I'm trying to solve the following problem using MATLAB but I faced multiple issues. The plot I obtained doesn't seem right even though I tried to obtain the steady-state solution, I got a plot that doesn't look steady.
The problem I'm trying to solve
The incorrect plot I got.
and here is the code
% system parameters
m=1; k=1; c=.1; wn=sqrt(k/m); z=c/2/sqrt(m*k); wd=wn*sqrt(1-z^2);
% initial conditions
x0=0; v0=0;
%% time
dt=.001; tMax=8*pi; t=0:(tMax-0)/999:tMax;
% input
A=1
omega=(2*pi)/10
F=A/2-(4*A/pi^2)*cos(omega*t); Fw=fft(F);
F=k*A*cos(omega*t); Fw=fft(F);
% normalize
y = F/m;
% compute coefficients proportional to the Fourier series coefficients
Yw = fft(y);
% setup the equations to solve the particular solution of the differential equation
% by the method of undetermined coefficients
N=1000
T=10
k = [0:N/2];
w = 2*pi*k/T;
A = wn*wn-w.*w;
B = 2*z*wn*w;
% solve the equation [A B;-B A][real(Xw); imag(Xw)] = [real(Yw); imag(Yw)] equation
% Note that solution can be obtained by writing [A B;-B A] as a scaling + rotation
% of a 2D vector, which we solve using complex number algebra
C = sqrt(A.*A+B.*B);
theta = acos(A./C);
Ywp = exp(j*theta)./C.*Yw([1:N/2+1]);
% build a hermitian-symmetric spectrum
Xw = [Ywp conj(fliplr(Ywp(2:end-1)))];
% bring back to time-domain (function synthesis from Fourier Series coefficients)
x = ifft(Xw);
figure()
plot(t,x)
Your forcing function doesn't look like the triangle wave in the problem. I edited the %% time section of your code into the following and appeared to give a steady state response.
%% time
TP = 10; % forcing time period (10 s)
dt=.001;
tMax= 3*TP; % needs to be multiple of the time period
t=0:(tMax-0)/999:tMax;
% input
A=1; % Forcing amplitude
omega=(2*pi)/TP;
% forcing is a triangle wave
% generate a triangle wave with min/max values of 0/1.
F = 0*t;
for i = 1:length(t)
if mod(t(i), TP) <= TP/2
F(i) = mod(t(i), TP)/(TP/2);
else
F(i) = 2 - mod(t(i), TP)/(TP/2);
end
end
F = F*A; % scale triangle wave by amplitude
% you can also use MATLAB's sawtooth() function if you have the signal
% processing toolbox
I am trying to determine the pose (x,y,theta) of a differential drive robot using ode45. The code I have below solves for the x position, but I am having an issue with the initial condition. I set it to 0 since at time 0 the robot is assumed to be at the origin, but I get an error. How do I set the initial condition for ode45 such that I get the expected output?
I tried to make the initial conditions vector of the same length as dxdt by setting initial conditions as a 41x1 zeros matrix, but I didn't understand the output and I don't believe I did it correctly.
% Given conditions for a differential drive robot:
B = 20; % (cm) distance along axle between centers of two wheels
r = 10; % (cm) diameter of both wheels
w_l = 5*sin(3*t); % (rad/s) angular rate of left wheel
w_r = 5*sin(3*t); % (rad/s) angular rate of right wheel
v_l = r*w_l; % (cm/s) velocity of left wheel
v_r = r*w_r; % (cm/s) velocity of right wheel
v = (v_r+v_l)/B; % (cm/s) velocity of robot
theta = 90; % constant orientation of robot since trajectory is straight
% Solve differential equation for x:
dxdt = v*cos(theta); % diff equaition for x
tspan = [0 20]; % time period to integrate over
x0 = 0; % initial condition since robot begins at origin
[t,x] = ode45(#(t,y) dxdt, tspan, x0);
I want to solve the differential equation dxdt for 0 to 20 seconds with an initial condition of 0. I expect the output to give me a vector of time from 0 to 20 and an array of for x. The problem I believe lies with the initial condition. MATLAB gives me an error in the live editor telling me, " #(t,y)dxdt returns a vector of length 69, but the length of initial conditions vector is 1. The vector returned by #(t,y)dxdt and the initial conditions vector must have the same number of elements."
The issue is not the initial condition. You need to define dxdt as a function i.e.
