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水箱的水流量估计

上传者:徐小白 |  格式:docx  |  页数:10 |  大小:0KB

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olum.m如下Рfunction yhat=volum(beta,x1) yhat=beta(1)-beta(2)*x1.^3+beta(3)*x1.^5-beta(4)*cos(0.1*x1)+beta(5)*sin(0.1*x1)Р输入程序:РclcРclearРx1=[0.46,1.38,2.4,3.41,4.43,5.44,6.46,7.47,8.45,9.48,10.46,10.94,11.49,12.22,13.14,14.43,15.44,16.37,17.38,18.49,19.5,20.4,21.43,22.49,23.42,24.44,25.45];Рy1=[13490,11620,10320,9760,9460,8710,9460,8920,10100,12751,15712,17117,18560,19710,18480,16140,16600,15400,14750,14620,14729,14264,14714,14490,14190 ,13410,11830];Рbeta0=[112000,200,1,10000,30000]';Р[beta,r ,J]=nlinfit(x1',y1','volum',beta0)Р[YY,delta]=nlpredci('volum',x1',beta,r ,J);Рplot(x1,y1,'k+',x1,YY,'r') %拟合图Рz=[YY-delta, -YY+delta]Рrcoplot(r,z) % 残差图Р附录三Р定积分求解РclearРsyms xРy=97566-16.8*x.^3+0.013*x.^5-83143*cos(0.1*x)-27478*sin(0.1*x);Рp=int(y,8.97,10.93);Рp1=vpa(p)Рq=int(y,20.84,22.96);Рp2=vpa(q)

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