I have a signal that is a periodic concatenation of truncated Lorentzians ($\frac{1}{1+x^2}$), like so:
The exact shape of the signal is unimportant I guess. I can approximate it with its first few Fourier components, so assume it is a single sine tone if it makes it any easier.
Let us name this signal $x(t)$. The frequency is $f_\mathrm{c} = 1\,\mathrm{kHz}$.
The signal above is now weakly phase-modulated by $\phi_\mathrm{PM}(t) = \Delta\phi \sin(2\pi f_\mathrm{PM}t)$. The resulting signal is then $x'(t) = x(t + \frac{\phi_\mathrm{PM}}{2\pi f_c})$. The modulation depth is small ($\Delta\phi\ll1$) and the modulation rate $f_\mathrm{PM}\approx 32\,\mathrm{kHz}$. The rate is known exactly, but it is not commensurate with $f_\mathrm{c}$. Note that the modulating signal is much faster than the signal being modulated.
I am interested in recovering the modulation depth $\Delta\phi$.
My gut tells me to look at the sum & difference frequencies, i.e. $f_\mathrm{PM} \pm f_\mathrm{c}$. (And, since $x(t)$ is not a pure tone, also $f_\mathrm{PM} \pm nf_\mathrm{c}$.)
Am I on the right track? If so, how does the amplitude of these sum & difference frequency components relate to $\Delta\phi$?
I feel thrown off by the modulation being so much faster. If $f_\mathrm{PM}\ll f_\mathrm{c}$, I would first down-convert by $f_c$, look at the phase, and down-convert that one by $f_\mathrm{PM}$. For a small modulation depth, I can skip computing the phase, and directly demodulate at $f_\mathrm{PM} + f_c$, hence my hunch above.