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  1. For flat fading, the bandwidth of the transmitted signal must be less than than the coherence bandwidth of the channel, and no intersymbol interference (ISI) occurs.
  2. Nyquist showed that the theoretical minimum system bandwidth required to detect R symbols/sec without ISI is R/2.

How can I relate the above two points? does this mean that for flat fading, the signal bandwidth must be less than the channel coherence bandwidth and more than or equal R/2?

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  • $\begingroup$ You can't relate these two points, since they have nothing to do with each other; what would the ADC "know" about the channel?. So, yes, you need to fulfill both criteria to have both flat fading, and a sufficient sampling. But: you're even putting Nyquist wrong. The sampling rate must be twice the signal bandwidth, not the other way around. $\endgroup$ – Marcus Müller Mar 14 at 12:49
  • $\begingroup$ @Marcus Müller I am talking about the Nyquist ISI criteria not the sampling theory. $\endgroup$ – Noha Mar 15 at 8:12
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The two guiding principles are:

  • In order to transmit at rate $R_p$ baud with no ISI, we require a bandwidth $B \geq R_p/2$, with equality only for sinc pulses. It follows that, given bandwidth $B$, the maximum rate is $R_p = 2B$.

  • A signal with bandwidth $B$ is subject to flat fading if and only if $B < B_c$, where $B_c$ is the channel's coherence bandwidth.

Then, we can draw two possible relationships between $R_p$ and $B_c$:

  1. For a given coherence bandwdith $B_c$, the maximum baud rate under flat fading is $R_{p,max} = 2B_c$. If you try to transmit faster than this, the channel becomes frequency-selective. This is an upper bound only: any baud rate below $R_{p,max}$ also sees flat fading.

  2. For a given rate $R_p$, the smallest coherence bandwidth that guarantees flat fading is $B_{c,min} = R_p/2$. If the coherence bandwidth $B_c$ is smaller than $B_{c,min}}, the signal will see frequency-selective fading.

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  • $\begingroup$ Thus, if we need to speak in terms of the minimum bandwidth required instead of the the maximum baud rate that can be transmitted, what will be the correct relation?according to Nyquist, the minimum bandwidth required to transmit R symbols/sec is R/2, and in case of flat fading with no ISI, the bandwidth is less than the coherence bandwidth Bc. My exact question now is: Does any value for the signal bandwidth less than Bc achieves flat fading? or we need the value of the bandwidth to be less than Bc and greater than or equal to R/2 according to Nyquist? @MBaz $\endgroup$ – Noha Mar 15 at 8:09
  • $\begingroup$ @Noha Please see the edited answer. $\endgroup$ – MBaz Mar 15 at 15:18
  • $\begingroup$ Thanks for the illustration $\endgroup$ – Noha Mar 15 at 19:01
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MBaz answer gives the limit but I thought to give another perspective. For #2, I am really thinking you might be talking about Nyquist ISI Criterion (https://en.wikipedia.org/wiki/Nyquist_ISI_criterion) .

$\frac{1}{T_s}\sum_{-\inf}^{+\inf}H(f - \frac{k}{T_s}) = 1$

The $H$ includes your channel as well as rrc filtering. So here, your transmit signal BW $\frac{1}{T_s}$ should be $\le 2B$. Since $R = \frac{1}{T_s}$, $R \le 2B$.

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