MATLAB: Getting extra parameters from ODE45 and the thestery transpose

MATLABode45

Hi,
I've been reading various posts on getting extra parameters from ODEs and not having much luck implementing them. Concerned that I'm trying something too complicated for my Matlab skills, I've gone back to a basic model to try and understand where I'm going wrong. This has raised a couple of questions.
First, the code, the majority of which is borrowed from others. This is the top half of a simple two degree of freedom mass-spring-damper model (without the plotting and analysis parts):
clear
clc
% define time span
t0 = 0; % s. Input signal start time
t1 = 5.115; % s. Input signal end time. 5.115 gives 1024 points; 61.435 gives 12,288 points
dt = 0.005; % time resolution
tvec = t0:dt:t1; % creates a horizontal vector between t0 and t1 that increments by dt
% define chirp input
A = 10; % mm. Input signal peak amplitude
f0 = 0.5; % Hz. Input signal start frequency
f1 = 20; % Hz. Input signal end frequency
g = (f1./f0).^(1./(t1-t0)); % Exponential growth of chirp frequency
i = A.*sin(f0.*(((g.^tvec)-1)./log(g)).*2.*pi); % Ground input displacement - exponential chirp signal
idot = A.*cos(f0.*(((g.^tvec)-1)/log(g)).*2.*pi).*(2.*pi.*f0.*g.^tvec); % Ground input velocity - exponential chirp signal
% set initial conditions
x0 = [0; 0; 0; 0];
% Sprung mass parameters
ms = 540.5; % kg

ks = 41; % N/mm

cs = 1.5; % Ns/mm

% Unsprung mass parameters
mu = 40; % kg
ku = 350; % N/mm
cu = 0.35; % Ns/mm
% Solve model
[T, x] = ode45(@(t,x) Two_DOF_QCM_Basic_ODE(t, x, i, idot, tvec, ms, ks, cs, mu, ku, cu), tvec, x0);
And the ODE function is:
function [dx, Fs] = Two_DOF_QCM_Basic_ODE(t, x, i, idot, tvec, ms, ks, cs, mu, ku, cu)
i = interp1(tvec, i, t); % this is interpolating i at t
idot = interp1(tvec, idot, t); % this is interpolating idot at t
% as Matlab is using unspecified time steps it needs a value of i for
% each t
% x(1) is sprung mass displacement
% x(2) is unsprung mass displacement
dx(1) = x(3); % sprung mass velocity. This is first column of dx
dx(2) = x(4); % unsprung mass velocity. This is second column of dx, and so on....
dx(3) = 1./ms.*(ks.*(x(2)-x(1)) + cs.*(x(4)-x(3))).*1000; % sprung mass acceleration

dx(4) = 1./mu.*((ku.*(i(1)-x(2))) + (cu.*(idot(1) - x(4))) - (ks.*(x(2)-x(1))) - (cs.*(x(4)-x(3)))).*1000; % unsprung mass acceleration

dx = dx'; % transpose results from horizontal to vertical
end
This runs fine but the first question is why the need to do the transpose at the bottom when I don't see that line in anyone else's examples? Without it, the code throws an error, and I'm aware that the ODE must return the results in columns. However, I'm confused as to why I don't see the transpose in the help pages or here on Answers.
I then adjusted the code to see if I could get an extra parameter out of the ODE – just a simple dummy parameter as an example. I inserted the following in the ODE:
Fs = ks.*(x(2)-x(1)) + cs.*(x(4)-x(3)); % sum of forces on sprung mass
And adjusted two of the ODE statements to accept the new parameter:
dx(3) = 1./ms.*(Fs).*1000; % sprung mass acceleration
dx(4) = 1./mu.*((ku.*(i(1)-x(2))) + (cu.*(idot(1) - x(4))) - Fs).*1000; % unsprung mass acceleration
I adjusted the function declaration to include the new parameter:
function [dx, Fs] = Two_DOF_QCM_Basic_ODE(t, x, i, idot, tvec, ms, ks, cs, mu, ku, cu)
I then asked for the parameter from the ODE:
[dx, Fs] = Two_DOF_QCM_Basic_ODE(T, x, i, idot, tvec, ms, ks, cs, mu, ku, cu);
It ran without error but I only got Fs back with a single value in it, rather than a value for each element of T. How do I get Fs back at all times in T?
Thanks, Simon.

Best Answer

Hi Simon, to your first question: When you assign values to dx(1), dx(2), and so on you are creating a row vector. However, you need for the ode45 function to return a column vector, hence you have to transpose. What is frequently done in the ode function is to intialize the vector, e.g.
dx = zeros(4,1); % this is a 4x1 column vector
as a column vector. Assigning values to the dx now preserves the column vector. The other approach (which I use frequently) is to assign dx as:
dx    = [x(3); ... % sprung mass velocity. This is first column of dx
         x(4); ... % unsprung mass velocity. This is second column of dx, and so on....
         1./ms.*(ks.*(x(2)-x(1)) + cs.*(x(4)-x(3))).*1000; ... % sprung mass acceleration
         1./mu.*((ku.*(i(1)-x(2))) + (cu.*(idot(1) - x(4))) - (ks.*(x(2)-x(1))) - (cs.*(x(4)-x(3)))).*1000]; % unsprung mass acceleration
which by design is a column vector. To your second question, since you have an equation for Fs you can simply calculate Fs after solving the ODE, i.e. with the return values from the ode45 call.
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