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 Commenti
John D'Errico
il 24 Giu 2018
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
il 24 Giu 2018
Torsten
il 25 Giu 2018
The variable you want to return from "objfun" is called "J", not "f".
christina
il 25 Giu 2018
Torsten
il 25 Giu 2018
J = @(theta) objfun(x_t,c_mn,theta )
instead of
J = @(theta) objfuntc(x_t,c_mn,theta )
christina
il 25 Giu 2018
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
il 25 Giu 2018
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
il 25 Giu 2018
Torsten
il 25 Giu 2018
I complemented my comment.
christina
il 25 Giu 2018
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