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Let us assume I have $k$ (a fixed number) sensors in a wireless sensor network with unknown channel statistics $\{h_1,h_2 \ldots, h_k\}$.

In my system model, each of these $k$ sensors has an independent $n$-length binary ID sequence $x^i=\{x_1^i,x_2^i,\ldots, x_n^i\}, \forall i \in \{1,2,\ldots k\}$. This sequence is made of $0's$ and $1's$ (On-Off keying). Moreover, during the initial access phase, all the devices transmit this ID sequence synchronously.

At the receiver, during each of the $n$ channel-uses, we receive $y_j= \sum_{i=1}^{n}x_j^ih_i + n_j, \forall j \in\{1,2,\ldots n\}.$ Here, $n \sim \mathcal{N}(0,1), \forall j$ is the additive white gaussian noise.

Question 1: Using this model, can I estimate the channel statistics $h_i$? My initial though is $Yes!$ However, I searched a lot in google scholar to find papers that use 0-1 binary sequences (On-Off keying pilot sequences) for channel estimation in wireless communication scenarios. I could not find any. There are some optical communication papers though.

Question 2: Why is on-off keying bad for using as pilot sequences? In my case, I have to transmit them in the initial access phase anyways. Hence, I am thinking that though this channel estimate may not be a good one, it could be a coarse estimate which is useful. Do you agree?

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  • $\begingroup$ What is the difference, if any, in your mind between the meaning of $k$ in the phrase "...each of these $k$ sensors..." (which suggests that $k$ is a fixed number, say $20$) and "...sequence $x^k=\{x_1^k,x_2^k,\ldots, x_n^k\}, \forall k. $.... (which suggests that $k$ is a variable)?? The transmission of these sequences might be synchronous, but the reception _ might not be: when they arrive at the receiver, differing path lengths (a.k.a. channel delays) will give rise to _asnychronous reception. I vote to close this question pending the OP's edits providing more details and clarity. $\endgroup$
    – Dilip Sarwate
    Dec 20, 2021 at 20:02
  • $\begingroup$ @DilipSarwate I have edited the question to give more clarity. Hope it is more clear now. $\endgroup$
    – wanderer
    Dec 20, 2021 at 20:49

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Q1: no, not in general. If the sum of multiple sequences is one of the sequences that are in your codebook $x^i$ (or a multiple), then this will become impossible to resolve. The sequences need to be mutually orthogonal, or the channels will "crosstalk".

Look up DSSS, this is exactly what you're doing.

Q2: because OOK doesn't use the channel half of the time. Generally, there's very few channels where OOK is a spectrally efficient solution, and flat channels like yours are not among them.

Whether or not a coarse estimate is "useful": I can't tell, I don't know what you need. What I can tell you is that if you'd use BPSK instead of OOK, you'd not have missed (on average) half of the chip times you could have used to estimate the channel. At the same average power, BPSK simply leads to higher "distance" between sequences, and thus to a clearer estimate in presence of noise.

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  • $\begingroup$ For question 1: My line of thought is that I could still use the pseudo-inverse and find a least square estimate as long as it has full rank. Also, random binary matrices tend to have full rank have as the dimensions grow. $\endgroup$
    – wanderer
    Dec 22, 2021 at 3:35
  • $\begingroup$ but two or more sequences adding up to a possible other sequence means exactly that the thing doesn't get full rank. And, yes, random binary matrices do, but they can have arbitrarily bad distance properties, which make your estimates unncessarily bad. $\endgroup$
    – Marcus Müller
    Dec 22, 2021 at 11:58
  • $\begingroup$ By bad distance do you mean that the probability that two random binary ID sequences being similar is very high? $\endgroup$
    – wanderer
    Dec 22, 2021 at 14:05
  • $\begingroup$ no, I mean distance in the metric sense. $\endgroup$
    – Marcus Müller
    Dec 22, 2021 at 14:07
  • $\begingroup$ I guess you are implying the matrix could be still ill-conditioned though full rank. (The phrase bad distance of not too clear for me .) $\endgroup$
    – wanderer
    Dec 22, 2021 at 23:34

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