2
$\begingroup$

In the literature, the instantaneous frequency for a signal $s(t)=e^{-j \phi(t)}$ is given by: $$f(t)=\tfrac{1}{2\pi}\frac{d\phi(t)}{dt}$$

This my problem: I have a signal $$u(t)=e^{-j \alpha_1 \phi(t)}+e^{-j \alpha_2 \phi(t)}+e^{-j \alpha_3 \phi(t)}$$ with $\alpha_1,\alpha_2,\alpha_3 \in \mathbb{Z} > 0$. How to find the instantaneous frequency for the signal $u(t)$?

$\endgroup$
  • 3
    $\begingroup$ if $\alpha_1$ != $\alpha_2$ != $\alpha_3$, the sum has no frequency. $\endgroup$ – Ben Nov 14 at 21:33
  • $\begingroup$ you should remove the $2 \pi$ factors in the complex sinusoids. $\endgroup$ – robert bristow-johnson Nov 15 at 2:38
  • 1
    $\begingroup$ you gotta figure out the overall real and imaginary parts and from that get the overall magnitude and phase, and take the derivative of the overall phase. $\endgroup$ – robert bristow-johnson Nov 15 at 2:42
3
$\begingroup$

Eventhough Laurent has given a broader sense of the answer, let me put here the communications theory sense ot it.

The concept of instantaneous freqency emerges when you consider Frequency Modulation or Phase Modulation systems, where the message is embedded into the change of the frequency or phase of a carrier signal.

This carrier is typically a single cosine wave or a complex exponential equivalent which posses a single well defined angle, whose time derivative gives you the instantaneous frequeny; hence the message. That's why they are also called angle modulation or exponential modulation systems.

The understanding is that there must exist a single carrier to represent the modulated message. Your example includes a general sum of three complex exponentials (unless you speciy those $\alpha$ pars) whose simplification into a single carrier seems not possible on the surface. Then there won't be a unique single instantaneous frequency definition for $u(t)$.

But you can consider three different instantaneous frequencies per complex exponential for your signal $u(t)$. Indeed this would be the case when multiple FM modulated signals (at different frequencies) were existing simultaneously as they are transmitted at different channels.

$\endgroup$
  • $\begingroup$ The communications sense your provided is essential, as I have close to none :) $\endgroup$ – Laurent Duval Nov 14 at 22:18
  • 1
    $\begingroup$ @LaurentDuval Depends on the perspective of the OP ;-) I'm also not so deep into it ;-) $\endgroup$ – Fat32 Nov 14 at 22:21
  • $\begingroup$ Your digital-communications tag ranking is way above mine. $\endgroup$ – Laurent Duval Nov 14 at 22:25
  • $\begingroup$ @LaurentDuval is there a tag ranking ? I didn't know that... :-) $\endgroup$ – Fat32 Nov 14 at 22:47
  • $\begingroup$ I shall say I do not fully understand. Fat32-tags $\endgroup$ – Laurent Duval Nov 14 at 22:56
3
$\begingroup$

My first inclination is to say this is a meaningless question. The concept of "instantaneous" frequency really only pertains to a single pure tone with a slightly varying frequency.

In this light, one may construct a definition saying "The instantaneous frequency at time $t$ is the same as that of a pure tone which matches the function (sum) in the immediate time neighborhood".

I think you have the signs wrong in your statement, and they are conceptual clutter, so I have removed them. The same solution can be done with them retained.

$$ A e^{j (\omega t + \theta) }=e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)} $$

Take the derivative of both sides.

$$ j \omega A e^{j (\omega t + \theta) }= j \left( \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} \right) \frac{d\phi}{dt} $$

Cancel the $j$s and divide both sides by the single tone definition.

$$ \omega = \frac{ \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} }{A e^{j (\omega t + \theta) }} \cdot \frac{d\phi}{dt} $$

Substitute it back in.

$$ \omega = \frac{ \alpha_1 e^{j \alpha_1 \phi(t)}+ \alpha_2 e^{j \alpha_2 \phi(t)}+ \alpha_3 e^{j \alpha_3 \phi(t)} }{e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)}} \cdot \frac{d\phi}{dt} $$

You can now solve for $\omega$. It may be complex. The real part represents the rotational frequency and the imaginary part the decay in amplitude coefficient.

