MATLAB HELP ME PLS
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close all
clear
clc
format long
a=0;
b=10;
function [Distance] = trapezoidal(Speed, a, b, n)
h = (b-a)/n;
result = 0.5*Speed*a + 0.5*Speed*b;
for i = 1:(n-1)
result = result + Speed*(a + i*h);
end
[Distance] = h*result;
fprintf('ans=%6.6ff\n',Distance)
end
답변 (1개)
TADA
2019년 7월 18일
편집: TADA
2019년 7월 18일
The trapezoidal integration method is a numeric calculation which approximates the area trapped between the curve and the x axis by dividing the curve to segments. the area of each segment is calculated by calculating the area of a trapeze entrapped by the coordinates of that segment.
function [auc, integration] = trapezoidal(x,y)
dx = diff(x);
integration = ((y(1:end-1)+y(2:end)) .* dx) / 2;
auc = sum(integration);
end
The accuracy of this method is limited mainly by the size of the segments. smaller dx means more accurate results
so define your time vector with more elements:
t = linspace(0,10,1000);
댓글 수: 1
TADA
2019년 7월 18일
to iteratively improve the approximation as required, I would start from a small number of segments, lets say 10 segments like you did
then check the error of the calculated distance
To calculate the actual distance you can integrate the polynomial:
% velocity polynomial coefficients
vp = [0.0011, -0.02928, 0.2807, -1.1837, -0.8283, 41.234, -3.3549];
tRange = [0 10];
nSegments = 10;
% call this in a loop that improves the precesion
while logical condition
% do something to improve precesion here
% calculate distance and error again
[d, err] = tryTrapz(tRange, vp, nSegments);
fprintf('Your message');
end
function [d, err] = tryTrapz(tRange, vp, nSegments)
t = linspace(tRange(1), tRange(2), nSegments);
d = trapz(t, polyval(vp, t));
% distance polynomial coefficients = integrated velocity
dp = polyint(vp);
theoreticalDist = polyval(dp, t(end));
err = 100*(theoreticalDist - d)/theoreticalDist;
end
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