matlab people helo hellp

i don't know what to say

[Stephen23]

Original question by Constantine Kenny retrieved from Google Cache:

https://web.archive.org/web/20211207072002/https://webcache.googleusercontent.com/search?q=cache%3AjX2SMniq6qwJ%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F1599824-convert-time-series-for-acceleration-to-displacement-using-ode45

https://web.archive.org/web/20211207072114/https://www.mathworks.com/matlabcentral/answers/1599824-matlab-people-helo-hellp

Convert time series for acceleration to displacement using ode45

I have data for ground motion acceleration as a function of time. I would like to convert this acceleration data to displacement data using matlab's ode45 function. Is this possible, and is this the right way to do so?

The differential equation is:

ẍ(t) + 2ζνẋ(t) + (v^2)*x(t) = -a(t)

where a(t) is the acceleration time series that I have.

Answer 1

[Star Strider]

It appears that ζ anc v are constants. Are they known and provided or are they to be estimated?

Assuming they are provided —

syms a(t) t v x(t) Y zeta 
D1x = diff(x);
D2x = diff(D1x);
DE = D2x + 2*zeta*v*D1x + v^2*x + a

DE(t) = 

[VF,Sbs] = odeToVectorField(DE)

VF = 

Sbs = 

odefcn = matlabFunction(VF, 'Vars',{'t','Y','v','zeta','a'})

Warning: Function 'a' not verified to be a valid MATLAB function.

odefcn = function_handle with value:
    @(t,Y,v,zeta,a)[Y(2);-a(t)-v.^2.*Y(1)-v.*zeta.*Y(2).*2.0]

v = 4.2;
zeta = 0.42;
t = linspace(0, 50, 15);
a = randn(size(t))*10;                                          % Acceleration Values Vector
aintp = @(tv) interp1(t, a, tv);                                % Interpolate Data
ic = [0 2];                                                     % Use Appropriate Values
[t,y] = ode45(@(t,y)odefcn(t,y,v,zeta,aintp), t, ic);

figure
plot(t, y)
grid
xlabel('t')
ylabel('Amplitude')
legend(string(Sbs), 'Location','best')

Seismology is not an area of my expertise, so make appropriate changes (and use the correct parameter values) to get the desired results.

.