In response to a comment I posted recently to What's behind light?, @Emilio Pisanty wrote a statement that effectively asserts that the Fourier transform of a pulse provides results that are unphysical. His argument had two sub-assertions offered as proof:

  1. If you add a sharp filter then you'll get a long quasi-monochromatic 'tail' behind the pulse, but that's not the pulse itself you're seeing, it's the inertia of the filter itself, which takes a long time (inversely proportional to its bandwidth) to wind down.

  2. no matter what you do, all physical filters are causal, which means that they will never produce light in front of the pulse.

Both are excellent points. However, they both suggest that Fourier transforms provide unphysical results - at least in this case. This is a bit distressing, because Fourier transforms are used throughout physics and it seems that they are considered to contain exactly the same information as, e.g., a straight temporal representation.

My question has two parts:

  1. When is it improper to consider Fourier components to be physically real?
  2. What experiment can demonstrate that, as Emilio said:

If you add a sharp filter then you'll get a long quasi-monochromatic 'tail' behind the pulse, but that's not the pulse itself you're seeing, it's the inertia of the filter itself, which takes a long time (inversely proportional to its bandwidth) to wind down.

Two specific cases come to mind, using femtosecond pulses which, as we know, can be filtered to produce longer, nearly-arbitrarily shaped pulses. In the first case, a diffraction grating spatially separates the femtosecond pulse into its spectral components. An aperture selects one or more sharp frequency bands which are focused by a lens back onto another identical diffraction grating. The components are all superimposed upon emerging from the diffraction grating, thereby re-forming the pulse, sans the frequency components that have been removed. The resulting pulse can be much longer than the original pulse.

The second case is the same as the first, but instead of diffraction gratings, prisms are used to separate the pulse into its frequency components and to recombine them into a resulting pulse after an intervening aperture removes some of the components.

The resultant lengthened pulse is identical in the two cases, if the aperture removes precisely the same frequency components. It should not matter how far the components are separated in the spatial domain, nor what material the apertures are made of, nor what material the gratings or prisms are made of, nor even what shapes the prism are or the pitch of the gratings, as long as the geometry of the setup is adjusted accordingly. The easiest way to understand both of those experiments is to imagine that the optics form a temporal Fourier transform, perform filtering in the Fourier domain, then form an inverse Fourier transform.

In order to bring "inertia of the filter itself" into the picture, it seems necessary to define carefully just what part(s) of each apparatus constitute the "filter", and explain why the filter's inertia is the same in the two cases.

  • 1
    $\begingroup$ Since you only have access to a finite, discretely sampled version of whatever it is you're measuring, a DFT assumes a periodic extension. Thus the need for tapers. Further it's not clear when you're talking about the FT and the periodogram. I assume you know about spectral leakage, variance-bias trade-off. What are you really asking? If you just look at the periodogram of a signal there will be lots of unphysical artifacts caused by the implicit box-car filter. $\endgroup$
    – JohnS
    Commented May 8, 2018 at 19:06
  • $\begingroup$ That's a good point: that if you sample a pulse for a finite length of time, you can't say for sure that the pulse doesn't repeat periodically so it's improper to assume it doesn't. Still, in a physical experiment it is relatively easy to ensure that the pulse cannot repeat periodically. One could ask what physical artifacts are created by diverting one pulse from a series of pulses emitted by a mode-locked laser. The Fourier transform of the solitary pulse (measured over a long time) contains spectral components that are not present in the series of pulses. $\endgroup$
    – S. McGrew
    Commented May 8, 2018 at 19:53
  • $\begingroup$ For example: Suppose we have a very long modelocked laser that emits a 5-femtosecond pulse every 10 nanoseconds. We can set up an electro-optical shutter that is open for anywhere from, say, 20 femtoseconds to 18 nanoseconds, timed so that the pulse passes through right in the middle of that interval. Does the pulse itself have different spectral components depending on the width of the interval? Could that be true if there is nothing between the pulses emitted by the modelocked laser? $\endgroup$
    – S. McGrew
    Commented May 8, 2018 at 20:14
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    $\begingroup$ "Both are excellent points. However, they both suggest that Fourier transforms provide unphysical results" That's really not what Emilio said. He said that there exist some filter functions you can write down but that are not physically realizable. That's not the same thing as saying that a "Fourier transforms provide unphysical results". $\endgroup$
    – DanielSank
    Commented May 8, 2018 at 20:45
  • 1
    $\begingroup$ ↑ Daniel's interpretation is much closer to the mark than yours, I think. $\endgroup$ Commented May 8, 2018 at 21:09

2 Answers 2


I assume you're talking about discrete Fourier Transforms. We never have access experimentally to the underlying continuous function (the pulse, say). So we can never know the power spectrum exactly. The artifacts you mention are not in the experiment, but arise from the discrete Fourier transform.

