ODE plot with 2nd order diff eqs
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I wanted to check to make sure I am creating the correct graph for problem 1 of the attached project file. It is similar to a previous problem I had to do , so I used that code and just changed the equation and conditions to match this problem.
Fp = @(t,x,w) [x(2); 10*cos(t.*w)-x(1).*4-x(1).*5]; % Spring ODE
ww = [0, 0.5, 1, 2, 4, 8, 16]; % ‘epsilon’ Vector
ts = linspace(0,80, 500); % Time Vector
for k1 = 1:length(ww)
[~, x{k1}]=ode45(@(t,x) Fp(t,x,ww(k1)), ts, [0,1]);
end
figure(1)
for k1 = 1:length(ww)
subplot(length(ww), 1, k1)
plot(ts,x{k1})
legend(sprintf('\\epsilon = %.1f', ww(k1)))
grid
axis([xlim -5 +5])
1 个评论
Torsten
2015-7-14
Your equation to be solved and your initial conditions are not consistent with project2.pdf.
Check again.
Best wishes
Torsten.
回答（1 个）
Star Strider
2015-7-14
The problem is that the loop code isn’t updated to match your new ODE. (Your ODE code and ode45 call are correct.)
Use this loop instead of the old one:
figure(1)
for k1 = 1:length(ww)
subplot(length(ww), 1, k1)
plot(ts,x{k1}(:,1))
hold on
legend(sprintf('\\omega = %.1f', ww(k1)))
grid
axis([xlim -2 +2])
xv = x{k1}(:,1);
zxix = find((xv.*circshift(xv, [1 0])) <= 0); % Approximate Zero-Crossing Indices
if length(zxix) > 2
ixrng = max(1,zxix(3)-1):min(zxix(3)+1,length(ts)); % Index Range
tzx{k1} = interp1(xv(ixrng), ts(ixrng), 0); % ‘tx’ Values At Zero-Crossings
plot(tzx{k1}, zeros(size(tzx{k1})), 'bp') % Plot Zero Crossings
end
hold off
end
xlabel('Time')
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2015-7-14
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