How to demodulate PM with signal faster than carrier?

Question

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.

Answer 1

Since the result is small angle FM modulation, for a single tone there will only be two significant sidebands each with magnitude close to $\beta/2$ (for the single sinusoidal modulation case with small $\beta$), where $\beta$ is the modulation index (peak angle). The two sidebands would be centered about the carrier spaced by the modulation rate - regardless of what frequency the carrier is.

When the carrier is smaller than the modulation, one of the two sidebands would extend into the negative frequency; but since the modulating signal and carrier is real (assuming that is the case) there will be a negative frequency carrier with two sidebands as well resulting in two real sidebands at $f_c + f_m$ and $f_m - f_c$ along with the dominant carrier at $f_c$ when $f_m > f_c$. Note how ultimately the spectrum is indifferent to what carrier frequency is used; the modulation index which is used to predict the sideband levels (which are spaced from any carrier by the modulation rate) is completely determined by the frequency deviation (peak instantaneous frequency of the modulated carrier) and modulation rate (for a single tone $\beta = f_m/f_{dev}$ where $f_m$ is the modulation frequency and $f_{dev}$ is the frequency deviation), but the actual carrier frequency is not a factor.

If we had a multi tone FM modulated signal that occupied the entire spectrum within a given bandwidth and the carrier was less than half that bandwidth, if the signal and carrier were real then we would see aliasing in this case consistent with what has been described above for the case of a single tone, while if the carrier were complex (as well as the resulting modulated waveform) then there would be no aliasing effect either.

As far as demodulation, this is possible for the case of a complex waveform where a complex carrier was used but would not be possible with a real carrier with an arbitrary unknown modulated multi-tone signal due to the aliasing. With a single tone modulation, the demodulation of the modulated tone can be easily determined. For example: consider a 1 Hz carrier with 5 Hz modulation and mod Index 0.2. Given the mod index this would produce two dominant sidebands 20 dB below the carrier (approximately), with the carrier at 1 Hz and the two sidebands at 6 Hz and 4 Hz. (Which is +6 Hz and -4 Hz for the positive frequency of the real carrier and -6 Hz and +4 Hz for the negative frequency of the real carrier). The levels and spacing of the sidebands when viewing the positive and negative frequencies are not affected by the low carrier but their spans overlap. For a single tone modulation we are able to extract all relevant information regardless but if multi-tones were present the overlap may result in aliasing effects that would be challenging to separate without knowledge of the modulated waveform itself (which if known completely, demodulation wouldn’t be necessary).

Answer 2

As per the question, the modulated signal is $x(t + \frac{\phi_{\mathrm{PM}}(t)}{2\pi f_c})$. In the limit of a weak phase deviation ($\Delta\phi\ll1$), we can expand to 1^st order:

$$ x(t + \frac{\Delta\phi \sin(2\pi f_{\mathrm{PM}}t)}{2\pi f_c}) \approx x(t) + \dot{x}(t) \frac{\Delta\phi}{2\pi f_c}\sin(2\pi f_{\mathrm{PM}}t)$$

The first term is simply the unmodulated signal. The second term is an AM of $\dot x(t)$ at the frequency $f_\mathrm{PM}$. It is indeed entirely irrelevant which of $f_c$ or $f_\mathrm{PM}$ is larger.

Now, $x(t)$ is not a single tone. It is periodic with period $1/f_c$ though, so we conveniently express it via its Fourier series:

$$x(t) = \sum_{k=-\infty}^{\infty} c_k \exp(2\pi i k f_c t)$$

For a Lorentzian, the Fourier coefficients are known, $c_k \propto \exp(-\alpha|k|)$, for some $\alpha$ relating to the linewidth. In my case, the Lorentzians are truncated. In frequency space, the truncation coresponds to a convolution with a sinc function. From here, it is straight forward to compute the Fourier representation of $\dot x(t)$.

The spectral components of $\dot x(t)$ (located at $kf_c$, i.e. 1 kHz, 2 kHz, etc.) each are mixed with $f_\mathrm{PM}$, hence showing in the spectrum at $f_\mathrm{PM} \pm k f_c$. Skipping the details here, the magnitude of the peaks each is $ \Delta \phi\, k c_k /2.$

_{(Measuring these individually, and combining them into an optimal estimate in the presence of noise is a different matter.)}

As mentioned by Dan Boschen, there is a risk that the spectra of $x(t)$ and $\dot x(t) \sin (2\pi f_\mathrm{PM}t)$ overlap, complicating the detection. It is something I'll have to keep in mind.