just want the magnitude of the random noise to increase.
Any advice on how to set up a better random force to the system of equations mentioned?
I am trying to solve two equations with complex coefficients using ode45.
But iam getting an error message as "Inputs must be floats, namely single or
double."
X = sym(['[',sprintf('X(%d) ',1:2),']']);
Eqns=[-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
f=#(t,X)[Eqns];
[t,Xabc]=ode45(f,[0 300*10^-6],[0 1])
How can i fix this ? Can somebody can help me ?
Per the MathWorks Support Team, the "ODE solvers in MATLAB 5 (R12) and later releases properly handle complex valued systems." So the complex numbers are the not the issue.
The error "Inputs must be floats, namely single or double." stems from your definition of f using Symbolic Variables that are, unlike complex numbers, not floats. The easiest way to get around this is to not use the Symbolic Toolbox at all; just makes Eqns an anonymous function:
Eqns= #(t,X) [-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
[t,Xabc]=ode45(Eqns,[0 300*10^-6],[0 1]);
That being said, I'd like to point out that numerically time integrating this system over 300 microseconds (I assume without units given) will take a long time since your coefficient matrix has imaginary eigenvalues on the order of 10E+10. The extremely short wavelength of those oscillations will more than likely be resolved by Matlab's adaptive methods, and that will take a while to solve for a time span just a few orders greater than the wavelength.
I'd, therefore, suggest an analytical approach to this problem; unless it is a stepping stone another problem that is non-analytically solvable.
Systems of ordinary differential equations of the form
,
which is a linear, homogenous system with a constant coefficient matrix, has the general solution
,
where the m-subscripted exponential function is the matrix exponential.
Therefore, the analytical solution to the system can be calculated exactly assuming the matrix exponential can be calculated.
In Matlab, the matrix exponential is calculate via the expm function.
The following code computes the analytical solution and compares it to the numerical one for a short time span:
% Set-up
A = [-23788605396486326904946699391889i/38685626227668133590597632,23788605396486326904946699391889i/38685626227668133590597632;...
-2500000+5223289665997855453060886952725538686654593059791i/324518553658426726783156020576256,23788605396486326904946699391889i/38685626227668133590597632];
Eqns = #(t,X) A*X;
X0 = [0;1];
% Numerical
options = odeset('RelTol',1E-8,'AbsTol',1E-8);
[t,Xabc]=ode45(Eqns,[0 1E-9],X0,options);
% Analytical
Xana = cell2mat(arrayfun(#(tk) expm(A*tk)*X0,t,'UniformOutput',false)')';
k = 1;
% Plots
figure(1);
subplot(3,1,1)
plot(t,abs(Xana(:,k)),t,abs(Xabc(:,k)),'--');
title('Magnitude');
subplot(3,1,2)
plot(t,real(Xana(:,k)),t,real(Xabc(:,k)),'--');
title('Real');
ylabel('Values');
subplot(3,1,3)
plot(t,imag(Xana(:,k)),t,imag(Xabc(:,k)),'--');
title('Imaginary');
xlabel('Time');
The comparison plot is:
The output of ode45 matches the magnitude and real parts of the solution very well, but the imaginary portion is out-of-phase by exactly π.
However, since ode45's error estimator only looks at norms, the phase difference is not noticed which may lead to problems depending on the application.
It will be noted that while the matrix exponential solution is far more costly than ode45 for the same number of time vector elements, the analytical solution will produce the exact solution for any time vector of any density given to it. So for long time solutions, the matrix exponential can be viewed as an improvement in some sense.
