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In OFDM, we usually insert the guard interval to avoid the inter-symbol interference at the expense of achieved data rate. that guard interval can be either zeros or CP guard interval, which it must be longer than the channel delay.

I am wondering, what's about using long symbol to avoid that guard interval ? for example instead of having number of subcarriers $N= 64$ or $N = 1024$, we can use $N = 5120$, what's the impact of using a high number of $N$ ??

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  • $\begingroup$ Why do you think that a larger $N$ helps to avoid the ISI? $\endgroup$
    – AlexTP
    Apr 8 at 16:26
  • $\begingroup$ @AlexTP because we will only have one symbol instead of many symbols, so we won' t have ISI. $\endgroup$
    – Fatima_Ali
    Apr 9 at 5:30
  • $\begingroup$ Yes, it is true that if you have only one symbol, the inter-symbol interference does not exist by definition. But a system with only one symbol does not make sense. Differently put, without guard interval, no ISI for one symbol for any N; but as soon as you have more than one symbol, always ISI for, again, any N. $\endgroup$
    – AlexTP
    Apr 9 at 9:07
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Sure, you can !

Practically, some systems are using OFDM system with number of subcarriers 4096, but the PAPR will be an issue in that case, and if your channel is fast-fading, the time-domain equalizer will be high-complex since one-tap frequency-domain equalizer will be useless.

Usually, choosing the length of the symbols depends on the channel delay, e.g., if your channel delay is 128, you must use at least length of symbol 1024 and so on.

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  • $\begingroup$ Can you elaborate how that works? I don't see at all how it's solving the ISI problem. $\endgroup$
    – mmmm
    Apr 9 at 7:15
  • $\begingroup$ I didn't say that solve the ISI, I said such practical systems whose $N = 4096$ are used. The ISI issue is something else. $\endgroup$
    – Zeyad_Zeyad
    Apr 9 at 7:35
  • $\begingroup$ But it's what Fatima asked about, no? I am wondering, what's about using long symbol to avoid that guard interval? not: can I generally use large symbols?. $\endgroup$
    – mmmm
    Apr 9 at 12:34
  • $\begingroup$ The answer is yes, one can use long long symbol to avoid the ISI since when having one symbol, it means there is not ISI. However, I don't that that will be logic $\endgroup$
    – Zeyad_Zeyad
    Apr 10 at 7:42
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I'm giving a new answer because I suspect my previous answer was based on a misunderstanding of your question, but I'm not sure (so I'm not deleting the previous question right away).

I think you meant to ask

What are the limits to increasing the subcarrier count in an OFDM system? It seems like as-long-as-possible OFDM symbols are advantageous.

And the answer to that is time-variability: If your channel impulse response changes during the length of a symbol, your OFDM system loses the "O" aspect, and even if it didn't, the subcarrier symbols would experience a loss of SNR due to changing channel.

So, the length of the DFT is inherently limited by your channel model not being stationary. It's very rare that a multipath channel is constant - if there are many things to reflect from, chances are high something changes. For example, You can use very normal 2.4 GHz WLAN OFDM signal reflections to sense the movement of a human chest while breathing; generally, OFDM radar very much is a thing, exactly because OFDM systems are exquisitely sensitive to changes in environment.

So, your OFDM symbol duration must be shorter than what you consider the coherence time of your channel.

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    $\begingroup$ I think you will have to point out that you are talking about keeping the bandwidth unchanged and, therefore, increasing DFT size is equivalent to reducing the subcarrier spacing that in turn increases the OFDM symbol duration. Another reason to not increase infinitely the DFT size in this setup is synchronization difficulty with such narrow subcarrier spacing. $\endgroup$
    – AlexTP
    Apr 9 at 9:23
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No, that can't work.

The guard interval has the purpose of "swallowing" the previous' symbol's impulse response.

Your longer DFT distributes energy all over the symbol duration. This doesn't solve the problem.

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  • $\begingroup$ Could you add more details? I didn't understand what you mean. $\endgroup$
    – Fatima_Ali
    Apr 8 at 13:41
  • $\begingroup$ You're proposing something that simply doesn't work. When you write down the DFT formula, your proposal simply doesn't address the problem that's the cause for having a guard interval. So, it doesn't work – I don't really know what more to explain! $\endgroup$
    – Marcus Müller
    Apr 8 at 15:24

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