I'm reading a the paper "Dual-Sideband Linear FMCW Lidar with Homodyne Detection for Application in 3D Imaging" (PDF) in which an in-phase/quadrature (I/Q) modulator is used in different modes: in single-sideband (SSB) mode, in which
$V_{I,SSB}(t)=V_D \cos(2\pi f_1 t + \pi \frac{f_2-f_1}{T}t^2), V_{Q,SSB}(t)=V_D \sin(2\pi f_1 t + \pi \frac{f_2-f_1}{T}t^2)$
and in double-sideband (DSB) mode, in which
$V_{I,DSB}(t)=V_D \cos(-2\pi f_2 t + 2\pi \frac{f_2}{T}t^2), V_{Q,DSB}(t)=V_D \sin(-2\pi f_2 t + 2\pi \frac{f_2}{T}t^2)$
where $V_D$ is a constant amplitude factor, $f_2-f_1$ is the bandwidth of the chirp, and $T$ is the period of the chirp.
I'm having trouble seeing where these expressions come from. In particular, further down in the article some examples are given why the thus obtained DSB signal has twice the bandwidth (resulting in twice the range resolution), but it is not at all clear to me why this is the case. Can someone explain this in simple terms?
The instantaneous frequency of the modulating signal is $f_1 + \frac{f_2 - f_1}{T}t$ in the SSB case, and $-f_2 + 2 f_2 \frac{t}{T}$ in the DSB case. Hence in the period from $0$ to $T$, the frequency ranges over the intervals $[f_1, f_2]$ and $[-f_2, f_2]$, respectively.
In the paper it is given that $f_1$ is 0.1 GHz and $f_2$ is 20 GHz, so for these particular parameters, in the second case the total frequency excursion is about twice as large.