Can a complex signal be one-sided (causal in time and positive only spectrum in frequency) in both domains?

I understand that a function can't have finite support in both domains, but what if both domains extend to infinity in just the positive direction?

The analytic signal $x_a = x + j \hat x$, where $\hat x$ is the Hilbert Transform of $x$, will be one-sided in the other domain. The impulse response of the Hilbert transform itself is non-causal and extends to $\pm \infty$, so even if $x$ is causal, the analytic signal itself cannot be causal. So we can rule out analytic signals, but these are not the only signals that can be one-sided in the other domain. (Pass any analytic signal through an all-pass signal and you will no longer have the Hilbert relationship between the real and imaginary components required of an analytic signal).

Generally any one-sided signal can be decomposed into even and odd components with symmetry such that when added only the positive half remains, and the even components if real will be real in the other domain, and the odd components if real will be imaginary in the other domain (and vice versa). So we seek to show that a more generalized complex one-side waveform which can be reduced into even and odd components does or can't possibly have transformed components that are also even and odd in the other domain. I can't yet get past this point to either find such a case that does exist, or clear proof that it can't possibly exist (with no restriction to real signals only etc).

Or is there a clearer proof or example of the ability to have a one-sided signal in both domains?

Clearly we can approximate obtaining single-side waveforms in both domains sufficiently for any practical use, to the same extent we are able to practically use Hilbert Transforms through delay and truncation and ignoring the error- so I'm not looking for that answer but the theoretical proof that it can't be done within the constraints of finite delay (or the example that can be).

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    $\begingroup$ I don't have an answer, but I'm wondering if one can be constructed using the functions that Clements & Pease used in their "On causal linear phase IIR digital filters" paper. The idea being that you start in the continuous domain and generate signals that, when sampled, have their zero crossings at the right place in the left half-plane in both domains. Not really an answer, so just putting it here as a comment. $\endgroup$
    – Peter K.
    Jun 14, 2021 at 14:06

4 Answers 4


The Fourier transform of a causal signal cannot be zero in any interval $[\omega_1,\omega_2]$ with $\omega_1<\omega_2$. This follows from the Paley-Wiener condition, which states that if $A(\omega)=|X(\omega)|$ is the magnitude of the Fourier transform $X(\omega)$ of a causal signal $x(t)$, then the integral

$$\int_{-\infty}^{\infty}\frac{\big|\ln A(\omega)\big|}{1+\omega^2}d\omega\tag{1}$$

must converge. Clearly, if $A(\omega)=0$ on some interval, the integral $(1)$ cannot be finite.

Note that the convergence of $(1)$ is a necessary and sufficient condition for $x(t)$ to be causal.

Consequently, there are no causal (one-sided) analytic signals, just as there are no causal ideal frequency-selective filters. Regarding the existence of causal signals with one-sided transforms, there is nothing special about analytic signals. As soon as the magnitude of the Fourier transform is required to be zero over some interval, it follows from $(1)$ that the corresponding signal cannot be causal. So we can generalize the terms causal and analytic to one-sided for both domains.

  • $\begingroup$ Matt I understood as I wrote that there are no causal analytic signals; I was looking for confirmation that this does (or doesn’t) extends beyond analytic signals (since it is not just analytic signals that have a one sided spectrum). Thanks for your insights. $\endgroup$ Jun 14, 2021 at 17:27
  • $\begingroup$ @DanBoschen Isn't a one-sided spectrum necessarily that of an analytic signal? Can always reflect etc. the spectrum to yield the signal's real part only. (If not, what's a counter-example?) $\endgroup$ Jun 14, 2021 at 17:47
  • $\begingroup$ @OverLordGoldDragon an analytic signal has a one-sided transform but not all one-sided signals are transforms of an analytic signal. I explain how/why in my question. (Basically the anayltic signal is the impulse response of a minimum phase system (for example) but we can have a non-minimum phase equivalent with the same magnitude response. $\endgroup$ Jun 14, 2021 at 17:50
  • $\begingroup$ (By simply cascading the system with an all pass that modifies phase and passes all magnitudes—- thus the imaginary term is no longer the Hilbert transform of the real term and therefore cannot be an analytic signal. $\endgroup$ Jun 14, 2021 at 17:52
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    $\begingroup$ @DanBoschen: Isolated zeros are generally not a problem for the convergence of the integral, and consequently, transforms of causal signal can have such zeros. However, the magnitude response cannot be zero over any interval of finite or infinite length. This precludes all one-sided transforms, no matter whether they are analytic or not. $\endgroup$
    – Matt L.
    Jun 15, 2021 at 12:25

I thought about this a bit, and I wondered whether something that numerically approximates the wished-for signal might be good. So I implemented a project onto convex sets (POCS) approach to synthesizing such a signal.

