How to ignore negative indices or assign it to zero in a loop

x=[1 2 -1 -3 -4]
h=[0 1 3]

n=length(x)
m=length(h)
y=zeros(n,m)
for i=1:n
  for  j=1:m
    
     if i-j>0 
          y(i,j)=h(j)*x(i-j+1)
     else
         y(i,j)=0 % My problem is here to assign if the element x(i-j) does not exist return zero
                  %but it is ignoring all the element in the matrix that is (i-j)< zero
     end
                    
  end
end

I have tried approach this problem by setting up a matrix, however i did not start loop from zero since it is same both way if i am right

Answer 1

[Stephan]

It is not needed to make the elements zero - they are already, because you preallocated y with zeros:

x=[1 2 -1 -3 -4]
h=[0 1 3]

n=length(x)
m=length(h)
y=zeros(n,m)
for i=1:n
  for  j=1:m
    
     if i-j>0 
          y(i,j)=h(j)*x(i-j+1)
     end
                    
  end
end

gives:

% alot of stuff
% ...
y =

     0     0     0
     0     0     0
     0     2     0
     0    -1     6
     0    -3    -3

[MD JAHIDUL HAQUE KHAN]

Hi stephan

But have you checked my code did not work for row 2 column 2 and row 3 column 3 , if you cross check both the analytical formula and my code. where sum of row 2 and sum of row 3 of the matrix does not justify the formula. i thought i could not assign that if x(i-j) is not in the rage of x it should consider zero.

regards

Jahidul

[Stephan]

use:

if i-j>=0 

instead of

if i-j>0 

[MD JAHIDUL HAQUE KHAN]

Thanks Stefan