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 $u(t)$?

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    $\begingroup$ if $\alpha_1$ != $\alpha_2$ != $\alpha_3$, the sum has no frequency. $\endgroup$ – Ben Nov 14 '19 at 21:33
  • $\begingroup$ you should remove the $2 \pi$ factors in the complex sinusoids. $\endgroup$ – robert bristow-johnson Nov 15 '19 at 2:38
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    $\begingroup$ 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 '19 at 2:42

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.

  • $\begingroup$ The communications sense your provided is essential, as I have close to none :) $\endgroup$ – Laurent Duval Nov 14 '19 at 22:18
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    $\begingroup$ @LaurentDuval Depends on the perspective of the OP ;-) I'm also not so deep into it ;-) $\endgroup$ – Fat32 Nov 14 '19 at 22:21
  • $\begingroup$ Your digital-communications tag ranking is way above mine. $\endgroup$ – Laurent Duval Nov 14 '19 at 22:25
  • $\begingroup$ @LaurentDuval is there a tag ranking ? I didn't know that... :-) $\endgroup$ – Fat32 Nov 14 '19 at 22:47
  • $\begingroup$ I shall say I do not fully understand. Fat32-tags $\endgroup$ – Laurent Duval Nov 14 '19 at 22:56

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.

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    $\begingroup$ This might have been an early upvote, but I have a guess why you're updating this slowly :) $\endgroup$ – Marcus Müller Nov 14 '19 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 '19 at 21:52
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    $\begingroup$ 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 '19 at 22:28
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    $\begingroup$ 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 '19 at 22:32
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    $\begingroup$ 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 '19 at 22:41

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.


$$ S = t^2 $$

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.

  • 1
    $\begingroup$ Nice work on solving for ω but to be clear ω is NOT the instantaneous frequency (see this post which details that trap: dsp.stackexchange.com/questions/31578/…), instantaneous frequency in signal processing applications has a well established and clearly defined mathematical construction. I added a polite down-vote until resolved as the answer as is may be misleading to others; certainly the question and its proper answer is not meaningless. $\endgroup$ – Dan Boschen May 29 at 15:29

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.


I work in the field of time and frequency in the design of atomic clocks, and there the notion of the instantaneous frequency for a sum of two tones is very clearly defined and useful. Fat32 touched on this with reference to communications (where I spent most of my career), but it extends far beyond that in more general cases starting with the clear definition of what instantaneous frequency is, without any limitation that the waveform be monochromatic or restrictions on the time duration, etc:

Instantaneous frequency is simply the time derivative of the phase of the analytic signal with respect to time:

$$f(t) = \frac{d\phi(t)}{dt}$$

Where $f(t)$ would be the instantaneous frequency in radians/sec with time $t$ in units of seconds. (Or divide by $2\pi$ to get in units of $Hz$ as the OP has done).

With the analytic signal of a real waveform $x(t)$ defined as $x(t)+ j \hat x(t)$ with $\hat x(t)$ as the Hilbert Transform. The analytic signal representation (which is a direct and reversible mapping from the real waveform by eliminating all the redundant negative frequency components) is convenient to us mathematically as it readily reveals the amplitude and phase components of the waveform that vary with time (there are no other constraints on this other than the validity of the analytic signal mathematically for a given waveform $x(t)$):

$$x_a(t) = A(t)e^{j\phi(t)} \tag{1} \label{1}$$

That said, consider the case of two cosine waves of equal amplitude of one and different frequencies for each cosine given as:

$$x(t) = \cos(\omega_1 t) + \cos(\omega_2 t)$$

With $\omega_1, \omega_2, t \in \mathbb{R}$

The analytic signal of $x(t)$ is:

$$\hat x(t) = e^{j \omega_1 t} + e^{j \omega_2 t} \tag{2} \label{2}$$

In this case, we can conveniently rearrange \ref{2} to put it directly in the form of \ref{1} as follows:

