# How to fill array with previous values?

2 ビュー (過去 30 日間)
George Francis 2018 年 6 月 11 日

Hello guys I have data which contains 2 columns and 600 rows. It basically contains inputs u and outputs y. And my question is how can I fill array 3x1 as follows:
z=[-u(n)-2*u(n-1)-u(n-2);y(n-2)-y(n-1);y(n)-y(n-1)];
I know that I have to make loop but my loop which is below isn't correct and I don't know how to fix it. So can you please help me?
N=length(y);
for n=1:N
for i=1:2
W(i) = y(n-i); %output
end
for i=1:2
V(i) = u(n-i+1); %input
end
z = [V';W'];
end

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### 回答 (2 件)

KSSV 2018 年 6 月 11 日
A = rand(600,2) ; % some random data
u = A(:,1) ; y = A(:,2) ;
n = 1:600 ;
z = zeros(600,3) ;
for n = 3:600
z(n,:) = [-u(n)-2*u(n-1)-u(n-2) y(n-2)-y(n-1) y(n)-y(n-1)];
end
z(1:2,:) = [] ;
Note, the above can be achieved without loop also.
##### 2 件のコメントなしを表示なしを非表示
George Francis 2018 年 6 月 11 日
the problem is z has to be array 3x1 when I tried your version z is 598x3.
KSSV 2018 年 6 月 11 日
As z is inside loop, every loop value has been saved..

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George Francis 2018 年 6 月 11 日
I'm making some other calculations from z so z has to be 3x1 in that loop. I actually tried something see code below. But I don't know if it's working correctly. I mean it works but I don't know if that is solution which I was looking for.
delta = 0.002;
sigma=1;
lamda=1;
C = sigma*eye(3);
P = [1;1;1];
N=length(y);
for n = 4:N
for i=1:3
W(i) = y(n-i);
end
y_n=W(1,1);
yn_1=W(1,2);
yn_2=W(1,3);
for i=1:3
D(i) = u(n-i);
end
u_n=D(1,1);
un_1=D(1,2);
un_2=D(1,3);
z = [-u_n-2*un_1-un_2;yn_2-yn_1;y_n-yn_1];
rk=-4*delta^2*yn_1;
C = (C - ( (C*z*z'*C)/( lamda+(z'*C*z) ) ))/lamda;
P = P + C*z*(rk - (z'*P))/(lamda+(z'*C*z));
end

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