Matrix double and complex double
이전 댓글 표시
Hello, i'm trying to calculate between
t1 = H*(t);
while the H is [32 128] and t is [4096 1] and i was searching for formula to make the matrix has same size and i found this
[m, n] = size(t);
[m1, n1] = size(H);
if m < m1
t(m1,:) = 0; % extend t with rows of zeros, up to the number of rows of H
elseif m > m1
H(m,:) = 0; % extend H with rows of zeros, up to the number of rows of t
end
if n < n1
t(:,n1) = 0; % extend t with columns of zeros, up to the number of columns of H
elseif n > n1
H(:,n) = 0; % extend H with columns of zeros, up to the number of columns of t
end
after i run it both matrix are [4096 128] but i still counldn't calculate it, any ide where the error come? btw the size of H is [4096 128] complex double while t is [4096 128] double, can it be the reason why i cant calculate the t1 = H*(t);? Thankyou!
채택된 답변
If you do a check before the if-elseif-else-end statement, the values of m, m1 and n , n1 are same
That leads to the if-elseif - end conditions not satisfied . Hence both matrices remain same despite using some modifications to the H & t matrixces through if-else statements
H = rand(4096,128)
t = rand(4096,128)
[m, n] = size(t)
[m1, n1] = size(H)
% check with this
isequal(m,m1)
% check the size
isequal(n,n1)
if m < m1
t(m1,:) = 0; % extend t with rows of zeros, up to the number of rows of H
elseif m > m1
H(m,:) = 0; % extend H with rows of zeros, up to the number of rows of t
end
if n < n1
t(:,n1) = 0; % extend t with columns of zeros, up to the number of columns of H
elseif n > n1
H(:,n) = 0; % extend H with columns of zeros, up to the number of columns of t
end
H, t
댓글 수: 1
Consider the first block of if-elseif-end statements, the if-condition is not satisfied howeever, the else-if condition is satisfied. So when that happens, it is replacing the old rows of H matrix which is m1 =32 with m = 4096 which comes from size of vector t. So essentially,size of H matrix now has become 4096 x 128 . Matlab dynamically resizes the matrix size by creating additional rows with zeros
% H matrix has 128 columns
H = rand(32,128);
% t vector has 4096 rows
t = rand(4096,1);
[m, n] = size(t)
[m1, n1] = size(H)
% if condition not satisfied
if m < m1
t(m1,:) = 0; % extend t with rows of zeros, up to the number of rows of H
% else-if condition satisfied
elseif m > m1
% H matrix now has 4096 rows and 128 columns
H(m,:) = 0; % extend H with rows of zeros, up to the number of rows of t
end
Now consider the second block of if-elseif-end statements, the if-condition is NOW satisfied howeever, the else-if condition is NOT satisfied. So when that happens, it is replacing the old columns of t vector which is having n = 1 with n1 = 128 which comes from size of Matrix H. So essentially,size of t vector has become 4096 x 128 Matlab dynamically resizes the matrix size by creating additional columns with zeros in this case
% if - condition is satisfied
if n < n1
t(:,n1) = 0; % extend t with columns of zeros, up to the number of columns of H
elseif n > n1
H(:,n) = 0; % extend H with columns of zeros, up to the number of columns of t
end
t1 = H.*t
H, t
Hence, thats the reason why both H and t will have same sizes now, after the execution of if-else-if-end statement blocks and not due to presence of any complex values in matrix. I hope this clarifies
추가 답변 (1개)
That algorithm would be for element-by-element multiplication. For * (matrix multiplication) you need to mentally transpose the second input -- and you only need to match one of the dimensions
%test data
H = rand(32,128);
t = rand(4096,1);
%the algorithm
[m, n] = size(t);
[m1, n1] = size(H);
if n1 < m
H(:,m) = 0; % extend H with columns of zeros, up to the number of rows of t
elseif n1 > m
t(n1,:) = 0; % extend t with rows of zeros, up to the number of columns of H
end
whos
t1 = H * t;
size(t1)
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