I would like to perform conditional simulations for Gaussian process (GP) models in Matlab. I have found a tutorial by Martin Kolář (http://mrmartin.net/?p=223).
sigma_f = 1.1251; %parameter of the squared exponential kernel
l = 0.90441; %parameter of the squared exponential kernel
kernel_function = #(x,x2) sigma_f^2*exp((x-x2)^2/(-2*l^2));
%This is one of many popular kernel functions, the squared exponential
%kernel. It favors smooth functions. (Here, it is defined here as an anonymous
%function handle)
% we can also define an error function, which models the observation noise
sigma_n = 0.1; %known noise on observed data
error_function = #(x,x2) sigma_n^2*(x==x2);
%this is just iid gaussian noise with mean 0 and variance sigma_n^2s
%kernel functions can be added together. Here, we add the error kernel to
%the squared exponential kernel)
k = #(x,x2) kernel_function(x,x2)+error_function(x,x2);
X_o = [-1.5 -1 -0.75 -0.4 -0.3 0]';
Y_o = [-1.6 -1.3 -0.5 0 0.3 0.6]';
prediction_x=-2:0.01:1;
K = zeros(length(X_o));
for i=1:length(X_o)
for j=1:length(X_o)
K(i,j)=k(X_o(i),X_o(j));
end
end
%% Demo #5.2 Sample from the Gaussian Process posterior
clearvars -except k prediction_x K X_o Y_o
%We can also sample from this posterior, the same way as we sampled before:
K_ss=zeros(length(prediction_x),length(prediction_x));
for i=1:length(prediction_x)
for j=i:length(prediction_x)%We only calculate the top half of the matrix. This an unnecessary speedup trick
K_ss(i,j)=k(prediction_x(i),prediction_x(j));
end
end
K_ss=K_ss+triu(K_ss,1)'; % We can use the upper half of the matrix and copy it to the
K_s=zeros(length(prediction_x),length(X_o));
for i=1:length(prediction_x)
for j=1:length(X_o)
K_s(i,j)=k(prediction_x(i),X_o(j));
end
end
[V,D]=eig(K_ss-K_s/K*K_s');
A=real(V*(D.^(1/2)));
for i=1:7
standard_random_vector = randn(length(A),1);
gaussian_process_sample(:,i) = A * standard_random_vector+K_s/K*Y_o;
end
hold on
plot(prediction_x,real(gaussian_process_sample))
set(plot(X_o,Y_o,'r.'),'MarkerSize',20)
The tutorial generates the conditional simulations using a direct simulation method based on covariance matrix decomposition. It is my understanding that there are several methods of generating conditional simulations that may be better when the number of simulation points is large such as conditioning by Kriging using a local neighborhood. I have found information regarding several methods in J.-P. Chilès and P. Delfiner, “Chapter 7 - Conditional Simulations,” in Geostatistics: Modeling Spatial Uncertainty, Second Edition, John Wiley & Sons, Inc., 2012, pp. 478–628.
Is there an existing Matlab toolbox that can be used for conditional simulations? I am aware of DACE, GPML, and mGstat (http://mgstat.sourceforge.net/). I believe only mGstat offers the capability to perform conditional simulations. However, mGstat also seems to be limited to only 3D models and I am interested in higher dimensional models.
Can anybody offer any advice on getting started performing conditional simulations with an existing toolbox such as GPML?
===================================================================
EDIT
I have found a few more Matlab toolboxes: STK, ScalaGauss, ooDACE
It appears STK is capable of conditional simulations using covariance matrix decomposition. However, is limited to a moderate number (maybe a few thousand?) of simulation points due to the Cholesky factorization.
I used the STK toolbox and I recommend it for others:
http://kriging.sourceforge.net/htmldoc/
I found that if you need conditional simulations at a large number of points then you might consider generating a conditional simulation at the points in a large design of experiment (DoE) and then simply relying on the mean prediction conditional on that DoE.
I'm trying to convolve a rectangular pulse with itself by taking the Fourier transform, squaring it, and then taking the inverse Fourier transform. I realize there is a conv() function but I would prefer to do it in the frequency domain for future, more complex problems. My problem is that when I do this, it does not produce a triangular function as expected. The code I'm using is below:
clc
clear all
x=-5:.01:5;
y=rectangularPulse(x);
Y=fft(y);
H=Y.^2;
h=ifft(H);
plot(x,h)
You need to zeropad in order to ensure that the convolution is linear. Currently you are performing a circular convolution. Try something like this:
y = ones(100,1);
N = length(y);
Nfft = 2*length(y) - 1;
Y=fft(y,Nfft);
H=Y.^2;
h=ifft(H);
plot(h);
I have a matrix z(x,y)
This is an NxN abitary pdf constructed from a unique Kernel density estimation (i.e. not a usual pdf and it doesn't have a function). It is multivariate and can't be separated and is discrete data.
