The signal processing is not only done in the baseband, but also in discrete time. This is because the digital chips ultimately work in the discrete time. However, the signal transmitted in the air is continuous-time.
How do we convert the discrete-time signal in devices to a continuous-time signal ready for transmission, and how we convert the continuous-time signal received to a discrete-time signal for processing in the receiver?
For any signal $s(t)$ with a bandwidth limited in $[-W/2, W/2]$, the sampling theorem asserts that we can sample it at a sampling frequency equal to its bandwidth $W$, namely at a sampling interval of $1/W$, and perfectly reconstruct it by
\begin{align} s(t) = \sum_{n} s[n] \text{sinc}(Wt-n), \end{align} where $s[n] = s(n/W)$ is the $n$-th sample, and $\text{sinc}(t)$ is defined as
\begin{align} \text{sinc}(t) \triangleq \frac{\sin(\pi t)}{\pi t}. \end{align}
We can see that to perfectly reconstruct the continuous-time signal from its samples, we must sample it at a frequency at least as large as its bandwidth. This is intuitive. A signal with a larger bandwidth contains components at higher frequencies, which translate to faster variations in the time domain. Therefore, we need to sample it at smaller intervals to avoid missing any information.
This is also part of the reason why we want to do signal processing in the baseband. The passband signal has the highest frequency up to $f_c+W/2$, requiring us to sample it at a sampling frequency of $2 f_c + W$, which is much higher than the sampling frequency of $W$ required for a baseband signal.
The sampling theorem also suggests how we should convert the discrete-time data stream $x[n]$ to the continuous-time baseband signal $x_b(t)$:
\begin{align} x_b(t) = \sum_{n} x[n] \text{sinc}(Wt-n) \end{align}
In other words, we pass the data stream $x[n]$ through a filter whose impulse response is the sinc function. We can also say that we “modulate” the data stream by the sinc function.
In practice, we usually use other functions (e.g., raised cosine) to modulate the discrete-time signals.
Now that we know how to convert between the discrete-time signal and the continuous-time signal, we can derive an equivalent input/output model of the channel in the baseband and also in discrete time. As you may have guessed, the equivalent channel model is also a FIR filter, but in discrete time.
Using the equivalent channel model in the baseband, we can write the baseband receive signal as
\[\begin{align} y_b(t) & = \sum_i a_i^b(t) x_b(t - \tau_i(t)) \notag\\ & = \sum_i a_i^b(t) \sum_{n} x[n] \text{sinc}\left[W(t-\tau_i(t))-n\right] \notag\\ & = \sum_{n} x[n] \sum_i a_i^b(t) \text{sinc}\left(Wt-W\tau_i(t)-n\right). \notag \end{align}\]The $m$-th sample of the baseband receive signal $y_b(t)$ is then
\begin{align} y[m] = y_b(m/W) = \sum_{n} x[n] \sum_i a_i^b(m/W) \text{sinc}\left[m-W\tau_i(m/W)-n\right] \notag \end{align}
Defining the delay $\ell \triangleq m-n$, we have
\begin{align} y[m] = \sum_{\ell} x[m-\ell] \sum_i a_i^b(m/W) \text{sinc}\left[\ell-\tau_i(m/W) W\right] \notag \end{align}
In conclusion, we can define the discrete-time baseband equivalent channel model as \[ h_\ell[m] = \sum_i a_i^b(m/W) \text{sinc}\left[\ell-W\tau_i(m/W)\right], \] so that the discrete-time baseband receive signal can be written as \(y[m] = \sum_{\ell} h_\ell[m] x[m-\ell].\)
Therefore, the discrete-time baseband equivalent chanenl model is also a complex-valued FIR filter!