One-sided waveforms in both time and frequency?

Question

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).

Answer 1

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.

Answer 2

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.

---

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):
        noise.append(random.gauss(0,1))
    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)

plt.plot(20*numpy.log10(abs(x)))
plt.plot(20*numpy.log10(abs(X)))

Answer 3

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 $t < 0$,

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

or

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

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.

Answer 4

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).