I wan't to construct a NxN matrix (F(x,y)) that is the cumulative distribution function in 2 dimensions of this pdf so that I can then randomly sample the F(x,y) = P(x < X ,y < Y);
Analytically I think the CDF of a multivariate function is the surface integral of the pdf.
What I have tried is using the cumsum function in order to calculate the surface integral and tested this with a multivariate normal against the analytical solution and there seems to be some discrepancy between the two:
% multivariate parameters
delta = 100;
mu = [1 1];
Sigma = [0.25 .3; .3 1];
x1 = linspace(-2,4,delta); x2 = linspace(-2,4,delta);
[X1,X2] = meshgrid(x1,x2);
% Calculate Normal multivariate pdf
F = mvnpdf([X1(:) X2(:)],mu,Sigma);
F = reshape(F,length(x2),length(x1));
% My attempt at a numerical surface integral
FN = cumsum(cumsum(F,1),2);
% Normalise the CDF
FN = FN./max(max(FN));
X = [X1(:) X2(:)];
% Analytic solution to a multivariate normal pdf
p = mvncdf(X,mu,Sigma);
p = reshape(p,delta,delta);
% Highlight the difference
dif = p - FN;
error = max(max(sqrt(dif.^2)));
% %% Plot
figure(1)
surf(x1,x2,F);
caxis([min(F(:))-.5*range(F(:)),max(F(:))]);
xlabel('x1'); ylabel('x2'); zlabel('Probability Density');
figure(2)
surf(X1,X2,FN);
xlabel('x1'); ylabel('x2');
figure(3);
surf(X1,X2,p);
xlabel('x1'); ylabel('x2');
figure(5)
surf(X1,X2,dif)
xlabel('x1'); ylabel('x2');
Particularly the error seems to be in the transition region which is the most important.
Does anyone have any better solution to this problem or see what I'm doing wrong??
Any help would be much appreciated!
EDIT: This is the desired outcome of the cumulative integration, The reason this function is of value to me is that when you randomly generate samples from this function on the closed interval [0,1] the higher weighted (i.e. the more likely) values appear more often in this way the samples converge on the expected value(s) (in the case of multiple peaks) this is desired outcome for algorithms such as particle filters, neural networks etc.
Think of the 1-dimensional case first. You have a function represented by a vector F and want to numerically integrate. cumsum(F) will do that, but it uses a poor form of numerical integration. Namely, it treats F as a step function. You could instead do a more accurate numerical integration using the Trapezoidal rule or Simpson's rule.
The 2-dimensional case is no different. Your use of cumsum(cumsum(F,1),2) is again treating F as a step function, and the numerical errors resulting from that assumption only get worse as the number of dimensions of integration increases. There exist 2-dimensional analogues of the Trapezoidal rule and Simpson's rule. Since there's a bit too much math to repeat here, take a look here:
http://onestopgate.com/gate-study-material/mathematics/numerical-analysis/numerical-integration/2d-trapezoidal.asp.
You DO NOT need to compute the 2-dimensional integral of the probability density function in order to sample from the distribution. If you are computing the 2-d integral, you are going about the problem incorrectly.
Here are two ways to approach the sampling problem.
(1) You write that you have a kernel density estimate. A kernel density estimate is a special case of a mixture density. Any mixture density can be sampled by first selecting one kernel (perhaps differently or equally weighted, same procedure applies), and then sampling from that kernel. (That applies in any number of dimensions.) Typically the kernels are some relatively simple distribution such as a Gaussian distribution so that it is easy to sample from it.
(2) Any joint density P(X, Y) is equal to P(X | Y) P(Y) (and equivalently P(Y | X) P(X)). Therefore you can sample from P(Y) (or P(X)) and then from P(X | Y). In order to sample from P(X | Y), you will need to integrate P(X, Y) along a line Y = y (where y is the sampled value of Y), but (this is crucial) you only need to integrate along that line; you don't need to integrate over all values of X and Y.
If you tell us more about your problem, I can help with the details.