linear least square method to fit the log data
显示 更早的评论
This is what I have so far. Im trying to fit the line (yp) to the data points plotted. Any suggestions of what Im doing wrong? Thanks!
function LogModel()
t=[1:1:48]'; %% data for time : column vector
y=[1.7039;5.9098;10.6923;15.6497;19.0261;23.1650;25.8011;26.2864;28.0410; 28.0070;28.8502;27.0665;27.4650;
25.9110; 23.3441;21.9954;20.7284;19.1973;17.3139;15.5076;13.7446;12.5922;11.4729;
10.4418;9.1933;7.6495;7.2157; 6.0907; 5.3670;4.6935;4.3370;3.6142; 3.1584;2.8272;
2.4244;2.0813; 1.8584; 1.5881; 1.4892;1.2528;1.1232; 1.0128;0.7994;0.7104;0.6527;0.4646;
0.4801;0.4537]; %% data for auto supply
n = length(t);
yln=log(y);
A=[ones(n,2),t, yln]; %% matrix formed by basis function 1, t at all time data
coef=inv(A'*A).*(A'*yln); %% least square solution
c1=exp(coef(1)); c2=exp(coef(2)); c3=coef(3);
tp=[0:0.01:50]; %% for plotting the model functiona f(t)
yp=c1.*tp.^(c2).*exp(c3.*tp);
m=1:n;
RMSE=norm((1./m).*sum(abs(y-yln).^2)).^(0.5);
fprintf('RMSE for log model = %12.5e\n', RMSE);
plot(t,y,'o', tp, yp, '-');
回答（1 个）
Star Strider
2021-3-18
Are you required to use a log fit for those data?
If not, try this:
objfcn = @(b,x) b(1).*x.*exp(b(2).*x) + b(3).*x.*exp(b(4).*x);
[B,resnrm] = fminsearch(@(b) norm(y - objfcn(b,t)), rand(4,1))
figure
plot(t, y, '.')
hold on
plot(t, objfcn(B,t), '-r')
hold off
grid
This parameter set provided an acceptable fit:
B =
-22.028821282813610
-0.268737534521761
21.509885470995002
-0.159842817422680
.
类别
在 帮助中心 和 File Exchange 中查找有关 Linear Predictive Coding 的更多信息
2021-3-18
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
