Coherence bandwidth/delay spread in OFDM

Question

Say we want to design an OFDM system and we have some parameters given to us, i.e. the coherence time $T_c$ and the max delay spread $T_m$.

It's clear to me why we need to ensure $T_s(N+N_{cp}) \ll T_c$ where $N_{cp}$ is the cyclic prefix in samples - we want the channel to stay coherent during the entire symbol. What is not clear is what happens to the constraint $T_s(N+N_{cp}) \gg T_m \Leftrightarrow B_N \ll B_c$. Apparently, in OFDM it suffices with $T_s N_{cp} \geq T_m$.

It makes intuitive sense in a way, but I'm still not sure exactly how the coherence bandwidth interacts with the subchannel bandwidths. Why don't we need to ensure $B_N \ll B_c$?

Answer 1

Indeed, the constraint $T_s(N+N_{cp}) \gg T_m$ should be derived as $T_s(N+N_{cp}) \gg T_s N_{cp} \geq T_m$.

The first inequality to make sure that the CP overhead is as small as possible. The second inquality is what you have said, to avoid ISI between OFDM symbols and to convert linear convolution to circular convolution so that we can use single-tap equalizer.

Thus, $B_N << B_c$ is a consequence of the prior design criterion. However, if subcarrier spacing $B_N$ is much smaller than coherence bandwidth $B_c$, the channel estimation is facilitated by adding several pilots in $B_c$. For info, in LTE, two pilot are separated by 45 kHz while the rms delay spread is about 1us gives $B_{c,50\%} \approx 200kHz$ and $B_{c,90\%} \approx 20kHz$; thus 45kHz pilot spacing is good compromise between pilot overhead and estimation

Answer 2

Well, the reason you do OFDM in the first place is that you can take a large piece of bandwidth, and instead of using it as a single channel that would be heavily frequency-selective, use it as a set of much narrower subchannels which can typically be considered to be represented by single tap channels.

Hence, coherence bandwidth isn't directly bounded by the overall OFDM bandwidth.