$$2\omega_m = \omega_2 - \omega_1$$

$$\omega_c = (\omega_1 + \omega_2)/2$$

$$\hat x(t) = e^{j(\omega_c+\omega_m)t} + e^{j(\omega_c-\omega_m)t}$$

$$\hat x(t) = (e^{j \omega_m t} + e^{-j \omega_m t})e^{\omega_c t}$$

$$ = 2\cos(\omega_m t)e^{\omega_c t} \tag{3} \label{3}$$

And thus we see for this case that $\phi(t) = \omega_c t$ and therefore the instantaneous frequency as $d\phi(t)/d(t)$ is the constant $\omega_c = (\omega_1 + \omega_2)/2$. We see too from $\ref{3}$ that this is an amplitude modulated waveform at a fixed carrier $\omega_c$: The carrier frequency does not change with time and only the amplitude goes up and down sinusoidally. This both makes sense and it quite meaningful.

In this case the instantaneous frequency was constant with time, but that needn't generally be the case; we can have arbitrary waveforms where both phase and amplitude vary with time. In this case with a time varying phase, the instantaneous frequency will be time varying and clearly defined for every "instant" in time. In practice we can only estimate the instantaneous frequency; such as one approach by getting the average of the instantaneous frequency between two time samples by the difference in phase over the difference in time. This is just like our ability to estimate velocity by measuring position over two samples in time.

Ultimately for any analytic signal with phase as a function of time, as long as we can determine the derivative of that phase, we can determine the instantaneous frequency.

For the specific question by the OP, given a $\phi(t)$ with integer coefficients, the process would be to determine the real and imaginary components of the sum from which the equation can be put in the form of $\ref{1}$ through conversion of cartesian to polar coordinates (providing the magnitude and phase versus time), and with that the instantaneous frequency would be time time derivative of the phase.

  • $\begingroup$ Interesting. I can follow the interpretation of an amplitude modulated carrier if $\omega_c >> \omega_m$ But what about a sum of 1 kHz and 2 kHz ? Following your argument you would claim that the instantaneous frequency would be a constant 1.5kHz. So it would be the same that of 1.5 kHz sine wave? $\endgroup$ – Hilmar May 29 at 14:06
  • $\begingroup$ @Hilmar Good one Hilmar and you must be recollecting that the analytic signal is only valid when $\omega_c > \omega_m$--- note why this is the case: The analytic signal has no negative frequency components so if we have a modulated signal spread over a given bandwidth, if $\omega_c < \omega_m$ it can no longer be analytic. With your new example here that is not the case so yes it would have an instantaneous frequency of 1.5 KHz to state it clearly. That said, we can't come up with two frequencies of individual sine waves where this relationship will ever fail (conveniently). Do you agree? $\endgroup$ – Dan Boschen May 29 at 14:29
  • $\begingroup$ Given that for any two sine waves, the $\omega_c$ will always be higher than $\omega_m$ since $\omega_c$ is the mean of the two frequencies and $\omega_m$ is half of the difference. So this can always be described/explained with analytic signals (which exist as a mathematical form because of the great convenience they give us to approach problems such as this) but we can come up with other cases that cannot be. $\endgroup$ – Dan Boschen May 29 at 14:33
  • $\begingroup$ "difference in phase over the difference in time" is only one method, limited to uni-modal signals. Sines' sum frequency depends on assumed definitions and isn't uniquely resolved for all. It's what time-frequency analysis studies - relevant literature. $\endgroup$ – OverLordGoldDragon May 29 at 14:44
  • $\begingroup$ @OverLordGoldDragon Yup it's only one method to estimate the frequency but I don't want to distract with various estimation techniques as it's off topic and the main point is Instantaneous Frequency on its own is a clearly defined term with a mathematical construct for the OP's formula / question. – but nice link on wavelets, thanks! $\endgroup$ – Dan Boschen May 29 at 15:16

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