MATLAB分段函数 [英] MATLAB Piecewise function

问题描述

我必须在MATLAB中构造以下函数，并且遇到了麻烦.

I have to construct the following function in MATLAB and am having trouble.

考虑在[0,4)中为t定义的函数s(t)，

Consider the function s(t) defined for t in [0,4) by

              { sin(pi*t/2)  , for t in [0,1)
       s(t) = { -(t-2)^3     , for t in [1,3)*
              { sin(pi*t/2)  , for t in [3,4)

(i)生成由512个均值组成的列向量 在[0,4)间隔内此函数的样本. (这 最好通过串联三个向量来完成.)

(i) Generate a column vector s consisting of 512 uniform samples of this function over the interval [0,4). (This is best done by concatenating three vectors.)

我知道它一定是某种形式.

I know it has to be something of the form.

N = 512;
s = sin(5 * t/N)." ;

N = 512;
s = sin(5*t/N).' ;

但是我需要s为分段函数，有人可以为此提供帮助吗?

But I need s to be the piecewise function, can someone provide assistance with this?

推荐答案

如果我理解正确，那么您正在尝试创建3个向量来计算所有t的特定函数输出，然后对每个向量进行切片并将它们连接起来取决于t的实际值.这是低效的，因为您初始化的向量是您实际想要的3倍(内存)，并且进行的计算是CPU的3倍，而其中大部分将被丢弃.最重要的是，如果您的t并非您期望的 (即单调递增)，则使用级联会有些棘手.这可能是不太可能的情况，但总的来说更好.

If I understand correctly, you're trying to create 3 vectors which calculate the specific function outputs for all t, then take slices of each and concatenate them depending on the actual value of t. This is inefficient as you're initialising 3 times as many vectors as you actually want (memory), and also making 3 times as many calculations (CPU), most of which will just be thrown away. To top it off, it'll be a bit tricky to use concatenate if your t is ever not as you expect (i.e. monotonically increasing). It might be an unlikely situation, but better to be general.

这里有两个选择，第一个是imho不错的Matlab方法，第二个是更常规的方法(如果您来自C ++或类似的东西，您可能会更习惯了，我使用了很长时间).

Here are two alternatives, the first is imho the nice Matlab way, the second is the more conventional way (you might be more used to that if you're coming from C++ or something, I was for a long time).

function example()
    t = linspace(0,4,513); % generate your time-trajectory
    t = t(1:end-1); % exclude final value which is 4

    tic
    traj1 = myFunc(t);
    toc

    tic
    traj2 = classicStyle(t);
    toc
end

function trajectory = myFunc(t)
    trajectory = zeros(size(t)); % since you know the size of your output, generate it at the beginning. More efficient than dynamically growing this.

    % you could put an assert for t>0 and t<3, otherwise you could end up with 0s wherever t is outside your expected range

    % find the indices for each piecewise segment you care about
    idx1 = find(t<1); 
    idx2 = find(t>=1 & t<3);
    idx3 = find(t>=3 & t<4);

    % now calculate each entry apprioriately
    trajectory(idx1) = sin(pi.*t(idx1)./2);
    trajectory(idx2) = -(t(idx2)-2).^3;
    trajectory(idx3) = sin(pi.*t(idx3)./2);
end

function trajectory = classicStyle(t)
    trajectory = zeros(size(t));

    % conventional way: loop over each t, and differentiate with if-else
    % works, but a lot more code and ugly
    for i=1:numel(t)
        if t(i)<1
            trajectory(i) = sin(pi*t(i)/2);
        elseif t(i)>=1 & t(i)<3
            trajectory(i) = -(t(i)-2)^3;
        elseif t(i)>=3 & t(i)<4
            trajectory(i) = sin(pi*t(i)/2);
        else
            error('t is beyond bounds!')
        end
    end     
end

请注意，当我尝试使用常规方式"时，有时对于您正在处理的采样大小而言会更快，尽管第一种方式(myFunc)确实会更快，因为您实际进行了很多扩展.无论如何，我建议采用第一种方法，因为它更易于阅读.

Note that when I tried it, the 'conventional way' is sometimes faster for the sampling size you're working on, although the first way (myFunc) is definitely faster as you scale up really a lot. In anycase I recommend the first approach, as it is much easier to read.

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