Second time derivative of function?
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I have defined a model of displacement 'x' and represented its first derivative with 'v' using the code below. However, when I want to differentiate 'v' using the same function it does not work. Is there a different approach to differentiating derivatives in MATLAB? here is the code:
h = 60 ;
b =130 ;
r = 28 ;
w = 2*pi; % angular velocity
t =0:(1/36):1 ;
x = b*(r*sin(w*t))./(h-r*cos(w*t));
v = diff(x)./diff(t) ; % derivative of x w.r.t t
a = diff(v)./diff(t) ; % derivative of v w.r.t t
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Aquatris
2018-7-25
It does not work, because the length of diff(x) and diff(t) are not the same. diff function takes the difference between each element. However returns 1 element less than the original vector since there is no way to find the difference for the last element.
Since you have fixed step size you can use
v = diff(x)./(1/36);
td = t(1:end-1); % time vector for plotting first derivative
a = diff(v)./(1/36) ;tdd = t(1:end-2)
tdd = t(1:end-2); % time vector for plotting second derivative
Alternatively you might try symbolic toolbox to derive the derivative of the expression symbolicly and then plug in numbers.
更多回答（1 个）
KSSV
2018-7-25
diff reduces the dimension of vector by one. You may use gradient
h = 60 ;
b =130 ;
r = 28 ;
w = 2*pi; % angular velocity
t =0:(1/36):1 ;
x = b*(r*sin(w*t))./(h-r*cos(w*t));
v = gradient(x)./gradient(t) ; % derivative of x w.r.t t
a = gradient(v)./gradient(t) ; % derivative of v w.r.t t
Or, reduce the time step by one and use _diff_
h = 60 ;
b =130 ;
r = 28 ;
w = 2*pi; % angular velocity
t =0:(1/36):1 ;
x = b*(r*sin(w*t))./(h-r*cos(w*t));
v = diff(x)./diff(t) ; % derivative of x w.r.t t
a = diff(v)./diff(t(1:end-1)) ; % derivative of v w.r.t t
1 个评论
Beau Olafson
2019-5-30
That gradient code 'literally' saved my life. I was so dang lost until then!
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