How to multiply function handles?

Question

Max 2016-1-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/264762-how-to-multiply-function-handles

评论： James Morrison 2023-2-6

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hello,

I would like to calculate and to plot the following equation:

f(t,x) = (beta_hat*s(t,x)*[I(t,x)]^(beta_hat-1))*exp(-[I(t,x)]^beta_hat)

where:

      s(t,x): step-function(330 for x=[0;2000])
                           (350 for x=[2000;3000])
                           (390 for x=[3000;4000])
     I(t,x): Integral(exp(-B_hat/x(u))/C_hat)) from x=0 to t
     beta_hat, C_hat and B_hat are known parameters

My MATLAB-script looks like that:

beta_hat = 4.2915822 
B_hat = 1861.6186657 
C_hat = 58.9848692  
%%Step-function x(t)
syms t 
y(t) = (exp(-B_hat/((heaviside(t)-heaviside(t-2000))*(330)+(heaviside(t-2000)-heaviside(t-3000))*(350)+...
       (heaviside(t-3000)-heaviside(t-4000))*(390))))/C_hat;  
fnum=matlabFunction(y);
Inum=@(x)integral(fnum,0,x);
f = @(x)(beta_hat*fnum(x).*Inum(x).^(beta_hat-1)).*exp(-(Inum(x).^beta_hat))
ezplot(f,[0,14000])

But if I plot my function f the figure isn´t right. There should be a smooth curve, similar to the bell-shaped curve.

Does anybody recognize my fault?

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Answer 1

Walter Roberson 2016-1-22

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https://ww2.mathworks.cn/matlabcentral/answers/264762-how-to-multiply-function-handles#answer_207044

What value are you assuming that heaviside(0) has?

As you are working with the Symbolic toolbox anyhow, have you considered doing a symbolic integration with int() instead of changing to a numeric function and using integral() ?

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Max 2016-1-22

Hi Walter,

thank you for your answer. I found found the mistake. It wasn´t a problem in MATLAB, but in a mathematical equation.

Walter Roberson 2016-1-22

heaviside(x) is defined to be 0 for x < 0,, and is defined to be 1 for x > 0. You cannot have heaviside(t=4000) be 390 .

The value of heaviside(x) for x = 0 is not precisely defined. In some systems, heaviside(0) is undefined. In some systems, heaviside(0) is 1/2 . In some systems, heaviside(0) is 0. In some systems, heaviside(0) is 1.

As of R2015a, you can use the new sympref() to control the value that is used for heaviside(0)

Remember, no matter what value you use, heaviside is discontinuous, so you should avoid using numeric integration with it, unless you split your integral up into disjoint ranges at each boundary, integrate each separately, and total the results.

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Answer 2

Tiago Araujo 2021-6-24

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/264762-how-to-multiply-function-handles#answer_732000

在 MATLAB Online 中打开

I need to do a integral with a function by combining two functions, something like this:

syms q r
x(q,r) =  q   + 2*r;
y(q,r) =  q*9 + r;
G = @(q,r) x+y
integral2(G,1,2,1,2)

It is only a example, my real problem is much more complex. How can i do that?

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Tiago Araujo 2021-6-24

在 MATLAB Online 中打开

Please, look at my real problem:

% Coordenadas no espaço X,Y,Z
P1 = [1 1 1];
P2 = [1 1 1.4];
P3 = [1.2 1 1];
P4 = [1.2 1 1.4];
P5 = [1 1.3 1];
P6 = [1 1.3 1.4];
P7 = [1.2 1.3 1];
P8 = [1.2 1.3 1.4];
% Transformação de Coordenadas
C1 = (1/8)*(1+q)*(1-r)*(1-t);
C2 = (1/8)*(1+q)*(1+r)*(1-t);
C3 = (1/8)*(1-q)*(1+r)*(1-t);
C4 = (1/8)*(1-q)*(1-r)*(1-t);
C5 = (1/8)*(1+q)*(1-r)*(1+t);
C6 = (1/8)*(1+q)*(1+r)*(1+t);
C7 = (1/8)*(1-q)*(1+r)*(1+t);
C8 = (1/8)*(1-q)*(1-r)*(1+t);
% Coordenadas em função de q,r e t
xi = C1*P1(:,1)+C2*P2(:,1)+C3*P3(:,1)+C4*P4(:,1)+...
    C5*P5(:,1)+C6*P6(:,1)+C7*P7(:,1)+C8*P8(:,1);
yi = C1*P1(:,2)+C2*P2(:,2)+C3*P3(:,2)+C4*P4(:,2)+...
    C5*P5(:,2)+C6*P6(:,2)+C7*P7(:,2)+C8*P8(:,2);
zi = C1*P1(:,3)+C2*P2(:,3)+C3*P3(:,3)+C4*P4(:,3)+...
    C5*P5(:,3)+C6*P6(:,3)+C7*P7(:,3)+C8*P8(:,3);
% Jacobiano
J =  [diff(xi,q) diff(yi,q) diff(zi,q)
      diff(xi,r) diff(yi,r) diff(zi,r)
      diff(xi,t) diff(yi,t) diff(zi,t)];
% Função G(q,r,t)
F = @(q,r,t) xi(q,r,t) .* yi(q,r,t)./zi(q,r,t)
G = det(J)*F
    
% Solução analítica
integral3(G,-1, 1, -1, 1, -1, 1)