Smoothing by Least Square Technique !!!
4 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Serhat
el 21 de Mayo de 2014
Respondida: Star Strider
el 21 de Mayo de 2014
Hi all,

We are trying to find coefficients ' a 's for a given 200x1 t and 200x1 r(t)
r(t) = [ 1530 1215 3243 1111 ..... ]' size: 200 x 1
t = [0:0.5:99.5] size: 200 x 1
N=200
Thanks :)
0 comentarios
Respuesta aceptada
Star Strider
el 21 de Mayo de 2014
Interesting problem. The system is essentially this matrix equation:
r = A*[t^n]
where r, t and n are defined by necessity as column vectors and A is the matrix of coefficients. This is the inverse of the usual least-squares problem:
t = 0:0.5:99.5; % Define ‘t’
n = 0:length(t)-1; % Define ‘n’
tn = t.^n; % Define ‘t^n’
r = [1530 1215 3243 1111]'; % Given ‘r’ vector
stn = tn(1:length(r))'; % Truncate length of ‘tv’ to match sample ‘r’
stn(1) = eps; % Replace zero with ‘eps’ in ‘stv’
stni = pinv(stn); % Take pseudo-inverse of ‘t^n’
A = r*stni % Calculate ‘A’ coefficient matrix
rt = A*stn % Verify ‘A’ calculation
At least for the data available, this works!
0 comentarios
Más respuestas (1)
Image Analyst
el 21 de Mayo de 2014
See my attached demo for polyfit.
2 comentarios
Image Analyst
el 21 de Mayo de 2014
I did not give constants for them. I computed all the coefficients (slope and intercept). Look again, specifically for these lines where I calculate them:
% Do the regression with polyfit
linearCoefficients = polyfit(x, y, 1)
Ver también
Categorías
Más información sobre Descriptive Statistics en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!