I have a signal that is a periodic concatenation of truncated Lorentzians ($\frac{1}{1+x^2}$), like so:

carrier signal

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.

  • $\begingroup$ Hi! I think this might not be hard, but I'm not 100% certain we have the same definition of "weakly phase-modulated by …": Could you write down as formula (or link to a definition) of what that means to you? $\endgroup$ Jul 9, 2021 at 18:08
  • $\begingroup$ @MarcusMüller Done. Does that help? I mean that the entire waveform wiggles back and forth in time. $\endgroup$
    – polwel
    Jul 9, 2021 at 18:17
  • $\begingroup$ Thanks! Yeah, that helps, but it also gives me stuff to think about... $\endgroup$ Jul 9, 2021 at 18:18
  • $\begingroup$ as a first thing: does high-pass filtering help divide $x$ from $\phi_{\text{PM}}$? $\endgroup$ Jul 9, 2021 at 18:20
  • $\begingroup$ I guess that's what I am asking about. Does the PM extend to low frequencies? Or even DC? My instinct says it should be localized around $f_\mathrm{PM}$, so yes, a high pass filter should nuke the non-PM part. $\endgroup$
    – polwel
    Jul 9, 2021 at 18:22

2 Answers 2


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).

  • $\begingroup$ Thanks, very helpful! This agrees well if my thinking. As far as the 'overlap' goes: luckily my 'carrier' is band-limited (the FT of a Lorentzian is a decaying exponential). Its 5th overtone (i.e 5 kHz) is already below the noise floor. I do not worry about the carrier overlapping with the modulation (the latter centered around 32 kHz). $\endgroup$
    – polwel
    Jul 10, 2021 at 14:49
  • $\begingroup$ Yes and prior to considering the frequency modulation itself, your frequency spectrum of the periodic Lorentzian would be samples of the Fourier Transform of the Lorenzian (which is continuous) spaced by that periodic rate in frequency. So harmonics at 32 KHz spacing so if the Lorenzian spectrum isn’t much larger than that then you would have one dominant tone. If the Lorenzian is wider such as to include more significant tones then each of those would be a modulation component as well. $\endgroup$ Jul 10, 2021 at 15:16
  • $\begingroup$ Why spaced at 32 kHz? The 32 kHz signal is a single tone, but the 1 kHz one is not. I'd expect only peaks spaced by 1 kHz around 32 kHz... Unless the PM is mixing with itself. $\endgroup$
    – polwel
    Jul 11, 2021 at 15:39
  • $\begingroup$ @polwel small signal FM and large carrier AM are similar in this regard: the modulation sidebands are spaced from the carrier by the modulation rate. If the carrier is a signal tone (sinusoidal) there is no reason we would get harmonics of the carrier. $\endgroup$ Jul 11, 2021 at 19:00
  • $\begingroup$ And as I described in my comments the harmonics of 1 KHz would sample the FT of the Lorentzian but that has nothing to do with that FM modulation, just harmonics of the carrier prior to considering the FM effects. Depending on the actual bandwidth of the Lorenzian in this case, those additional components may be insignificant or at least significantly lower resulting in one dominant carrier as the OP wanted to proceed with. $\endgroup$ Jul 11, 2021 at 19:03

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 1st 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.

  • $\begingroup$ @polwei I don't think your relationship for x(t) as a phase modulation is correct. Delay modulation and phase modulation are not the same (although for narrow bandwidth relative to the carrier will be approximately so). FM and PM are constant envelope modulations with no AM components. $\endgroup$ Jul 13, 2021 at 4:58
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    $\begingroup$ Isn't that because I truncated the Taylor expansion? For instance, the second order term ($\sim \sin^2$) will have a DC term that to an extent balances the first order term. The original expression $x(t + ...)$ clearly preserves the envelope. $\endgroup$
    – polwel
    Jul 13, 2021 at 7:16
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    $\begingroup$ I follow now and that all makes sense! That is consistent with the spectrums shown in this post dsp.stackexchange.com/questions/72533/spectrum-of-fm-signal/… $\endgroup$ Jul 13, 2021 at 11:58

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