Why does every single band-limited signal in frequency have an infinite time domain and vice versa (As it's a symmetric relation, inf in one is finite in the other).

I understand how a digital signal with an infinite slope needs infinite amount of sine waves to approximate or achieve, hence it being infinite in the frequency domain. However, what about any terminating curve with finite slope throughout?

Either an intuitive or a mathematical understanding would be helpful

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    \$\begingroup\$ I think this might be better answered on the Math SE. You'll want to verify this, but if a single frequency sinusoid is truly periodic and therefore lasting for all time, the only single frequency that could cancel it out is the same sinusoid 180 degrees out of phase. But obviously that won't work since you need parts of it to be non-cancelling to contribute to your waveform existing for a finite amount of time. So it could be that you need a great many other sinusoids to cancel out that first sinusoid, though you don't want perfect cancellation; just perfect outside of a time interval. \$\endgroup\$
    – DKNguyen
    Commented Aug 8, 2021 at 17:44
  • \$\begingroup\$ Ideal impulse contains all frequencies, doesn't it? Though such a signal is not physically realizable. \$\endgroup\$
    – SteveSh
    Commented Aug 8, 2021 at 18:07
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    \$\begingroup\$ "inf in one is finite in the other" Not necessarily. There are signals that are infinite in both domains. For example, a Gaussian pulse is its own Fourier transform. The only thing you can say is that there are no signals that are finite in both domains. \$\endgroup\$
    – Dave Tweed
    Commented Aug 8, 2021 at 18:33
  • \$\begingroup\$ I would think that all signals are time and frequency limited, which is easily demonstrated and measured. Also, we cannot prove anything is infinite. But we can say that in theory, just as the case just as a singularity is infinite bandwidth and amplitude in zero time exists but we cannot demonstrate. \$\endgroup\$ Commented Aug 8, 2021 at 18:43
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    \$\begingroup\$ When you connect this statement with reality, it's not really true in any practical sense. A gaussian has a Fourier transform that is also a gaussian. Although a gaussian is in principle a function that extends to infinity on both sides, in reality its tails fall off extremely fast, and are therefore undetectable past a certain point. So you can certainly have a signal that is finite in the time domain for all practical purposes, and also finite in the frequency domain for all practical purposes. \$\endgroup\$
    – user46399
    Commented Aug 9, 2021 at 2:18

3 Answers 3


Many texts prove that a signal cannot be both time limited and bandlimited. It is quite a deep result and depends on complex analysis, but the shortest proof I know starts with a bandlimited signal \$f(t)\$. It is straightforward to show from the fourier transform that being bandlimited means that \$f(t)\$ is analytic over the entire complex plane, hence if it vanishes on any interval (e.g. \$f(t) = 0 \mathrm{~for~} t>T\$), then it vanishes everywhere. Hence a bandlimited signal cannot be time-limited. The converse is identical. You will find this stuff in more rigorous texts (e.g. Papoulis).

A more handwaving argument would be that if it is bandlimited then you don't change it by multiplying the spectrum by a rectangular window, hence you don't change it in the time domain by convolving with a \$\mathrm{sinc}\$ function - this tends to spread it outside any time window you assume it is contained in.

Your question also touches on discontinuities and slopes. There are a lot of useful results about the rate of roll off depending on the level of discontinuity. From memory, a function with step discontinuities has a spectrum that falls off at \$\frac{1}{f}\$, a continuous function with step discontinuities in the first derivative goes as \$\frac{1}{f^2}\$ and so on. The smoother the function (more continuous derivatives) the more rapid the fall off of the spectrum, but also the more spread out on time.

Some problems can benefit from some of the serious maths out there. For example, given that a signal must be constrained to a certain bandwidth, what waveform concentrates most of the signal energy into a certain time interval? The solution to this is the prolate spheroidal functions — see Papoulis's book Signal Analysis.

