# Getting extra parameters from ODE45 and the mystery transpose

조회 수: 28 (최근 30일)
Simon Aldworth 2021년 1월 19일
편집: Stephen23 2021년 1월 25일
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.

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### 채택된 답변

Mischa Kim 2021년 1월 19일
편집: Mischa Kim 2021년 1월 19일
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.
##### 댓글 수: 8이전 댓글 6개 표시이전 댓글 6개 숨기기
Walter Roberson 2021년 1월 24일
편집: Walter Roberson 2021년 1월 24일
It is common for people to think that they would like to record all calculated intermediate values such as Fs, thinking that this will give a record of the Fs values that were used for each integration step. However, that does not work in practice.
Each "step" for ode45 involves evaluating the given function with 6 different boundary conditions including a mix of different times on the 6 different calls. If the error estimate between the 5th and 6th evaluation is satisfactory, then the step is "accepted". If the error estimate is not satisfactory, then the step is rejected and the same base point is used but with a smaller step size, again making 6 calls. A number of rejections can occur in a row, until it has found a step size that is small enough to accomodate whatever rapid changes in the function are happening.
From time to time, results are output, but not necessarily at any point that was evaluated -- for example instead of evaluating at time 2.0 exactly, it might synthesize and output based upon its calculations at 1.97 and 2.22 and 2.05, extrapolating what the readings for 2.0 would have been.
Now, it is unlikely that you want the value of intermediate variables during calculations for steps that end up being rejected --- the evaluations might even have been at locations that turn out to be outside boundaries (according to event functions.)
And you probably do not want the value of intermediate variables during the 6 different calculations that go into making up one cycle (though sometimes you do). Most of the time, what you want is the value of the intermediate variables associated with the locations actually output... but as I indicated earlier, ode*() might not have evaluated at that exact location!
Because of these factors, most of the time it is better to take the output locations and invoke a function that calculates just the intermediate values for those locations. It could be a function that you would otherwise invoke from inside the ode function.
Simon Aldworth 2021년 1월 25일
@Mischa Kim, many thanks - that's working fine; and thanks for the lessons. I have also managed to get the alternative method working, whereby the solver is called to complete the integration and then ode is called a second time to provide addtional output at the solved time steps. Thanks again.
@Walter Roberson. Thank you for your explanation; I understand the concept that the integration time steps do not necessarily align to the output steps. This is why my original post was written the way it was. However, so that I fully understand your reasons for posting: are you saying that something we are doing with my code is wrong? Should I be doing something differently? Regards.
Simon.

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### 추가 답변 (1개)

Stephen23 2021년 1월 25일
편집: Stephen23 2021년 1월 25일
Using the OutputFcn is a complex way to get the Fs values.
The simpler approach is to run the ODE solver normally, and then run the function with the values returned by the ODE solver, which automatically calculates Fs values which correspond exactly to the T and x values returned by the ODE solver. This just takes two lines of code, as shown below:
% 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
fun = @(t,x) Two_DOF_QCM_Basic_ODE(t, x, i, idot, tvec, ms, ks, cs, mu, ku, cu);
[T, x] = ode45(fun, tvec, x0) % solve
T = 1024×1
0 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450
x = 1024×4
0 0 0 0 0.0000 0.0085 0.0244 4.3093 0.0004 0.0502 0.1487 12.8568 0.0018 0.1397 0.4273 22.9849 0.0050 0.2786 0.8805 32.2528 0.0109 0.4578 1.4973 38.9086 0.0202 0.6619 2.2440 42.1296 0.0334 0.8734 3.0753 41.9685 0.0510 1.0774 3.9473 39.2069 0.0729 1.2631 4.8229 35.0102
[~,Fs] = cellfun(fun,num2cell(T),num2cell(x,2),'uni',0); % This is all you need...
Fs = cell2mat(Fs) % ... and the Fs values correspond exactly to the T and x values.
Fs = 1024×1
0 6.7739 21.1026 39.4898 58.2753 74.4413 86.1407 92.7790 94.9738 94.0782
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
Fs = ks.*(x(2)-x(1)) + cs.*(x(4)-x(3)); % sum of forces on sprung mass
dx = nan(4,1); % !!! preallocate column output !!!
% 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.*(Fs).*1000; % sprung mass acceleration
dx(4) = 1./mu.*((ku.*(i(1)-x(2))) + (cu.*(idot(1) - x(4))) - Fs).*1000; % unsprung mass acceleration
end
##### 댓글 수: 2없음 표시없음 숨기기
Simon Aldworth 2021년 1월 25일
Many thanks for this Stephen, I'll study this in more detail later. For now, though, I'm a bit confused about your statement that the outputfcn approach collects all intermediate points, including when the ODE steps backwards or takes tiny steps. The help page for odeset says that "the ODE solver calls the output function after each successful time step", which to me implies that since I've called the ode with the tspan option, a successful time step will be at each element of tspan.
That has been the premise of my enquiry from the beginning - that I should only get Fs for each successful integration time step and that those steps are defined by tspan.
Regards.
Stephen23 2021년 1월 25일
편집: Stephen23 2021년 1월 25일
"...l'm a bit confused about your statement that the outputfcn approach collects all intermediate points..."
Sorry, that was my mistake (since removed from my answer): I confused the common attempt of using persistent variables or similar to store all intermediate calculations, including those backward steps, etc., with the use of outputfcn.
I would have to check the outputFcn documentation, and probably try some examples.
In any case, it is still the more complex approach, if your goal is just to get the Fs values.

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