Using for loop for midpoint method
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hi guys, this is my code
m = 1000;
f = 75000;
v0 = 0;
h = 0.1;
tspan = [0 10];
c = [50,100,500,1000,2000];
dvdt = cell(1,length(c));
for i = 1:length(c);
dvdt{i} = @(t,v) (f-c{i}*v)/m;
end
t =cell(1,length(c));
v =cell(1,length(c));
for k = 1:length(c)
[t{k},v{k}]=midpoint(dvdt{1,k},tspan,v0,h)
end
%% Question
I want to use midpoint to calculate the t and v, and my dvdt is relied on c(constant values run from 50 to 2000). The code works, but t and v are released just only based on c = 50, which means my for loop does not work. Can someone please help !!!
Thanks in advance
0 comentarios
Respuestas (1)
Hitesh
el 7 de Abr. de 2025
Hi Khang,
The error "Too many output arguments" indicates that the "midpoint" function did not return two outputs. You need to initializes time and velocity vectors and applies the midpoint method to solve the differential equation. Kindly refer to the folllowing revised code for better understanding:
function [t, v] = midpoint(dvdt, tspan, v0, h)
% Initialize time and solution vectors
t = tspan(1):h:tspan(2);
v = zeros(size(t));
v(1) = v0;
% Midpoint method loop
for n = 1:(length(t) - 1)
tn = t(n);
vn = v(n);
k1 = h * dvdt(tn, vn);
k2 = h * dvdt(tn + h/2, vn + k1/2);
v(n+1) = vn + k2;
end
end
m = 1000;
f = 75000;
v0 = 0;
h = 0.1;
tspan = [0 10];
c = [50, 100, 500, 1000, 2000];
dvdt = cell(1, length(c));
% Create the differential equations for each value of c
for i = 1:length(c)
dvdt{i} = @(t, v) (f - c(i) * v) / m;
end
t = cell(1, length(c));
v = cell(1, length(c));
% Solve using the midpoint method for each value of c
for k = 1:length(c)
[t{k}, v{k}] = midpoint(dvdt{k}, tspan, v0, h);
end
0 comentarios
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!