An inherent property of discrete Fourier transforms is that the computation actually yields the Fourier amplitudes for a time series of infinite length consisting of repeated copies of the observed time series, suitably normalized to make the results bounded. Thus the Fourier transform we actually compute is for the series which has repeated discontinuities at the points where the original, finite-length series repeats. These jumps are unphysical: they are a consequence of the DFT algorithm. In addition to the spectral leakage, the periodogram is a biased estimate of the underlying power spectrum. It's at one end of the variance-bias tradeoff. For more see the first two references.

The first figure shows an example of two power spectral estimates from a noisy interferometer measurements of an oscillating target: the periodogram and the result of Thomson's multi-taper approach. Thomson's method lends itself readily to computing confidence intervals. So you can test the hypothesis that a certain peak is real or not. Clearly the multi-taper estimate has a lot less variance.

periodogram vs multi-taper

The second big point is that naive filters can easily violate causality. The second figure is based on an example by Toll (reference below). It shows at top a causal wavepacket (source at t=0, say) made via a weighted sum of sinusoids all of which extend from minus infinity to infinity. If you remove a single component (middle figure) from this sum without phase shifting the other components you get the non-causal result at the bottom. The moral to the story is that you can't have attenuation without dispersion. Moreover, the converse is true as well; namely, a phase shift of one frequency is necessarily accompanied by an absorption at other frequencies.

Toll causality

Thomson, D. J. (1982) "Spectrum estimation and harmonic analysis." Proceedings of the IEEE, 70, 1055–1096

Percival, D. B., and A. T. Walden. Spectral Analysis for Physical Applications: Multitaper and Conventional Univariate Techniques. Cambridge: Cambridge University Press, 1993.

Toll J S 1956 Causality and the dispersion relation: logical foundations Phys. Rev. 104 1760–70

  • $\begingroup$ I assume that by "naive filters" you mean filters that are not physically possible. Your statement, "If you remove a single component (middle figure) from this sum without phase shifting the other components you get the non-causal result at the bottom" confuses me. Are you saying that in a spectral filter that passes all but a very narrow band of frequencies, all the frequencies that are passed inevitably end up getting phase shifted? $\endgroup$
    – S. McGrew
    Commented May 9, 2018 at 4:07
  • $\begingroup$ @S.McGrew I didn't say all the frequencies would be shifted but some must be or you'll violate Kramers-Kronig. By naive I meant literally that. I've seen people arbitrarily assign an absorption based on some simple idea, only to discover later via KK that their simple idea violates causality. $\endgroup$
    – JohnS
    Commented May 9, 2018 at 18:54
  • $\begingroup$ I'm imagining a spectrometer consisting of a grating and a lens. At the focal plane there is an aperture mask (e.g., lamp black on glass). Following that there is a thin retroreflector array that reverses all beam paths. The filtered beam is separated from the input beam using a beamsplitter. I'm not sure if the retroreflector array would be an issue, but am guessing that scaling the system larger will reduce effects of the array elements' sizes and arrangement. Not expecting anything acausal; just wondering how to analyze it without invoking non-physical math objects. $\endgroup$
    – S. McGrew
    Commented May 9, 2018 at 20:28

Indeed, not all possible functions $\tilde{f}(\omega)$ correspond to Fourier transforms of functions that respect causality. The Kramers-Kronig relations impose constraints on the set of such Fourier-transformed functions.

  • $\begingroup$ Signals themselves are blind to causality - they're just given, and that's it. It is transfer functions that really matter, as they represent actual physical transformations, and that are required to be causal (and which are thus restricted in the frequency-domain forms they can take). Most of OP's concern (and particularly in the linked question), as I read it, is with signals, rather than transfer functions. In either case, it's worth drawing up the distinction with a bit more care. $\endgroup$ Commented May 8, 2018 at 21:17

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