A frequency selective wireless channel is necessarily a multi-tap channel, in the tap-delay line model, as the delay spread is significantly larger than the symbol duration. And a single-tap channel is necessarily flat. But the converse is not necessarily true, is it? In particular, a flat fading channel can also be a multi-tap channel, can't it?

According to Tse & Viswanath's book (Ch2), the $l$-th tap of the channel filter is given by equation (2.36)

$$h_l=\sum_i a_i^b \mathrm{sinc}(l-W\tau_i)=\sum_i a_i^b \mathrm{sinc}(l-\frac{\tau_i}T)$$

where $a_i^b$ and $\tau_i$ are the gain and the delay of the $i$-th path, respectively, $W$ is the channel bandwidth and $T=1/W$ is the symbol duration. Consider a channel with a single path (or ray), whose delay is $\tau_0 = T/2$. This channel is of course flat. But the channel filter has infinite taps, doesn't it? (c.f. $h_l=a_0^b \mathrm{sinc}(l-0.5)$)

Edits 2024/5/3

This paragraph is added to clarify the discretization of the (baseband) multipath channel, and the derivation of $h_l$. Let $x_b(t)$ be a (base-band) signal, band-limited to $[-W/2, W/2]$. So $x_b(t)$ may be represented by

$$x_b(t)=\sum_{n=-\infty}^\infty x_n\mathrm{sinc}\left(\frac{t}T-n\right)$$ where $T=1/W$ and $x_n\triangleq x_b(nT)$. The received signal (over the multipath channel) is $$y_b(t)=\sum_i a_i^b x_b(t-\tau_i)$$ where $i$ is the index the multipath. Next, we sample $y_b(t)$:

$$y_m\triangleq y_b(mT)=\sum_i a_i^b\sum_{n=-\infty}^\infty x_n\mathrm{sinc}\left(m-n-\frac{\tau_i}T\right)=\sum_{l=-\infty}^\infty h_l x_{m-l}$$

where $\displaystyle h_l\triangleq \sum_i a_i^b \mathrm{sinc}(l-\frac{\tau_i}T)$. The last equality of the equation above follows from swapping the order of the 2 summations and change of variables, $l=m-n$.


1 Answer 1


But the converse is not necessarily true, is it?

Ah, it usually is.

Here's two aspects we need to address:

  1. Is every multi-tap channel frequency selective?
  2. We need to differentiate between the channel (a random, unknown thing happening to your signal) and an realization of the channel (a fix, but potentially unknown channel)

1. Is every multi-tap channel frequency selective?

The trivial answer is "no", because $h = (0, 5, 0)$ has three taps and is still flat, clearly. However, nobody would mean "channel with multiple taps, of which all but one tap are zero" when they say "multi-tap channel". Yet, $h=(0,5,0)$ is a perfectly valid realization of a three-tap channel. It might just happen!

If we restrict us to channels where actually multiple taps are non-zero, then the only non-frequency-selective channel must be an all-pass filter (because "linear convolutional operation that leaves the amplitude of all frequencies alike" is the definition of all-pass filters!). However, there's a difference between "flat" and "all-pass": When communications engineers say "flat", they also mean that the phase stays constant (or at worst, is a linear function of frequency), whereas an all-pass filter might change phase, arbitrarily rapidly.

So, we can approach this from two perspectives:

  • Either, we show that the inverse DTFT of a constant (flat response) must be a single "tap" (i.e., a Kronecker delta),
  • Or, we show that the only All-Pass filter has exactly one pole and one zero, both at zero, and hence is "equivalent" to the Dirac delta.

Since the first one is less work to write down, the definition of the DTFT is

$$h[n] = T \int\limits_{a}^{a+\frac{1}{T}} H(f) \cdot e^{j2\pi fnT} \,\mathrm{d}f\quad$$

In other words: you integrate the (discrete-time) spectrum of your channel realization over any bandwidth of the "tap rate" (the inverse of the tap spacing) multiplied with a "Fourier transform kernel", and you get the $n$th tap.