% Define differential equation
function derivative = dxdt(t,y)
% Given conditions for a differential drive robot:
B = 20; % (cm) distance along axle between centers of two wheels
r = 10; % (cm) diameter of both wheels
w_l = 5*sin(3*t); % (rad/s) angular rate of left wheel
w_r = 5*sin(3*t); % (rad/s) angular rate of right wheel
v_l = r*w_l; % (cm/s) velocity of left wheel
v_r = r*w_r; % (cm/s) velocity of right wheel
v = (v_r+v_l)/B; % (cm/s) velocity of robot
theta = 90; % constant orientation of robot since trajectory is straight
derivative = v*cos(theta); % diff equation for x
end
Then when you use ode45 you should tell it to pass the t and y variables as arguments to dxdt like
[t,x] = ode45(#(t,y) dxdt(t,y), tspan, x0);
This should then work. In this case, as dxdt only takes the default arguments, you could also write
[t,x] = ode45(#dxdt, tspan, x0);
The error you got indicates that at some point you made dxdt into a vector of length 69, whilst MATLAB was expecting to get back 1 value for dxdt when it passed one t and one y to your dxdt 'function'. Whenever you get errors like this, I'd recommend putting
clear all
`clearvars` % better than clear all - see am304's comment below
at the top of your script to avoid polluting your workspace with previously defined variables.
I have a question about the use of Matlab to compute solution of stochastic differentials equations. The equations are the 2.2a,b, page 3, in this paper (PDF).
My professor suggested using ode45 with a small time step, but the results do not match with those in the article. In particular the time series and the pdf. I also have a doubt about the definition of the white noise in the function.
Here the code for the integration function:
function dVdt = R_Lang( t,V )
global sigma lambda alpha
W1=sigma*randn(1,1);
W2=sigma*randn(1,1);
dVdt=[alpha*V(1)+lambda*V(1)^3+1/V(1)*0.5*sigma^2+W1;
sigma/V(1)*W2];
end
Main script:
clear variables
close all
global sigma lambda alpha
sigma=sqrt(2*0.0028);
alpha=3.81;
lambda=-5604;
tspan=[0,10];
options = odeset('RelTol',1E-6,'AbsTol',1E-6,'MaxStep',0.05);
A0=random('norm',0,0.5,[2,1]);
[t,L]=ode45(#(t,L) R_Lang(t,L),tspan,A0,options);
If you have any suggestions I'd be grateful.
Here the new code to confront my EM method and 'sde_euler'.
lambda = -5604;
sigma=sqrt(2*0.0028) ;
Rzero = 0.03; % problem parameters
phizero=-1;
dt=1e-5;
T = 0:dt:10;
N=length(T);
Xi1 = sigma*randn(1,N); % Gaussian Noise with variance=sigma^2
Xi2 = sigma*randn(1,N);
alpha=3.81;
Rem = zeros(1,N); % preallocate for efficiency
Rtemp = Rzero;
phiem = zeros(1,N); % preallocate for efficiency
phitemp = phizero;
for j = 1:N
Rtemp = Rtemp + dt*(alpha*Rtemp+lambda*Rtemp^3+sigma^2/(2*Rtemp)) + sigma*Xi1(j);
phitemp=phitemp+sigma/Rtemp*Xi2(j);
phiem(j)=phitemp;
Rem(j) = Rtemp;
end
f = #(t,V)[alpha*V(1)+lambda*V(1)^3+0.5*sigma^2/V(1)/2;
0]; % Drift function
g = #(t,V)[sigma;
sigma/V(1)]; % Diffusion function
A0 = [0.03;0]; % 2-by-1 initial condition
opts = sdeset('RandSeed',1,'SDEType','Ito'); % Set random seed, use Ito formulation
L = sde_euler(f,g,T,A0,opts);
plot(T,Rem,'r')
hold on
plot(T,L(:,1),'b')
Thanks again for the help !
ODEs and SDEs are very different and one should not use tools for ODEs, like ode45, to try to solve SDEs. Looking at the paper you linked to, they used a basic Euler-Maruyama scheme to integrate the system. This a very simple solver to implement yourself.
Before proceeding, you (and your professor!) should take some time to read up on SDEs and how to solve them numerically. I recommend this paper, which includes many Matlab examples:
Desmond J. Higham, 2001, An Algorithmic Introduction to Numerical Simulation of Stochastic Differential Equations, SIAM Rev. (Educ. Sect.), 43 525–46. http://dx.doi.org/10.1137/S0036144500378302
The URL to the Matlab files in the paper won't work; use this one. Note, that as this a 15-year old paper, some of the code related to random number generation is out of date (use rng(1) instead of randn('state',1) to seed the generator).
If you are familiar with ode45 you might look at my SDETools Matlab toolbox on GitHub. It was designed to be fast and has an interface that works very similarly to Matlab's ODE suite. Here is how you might code up your example using the Euler-Maruyma solver:
sigma = 1e-1*sqrt(2*0.0028);
lambda = -5604;
alpha = 3.81;
f = #(t,V)[alpha*V(1)+lambda*V(1)^3+0.5*sigma^2/V(1);
0]; % Drift function
g = #(t,V)[sigma;
sigma/V(1)]; % Diffusion function
dt = 1e-3; % Time step
t = 0:dt:10; % Time vector
A0 = [0.03;-2]; % 2-by-1 initial condition
opts = sdeset('RandSeed',1,'SDEType','Ito'); % Set random seed, use Ito formulation
L = sde_euler(f,g,t,A0,opts); % Integrate
figure;
subplot(211);
plot(t,L(:,2));
ylabel('\phi');
subplot(212);
plot(t,L(:,1));
ylabel('r');
xlabel('t');
I had to reduce the size of sigma or the noise was so large that it could cause the radius variable to go negative. I'm not sure if the paper discusses how they handle this singularity. You can try the 'NonNegative' option within sdeset to try to handle this or you may need to construct your own solver. I also couldn't find what integration time step the paper used. You should also consider contacting the authors of the paper directly.
