1
$\begingroup$

As we know that Zadoff-Chu matrix is similar to Walsh-Hadamard matrix where every columns in those matrices is orthogonal with the any other column.

For Walsh-Hadmard matrix, the orthogonality is clear and can be demonstrated easily. But regarding the Zadoff-Chu matrix, I don't see that orthogonality is right as shown below:

a = (sqrt(2) + 1j*sqrt(2))/2;
W = [a 1 a -1; -1 a 1 a ; a -1 a 1; 1 a -1 a];  %%Zadoff-Chu matrix 

C1 = W(:,1);     %% Take the first column

orth_results = []; 
for j = 1 : size(W,2)
    results = (abs(sum(W(:,j).*C1)).^2);  %Project the first column to every column in matrix W
    orth_results = [orth_results results];
end 

Following the condition of orthogonality between columns, orth_results should be equals to 0 except in the fist value, but what I get is orth_results = [8.0000 0 8.0000 0];

What's the problem of that? Is there any issue in the matrix itself ?

$\endgroup$
1
$\begingroup$

For complex matrices, the concept of orthogonality is replaced by unitarity. If $^H$ denotes the conjugate transpose of a matrix, we should check that $W$ is a unitary matrix (up to a scaling factor) via equations:

$$W^* W = WW^* =\lambda I$$

with $\lambda\neq 0$, which you can test positively with:

W'*W
W*W'

What is missing in your loop, and especially in the complex scalar product, is the conjugation on one of the complex vectors. This is explained for instance in Wikipedia/Dot product for Complex vectors.

You could for instance replace C1 by conj(C1).

| improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, that's condition is verified, it's OK. but Why when can't I proof the orthogonality of each columns by projecting it on the other columns? $\endgroup$ – New_student Jun 12 at 15:24
  • $\begingroup$ The conjugation is missing in the scalar product. Try conj(C1) instead of C1. Conjugation is implicit in the matrix transpose operator $\endgroup$ – Laurent Duval Jun 12 at 15:30
  • 1
    $\begingroup$ Thank you so much .. I got it. $\endgroup$ – New_student Jun 12 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.