Let's try that with our constant response $H(f)\equiv 1$ and $a=0$:

$$h[n] = T \int\limits_{0}^{\frac{1}{T}} e^{j2\pi fnT} \,\mathrm{d}f\quad$$

Huh, here it says "integrate a complex sinusoid of frequency $nT$ over $n$ periods. Integrating a sinusoid over a period is always going to be zero, so this formula will yield exactly $0$ for all $h[n\ne0]$, and $h[0] = 1$.

So, there's that: the iDTFT of a constant-1-channel is unambiguously $[1,0,0…]$, a single-tap channel. Since the iDTFT is a linear operation, that's true for any other constant channel, as well.

So, yeah! Every truly constant-in-frequency channel is a single-tap channel. It might have infinitely many zero-valued taps. There's the extension that a flat channel might also shift the signal in time (in which case, the spectrum is just a constant multiplied with a complex sinusoid, i.e., phase is exactly linear to frequency), but that's really just a shifting of the position of your single tap.

The example you cite is just exactly that: The sinc function has zeros in every place your formula evaluates it, with the exception of $l- \frac{\tau_i}{T} = 0$, which is equivalent to $l=\frac{\tau_i}{T}$; if your channel needs to be flat, then all but one $a_i$ need to be zero (see above) and that delay needs to be a multiple of $T$.

Difference between Channel and Channel Realization

As I mentioned quite in the beginning of the previous section, $h=(0,5,0)$ might be a perfectly valid realization of a multi-path channel. And that happens! Sometimes you're in luck, and the thing you needed to assume to be frequency-selective isn't very frequency-selective.

But you typically build communication systems to work over a range of different, unknown channels. You set up a stochastic model of what channels can occur, and what the probabilities of different types (ranges) of channels are. And we've identified the multi-path channel as a useful model of these kinds of channels, so, if we say

this is a flat channel

what we often mean is

the model, who gives us a whole range of possible channel realizations, will only give us channel realizations that fulfill the requirements of flatness

So, when someone says, for example, "fast fading channel with heavy Doppler", unless they explicitly say that's not the case (or there's physical reason to assume that's not the case), the "stationary, direct-line-of-sight flat channel" is just a special case of that, and might happen to occur at times! It's just that it's not the only thing that might occur.

  • $\begingroup$ I appreciate the detailed answer, but I still don't think the converse (i.e. a multi-tap channel must be frequency selective) is necessarily true, for two reasons. First of all, I wasn't clear enough in the question. When I said "flat fading", I meant "frequency non-selective" fading. I.e. when the coherent bandwidth is significantly larger than the channel bandwidth, or equivalently when the delay spread is much shorter than the symbol duration. Secondly, the effects of sampling which discretizes the channel. $\endgroup$
    – syeh_106
    Commented May 3 at 3:38
  • $\begingroup$ A frequency non-selective fading channel doesn't necessarily need to have constant magnitude over the channel bandwidth. The frequency response of the channel just need to be highly correlated over the channel bandwidth. $\endgroup$
    – syeh_106
    Commented May 3 at 3:38
  • $\begingroup$ The effects of sampling is hard to clarify in comments. I've edited the question. $\endgroup$
    – syeh_106
    Commented May 3 at 3:39
  • $\begingroup$ Consider the extreme 1-path (ray) channel again, say $a^b_0=1$ and $a^b_i=0$ for any other $i$, and $\tau_0=T/2$. Then $y_b(t)=x_b(t−T/2)$. So the CIR is $h_b(t)=\delta(t−T/2)$ and the channel is clearly frequency non-selective, with CFR being $H_b(f)=e^{−j\pi fT}$. $\endgroup$
    – syeh_106
    Commented May 3 at 3:41
  • $\begingroup$ However, after discretization, the DT CIR $h_l=\mathrm{sinc}(l−0.5)$ has infinitely many nonzero taps. And the DT CFR has nearly constant magnitude over most of the channel bandwidth, e.g. with 1000 taps. Even for this DT channel, it is practically frequency-nonselective, isn't it? $\endgroup$
    – syeh_106
    Commented May 3 at 3:46

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