Why multi-path channel has linear phase within the coherence bandwidth?
For the classical baseband channel model for multipath, $h(t)=\sum_ka_k\delta(t-\tau_k)$, the frequency response $H(f)=\sum_k a_ke^{-j2\pi f\tau_k}$. If $K=1$, single tap the phase would have been linear $\phi(f)=2\pi f\tau_k$. But for different path incident on the receiver, there is no guarantee that phase is linear. But let us see how phase changes between first path $k$ and last path $p$. For these incident paths, $2\pi f\tau_k-2\pi f\tau_p \approx \pi$ would cause fading. So if your frequency $f$ changes from $f_0$ to $f_0 +\frac{1}{2(\tau_k-\tau_p)}$, you will encounter fading. Hence $f_w=\frac{1}{2(\tau_k-\tau_p)}$ is called Coherence Bandwidth. It is the amount of bandwidth for which frequency response is 'flat' or constant. If the delay spread $\Delta \tau = (\tau_k-\tau_p)$ was less, $H(f)$ would be varying less over frequencies. If the difference in delay of paths was much much smaller than symbol time, that is, $\Delta \tau &lt;&lt; T$, the width of the channel would be much much greater than width of symbols. That is $\frac{1}{\Delta\tau} &gt;&gt; \frac{1}{T}$. So your transmission data would see a 'flat' or constant channel in terms of frequency response. A constant channel in frequency corresponds to a single tap as explained earlier. It's phase response is linear $\phi(f)=2\pi f\tau_k$.