The idea is:

  1. Initialize with a random signal.
  2. Take the FFT.
  3. Zero out the upper coefficients.
  4. Take the inverse FFT.
  5. Zero out the upper coefficients.
  6. Repeat steps 2 to 5 as many times as you want (I chose 100).

One example of the absolute value of the resulting signal is plotted below (along with the absolute value of its FFT) on a decibel scale.

Absolute value of resulting signal and its FFT, decibel scale.

Code Only Below

import numpy
import matplotlib.pyplot as plt
from scipy.fft import fft
from scipy.fft import ifft
import random 

def get_noise(N):
    noise = []
    for i in range(N):
    return noise

def project(x):
    N = len(x)
    high_half = numpy.arange(N/2+1,N).astype(int)
    X = fft(x)
    X[high_half] = 0
    x = ifft(X)
    x[high_half] = 0
    return x

N = 1024
x_init = get_noise(N)
x = x_init
for i in numpy.arange(1,100):
    x = project(x)

X = fft(x)

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    $\begingroup$ Nice Peter- this made me think if there then can actually be a discrete case since we can come up with cases for that which would still comply with Matt’s well stated Paley-Wiener condition. Which leases to the next and related question I will post! $\endgroup$ Jun 15, 2021 at 16:02
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    $\begingroup$ Ha, I also used that idea in my answer to the discrete case question and earlier for eigenvectors of DFT. Thanks for the POCS pointer. I only got it to converge in reasonable time for DFT sizes of up to 20 or so. $\endgroup$ Feb 27, 2023 at 20:19
  • $\begingroup$ @OlliNiemitalo POCS is a nice approach when nothing much else works or where you just have easily enforceable constraints. $\endgroup$
    – Peter K.
    Feb 28, 2023 at 1:12

Disclaimer: I understand the question as "is there an $x(t)$ that's $x(t<0)=0$ with $X(\omega < 0) = 0$". I present some ideas rather than proofs.

Approach 1

I'll answer in terms of the inverse Fourier transform.

We seek some $X(\omega)$ such that the complex sinusoids it weighs, in summation (integration), cancel to zero for $t < 0$ - but also $X(\omega < 0) = 0$. That is, both real and imaginary components of this sum are zero.

Let's focus on imaginary. Parts of cisoid that sum to imaginary part are the sine for real part of $X(\omega)$, and cosine for imaginary part of $X(\omega)$ - call them $R(\omega), I(\omega)$. Then we seek some (real) $R$ and $I$ such that, for $t < 0$,

$$ \int_{0}^{\infty} \left(R(\omega)\sin(\omega t) + I(\omega)\cos(\omega t)\right) d\omega = 0 \tag{1} $$


$$ \int_{0}^{\infty} R(\omega)\sin(\omega t)d\omega = - \int_{0}^{\infty} I(\omega)\cos(\omega t) d\omega. \tag{2} $$

Since sines and cosines are orthogonal, there can be no $R, I$ that yield $(2)$ over any interval $(t_0, t_1)$.

We also cannot have

$$ \int_0^\infty R(\omega) \sin(\omega t) d\omega = 0 $$

over any interval $(t_0, t_1)$ since sine over any frequency interval (/differential) $(\omega_0, \omega_1)$ is orthogonal to sine over any other interval (for $\omega > 0$), i.e. different frequencies cannot cancel each other over a continuous interval. What allows FT to ever be continuously zero in the first place is the negative frequencies.

In summary: no, because cosines and sines of same frequency, or sines of different frequencies, cannot cancel over a continuous interval - thus neither can any their possible sum. This argument repeats for the real part, and for forward FT - hence concludes the 'proof'.

Approach 2

Such an $x(t)$ would be equivalent to multiplying in time by a complex unit step, whose FT is

$$ U(\omega) = (1 + 1j) \cdot (\pi \delta (\omega) + 1/j\omega) $$

So we seek some $F(\omega)$ such that, when convolved with $U(\omega)$, is $0$ for all $\omega < 0$. The infinite tail doesn't make it easy... but I can't tell if such a $F(\omega)$ cannot exist.

Approach 3

Why not just make a causal real signal analytic? Suppose

$$ x(t) = e^{-t} u(t) \Leftrightarrow X(\omega) = \frac{1}{1 + j\omega} $$

then ... (ignoring $\omega = 0$ for now)

$$ X_a(\omega) = \frac{2}{1 + j\omega} U(\omega) $$

What's the inverse FT of this? Wolfram won't tell.