The answer is then $\frac{1}{2\pi}\Re(\omega$), where the fraction is the conversion factor between radians per unit time and cycles per unit time. $2\pi$ has units of radians per cycle.


After thoughts:

I think this answer qualifies under R B-J's "overall" categorization.

The constraint that the $\alpha$'s be integer value seems superfluous. All that does is ensure that the overall signal is periodic. The same would be true for rational $\alpha$ values too. This has nothing to do with local behavior.

Also, if you plug in $\alpha_1=-1,\alpha_2=0,\alpha_3=0$, you can see the OP's original assertion has a sign error, otherwise, they are in agreement.

You might find the formulas I derive in these articles interesting and pertinent.

These formulas are derived based on the assumption of a single pure tone which varies only in frequency. Surprisingly, the same formulas work for both the real and complex tone cases. Obviously, being at a peak or a zero crossing is only relevant in the real tone case.

Since the assumption of my solution is that the sum is acting like a single pure tone in a small neighborhood, the article formulas, based on a few neighboring points, should return close to the same results as the derived solution in my answer, based on the signal definition.


Per LD, the "close to a kind of an average of the base frequencies" formulation:

$$ S = e^{j \alpha_1 \phi(t)}+e^{j \alpha_2 \phi(t)}+e^{j \alpha_3 \phi(t)} $$

$$ \omega = \left[ \left( \frac{e^{j \alpha_1 \phi(t)}}{S} \right) \cdot \alpha_1 + \left( \frac{e^{j \alpha_2 \phi(t)}}{S} \right) \cdot \alpha_2 + \left( \frac{e^{j \alpha_3 \phi(t)}}{S} \right)\cdot \alpha_3 \right] \cdot \frac{d\phi}{dt} $$


Instead of matching the signal locally to a complex pure tone, it can instead be matched to a real pure tone.

$$ A \cos( \omega t + \phi ) = S $$

$$ -\omega A \sin( \omega t + \phi ) = S' $$

$$ -\omega^2 A \cos( \omega t + \phi ) = S'' $$

$$ \omega^2 = -\frac{S''}{S} $$

$$ \omega = j\sqrt{\frac{S''}{S}} $$

The reader is invited to plug the definition of $S$ in from above.

If you do plug in the the original $S$ in the real tone matching case, you will get an RMS like formula which is also a kind of weighted average of the base frequencies.

Note too, with these formulas, it is possible for the denominator to become zero while the numerator is non-zero. Thus the instantaneous frequency is supposedly infinite (loosely speaking). That doesn't make sense either.

To see that the two approaches don't necessarily reach the same value it is easier to use a simpler function.

Suppose

$$ S = t^2 $$

The two different approaches yield different results:

$$ \omega_{1} = -j\frac{S'}{S} = \frac{-2}{t}j $$

$$ \omega_{2} = j\sqrt{\frac{2}{t^2}} = \frac{\sqrt{2}}{|t|}j $$

There is nothing that says that either of these methods is any more or less valid than the other.

Taking a totally different tack, let's look at the specific case of

$$ \alpha_1 = 3, \alpha_2 = 7,\alpha_3 = 0, \phi(t)=t $$

$$ \begin{align} S &= e^{j3t} + e^{j7t} \\ &= e^{j5t} \left( e^{-j2t} + e^{j2t} \right) \\ &= 2 e^{j5t} \cos(2t) \end{align} $$

This specific example has an unwavering rotational frequency of 5 radians per unit time and an unwavering amplitude oscillation frequency of 2 radians per unit time. This is the complex analog of the beat phenomenon. Since both are unwavering, the "instantaneous" values should be the same.

On the other hand, the signal repeats itself every $2\pi$ units of time, so its overall global frequency is 1 radian per unit time. It is an example of a periodic signal missing its fundamental.

So, what is the actual frequency?

I return to my original inclination, this is a meaningless question.

