The uncertainty principle states that if you have a signal which is very concentrated in time, then its Fourier transform will be rather outspread and vice versa. However, I don't really understand how this implies that in general we can't find a function that gives us the frequency spectrum for a particular point in time.

I don't even know what exactly is meant with determining the frequency spectrum at a specific point in time. If I wanted to localise the frequency better (e.g. at zero), I would multiply by a corresponding indicator function and consider the fourier transform, i.e. consider $$ \begin{aligned} \int_{ \mathbb{R}} f( x) {1}_{ [ - \epsilon , \epsilon ] } ( x) e^{ -2\pi i\xi x}\, dx .\end{aligned} $$ However, with this interpretation the frequency spectrum at $ x = 0$ doesn't really make sense, since for $ \epsilon \to 0$ this integral will vanish. Can someone clarify this a bit?

  • 1
    $\begingroup$ The uncertainty principle is a property of bilinear (i.e. "inner product") time-frequency distributions. Such distributions give you information about linear compositions and also allow you to decompose the signal by linear superposition. That is not the only way to model a signal, and different models define frequency in ways different from the Fourier transform. Then the uncertainty principle might not apply. $\endgroup$
    – Jazzmaniac
    Commented Jul 4, 2023 at 15:38
  • $\begingroup$ @Jazzmaniac but could you still try to explain me what people mean when saying: We can't get both, "perfect" information about time and frequency at the same time? $\endgroup$
    – Richard
    Commented Jul 4, 2023 at 16:08
  • $\begingroup$ @Jazzmaniac Is my following interpretation correct: If you choose a very narrow window in the windowed Fourier transform, due to the Heisenberg uncertainty relation the corresponding "frequency representation" will look very "flat", meaning we can't really point out a single frequency which is contained "the most" in our signal. $\endgroup$
    – Richard
    Commented Jul 4, 2023 at 16:29
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    $\begingroup$ In simple terms: the smaller fraction of a period you see, the less you are able to pin down the period length or the frequency. $\endgroup$ Commented Jul 5, 2023 at 1:39

2 Answers 2


The Fourier Uncertainty Principle states that an arbitrary function cannot be compact both in time and frequency. The Fourier Time Scaling Property demonstrates this mathematically and is given as:

$$x(at) = \frac{1}{|a|}X(\omega/a)$$

Note how when we stretch the horizontal axis $t$ in the time domain signal to $at$, the horizontal axis $\omega$ in the frequency domain signal is compressed by $\omega/a$.

For continuous-time systems, a Gaussian pulse reaches the lower bound of what is possible in minimizing the joint localization in time and frequency, referred to as the Time-Bandwidth product. For discrete time systems the Slepian pulse reaches this limit (and Kaiser pulses are a good approximation of a Slepian).

Below I show two (truncated) Gaussian pulses in time and their Discrete Time Fourier Transform, where we see that the pulse that is localized tighter in time (blue trace) is distributed wider in frequency (blue trace) and vice versa. If we were to continue to localize time wider, the frequency distribution related to that would get wider. In the extreme, a single impulse in time would be constant for all frequencies.

Gaussian Pulses

If the time pulses given above were observations of a frequency (and as such would be a window multiplied by the underlying continuous stationary frequency signal), the resulting frequency in the Fourier Transform would have the same shape. Note as we make the time window narrower, the frequency result gets wider, and in the extreme if we were to select only a single point in time (select with an impulse), the frequency would be constant and we would have no indication of the original actual frequency that we selected the sample from:

Gaussian pulse for observing time and frequency


Why does it imply that we can't localise

It doesn't. It says that linear methods cannot resolve past its limit.

Perfectly jointly localizing

Single-component signals can be localized perfectly in both time and frequency. For a pure sine, as long as sampling above Nyquist, two points are sufficient to perfectly recover its frequency, phase, and amplitude.

(Both above are of course limited by numeric precision). For variable frequency/amplitude, time-frequency reassignment works. Synchrosqueezing examples:

Rather than perfect, these examples are challenging and feature more than one "component", just to show that performance is good even with >1 component. SSQ_STFT should${}^1$ handle a (single) linear chirp perfectly, and SSQ_CWT exponential chirp, though there's fundamental limitations with finite inputs near boundaries (information discontinuity). Note, SSQ's performance on single-component is very impressive, even handling pure noise as the frequency (code in source post):

The mapping may be imperfect, it's also stretched (log-scaled) since it's CWT.

Another way to perfectly map a single-component AM-FM may be using unwindowed STFT with size-2 DFT, via the sine inverse solution. In addition to the Nyquist rate, the sampling rate is constrained by $f'(t)$, i.e. frequency derivative, e.g. we obviously can't losslessly map a jump from $f=2$ to $f=4$ that happens between samples, even if $f=4$ is below Nyquist. This constraint is formalized for synchrosqueezing (see linked post).

