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?

댓글 수: 2

Wonsang You
Wonsang You 2017년 9월 4일
The question is unclear. What are w1 ans w2? Are they some variables?
dpb
dpb 2017년 9월 4일
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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 채택된 답변

Walter Roberson
Walter Roberson 2017년 9월 4일

1 개 추천

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

댓글 수: 3

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?
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
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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