4-point first derivative

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alburary daniel
alburary daniel 2018-8-3
评论： Aquatris 2019-3-13
I am given data t=[0 1 2 3 4 5 6 7 8 9 10] and
y(t)=[1 2.7 5.8 6.6 7.5 9.9 10.2 11.2 12.6 13.6 15.8]
and have to evaluate the derivative of y at each given t value
using the following finite difference schemes
at 4-point first derivative
My code at finite difference is
t= 0: 1: 10;
y= [1 2.7 5.8 6.6 7.5 9.9 10.2 11.2 12.6 13.6 15.8];
n=length(y);
dfdx=zeros(n,1);
dfdx(t)=(y(2)-y(1))/(t(2)-t(1));
for i=2:n-1
dfdx(1)=(y(i+1)-y(i-1))/(t(i+1)-t(i-1));
end
dfdx(n)=(y(n)-y(n-1))/(t(n)-t(n-1));
but I have interesting of method of 4-point first derivative for more accuracy thanks in avance!
  2 个评论
Aquatris
Aquatris 2018-8-3
Please format the question properly and include the "following scheme" for us to be able to help.

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回答（1 个）

Aquatris
Aquatris 2018-8-3
I found a source where the equations for the differentiation are shown, with some typos. Here is an example code, where d4 is 4 point d5 is 5 point differentiation.
dt = 1e-3;
t = 0:0.001:20;
x = sin(0.4*t)+exp(-0.9*t); % sample signal
xd = 0.4*cos(0.4*t)-0.9*exp(-0.9*t);% sample signals exact derivative
n=length(x);
d4=zeros(size(x));
d5=zeros(size(x));
for j = 3:n-2;
d4(j) = -1/6/dt*(-2*x(j+1)-3*x(j)+6*x(j-1)-x(j-2));
d5(j)= 1/12/dt*(x(j-2) - 8.*x(j-1) + 8.*x(j+1) - x(j+2));
end
d4(1:2)=d4(3);
d4(n-1:n)=d4(n-2);
d5(1:2)=d5(3);
d5(n-1:n)=d5(n-2);
plot(t,xd,t,d4,'r--',t,d5,'m-.') % comparison plot
  6 个评论
显示 4更早的评论
Mohammad Ezzad Hamdan
Mohammad Ezzad Hamdan 2019-3-13
What do you mean by the terms below;
d5(1:2)=d5(3);
d5(n-1:n)=d5(n-2);
Aquatris
Aquatris 2019-3-13
It is not possible to calculate the first 2 or last 2 elements for the d5, I just equate them to some value. For the last 2 elements I hold the last calculated value of d5 and for the first 2 elements I hold the first calculated value. Depending on the application this might be acceptable.

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