What's "single-component"? In short, it's controlled by the time-frequency kernel (e.g. wavelet/kernel), or the resolution of the linear method, but basically a low-frequency amplitude modulating a relatively high variable-frequency carrier is single-component - some details.

1: I'm not certain, I've not studied the math, but it seems very likely in continuous time, else near-perfect. What's for certain, the Heisenberg limit goes out the window for single-component.

"Resolution" means "ability to resolve"

If there's one sine, there's nothing to resolve. The Fourier transform is an imperfect tool for handling one sine. It is, however, a perfect (in a sense) linear tool for >1 sines.

The FT works by taking an inner product (similarity/dot product) of every possible sine against the input. What one learns when there's two sines, is that traditional limitations kick in if the sines aren't infinitely long - i.e. observation interval. With DFT, one can observe this by taking shorter and shorter durations of same frequency sines - the "correlation machine" that is the DFT will eventually fail to produce two distinct peaks.

Localizing general signals

Firstly, localization is non-unique:

$$ \cos(A)\cos(B) = .5[\cos(A+B) + \cos (A - B)] $$

Second, not every signal has a meaningful AM-FM representation - e.g. unit impulse, rectangle function.

This aside, AM-FM localization of general signals amounts to solving the cocktail party problem - i.e. unmixing. Linear methods are incapable of perfectly (or, most of the time, even satisfactorily) unmixing. Yet, fundamentally, it's critical to acknowledge, that the case here is not same as the physics uncertainty principle: here, we have perfect information (the signal is exactly determined) - in physics, the problem is that we can't measure exactly to begin with. Obviously every real-world signal is subject to this limitation (plus general noise which far outweighs the quantum limit), but that's separate from the mathematical problem.

Thus, nothing says perfect unmixing is impossible. Except, noise (mathematically), due to non-uniqueness.

The most successful methods for joint localizers are deep neural networks: they have excellent performance in e.g. voice separation, meaning their latent representations necessarily break the Heisenberg limit. Obtaining AM-FM mappings from such outputs, or from latent representations, however, is another story - the separated voices are still very much multi-component.

Proof idea: "Linear = Heisenberg-bound"

For any sequence of operations, as long as the net-operation is linear

$$ \text{op}(c_0x_0 + c_1x_1) = c_0\text{op}(x_0) + c_1\text{op}(x_1) $$

the operation has a Fourier transform${^2}$. Then, it suffices to prove that there cannot be a function $x(t)$ that is time-limited and also band-limited, i.e. $=0$ outside of $[a, b]$ in time and $[c, d]$ in freq. This is done in Wavelet Tour - full proof in screenshot. Hence, no linear operation can yield $a=b < \infty$ and $c=d<\infty$ at once.

2: I don't really know this. It makes sense to me, but I'm just regurgitating authoritative sources on the conclusion "Linear = Heisenberg-bound", which also may be a simplification; @Jazzmaniac:

The uncertainty principle is a property of bilinear (i.e. "inner product") time-frequency distributions.

The idea of using the invoked theorem as "proof" is also mine, I've no clue what the formal justification is. That all linear operators can be expressed as functions at all seems to be untrue (I'm unsure how to exactly interpret the Q&A), which'd make the statement definitively false.

Re: OP, what's "instantaneous frequency"?

Here I address the asker instead of just the title.

Can someone clarify this a bit?

referring to making sense of "instantaneous frequency". The field that studies this is called "time-frequency analysis":

OP's integral has the right idea concerning "instantaneous", but wrongly implemented. A universal definition of instantaneous frequency is

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

where $x(t) = A(t)e^{j\phi(t)}$, and $x(t)$ can be made real by adding complex conjugate (doesn't change $f(t)$). The $A, \phi$ immediately run into uniqueness problems, and such $f(t)$ has limited "meaningfulness" beyond single-component signals. But, it's great for making sense of inst freq: once understanding the rotational nature of complex numbers, it's simply the speed of a phasor's rotation. One can also define a non-vanishing $\text{lim}$, but differently.

  • $\begingroup$ For a pure sine, as long as sampling above Nyquist, two points are sufficient to perfectly recover its frequency, phase, and amplitude. --- Is that true? If the two points were half a period apart, then I don't think the amplitude and phase are recoverable uniquely? $\endgroup$
    – Peter K.
    Commented Jul 20, 2023 at 23:34
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    $\begingroup$ That'd put them at Nyquist. $\endgroup$ Commented Jul 20, 2023 at 23:40
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    $\begingroup$ True! OK, so I'm still concerned that two data points can infer three values. That seems like an under-determined system? $\endgroup$
    – Peter K.
    Commented Jul 20, 2023 at 23:55
  • $\begingroup$ Indeed I remained unsure on that - counter. And the two bin solution divides by zero for N=2. But N=3 works fine, as I've linked, in my virtually exhaustive test. Complex sines still should work for N=2, as that's 4 data points. All this likely also has time-domain proofs. Thanks for pointing out. $\endgroup$ Commented Jul 22, 2023 at 21:57

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