Matlab ode45检索参数 [英] matlab ode45 retrieving parameters

问题描述

我正在Matlab中试用ode45.我已经学习了如何将参数传递给ode函数，但是我仍然有一个问题.假设我想计算汽车的轨迹(速度曲线)，并且有一个函数，例如getAcceleration，它给了我汽车的加速度，但也给了我正确的档位:[acceleration, gear] = getAcceleration(speed,modelStructure)其中，modelStructure代表汽车的型号.

I'm experimenting with ode45 in Matlab. I've learned how to pass parameters to the ode function but I still have a question. Let's suppose that I want to compute the trajectory (speed profile) of a Car and I have a function, e.g. getAcceleration, that gives me the acceleration of the car but also the right gear: [acceleration, gear] = getAcceleration(speed,modelStructure) where modelStructure represents the model of the car.

ode函数为:

function [dy] = car(t,y,modelStructure)

dy           = zeros(2,1);
dy(1)        = y(2);
[dy(2),gear] = getAcceleration(y(1),modelStructure);

然后我以这种方式称呼Ode45集成商:

Then I call the Ode45 integrator in this way:

tInit = 0;
tEnd  = 5,
[t,y] = ode45(@car,[tInit tEnd], [speedInitial,accelerationInitial],options,modelStructure);

问题是:如何获得向量存储齿轮?我应该有类似[t,y,gear]=ode45(....)的东西还是gear应该在y向量之内?

The problem is: how do I get the vector storing gears? Should I have something like [t,y,gear]=ode45(....) or should gear be within the y vector?

我一直在编写代码，并使用事件功能，现在可以获取汽车的齿轮"变化(作为事件). 现在，我有一个与相同代码有关的新问题. 想象一下，当我评估"dy"向量时，我可以得到另一个值Z，这使我可以极大地提高加速度计算(getAcceleration)的速度:

I've been working on my code and using the events function I'm now able to get the car 'gears' changes (as events). Now I have a new problem related to the same code. Imagine that when I evaluate de 'dy' vector I'm able to get a further value Z which let me to have a massive speed up calling the acceleration computation (getAcceleration):

function [dy] = car(t,y,modelStructure)

dy           = zeros(2,1);
dy(1)        = y(2);
[dy(2),Z(t)] = getAcceleration(y(1),modelStructure,Z(t-1)); 

并假设我也能够在初始条件下计算Z.问题是我无法计算Z导数.

and suppose that I'm also able to compute Z at the initial condition. The problem is that I'm not able to compute the Z derivative.

有没有一种方法可以传递Z值而不整合步进?

Is there a way to pass Z value throw the stepping without integrating it?

谢谢大家.

推荐答案

首先:为什么微分方程的初始值是初始速度(speedInitial)和初始加速度(accelerationInitial)?这意味着，微分方程car将在每次t处计算加速度(y(1))和加速度率(y(2))，即加速度的时间导数.这似乎是不正确的...我想说初始值应该是初始位置(positionInitial)和初始速度(speedInitial).但是，我不知道您的模型，我可能是错的.

First off: why are the initial values to the differential equation the initial speed (speedInitial) and the initial acceleration (accelerationInitial)? That means that the differential equation car will be computing the acceleration (y(1)) and the jerk (y(2)), the time-derivative of the acceleration, at each time t. That seems incorrect...I would say the initial values should be the initial position (positionInitial) and the initial speed (speedInitial). But, I don't know your model, I could be wrong.

现在，直接在解决方案中获得gear:您不能，除非没有破解ode45.这也是合乎逻辑的；您也不能一直都直接获取dy，可以吗?但这不是设置ode45的方式.

Now, getting the gear in the solution directlty: you can't, not without hacking ode45. This is also logical; you also cannot get dy at all times directly, can you? That's just not how ode45 is set up.

我在这里看到两种解决方法:

There's two ways out I see here:

免责声明 :请勿使用此方法.只是在这里显示大多数人会做的第一次尝试.

您可以将gear存储在全局变量中.它可能是最少的编码量，但也是最不方便的结果:

You can store gear in a global variable. It's probably the least amount of coding, but also the least convenient outcome:

global ts gear ii

ii    = 1;
tInit = 0;
tEnd  = 5,
[t,y] = ode45(...
    @(t,y) car(t,y,modelStructure), ...
    [tInit tEnd], ...
    [speedInitial, accelerationInitial], options);

...

function [dy] = car(t,y,modelStructure)
global ts gear ii

dy    = zeros(2,1);
dy(1) = y(2);
[dy(2),gear(ii)] = getAcceleration(y(1),modelStructure);

ts(ii) = t;
ii = ii + 1;

但是，由于ode45的性质，这将为您提供一系列时间ts和关联的gear，其中包含中间点和/或被ode45拒绝的点.因此，您之后必须对这些内容进行过滤:

But, due to the nature of ode45, this will get you an array of times ts and associated gear which contains intermediate points and/or points that got rejected by ode45. So, you'll have to filter for those afterwards:

ts( ~ismember(ts, t) ) = [];

我再说一遍:这是我推荐的方法 不 .仅在测试或执行一些快速处理工作时才使用全局变量，但始终非常迅速地转向其他解决方案.同样，全局变量在ode45的每个(子)重复项处增长，这是不可接受的性能损失.

I'll say it again: this is NOT the method I'd recommend. Only use global variables when testing or doing some quick-n-dirty stuff, but always very quickly shift towards other solutions. Also, the global variables grow at each (sub-)iteration of ode45, which is an unacceptable performance penalty.

最好使用下一种方法:

对于您的情况，这也不是一件难事，也是我建议您采取的方式.首先，如下修改微分方程，然后按正常方式求解:

This is also not too hard for your case, and the way I'd recommend you to go. First, modify the differential equation as below, and solve as normal:

tInit = 0;
tEnd  = 5,
[t,y] = ode45(...
    @(t,y) car(t,y,modelStructure), ...
    [tInit tEnd], ...
    [speedInitial, accelerationInitial], options);

...

function [dy, gear] = car(t,y,modelStructure)    

dy    = [0;0];
dy(1) = y(2);
[dy(2),gear] = getAcceleration(y(1),modelStructure);

，然后在ode45完成后，执行以下操作:

and then after ode45 completes, do this:

gear = zeros(size(t));
for ii = 1:numel(t)
    [~, gear(ii)] = car(t(ii), y(ii,:).', modelStructure); 
end

这有时会为您提供汽车所需的所有挡位(c11).

That will get you all the gears the car would have at times t.

我在这里看到的唯一缺点是，与ode45单独使用相比，您对car进行的函数评估要多得多.但是，如果每个car评估时间都在几秒或更长时间内，这只是一个实际问题，我怀疑在您的设置中情况并非如此.

The only drawback that I can see here is that you'll have many more function evaluations of car than ode45 would use by itself. But this is only a real problem if each evaluation of car takes in the order of seconds or longer, which I suspect is not the case in your setup.

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