How to multiply function handles?

Max
2016 1 22
2 个回答
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更新时间：2023 2 6
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0 个投票

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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 采纳的回答

Walter Roberson
Walter Roberson 2016-1-22

0 个投票

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
Max 2016-1-22
Hello Walter,
heaviside(t=0) should be 330 and heaviside(t=4000) should be 390. But I didn´t know how to realize that step-function in MATLAB which I could use for calculation the Integral.
I´ve also tried that: I=int(y); f_new = (beta_hat*y.*(I)^(beta_hat-1)).*exp(-((I).^beta_hat)) ezplot(f_new) But,I couldn´t plot that expression.
Max
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
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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更多回答（1 个）

Tiago Araujo
Tiago Araujo 2021-6-24

1 个投票

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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Steven Lord
Steven Lord 2021-6-24
Ran in:
Ran in:
The only function handle you have in your code is G.
But if x and y were defined as function handles, you can't multiply function handles. What you can do is multiply the values obtained by evaluating the function handles.
x = @(q, r) q + 2*r;
y = @(q, r) 9*q + r;
G = @(q, r) x(q, r) .* y(q, r); % let's multiply rather than add
integral2(G, 1, 2, 1, 2)
ans = 68.4167
Tiago Araujo
Tiago Araujo 2021-6-24
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)
It doesnt work.
Walter Roberson
Walter Roberson 2021-6-24
You have F as a function handle. You cannot multiply a function handle by anything.
You are also mixing symbolic expressions and function handles.
Stick with symbolic all the way through to creating G. Then use matlabFunction. Be sure to use the 'vars' option to control the order of variables.
Tiago Araujo
Tiago Araujo 2021-6-24
You're great! Thanks, matlabFunction solved my problem! I didnt know this function.
James Morrison
James Morrison 2023-2-6
Thank you @Steven Lord! worked perfectly :)

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