# Does the phrase “instantaneous frequency” make sense?

I had always thought of time and frequency as being two different (yet complete) descriptions of the same system, so to me, the phrase "instantaneous frequency" didn't make sense -- frequency is a global description of a function, whereas time is local description of a function.

However, I just got confused by something.

Let's say I have the function $x(t) = \sin(t^2)$. I believe I can say that the frequency of this function increases over time, and, at least intuitively, this makes sense.

But mathematically, I can't make sense of this. Sure, I could take a limit, but I feel like that would be changing the meaning of "frequency" altogether, since it's no longer a description of the actual function.

Is it actually sensible to say that a function's frequency "changes" over time? What is the proper way to state something like this more mathematically?

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When you ask two such closely related questions, it makes sense to cross-link them so that people don't waste their time answering overlapping parts of them separately. –  joriki Nov 24 '11 at 22:55
@joriki: I didn't feel that the answers were related, but thanks for linking it. –  Mehrdad Nov 24 '11 at 22:56
They're quite close related -- I had just started answering the other question and started writing something about the meaning or lack thereof of a frequency changing over time, when I saw this question. –  joriki Nov 24 '11 at 22:58
@joriki: Ah okay, thanks for letting me know! –  Mehrdad Nov 24 '11 at 22:59
Well, they were even more closely related before you edited this one to use a different function than the other one... –  joriki Nov 24 '11 at 23:10

Yes, you can make sense of the concept of instantaneous frequency. Imagine a time-dependent phase $\mathrm e^{i\phi(t)}$. For $\phi=\omega t$, we would call $\omega$ the angular frequency, and this is $\mathrm d\phi/\mathrm dt$. Thus we can regard $\mathrm d\phi/\mathrm dt$, the rate of change of the phase angle with time, as an instantaneous frequency. We can still do this if the amplitude is also time-dependent, by ignoring the amplitude and focussing on the rate of change of the phase.

If we only have a real signal, we would in general have to make an arbitrary decision what part of its change is due to a change in "amplitude" and what part is due to a change in "phase", since taking the real part of different complex signals can lead to the same real signal. However, there may be situations in which it makes sense to regard the signal as having a constant amplitude, and to regard all changes as changes in phase. The example you give is such a case, since the signal oscillates between $+1$ and $-1$. In such a case, we can consider the signal as the real part of a complex signal of constant amplitude.

That is, given a real signal $f(t)$ and a constant amplitude $A$, we can consider the complex signal $f(t)\pm\mathrm i\sqrt{A^2-f(t)^2}$, where the sign changes whenever the signal reaches an extremum with $f(t)=\pm A$. Then we can calculate the phase as $\phi(t)=\arccos(f(t)/A)$, where we let the phase wrap around and increase beyond the usual range $[0,2\pi]$ to avoid discontinuities. Thus we can define an instantaneous angular frequency as $\omega(t)=\mathrm d\phi(t)/\mathrm dt=\mathrm d/\mathrm dt(\arccos(f(t)/A)$. In your case, the result would be $\omega(t)=\mathrm d/\mathrm dt(t^2)=2t$.

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I think the frequency is only defined up to a sign change (because $A \sin(\omega t) = -A \sin(-\omega t)$). So maybe it would make more sense to take the absolute value of your formula. In that case, I think it simplifies to $\frac{|f'|}{\sqrt{A^2 - f^2}}$. This is an interesting idea, but I does not always coincide with the usual frequency. For example I think the frequency of $\frac{1+\cos(x)}{2}$ will not be constant (maybe somehow removing the "mean value", assuming we can define one, could solve the problem). –  Joel Cohen Nov 25 '11 at 3:29
Cool, thanks for the explanation. –  Mehrdad Nov 26 '11 at 11:04

To add a slightly different viewpoint to @Joriki's excellent answer, communications engineers often restrict the notion of instantaneous frequency to (real-valued) signals that can be expressed as $\text{Re}\{a(t)\exp(j\phi(t))\}$ where $a(t)$ is a (possibly complex-valued) continuous signal that is varying much more slowly with $t$ than $\phi(t)$ is, and $\phi(t)$ is a continuous function of $t$ that is differentiable for all $t$ except possibly for a discrete set of time instants. Then the instantaneous (radian or angular) frequency is the derivative of the phase. For the OP's problem, we have $a(t) = -j, \phi(t) = t^2$ and everyone is in agreement that the instantaneous frequency of $\sin(t^2)$ is $2t$.

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To add to a previous reply: when a signal is not expressible as a sine/cosine or complex exponential of a function of time, it may still be decomposable into a superposition of such signals. In that case, the instantaneous frequency concept would apply to each component separately, thus resulting in a spectrum concentrated on a region shaped like a web, whose lines, in fact, are called "partials".

To obtain the spectrum, you'll first have to segregate the components. The simplest way to do that is to rely on the assumption that the different components are not crowding the same area of frequency space too much, to apply a time-frequency transform (e.g. windowed fourier transform, or wavelet or S-transform) and then apply the instantaneous frequency to each frequency channel of the resulting spectrum. If the components are well-enough spaced apart, then each one will register over a given frequency range in the spectrum -- namely the range that is "tuned into" that signal component.

I illustrate this here http://www.youtube.com/watch?v=iVO9-Y_Dgxw in a side-by-side contrast with a readout from a usual fourier-based spectrogram. The 3 spectra on the right in the view are of the same signal at three different resolutions (showing that the particular method can be applied at multiple resolutions, leading to equally sharp results) and the phase is color-coded in each one. This YouTube demo shows what partials looks like at a resolution of 5280 pixels/second (again the phase is color coded) http://www.youtube.com/watch?v=gDMixb7-_LM

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I just thought I'd post this link I came across, while looking for something else -- apparently a signal whose frequency changes with time is not only well-defined, but it is also called a chirp. :)

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Hopefully that somewhat clarifies that algorithm I told you about a while back... :) –  Ｊ. Ｍ. Nov 26 '11 at 11:11
@Ｊ.Ｍ.: I entirely missed your comment until now (which is 3 years later)! Yes it does! haha :) –  Mehrdad Jul 7 at 11:13