UPDATE
Here's an Euler-Maruyama implementation that matches the sde_euler code above:
sigma = 1e-1*sqrt(2*0.0028);
lambda = -5604;
alpha = 3.81;
f = #(t,V)[alpha*V(1)+lambda*V(1)^3+0.5*sigma^2/V(1);
0]; % Drift function
g = #(t,V)[sigma;
sigma/V(1)]; % Diffusion function
dt = 1e-3; % Time step
t = 0:dt:10; % Time vector
A0 = [0.03;-2]; % 2-by-1 initial condition
% Create and initialize state vector (L here is transposed relative to sde_euler output)
lt = length(t);
n = length(A0);
L = zeros(n,lt);
L(:,1) = A0;
% Set seed and pre-calculate Wiener increments with order matching sde_euler
rng(1);
r = sqrt(dt)*randn(lt-1,n).';
% General Euler-Maruyama integration loop
for i = 1:lt-1
L(:,i+1) = L(:,i)+f(t(i),L(:,i))*dt+r(:,i).*g(t(i),L(:,i));
end
figure;
subplot(211);
plot(t,L(2,:));
ylabel('\phi');
subplot(212);
plot(t,L(1,:));
ylabel('r');
xlabel('t');
I have here an ODE to solve that has a test parameter a.
I would like a to be a function of temperature T (in this code it is solved as X(2) since it varies with time).
I would like to base it off experimental values of a and T.
Hence, I am thinking of doing it along the lines of interp1.
Here is my test dummy ODE.
function xprime = RabbitTemp(t,X)
% Model of Rabbit Population
% where,
% Xo = Initial Population of Rabbits
% X(1) = Population density of Rabbit
% X(2) = Temperature T (that varies with time)
% a = test parameter
% Interpolate here
Texptdata = [1 2 3 4];
aexptdata =[10 14 19 30];
a= interp1(Texptdata,aexptdata,X(2),'spline');
% ODE
dx = [0 0];
dx(1) = (X(1))*(1 - X(1)*a - 3*(X(2))));
dx(2) = sin(t);
xprime = [dx(1) dx(2)]';
end
I've tried running the code with my solver and it does compile fine.
However I am just wondering if the ODE solver is calling the correct interpolated a corresponding to the correct corresponding value of X(2). I cannot seem to figure out how to verify this.
Can anyone help me verify this or point me to how I can go about doing it? Much appreciated!
Lets say I have a simple logistic equation
dx/dt = 2ax(1 - x/N)
where N is the carrying capacity, a is some growth rate, and both a and N are parameters I'd like to vary.
So what I want to do is to plot a 3D graph of my fixed point and the two parameters.
I understand how to find a fixed point of a single parameter.
Here is my sample code
function xprime = MyLogisticFunction(t,X) %% The ODE
% Parameters
N = 10 % Carrying Capacity
a = 0.5 % Growth Rate
x1prime = 2*a*X(1)*(1 - X(1)/N );
xprime = [x1prime ]';
end
Next my solver
% Initial Number
x0 = 0.4;
%Time Window
tspan=[0 100];
[t,x]=ode45(#MyLogisticFunction,tspan,x0);
clf
x(end,1) % This gives me the fixed point for the parameters above.
So my real question is, how do I put a for loop across two functions, that allows me to vary a and N, so that I can plot out a 3D graph of a and N and my fixed point x*.
I've tried combining both functions into one .m file but it does not seem to work
You need to pass the parameters to your function:
function xprime = MyLogisticFunction(t,X,a,N) %% The ODE
% Parameters (passed as function arguments)
% N = 10 % Carrying Capacity
% a = 0.5 % Growth Rate
x1prime = 2*a*X(1)*(1 - X(1)/N );
xprime = [x1prime ]';
end
and then when you call the ode solver:
% Initial Number
x0 = 0.4;
%Time Window
tspan=[0 100];
a = 0.1:0.1:1; % or whatever
N = 1:10; % or whatever
x_end = zeros(length(a),length(N));
for ii = 1:length(a)
for jj = 1:length(N)
[t,x]=ode45(#(t,X)MyLogisticFunction(t,X,a(ii),N(jj)),tspan,x0);
x_end(ii,jj) = x(end,1);
end
end