load('S_chest.mat');
load('S_abdomen.mat');
x= S_chest;
y= S_abdomen;
A=zeros(5955,2);
A(1,1)=x(1);
for n= 2:5953
A(n,1)= x(n);
A(n,2)= x(n-1);
end
B=zeros(5955,2);
B(1,1)=y(1);
for i=2:5953
B(i,1)=y(i);
end
w= A/B;
w(w==0)= [];
disp(w);
I need to obtain w1, w2. Matlab comes out of many results but I only need two. How am I going to get them?

2 件のコメント

Wonsang You
Wonsang You 2017 年 9 月 4 日
The question is unclear. What are w1 ans w2? Are they some variables?
dpb
dpb 2017 年 9 月 4 日
I'm guessing OP is trying to fit some model but isn't clear what model is wanted. Knowing the size and what are the contents of the two data files might help some but as you point out, a clear problem statement would be better still.

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 採用された回答

Walter Roberson
Walter Roberson 2017 年 9 月 4 日

1 投票

My guess is that you want
w = A\B(:,1);

3 件のコメント

Stelios Fanourakis
Stelios Fanourakis 2017 年 9 月 5 日
You are right. Since my equation is y(n)=x(n)*w1+x(n-1)*w2 and I need to find out the w1,w2 variables. I have assigned table A as my x(n) and x(n-1) and B as my y(n). Although, mathematically correct would be w=B\A; my teacher claims to be w=A\B; correct. Can you figure out why?
Stelios Fanourakis
Stelios Fanourakis 2017 年 9 月 5 日
Results are different for w1,w2 whether A\B or B\A. Mathematically as we shift x to the opposite side of equation comes B\A
Walter Roberson
Walter Roberson 2017 年 9 月 5 日
Vec = @(M) M(:);
A = [ Vec(x(2:end)), Vec(x(1:end-1))];
Now, A*x = y .
Pretend for a moment that A is a square matrix (so inv(A) exists). Pre-multiply by inv(A), getting
inv(A)*A*x = inv(A)*y
inv(A)*A would be the identity matrix, so you would have
x = inv(A)*y
which is the operation A\y

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