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Assume that we have a complex signal $z(t) = x(t) + i y(t)$ which is band-limited ($x(t), y(t) \in \mathbb R$).

So, band-limited:
there exists a finite $B\in \mathbb R$ such that $\forall \omega \in (-\infty, -B) \cup (B, \infty)$, $X(\omega) = 0$ and $Y(\omega) = 0$ where $X(\omega)$ and $Y(\omega)$ are the Fourier transforms of $x(t)$ and $y(t)$ respectively.
In short, the Fourier transform of $z(t)$ is zero for all $\omega$ s.t. $\vert \omega \vert > 0$.

Now we can write $z(t) = r(t) e^{i \phi(t)}$ with $r(t), \phi(t) \in \mathbb R$ and $r(t) \ge 0$ ($\phi(t)$ doesn't have to be in an interval with the length $2\pi$) since $z(t)$ is complex.

For any band-limited signal $z(t)$, are both $r(t)$ and $\phi(t)$ band-limited
(are the Fourier transforms of $r(t)$ and $\phi(t)$ are zero for all $\omega$ s.t. $\vert \omega \vert > 0 $)?

Naetmul
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  • Isn't there a theorem stating that a band-limited signal must be smooth? If $z(t)=\sin t$ then $r(t)=\lvert\sin t\rvert$ is not smooth. –  Sep 20 '18 at 06:34
  • @Rahul A band-limited signal must be smooth. Yes, then the question's statement is false by the condition $r(t) \ge 0$. Thanks. – Naetmul Sep 20 '18 at 06:40

1 Answers1

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The absolute value of a band-limited function needs not be band-limited.

The reason for this is that a band-limited function is necessarily analytic, in the sense that it can be extended to an analytic function in a complex strip containing the real axis. But the absolute value of an analytic function is not necessarily an analytic function.

To make a concrete example, consider $z(t)=\cos t$. Its Fourier transform is $$ Z(\tau)=\frac12\left( \delta(\tau-1)+\delta(\tau+1)\right), $$ up to irrelevant factors of $2\pi$, that depend on the normalization you choose. So, $z$ is band-limited. However, $|z(t)|$ is not, because as you see here, its Fourier series contains infinite terms. Indeed, the Fourier transform of $|z(t)|$ is a Dirac comb $$ \sum_{n=-\infty}^\infty c_n\delta(\tau-n), $$ where $c_n$ are the terms of the Fourier series of $|z(t)|$. Since infinitely many of the $c_n$ are nonzero, $|z(t)|$ is not band-limited.

The same reasoning shows that the argument of a band-limited function needs not be band-limited. In the above example, the argument of $z(t)$ is the square wave $$ \phi(t)=\begin{cases} 0, & \cos t> 0, \\ \pi, & \cos t<0, \end{cases}$$ which is again a periodic signal with an infinite number of Fourier series terms. Reasoning as above, we see that $\phi$ is not band-limited.

Giuseppe Negro
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