As known, the pilot tones are transmitted via subcarriers predefined and known in the transmitter and the receiver. The pilot spacing is repeated with a certain rate that depends on how fast the channel changes.

I am talking about comb-type channel estimation. The distance between every two pilot's sub-carrier is called, here, pilot's spacing.

If we have number of sub-carriers $N$ = 64; in the first case let's take 8 sub-carriers to transmit pilot's tones defined in locations 1,9,17...; with pilot's spacing $j_1 = \frac{64}{8} = 8$. The second case let's take 16 sub-carriers defined in locations 1,5,9,13...; with pilot's spacing $j_2 =\frac{64}{16} = 4$.

It's clear that second case pilots can track the time-varied channel better. But, with an OFDM system, the symbol data is converted into time domain using $iFFT$, and every subcarrier is spread over all the other sub-carriers including pilot's sub-carriers. How does the pilot's spacing affect the performance?


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I don't know whether I'd call it "artifacts", but of course your pilots are part of your signal, and taking $\frac1{j_1}N$ or $\frac1{j_2}N$ of the available subcarriers for pilots reduces the number of available subcarriers for data symbols by exactly that amount.

every subcarrier is spread over all the other sub-carriers including pilot's sub-carriers

No. This is important: the subcarriers are orthogonal, hence the O in OFDM.

They are not overlapping into each other; they remain orthogonal, don't influence each other. It's just that in time domain, you can't see that.

Remember (please), that OFDM is just the IDFT/DFT in its core: that's just an invertible transform between $N$-element vectors and $N$-element vectors; a base transform! In fact, it's even an orthogonal base transform. So, assuming you're synchronized, in OFDM, subcarriers do not spread into others. That is the whole point!

Pilot symbols aren't "special" in any way: you usually can't tell from the time domain signal whether and how many pilots there are.


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