9
$\begingroup$

Suppose there is a DFT vector $\mathbf{X}$ with length N, which presents complex conjugate symmetry around its middle point, i.e., $X(1) = X(N-1)^*$, $X(2) = X(N - 2)^*$ and so forth. $X(0)$ and $X(N/2)$ are the DC and Nyquist frequency respectively, therefore are real numbers. The remaining elements are complex.

Now, suppose there is a matrix $\mathbf{T}$, with size $N \times N$, which multiplies vector X.

\begin{align} \mathbf{Y} = \mathbf{T}\mathbf{X} \end{align}

The question is:

In what conditions, for matrix $\mathbf{T}$, the complex conjugte symmetry around the middle point of the resulting vector $\mathbf{Y}$ is preserved?

The motivation for this question is trying to come up with a precoder matrix $\mathbf{T}$ that results in a precoded (pre-equalized) symbol $\mathbf{Y}$ whose IFFT is real.

EDIT:

Thanks @MattL. and @niaren. The difficulty about this question is to find necessary conditions. Matt's answer is indeed sufficient. It is also sufficient to make the following modifications:

The first row and the first column do not need to be zero. Instead, they could be non-zero, as long as its values present a complex conjugate symmetry around the middle point, its first value is real and its $(N/2+1)$-th value is real, just like the symbol. The same could be stated for the $(N/2+1)$-th column, the $(N/2+1)$-th row, and the main diagonal.

Secondly, the same correspondence between the matrix in the upper left corner and lower right corner could be made between the upper right corner and the lower left corner, that is, chose an $(N/2 -1)\times(N/2-1)$ matrix starting from $t_{2,N/2 + 2}$ to $t_{N/2,N}$, flip from left to right, flip upside down and take the conjugate, then put at the lower left corner. On MATLAB, that would be:

T(N/2+2:N,2:N/2) = conj(fliplr(flipud(Tisi(2:(N/2),N/2+2:N))))

This structure is similar to the structure of the DFT matrix. Would that be a necessary condition?

EDIT(2):

The following code implements such a valid operator for any real-valued $N \times N$ matrix $\mathbf{A}$:

N = 8;  
A = rand(N,N); %must be real-valued  
w = exp(-1j*2*pi/N); % twiddle factor  
W = w.^(repmat(0:N-1,N,1).*repmat(0:N-1,N,1).'); % DFT matrix  
T = W*A*W'

EDIT(3):

It is also interesting to note the $\mathbf{T}^{-1}$ presents the sufficient condition too. This comes from the fact that:

\begin{align} \mathbf{T}^{-1} &= \mathbf{\left(W A W^{H}\right)}^{-1}\\ &= \mathbf{\left(W^{H}\right)}^{-1} \mathbf{A}^{-1} \mathbf{W}^{-1} \end{align} where $\mathbf{W}$ is the DFT matrix.

Since $\mathbf{W^{H}} = N\mathbf{W}^{-1}$. This equation becomes:

\begin{align} \mathbf{T}^{-1} &= \left(N\mathbf{W}^{-1}\right)^{-1} \mathbf{A}^{-1} \frac{1}{N}\mathbf{W^{H}}\\ &= \mathbf{W} \mathbf{A}^{-1} \mathbf{W^{H}} \end{align}

Finally, since $\mathbf{A}^{-1}$ is real-valued, provided that $\mathbf{A}$ is full rank,$\mathbf{T}^{-1}$ is sufficient.

$\endgroup$
  • $\begingroup$ I'll sleep over it before I might go into more details, but just for you to consider: even though the restriction of a diagonal matrix $\mathbf{T}$ is not necessary, it can be done without loss of generality, because all possible vectors $\mathbf{Y}$ can be generated. Do you agree? $\endgroup$ – Matt L. May 17 '13 at 22:23
  • $\begingroup$ Sure, I agree with that. $\endgroup$ – igorauad May 17 '13 at 22:37
1
$\begingroup$

I think that the entries in your matrix $\bf{T}$ must obey $a_{N-n+1,N-m+1} = a^*_{n,m}$. This is saying that the entries in row $N-n+1$ are the same as the coefficients in row n but where the coefficients are conjugated and reversed. The pattern in $\bf{T}$ for $N=4$ is

$T_4 = \left[ \begin{smallmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24} \\ a^*_{24}&a^*_{23}&a^*_{22}&a^*_{21} \\ a^*_{14}&a^*_{13}&a^*_{12}&a^*_{11}\end{smallmatrix} \right]$

I'm sure someone will come up with a better and more precise answer.

$\endgroup$
  • $\begingroup$ What about the DC component? The DC component of $\mathbf{Y}$ is the inner product of the first row of $\mathbf{T}$ with the (complex) vector $\mathbf{X}$. How is this going to be real-valued? $\endgroup$ – Matt L. May 17 '13 at 18:02
  • 1
    $\begingroup$ I left that as an exercise to the OP to stuff those two rows in cough. But I don't see how you come to the conclusion that only a diagonal matrix will work (not saying that you are wrong). $\endgroup$ – niaren May 17 '13 at 18:12
  • $\begingroup$ I might be wrong indeed. When I have more time I'll think about it again ... Let's put it like this: a diagonal matrix (with conjugate symmetry) will work in any case. $\endgroup$ – Matt L. May 17 '13 at 18:36
-1
$\begingroup$

If I'm not mistaken the only solution for $\mathbf{T}$ which is independent of the vector $\mathbf{X}$ is a diagonal (complex) matrix, where the diagonal satisfies the complex conjugate symmetry.

EDIT: OK, I was mistaken. Diagonal is fine, but it's not necessary. The matrix $\mathbf{T}$ must have the following general structure: elements $t_{11}$ and $t_{N/2+1,N/2+1}$ must be real-valued (they correspond to DC and Nyquist). Apart from $t_{11}$ the first row and column contains only zeros. For elements $t_{22}$ to $t_{N/2,N/2}$ chose an arbitray $(N/2-1)\times (N/2-1)$ matrix. Then use this arbitray matrix to form a new matrix by swapping all rows (the first row becomes the last one, the second row becomes the second last one, etc.), by flipping the rows from left to right, and by conjugating. Then put this submatrix in the lower right corner of the total matrix $\mathbf{T}$. All other elements of $\mathbf{T}$ must be zero. I'm aware that this is a bit tough to understand without a visualization, so I'll add one later on when I have more time.

$\endgroup$

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.