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2011207156

金虫 (小有名气)

[求助] matlab拟合动力学参数,非常谢谢~

function butongshike201506171
clear all;clc
format long
tspan=(0:10:390);
yexp=[0.046125761496241
   0.078139373146412
   0.117158327134899
   0.159846402230616
   0.199412869433285
   0.231111460229405
   0.259814231746068
   0.282697037233637
   0.306493661349134
   0.326396561089024
   0.345295684999190
   0.360444423309371
   0.377418750129898
   0.395250807577438
   0.414965633800510
   0.430017989495138
   0.445923212579888
   0.457995753035510
   0.469186259425871
   0.480284344291621
   0.491524402631167
   0.501658351934599
   0.510935126510884
   0.520202070112304
   0.531309455870521
   0.537782255717000
   0.547944130180314
   0.558094213007469
   0.565468701311900
   0.573757577102420
   0.581007002009939
   0.589392949772071
   0.597541754780127
   0.607751809867969
   0.613129267171725
   0.620580135887054
   0.626072239599544
   0.632475341850705
   0.638742438783615];
beta0=[67971 0.000839];
y0=0;
lb=[-1 -1]*1e6;
ub=[1e9 1e9];   
yy=[0 yexp'];
k0=beta0;
[k,resnorm,residual,exitflag,output,lambda,jacobian] = ...
    lsqnonlin(@ObjFunc,k0,lb,ub,[],tspan,y0,yexp);
fprintf('\n\n使用函数lsqnonlin()估计得到的参数值为:\n')
fprintf('\t待拟合参数 k1 = %.6f\n',k(1))
fprintf('\t待拟合参数 k2= %.6f\n',k(2))
fprintf('  残差平方和= %.6f\n\n',resnorm)
ts=0:10:max(tspan);
[ts ys]=ode45(@KineticsEqs,ts,y0,[],k)
[ttt XXsim] = ode45(@KineticsEqs,tspan,y0,[],k);
y=XXsim(2:end);
xexp=yexp;
R2=1-sum((xexp-y).^2)./sum((xexp-mean(y)).^2);
fprintf('\n\t决定系数R-Square = %.6f',R2);
figure(1)
plot(ts,ys,'b',tspan,yy,'or'),legend('计算值','实验值','Location','best');
xlabel('x'),ylabel('y')


%---------------------------------------------------------
function f = ObjFunc(k,tspan,y0,yexp)           % 目标函数
[t Xsim] = ode45(@KineticsEqs,tspan,y0,[],k) ;
ysim = Xsim(2:end);
size(ysim);
size(yexp);
f=ysim-yexp;
%----------------------------------------------------------

function dydt = KineticsEqs(t,y,k)
beta(1)=k(1);
beta(2)=k(2);
tspan=(0:10:390);
h=0.068031;
p=[3.472000000000000
   3.519000000000000
   3.548000000000000
   3.589000000000000
   3.634000000000000
   3.676000000000000
   3.710000000000000
   3.740000000000000
   3.765000000000000
   3.789000000000000
   3.811000000000000
   3.832000000000000
   3.847000000000000
   3.865000000000000
   3.885000000000000
   3.906000000000000
   3.921000000000000
   3.938000000000000
   3.952000000000000
   3.964000000000000
   3.977000000000000
   3.988000000000000
   4.000000000000000
   4.010000000000000
   4.020000000000000
   4.032000000000000
   4.039000000000000
   4.050000000000000
   4.061000000000000
   4.069000000000000
   4.078000000000000
   4.087000000000000
   4.095000000000000
   4.105000000000000
   4.115000000000000
   4.122000000000000
   4.129000000000000
   4.135000000000000
   4.142000000000000
   4.150000000000000]';
D=p*h;
sp=csaps(tspan,D,1);
pp=fnder(sp);
dD=fnval(pp,tspan);
dydt= (83.38-0.0014*exp(1600*(1/277.55-1/298.15))...
*interp1(tspan,p,t,'pchip')*10000-1/beta(2)...
*interp1(tspan,D,t,'pchip')*4.229*(0.0598...
-(y*0.0150*(54.2+17.68)+(0.236204-interp1(tspan,D,t,'pchip'))...
*0.0598*4.229)/(45.07-interp1(tspan,D,t,'pchip')*4.229))...
*interp1(tspan,dD,t,'pchip')/(-45.07+interp1(tspan,D,t,'pchip')...
*4.229)/0.0150-(0.0598-(y*0.0150*(54.2+17.68)+(0.236204-...
interp1(tspan,D,t,'pchip'))*0.0598*4.229)/...
(45.07-interp1(tspan,D,t,'pchip')*4.229))*...
interp1(tspan,dD,t,'pchip')/0.0150/beta(1))/...
(1/beta(1)+1/beta(2)-1/beta(2)*interp1(tspan,D,t,'pchip')...
*71.88/(-45.07+4.229*interp1(tspan,D,t,'pchip')));
以上是求动力学参数beta(1)和beta(2)的代码,p,D,和dD是随时间变化的已知量,不知以上程序运行时interp1(tspan,D,t,'pchip')能否对应上不同时刻的D。非常谢谢。
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