Though it won't guarantee the imaginary part to be zero, the real has to be (invalidaing Approach 1). An idea I've had, if it's possible to force one component to zero, then it'll take both $R$ and $I$ to do so, exhausting degrees of freedom to also force the other component to zero.

  • $\begingroup$ I like your approach. But it seems this would disprove the existence of any causal signals in time, following the same logic yet we know they exist and easily proved through the decomposition into the sum of even and odd non-causal functions in time (which leads us to the analytic signal in frequency: analytic signal in one domain = single-sided in the other domain). What's different about an analytic signal in frequency that makes this logical process not apply? $\endgroup$ Jun 14, 2021 at 21:59
  • $\begingroup$ (To be clearer- it appears your logical flow would disprove the existence of any one-sided waveforms in time even if we leave out the "but also $X(\omega<0)=0$, but we know this is not the case and can easily come up with many counter-examples; the set of all causal waveforms in time. I'm sure I missing something simple to come to that conclusion, just trying to see that) $\endgroup$ Jun 14, 2021 at 22:50
  • $\begingroup$ @DanBoschen But an analytic signal is infinite, not one sided? Also I think my answer is wrong; can't we make an "analytic square wave"? Then it's possible to get a "continuous zero", though I'm validated in the sense that it's an "energy zero" (Gibbs effect). This made me realize another perspective, added. $\endgroup$ Jun 14, 2021 at 22:52
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    $\begingroup$ But read my initial question carefully where I suggest the pitfall with that approach (at least as far as I could take it). It seems as soon as you make a signal analytic, it can never be causal. When we do this in practice we only approximate it through delay and truncation. This is why I said theoretically rather than practically (I can work with waveforms of infinite support as there values become insignificant but that is not what I am referring to here, but rather a strict theoretical / mathematical existence or proven that it is not possible). $\endgroup$ Jun 14, 2021 at 22:57
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    $\begingroup$ I really appreciate the thought digging. I suspect it's not possible but couldn't find this bottom lined anywhere and couldn't find my way to a conclusion (going down similar paths). Certainly it is equally impossible to even take the Hilbert Transform so an answer may be to the extent we say we can create analytic waveforms (which we do all the time in practice), we can create causal waveforms in time and frequency. This may be an important practical vs theoretical distinction, but I am now looking for the theoretical foundation similar to finite in one domain must be infinite in the other. $\endgroup$ Jun 14, 2021 at 23:07

This is not an answer, rather a rationale I am following.
(I know there is the "comment" functionality, but I don't have unlocked it yet)

Anyway, I would shortly describe my observation as: your answer reduces in determining whether a generic signal (not necessarily an analytic one) with finite support in time/frequency can have one-sided infinite support in frequency/time.

Let me clarify. Consider a time domain signal $x(t)$, even in time, centered at the origin and with finite duration $2T$, then its one-sided version in time (namely, causal) would be $$x(t-\tau), \; \tau>T.$$ We all know, from the FT's properties, that $$x(t-\tau) \Longleftrightarrow X(f) e^{-j2\pi f\tau}.$$ The complex exponential has infinite support extending from $-\infty$ to $+\infty$, hence the question reduces to the support of $X(f) = \mathfrak{F}\left\{\,x(t)\,\right\}$. As far as I know, being $x(t)$ a windowed version of some signal extending for the whole $t$-axis, the convolution with the $\mathrm{sinc}$ (considering a $\mathrm{rect}$ window) in the frequency domain would result in a support extending from $-\infty$ to $+\infty$. Clearly this is not the theoretical proof, but my aim was to point out that the kind of proof you are looking for may be different from the one you initially thought.

I don't expect the relaxation of some of my hypothesis on $x(t)$ (e.g., time-even) to have an impact on the core issue.

As usual, things should work also the other way (i.e., considering $X(f-\xi)$ as starting point).

  • $\begingroup$ Yes agreed, a delay in time simply adds a linear phase in frequency but doesn't modify the position in frequency of the waveform or the magnitude response or create new frequencies (so just adds phase shift). So I am with you through that point. However after that the explanation assumes a rectangular window (as it states) but we're left with the question or proof that there cannot be any window in time that can be one sided in frequency. Well that window and showing that it exists or doesn't can also be complex with arbitrary magnitude and you see we are right back to the first question! $\endgroup$ Jun 15, 2021 at 12:11

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