Error using function matlab

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Nikki Dobson
Nikki Dobson 2020-11-9
评论： Nikki Dobson 2020-11-9
Hi, I’m trying to use the bisection method to find x. Think there’s a problem using the function. If you have any hints to help me fix it, it would be much appreciated. Thanks
clc
k=1007.564; %stiffness (N/m)
m=98.668; %mass (kg)
F_0=108.764; %applied force magnitude (N)
x_0=-0.02; %initial displacement (m)
v_0=0.1; %initial velocity (m/s)
t=4.54; %time (s)
x_1=0.0386; %displacement (m)
om_n=sqrt(k/m);
f_0=F_0/m;
x=0; %omega
function bisection
n = 20;
% initial interval
a = 5.12;
b = 6.45;
fprintf('\n initial interval [%g, %g] \n total bisection steps %d\n', a,b,n);
%initialise and check that there is a root in the prescribed interval
x_left = a;
x_right = b;
f_left = f(x_left);
f_right = f(x_right);
if f_left*f_right > 0
error('ERROR: no root in the specified interval');
end
%the bisection method
for i=1:n
if f_left == 0
%exact root is reached by the left bound of the interval
fprintf('\n stage %g root %g with zero absolute error \n',i,x_left);
return;
end
if f_right==0
%exact root is reached by the right bound of the interval
fprintf('\n stage %g root %g with zero absolute error \n',i,x_right);
return
end
%the bisection process
x_mid = (x_left+x_right)/2.0;
f_mid = f(x_mid);
if f_left*f_mid <= 0
%there is a root in [x_left,x_mid]
x_right = x_mid;
f_right = f_mid;
end
if f_left*f_mid > 0
% there is a root in [x_mid,x_right]
x_left = x_mid;
f_left = f_mid;
end
%compute the approximate rood for the current bisection step
root = (x_left+x_right)/2.0;
%compute the absolute error for the current bisection step
abs_err=(x_right-x_left)/2.0;
fprintf('\n stage %g root %g absolute error < %g \n',i,root,abs_err);
end
%check satisfaction of equation at end of process
residual = f(root);
fprintf('\n final residual = %g \n',residual);
end
%Subfunction defines the equation f(x) = 0
function f_value = f(x) %x=omega
f_value = (v_0/om_n)*sin(om_n*t) + (x_0-((f_0)/(om_n)^2-(x)^2))*cos(om_n*t) + ((f_0)/(om_n)^2-(x)^2)*cos(x*t);
end

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采纳的回答

Walter Roberson
Walter Roberson 2020-11-9
You do not call the function bisection
Your function f does not have access to the variables defined in the script. f would need to be a nested function using shared variables -- and you need to nest inside a function, not inside a script.
For example,
function driver
k=1007.564; %stiffness (N/m)
m=98.668; %mass (kg)
F_0=108.764; %applied force magnitude (N)
bisection(@f); %call your function
function f_value = f(x) %x=omega
%this is a nested function that shares variables
f_value = (v_0/om_n)*sin(om_n*t) + (x_0-((f_0)/(om_n)^2-(x)^2))*cos(om_n*t) + ((f_0)/(om_n)^2-(x)^2)*cos(x*t);
end
end
function bisection(f)
stuff
end

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