Get all graphs from ODE23

Hello all,

I am trying to write a code using the ode23 to solve for a free vibration damped SDOF. I wrote the following code without using the ode23 function:

function f=free_viscous(m,c,k,x0,v0,n,tspan);
t=[0:tspan:tspan*n];
wn=sqrt(k/m);
cc=2*m*wn;
zeta=c/cc;
phi=atan((v0+zeta*wn*x0)/(x0*wn*sqrt(1-zeta^2)));
X=sqrt(x0^2*wn^2+v0^2+2*x0*v0*zeta*wn)/(wn*sqrt(1-zeta^2))
syms d;
x=X*cos(wn*d*sqrt(1-zeta^2)-phi)*exp(-zeta*wn*d);
v=diff(x,d)
a=diff(v,d)
for i=1:101    
    x(i)=X*cos(wn*t(i)*sqrt(1-zeta^2)-phi)*exp(-zeta*wn*t(i));
    v(i)=X*wn*exp(-t(i)*wn*zeta)*sin(phi - t(i)*wn*(1 - zeta^2)^(1/2))*(1 - zeta^2)^(1/2) - X*wn*zeta*exp(-t(i)*wn*zeta)*cos(phi - t(i)*wn*(1 - zeta^2)^(1/2)); 
    a(i)=X*wn^2*exp(-t(i)*wn*zeta)*cos(phi - t(i)*wn*(1 - zeta^2)^(1/2))*(zeta^2 - 1) + X*wn^2*zeta^2*exp(-t(i)*wn*zeta)*cos(phi - t(i)*wn*(1 - zeta^2)^(1/2)) - 2*X*wn^2*zeta*exp(-t(i)*wn*zeta)*sin(phi - t(i)*wn*(1 - zeta^2)^(1/2))*(1 - zeta^2)^(1/2);
 end    
plot(t,x,t,v,t,a);

Running code for the previous is(i was able to get all of my graphs):

clc
clear all
free_viscous(450,1000,26519.2,0.539657,1,100,0.025)

I was able to write the ode23 code but i didnt know how to use MATLAB to plot the acceleration. I managed only the displacement and velocity:

function f=dfunc2(t,x);
m=450;
c=1000;
k=26519.2;
f=zeros(2,1);
f(1)=x(2);
f(2)=(-c/m)*x(2)-(k/m)*x(1);

the main code is:

clc
clear all
tspan=[0:0.025:2.5];
x0=[0.539657;1];
[t,x]=ode23('dfunc2',tspan,x0);
plot(t,x(:,1),t,x(:,2));

Can anyone please help? this is a personal endeavour and not related to any homework.

Thanks!

Answer 1

[Star Strider]

Try this approach:

function f=dfunc2(t,x);
m=450;
c=1000;
k=26519.2;
f=zeros(2,1);
f(1)=x(2);
f(2)=(-c/m)*x(2)-(k/m)*x(1);
end

tspan=[0:0.025:2.5];
x0=[0.539657;1];
[t,x]=ode23(@dfunc2,tspan,x0);
acc = gradient(x(:,2))./gradient(t);                % Calculate Acceleration By Taking The Numerical Derivative Of Velocity

figure
plot(t,x(:,1),t,x(:,2), t,acc);
grid

.

Answer 2

[Jan]

Vectorize the function to be integrated: Then it accepts a matrix with different X values also:

tspan  = 0:0.025:2.5;
x0     = [0.539657; 1];
[t, x] = ode23(@dfunc2, tspan, x0);

% Call dfunc2 to get the accelerations:
a = dfunc2(t, x.').';  % x must be transposed twice

figure;
plot(t, x(:,1), t, x(:,2), t, a(:, 2));
end

function f = dfunc2(t, x)
m = 450;
c = 1000;
k = 26519.2;
f = [x(2, :); ...
     (-c/m) * x(2, :) - (k/m) * x(1, :)];
end

a(:, 1) is the same as x(:, 2) as expected.

[Ali Awada]

Hey Jan,

it works thank you. Can you explain trasnposing x twice?

[Jan]

The function to be integrated is called from the integrator with the input x as a column vector. In the output of the integrator the trajectory is saved as row vectors for each time point. To use the output of the integrator as input for the function to be integrated, it must be transposed. To have the output of the acceleration in the same format as the trajectory, you have to transpose again.

Obtaining the derivative directly is more accurate than a numerical differentiation by gradient.

[Ali Awada]

Thanks for the detailed explanation. Definitely a better approach!