在线时间:8:00-16:00
迪恩网络APP
随时随地掌握行业动态
扫描二维码
关注迪恩网络微信公众号
如何用matlab来计算某地的面积? 但是要处理成这样: X=[0 1 2 3 4 5 6 7 8 9 10 11 12 13 13.6087]; Y1=[0 1.5652 1.6425 2.8696 3.0343 3.7391 3.9130 3.9130 3.0343 2.6957 2.4348 2.1739 1.5652 0.8696 0]; Y2=[0 -0.6087 -0.3636 -0.3478 -0.4545 -1.0435 -1.3913 -1.3913 -1.3913 -1.3913 -1.3913 -1.1304 -1.7391 -1.4783 0]; 方法一:先拟合后求积分 X=[0 1 2 3 4 5 6 7 8 9 10 11 12 13 Y1=[0 1.5652 1.6425 2.8696 3.0343 3.7391 Y2=[0 -0.6087 -0.3636 -0.3478 -0.4545 C1=polyfit(X,Y1,7) C2=polyfit(X,Y2,7) Z1=polyval(C1,X); Z2=polyval(C2,X); plot(X,Y1,‘r*’,X,Z1,‘b’);%n=2阶显示拟合结果 hold plot(X,Y2,‘r*’,X,Z2,‘b’); A1=polyarea(X,Z1); A2=abs(polyarea(X,Z2)); A=A1+A2 A = 46.4513 方法二: X=[0 1 2 3 4 5 6 7 8 9 10 11 12 13 Y1=[0 1.5652 1.6425 2.8696 3.0343 3.7391 Y2=[0 -0.6087 -0.3636 -0.3478 -0.4545 plot(X,Y1,X,Y2) P(1)=spline(X,Y1); P(2)=spline(X,Y2); N=100000;x=unifrnd(0,14,[1,N]); y=unifrnd(0,4,[1,N]); count1=sum(ppval(P(1),x)>y); integral1=414count1/N; x=unifrnd(0,14,[1,N]); y=unifrnd(-2,0,[1,N]);count1=sum(ppval(P(2),x)>y); count2=sum(ppval(P(2),x)>y); N=100000;x=unifrnd(0,14,[1,N]); y=unifrnd(-2,0,[1,N]); count2=sum(ppval(P(2),x)<y); integral2=214count2/N;
S = 48.0771 |
2023-10-27
2022-08-15
2022-08-17
2022-09-23
2022-08-13
请发表评论