Computing the overlapping area of curves

조회 수: 3 (최근 30일)
Deepa Maheshvare
Deepa Maheshvare 2022년 7월 3일
편집: Chunru 2022년 7월 4일
I have a set of datapoints corresponding to 5 curves.
scale = 1.5;
x1 = [0,4,6,10,15,20]*scale;
y1 = [18,17.5,13,12,8,10];
x2 = [0,10.5,28]*scale;
y2= [18.2,10.6,10.3];
x3 = [0,4,6,10,15,20]*scale;
y3 = [18,13,15,12,11,9.6];
x4 = [9,17,28]*scale;
y4 = [5,5.5,7];
x5 = [1,10,20]*scale;
y5 = [3,0.8,2];
plot(x1,y1, '*-', x2, y2, '*-', x3, y3, '*-', x4, y4, '*-', x5, y5, '*-')
I want to find the overlapping area under these curves.
Suggestions on the functions that could be uded to find the intersecting area and the x limits of the intersecting area will be really helpful.
  댓글 수: 3
Chunru
Chunru 2022년 7월 4일
Which curves do you mean? How the overlapping area is defined?
Deepa Maheshvare
Deepa Maheshvare 2022년 7월 4일
Thank you for the reply. I am interested in finding the area under each of the curve and visualize the area that overlaps.

댓글을 달려면 로그인하십시오.

채택된 답변

Chunru
Chunru 2022년 7월 4일
편집: Chunru 2022년 7월 4일
scale = 1.5;
x{1} = [0,4,6,10,15,20]*scale;
y{1} = [18,17.5,13,12,8,10];
x{2} = [0,10.5,28]*scale;
y{2}= [18.2,10.6,10.3];
x{3} = [0,4,6,10,15,20]*scale;
y{3} = [18,13,15,12,11,9.6];
x{4} = [9,17,28]*scale;
y{4} = [5,5.5,7];
x{5} = [1,10,20]*scale;
y{5} = [3,0.8,2];
figure; hold on
xmin = -inf; xmax=inf;
for i=1:length(x)
area(x{i}, y{i}, 'FaceAlpha', 0.5); % visualize area with transparancy
a(i) = trapz(x{i}, y{i}); % compute area
xmin = max(xmin, min(x{i}));
xmax = min(xmax, max(x{i}));
end
xline([xmin xmax], 'LineWidth', 2)
legend
a
a = 1×5
369.7500 501.1125 379.5000 166.1250 46.6500
  댓글 수: 2
Deepa Maheshvare
Deepa Maheshvare 2022년 7월 4일
Thanks so much. Could we find/highlight the area that is intersected by all curves?
Chunru
Chunru 2022년 7월 4일
See the update.

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 2-D and 3-D Plots에 대해 자세히 알아보기

태그

제품


릴리스

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by