Optimization problem using Quasi Newton method
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Given x_t and c_mn, the objective function is defined as:
I am trying to solve the above objective function for theta using quasi newton method. But there seem to be some problem with my code as it's not working. Could somebody please help me with this?
Following is my matlab code:
Code for defining objective function:
function f = objfuntc(x_t,c_mn,theta )
L = length(x_t );
[M,N] =size(c_mn);
f = 0 ;
for t = 1:L
xs = 0 ;
for n = 1:N
for m=1:M
xs = xs + c_mn(m,n)*exp(1i*2*pi*m*(t - n -2) + theta);
end
end
f = f + (x_t(t) - xs)^2 ;
end
Code for calling objective function:
x_t = rand(16,1 );
L = length(x_t);
a = 1;
M = L;
c = rand(M*L/a,1 );
c_mn=reshape(c,[M,L/a]);
J = @(theta) objfuntc(x_t,c_mn,theta )
theta0 = 10 ;
options = optimoptions('fminunc','Algorithm','quasi-newton ');
[theta, thetaval] = fminunc(J,theta0,options)
12 个评论
John D'Errico
2018-6-24
Why do you want to write your own optimizer in MATLAB? That is generally a bad idea, when optimization tools written by professionals are readily available.
christina
2018-6-24
Torsten
2018-6-25
The variable you want to return from "objfun" is called "J", not "f".
christina
2018-6-25
Torsten
2018-6-25
J = @(theta) objfun(x_t,c_mn,theta )
instead of
J = @(theta) objfuntc(x_t,c_mn,theta )
christina
2018-6-25
Put this code in one file, name it "main.m", open it in MATLAB and run it.
function main
x_t = rand(16,1 );
L = length(x_t);
a = 1;
M = L;
c = rand(M*L/a,1 );
c_mn=reshape(c,[M,L/a]);
J = @(theta) objfuntc(x_t,c_mn,theta )
theta0 = 10 ;
options = optimoptions('fminunc','Algorithm','quasi-newton ');
[theta, thetaval] = fminunc(J,theta0,options)
end
function f = objfuntc(x_t,c_mn,theta )
L = length(x_t );
[M,N] =size(c_mn);
f = 0 ;
for t = 1:L
xs = 0 ;
for n = 1:N
for m=1:M
xs = xs + c_mn(m,n)*sin(2*pi*m*(t - n -2) + theta);
end
end
f = f + (x_t(t) - xs)^2 ;
end
end
christina
2018-6-25
x_t and c_mn can be complex-valued ?
You will have to use
f = f + (x_t(t) - xs)*conj(x_t(t) - xs) ;
instead of
f = f + (x_t(t) - xs)^2 ;
Best wishes
Torsten.
christina
2018-6-25
Torsten
2018-6-25
I complemented my comment.
christina
2018-6-25
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