$\endgroup$
2
$\begingroup$

The notion of instantaneous frequency is (hopefully) consistent with the monochromatic wave model:

$$x(t)=a \cos 2\pi \nu t\,,$$

where $a$ is the amplitude and $ \nu$ the frequency. It would be tempting to compute a similar formula for evolving amplitude and frequency cases, something like:

$$x(t)=a(t) \cos \left(\phi(t)\right)\,.$$

However, this is not well-defined, as an infinity of pairs of $a(t)$ and $\phi( t)$ would satisfy the same equation. It suffises to choose a function $\alpha(t)$ with $0<\alpha(t)<1$, and write

$$x(t)=\frac{a(t)}{\alpha(t)} \times \cos \left(\arccos\left( \alpha(t) \times \cos \left( \phi( t)\right)\right)\right)\,,$$ to get (any) other result.

One solution for the instantaneous frequency is thus to define it through the analytic signal, which is readily given in your multitone case:

$$ \begin{align}2u(t) &= \ \ \left(\cos(\alpha_1\phi(t))+\cos(\alpha_2\phi(t))+\cos(\alpha_3\phi(t))\right)\\ & \quad + \ i\left(\sin(\alpha_1\phi(t))+\sin(\alpha_2\phi(t))+\sin(\alpha_3\phi(t))\right) \end{align}$$

and then differentiate the phase argument of it. Which would be the $\arctan$ (plus phase unwrapping) of the imaginary part divided by the real part. Tonight, this seems very tedious to me. Indeed, in the case of the simple case of to sines with frequencies $\nu_1$ and $\nu_2$, then $y_H(t) = e^{-2\pi i \nu_1t}+e^{-2\pi i \nu_2t} $ rewrites, with instantaneous frequencies (see P. Flandrin, Time-Frequency Time-Scale Analysis):

$$ y_H(t) = 2|\cos\pi(\nu_2-\nu_1)t|\exp\left(-2\pi i \frac{\nu_1+\nu_2}{2}t+\pi i \mathrm{sign}(\cos\pi (\nu_2-\nu_1)t)\right)$$

In other words, the "instantaneous" frequency would close to a kind of average of the base frequencies. This corresponds to the notion of "beating" if they are close, but has less meaning when the frequencies are far apart. And I am not surprised: even though the instantaneous frequency is believed "local", computing it through a slow-decaying Hilbert transform makes it very global indeed.

I do suspect that, in your three sine sum model, the expression will be even more tedious, with limited physical meaning as well without knowledge on the $\alpha_k$ and $\phi(t)$. Since the $\alpha_k$ are integers, I doubt that the frequencies of each component could be considered as "close" in general.

Thus, people have been trying to define other concepts for "instantaneous frequencies", notably based on time-frequency transformations. I am not sure that people have agreed on one definition as of today.

$\endgroup$
  • 2
    $\begingroup$ This might have been an early upvote, but I have a guess why you're updating this slowly :) $\endgroup$ – Marcus Müller Nov 14 at 21:24
  • $\begingroup$ The upvote was clearly not deserved. Please do reconsider, though I thank you for your trust $\endgroup$ – Laurent Duval Nov 14 at 21:52
  • 1
    $\begingroup$ I must admit I was expecting something else (linearization of the (complex) $\ln$ to find an approximated instantaneous phase you could time-derive), but this is way better: You discuss the problem of using the notion of frequency to a multi-tone signal. So yeah, the upvote stays where it is :) $\endgroup$ – Marcus Müller Nov 14 at 22:28
  • 1
    $\begingroup$ This was indeed my first move, but I felt I had to reconsider it because (?) of the integer nature of the $\alpha_k$ (and maybe a too good white wine tonight) $\endgroup$ – Laurent Duval Nov 14 at 22:32
  • 1
    $\begingroup$ Your statement in bold has been bothering me. "Instantaneous" is "local" by definition, no faith required. A global solution is only possible if the signal is defined globally. Here it is, but having a global definition is not a requirement in general. $\endgroup$ – Cedron Dawg Nov 16 at 22:41
0
$\begingroup$

From an audio point of view, I would determine if the reciprocal of the least common multiple of a1,a2,a2 were within the range perceivable as pitch to a human; and, if so and stationary for about 6 periods or more, call that reciprocal the instantaneous pitch frequency.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.