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Given a function of the form: $$ f(x) = e^{i\phi(x)} | \phi(x)\in\Re $$ What are the properties of its Fourier transform? For instance, purely real functions have Fourier transforms with symmetric properties. Can anything similar be said about Fourier transforms of phase-only functions?

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  • $\begingroup$ Please precise the scope of work. Take $\phi(x)=0$, then the Fourier transform of $f$ is (up to a multiplicative constant) Dirac delta, so we work with distributions and not fnctions. $\endgroup$ – TZakrevskiy Nov 4 '13 at 11:00
  • $\begingroup$ This is primarily for engineering applications regarding "phase-only" signals. You can assume phi(x) being a continuous, "well-behaved" arbitrary signal. $\endgroup$ – David Nov 4 '13 at 11:13
  • $\begingroup$ The answer is no. $\endgroup$ – reuns Jun 23 at 2:35
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A "phase-only function" $f(x)$ as you call it can be equivalently expressed as a function for which $|f(x)|=1$ for all $x$. Or in other words for all $x\in\mathbb R$, $$ f(x)\cdot \overline{f(x)}=1.\qquad(1) $$ Let $g(y)$ denote the Fourier transform of $f(x)$, which is actually a tempered distribution for this class of functions (we will ignore this distinction for the rest of this paragraph). Since Fourier transforms interchange multiplication with convolution, intuitively $(1)$ should transform into $$ \delta(y)=g(y)*\overline{g(-y)}, $$ which can be interpreted (by definition of convolution) as saying that $g$ is orthogonal to all of its non-zero translates. Indeed, intuitively one has that $$ g(y)*\overline{g(-y)}=\int_{\mathbb R}g(t)\overline{g(t-y)}\ dt=\langle g(t),g(t-y)\rangle. $$ For this to "equal" the delta "function" means that, for $y\not=0$, the function $g(t)$ is orthogonal to its translate by $y$.

Now that we have the intuitive picture, we can proceed to formalize it. Let $\tau_z$ denote the operator that shifts a tempered distribution by $z$. (See this post by Terry Tao for a precise definition of what this means, along with a good outline of formal properties for working with tempered distributions.)

Claim. The Fourier transform of a "phase-only function" is a tempered distribution $g$ with the property that for all $z\in\mathbb R\setminus \{0\}$, there exists a sequence of Schwarz functions $g_n$ (possibly depending on $z$) converging to $g$ in the sense of tempered distributions, with the property that $$\lim_{n\to\infty}\langle \tau_z g,g_n\rangle=0.$$


Explicit examples.

  1. In the case of a linear phase $f(x)=e^{i(ax+b)}$, the Fourier transform is a multiple of a dirac spike, which obviously satisfies the condition (since any non-zero translate has the spike somewhere else, meaning that it is orthogonal to the first spike).

  2. In the case of a quadratic phase $f(x)=e^{-ix^2/2}$, the Fourier transform is a constant times $e^{iy^2/2}$, and we can see that it is orthogonal to its translate by $z\in\mathbb R\setminus \{0\}$ since we can take a sequence of functions $g_n$ by restricting $e^{iy^2/2}$ to the interval $[-\frac{2\pi n}{|z|},\frac{2\pi n}{|z|}]$ and obtaining $$ \langle e^{i(y+z)^2/2},g_n\rangle=\int_{-2\pi n/|z|}^{2\pi n/|z|}e^{i(y+z)^2/2}e^{-iy^2/2}\ dy=0, $$ since the integrand is a constant multiple of $e^{iyz}$, which has period $2\pi/|z|$.

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There are a class of filters called "All-Pass" filters that have unit magnitude and non-zero phase. They are often used to correct phase distortion introduced in other parts of the signal processing chain.

Due to the Fourier shift theorem, you are (in the time-domain) shifting the different frequency components. For example, if the phase is linear, you'll be shifting the frequencies with higher magnitude more.

There are numerous applications of these filters. You can read more about them on Wikipedia here: https://en.wikipedia.org/wiki/All-pass_filter

(I realize that these aren't mathematical properties, like symmetry, that you were requesting. But I thought it would be interesting to you nonetheless.)

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