# bit reversed order of FFT matrix for channel estimation

Although the question seems similar to this one HERE, but what I want to ask about is little bit different.

Assume we have a signal $$X$$ whose length is $$N$$x$$1$$, we convert that signal into time domain using bit reversed order FFT as: $$x = F^{'b}X = p*F'*X$$ , where $$F^{'b}$$ is the inversed bit-reversed order FFT matrix ($$F^{b}$$). $$p$$ is row permucation matrix and $$F'$$ is the inversed DFT matrix.

A cyclic prefix is added into the signal $$x$$ having $$x_{cp} = [cp; x]$$. Then that signal is transmitted over a frequency selective channel $$h$$ resulting $$y_{cp} = h*x_{cp}$$ where $$*$$ denotes to the circular convolution operation.

The CP is first removed from the signal $$y_{cp}$$ resulting the signal $$y$$.

Based on my analysis, when transforming the signal $$y$$ into frequency domain using $$F^{b}$$, as following $$Y = F^{b}y = p'*F*y$$, the channel is diagonalized where I can get the channel $$h = F^{'b}(Y./X)$$ it means in that case I use the whole signal $$X$$ as pilots.

The problem I am facing, when using such pilots from $$X$$, let's be $$X_{1:4:end}$$, and then perform the same steps, the results of $$h = F_2^{'b}(Y_{1:4:end} ./ X_{1:4:end})$$ is not equivalent to the channel !! where $$F_2^{'b}$$ is similar to $$F^{'b}$$ but with size similar to $$(Y_{1:4:end}/X_{1:4:end})$$ and $$./$$ is element wise division.

$$NP$$: When performing these steps with the normal $$FFT$$ matrix, everything is working well and I can recover the exact channel using the pilots data.

• I think this will be clearer for you if you write your $F^b$ (and $F^{'b}$) as combination of DFT matrix (or IDFT matrix) and permutation matrix! Mar 31 at 9:24
• I modified the question however I think it became little bit complicated to understand it now since more matrices have been added into it. Mar 31 at 9:40
• writing things down as matrix operations makes this a lot easier, not harder! Mar 31 at 9:40

So, writing your $$F^b$$ and $$F^{'b}$$ as permutations $$p, p^{-1}$$ of the DFT matrix $$F$$ , IDFT $$F'$$, and the cyclic prefixing and removal also as Matrix operation $$P$$, $$R$$ (which boils down to cyclically extended or truncated identity matrices), and the circular convolution with the channel impulse response as circulant matrix $$C$$,

\begin{align} x&=F^{'b}X=p F' X\\ x_{cp} &= PpF' X\\ y_{cp} &= y*x_{cp} \\ y&=R(y*x_{cp})= CpF'X\\ Y&=F^By \\ &= p'FCpF'X & F, F' \text{ are symmetric matrices!}\\ &=p'FCF'pX\\ &=p'\underbrace{FCF'}_{\text{IDFT-Channel-DFT}}\underbrace{pX}_{\text{permuted symbols}} \end{align}

So, all that's really happening here mathematically is that you permute the symbols before transmission, then de-permute them at the receiver. In between, you get normal OFDM. There's no advantage or disadvantage to the permutation w.r.t. to channel diagonalizability, but we've arrived here in your previous question already.

Now, for channel estimation, you'll want to insert pilots such that they make sense for the central $$FCF'$$ part – in other words, you need to think about where you'd want your normal OFDM system to have these pilots, and then shift these positions according to your $$p$$.

Of course, the channel you estimate left of your final reordering isn't the same as the channel you estimate with OFDM, it's a permutation. As shown above, it's however "just" a permutation, there's nothing in the maths here that changes anything about the channels properties at all, so if your results aren't the same as for straightforward OFDM, then you've got an implementation bug somewhere.

• thank you so much, but I have a question, how did you get $p'FCF'pX$ from $p'FCpF'X$ ? Do you mean that $F'p$ = $pF'$ ? Mar 31 at 16:32
• Because the matrix is symmetric – it doesn't matter whether we permute rows or columns. Mar 31 at 23:25
• Yes, but I get that if we have permutation matrix $p = [1 ,0, 0, 0; 0, 0, 1, 0; 0, 1, 0, 0; 0, 0, 0, 1];$ and DFT matrix $F = dftmtx(4)/sqrt(4);$ , why $Fp - pF$ is not zero ? Apr 1 at 2:09
• if your DFT matrix isn't symmetric, then you've got a bug in your DFT matrix. Permuting columns and permuting rows is the same in a symmetric matrix. Apr 1 at 9:31
• So, the same question will be raised again, how did you get $p'FCF'pX$ from $p'FCpF'X$ ? Apr 1 at 12:56