  • \$\begingroup\$ +1! Thanks. The uniform convergence of continuous functions must be continuous, but there's no requirement that this uniform limit of differentiable functions be differentiable. The canonical Weierstrass function is one illustration where the derivatives aren't close to uniform convergence. Denise Nocoletti discusses its transform in Properties of the Weierstrass Function in the Time and Frequency Domains. \$\endgroup\$
    – jonk
    Commented Aug 8, 2021 at 22:15
  • \$\begingroup\$ Well, crap. It's still better now. Thanks again. \$\endgroup\$
    – jonk
    Commented Aug 9, 2021 at 0:25
  • \$\begingroup\$ Why is this a handwavy argument? -> "A more handwaving argument would be that if it is bandlimited then you don't change it by multiplying the spectrum by a rectangular window, hence you don't change it in the time domain by convolving with a sinc function - this tends to spread it outside any time window you assume it is contained in." \$\endgroup\$ Commented Aug 9, 2021 at 10:00
  • \$\begingroup\$ @aravindh-krishnamoorthy: it sounds reasonable, but can you prove it is true? We know it is true from the first para, but can you prove that it is true using simpler arguments? Can you guarantee that some mathematician can't come up with some pathological function (like the sorts of things jonk threw in), that breaks it? I'm not saying you can't, maths is full of proofs that favour simplification over elucidation, I'm just offering the types of proof you will find in rigorous texts. \$\endgroup\$
    – Tesla23
    Commented Aug 9, 2021 at 10:29
  • \$\begingroup\$ @Tesla23 Thank you. I agree with your sentiment -- the second statement is easier to understand but the proof is no simpler. Nevertheless, in this particular case, both statements are equivalent and can be proven to be so. \$\endgroup\$ Commented Aug 9, 2021 at 10:36

why does a single band-limited signal in frequency have an infinite time domain

The assumption here is that if the time was limited it might have a discontinuity and thus an infinite rise time if the samples are ideal.

But this is not the case in real systems with BW limited, so it is assumed to be "steady-state".

Therefore the normal case when a system is analyzed with time-boundaries and bandwidth limits we ignore any discontinuities at the end of the "steady-state"

This is analogous to a simple low pass filter with a unit step function applied. In theory the step can be infinite or finite rise time and the exponential never reaches the unit voltage, but in practical terms with tolerances, the experiment duration can be stopped at 10 T=10RC.

At this point=10T, the residual error is about 144 PPM and the dV/dt has reduced the risetime to and spectrum BW or the peak risetime t=0.115% so could be captured with high accuracy with ~142x the -3dB BW.

So in theory yes, you cannot have simultaneous limited time and spectral Fourier BW but if you have an error tolerance, you can have both.

  • \$\begingroup\$ That's a good point. If the signal truly ended then there should be points at the start and end of the signal where it goes from "some signal" to literally zero signal and if you look closely enough there's a sharp discontinuity there...I think...I don't know if that applies to a continuous-time signal that grows and then decays asymptotically. It certainly makes sense for sampled signals though. \$\endgroup\$
    – DKNguyen
    Commented Aug 8, 2021 at 17:51
  • \$\begingroup\$ I don't quite understand - do you mean then that if there is no discontinuity in time, a signal could be both time and band limited? Also, could you expand on the latter half of your answer about BW limited real systems and steady state? \$\endgroup\$ Commented Aug 8, 2021 at 18:00
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    \$\begingroup\$ @DKNguyen It does not apply to continuous-time signals in general, as bump functions exist. Bump functions by definition go to zero quite smoothly (they're continuous, and so are all of their derivatives) and remain zero out to infinity, so one of those as your signal would have no such discontinuity. \$\endgroup\$
    – Hearth
    Commented Aug 8, 2021 at 20:11

Just to add to other answers here, a concise way to precisely mathematically state this property is to say that "the Fourier transform of a compactly supported L^2 function on R^n is holomorphic, so---if nonzero---is never compactly supported". This is a part of the Paley–Wiener theorem in functional analysis, which you can look up for a proof.


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