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Is it possible to use the bit-reverse order of FFT instead of FFT to estimate the channel?

Based on my reading, I found that bit reversed order of FFT matrix can diagonalize the channel, but unfortunately it can't estimate the channel based on pilot data. It means, that when having $y = h*x$, where $h$ is a channel, $*$ is convolution operation, and $x$ is a signal. Therefore, in order to get the frequency signal based on bit-reversed order FFT matrix, $Y = F^b y = F^b (h*x)$. Therefore, the signal $Y = HX$ where the $H$ and $X$ are the frequency-domain channel response and signal gotten using $F^b$, respectively. Based on my analysis, I got that $H = Y/X$, but when estimating $H$ based on pilot data taken from $X$. For example, $X_p = X_{1:4:end}$, the channel response $H$ cannot be estimated using $H = Y/X_p$. Why ? and is it possible to estimate $H$ based on $F^b$ and $X_p$ ?

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  • $\begingroup$ could you refer us to some literature on "bit-reversed order FFT diagonalizing the channel"? What kind of channel? "Diagonalizing" is something you can do to a matrix, so which matrix are we referring to then here? Rule of thumb: whenever you feel like writing "I read somewhere that...", simply cite that source – so that people can look it up. $\endgroup$ Mar 29 '21 at 12:34
  • $\begingroup$ As mentioned in the question, bit-reversed order FFT diagonalizing the channel is something I am studying and checking. I don't think that someone did it in literature. The term diagonalizing I used above is for the diagonal matrix $H$. thank you $\endgroup$ Mar 29 '21 at 12:42
  • $\begingroup$ but for an LTI channel, $H$ is already diagonal? So what's there to diagonalize? Oh wait $\endgroup$ Mar 29 '21 at 12:49
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So, according to your text,

$$ H := F^bh,$$

i.e. what you call the bit-reversed order FFT of the channel impulse response.

Since "bit-reversed order" just implies you're permuting outputs of the FFT, $F^b$ is just a row permutation $\Pi_{BR}$ of the standard DFT matrix $D$.

Since the "proper" DFT actually leads to a diagonal $H$ (as a linear channel never "mixes" any other frequencies in), you already know exactly what to do to convert your $H$ to a diagonal matrix:

Reverse the reverse bit ordering:

$$\Pi^{-1}_{BR}H=\Pi^{-1}_{BR}\left(F^bh\right) = \Pi^{-1}_{BR}\left(\Pi_{BR}Dh\right) = Dh,$$

which, in best classical OFDM nature, diagonalizes the channel.

(Using the full rank of any permutation matrix, you can also show this is the only way to diagonalize the result of the bit-reverse ordered FFT; putting this into a ray of optimism: Those who set out to invent OFDM will in expectation invent OFDM!.)

Channel estimation relates to the way you want to correct the channel influence. Hence, once you diagonalized your system, it becomes "normal" OFDM, and all the normal OFDM methods applies.

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  • $\begingroup$ Thank you so much, I will check if that achieve also the channel estimation based on pilots data or no. Then, I'll feedback you. $\endgroup$ Mar 30 '21 at 5:51
  • $\begingroup$ as said, as soon as you do the permutation, you're doing normal OFDM; there's nothing special about your system! $\endgroup$ Mar 30 '21 at 6:25
  • $\begingroup$ I think my question wasn't clear enough. Anyway, I will raise another question with more details. thanks again $\endgroup$ Mar 30 '21 at 10:25
  • $\begingroup$ not quite sure what you mean, but if you can ask it more clearly, I'm interested in reading it! Please don't be disappointed, however, if the solution stays "reorder to normal order". $\endgroup$ Mar 30 '21 at 18:09
  • $\begingroup$ I added the question into here dsp.stackexchange.com/questions/74164/… $\endgroup$ Mar 31 '21 at 9:04

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