3
$\begingroup$

If $\mathcal{H}$ denotes the Hilbert transform, the analytic signal of a signal $x(t)$ is $$x_a(t) = x(t) +\imath \mathcal{H}(x(t))\,.$$ The real and imaginary parts form Hilbert pairs.

Are there natural (and possibly useful) extensions, with another transform $\mathcal{T}$, which possesses involution, anti-involution or even semi-involution properties, see eg Is there a name for a function whose square is an involution?

A shape can be (for instance): $$x_b(t) = x(t) +\jmath \mathcal{T}(x(t)) +\jmath^2 \mathcal{T}^2(x(t))$$ of "Hilbert-like" triples? And $n$-uples? Or, if any, related to higher order algebras (quaternions, octonions). What properties can they reveal (generalized instantaneous phase, envelopes?)

I am mainly interested in constructions that, as least in some types of signals, in at least one application, have proven more efficient that state of the art.

$\endgroup$
  • $\begingroup$ More efficient for what application? $\endgroup$ – geometrikal May 13 '16 at 6:40
  • 1
    $\begingroup$ The paper Fourier Transform by Terence Tao might give you some leads (cited at the end of Harmonic_analysis) . $\endgroup$ – denis Nov 19 '17 at 11:21
3
$\begingroup$

The generalisation of the concept of an analytic signal is not straight forward. I'm quite certain however that looking for such a generalisation with quarternions (or even octonions) will not turn out fruitful. Those generalise complex numbers primarily algebraically, attempting to preserve as much of the field structure as possible, and not so much as a smooth geometric structure.

In my experience, the most useful generalisations of the complex numbers are usually either unitary Lie groups or Clifford algebras.

That brings me to my example, which I encountered in my work ten years ago. The mathematical background is unfortunately too vast to be laid out here, but I will attempt a sketch of the problem and explain how it relates to your question.

One of the key motivations of finding an analytic signal is the ability to extract instantaneous phase and amplitude. I will take this as the guiding theme for generalisation. In the following, I will refer to the Hilbert transform as the map from a real valued signal to its analytic counterpart and not the quadrature.

Consider the signal space $L^2(A,B)$ of square integrable functions from $A$ to $B$. The usual Hilbert transform is then a continuous (with respect to the natural topology) $\mathbb{R}$-linear shift invariant map $$H:L^2(\mathbb{R},\mathbb{R})\to L^2(\mathbb{R},\mathbb{C})$$

so that $\cos(\omega t)\mapsto \exp(i|\omega| t)$ for $\omega \in \mathbb{R}$.

We can slightly reformulate this by replacing $\mathbb{C}$ with $\mathbb{R}^+_0 \times \mathrm{U}(1)$, where $\mathrm{U}(1)$ is the unitary group of dimension 1. To be able to square-integrate that, we define the square operator $\square^2:\mathbb{R}^+_0 \times \mathrm{U}(1)\to\mathbb{R}$ so that $(a,\Phi)^2\mapsto a^2$ and $L^2(A,\mathbb{R}^+_0 \times \mathrm{U}(1))$ is well defined.

This is of course just the polar representation of the complex numbers and the Hilbert transform is still the same transform. However, with this reformulation, there is a nearly obvious generalisation. Replacing $\mathrm{U}(1)$ by a more general unitary group allows us to keep all the structure in place and have a sensible generalisation.

If we take $\mathrm{SU}(n)$ for the general case we also require an $n$-dimensional vector space for a faithful irreducible representation, so that we will allow signals in $L^2(\mathbb{R},\mathbb{R}^n)$ and the Hilbert transform is a continuous linear shift invariant map $$H:L^2(\mathbb{R},\mathbb{R}^n) \to L^2(\mathbb{R},\mathbb{R}^+_0\times \mathrm{SU}(n) )$$ The actual map however is not obvious. It is easier to first generalise how it will be undone. The original Hilbert transform can be undone by taking the real part of the analytic result. For the generalisation, we can simply apply the group element to a unit vector, multiply the result with the amplitude and take the real part. So we map $$(a \in \mathbb{R}^+_0,\Phi \in \mathrm{SU}(n)) \mapsto a\cdot\Re\left(\Phi(\mathbf{e}_1)\right)$$ with a canonical basis vector $\mathbf{e}_1\in\mathbb{C}^n$. It is easily verified that this reduces to the standard case for $\mathrm{U}(1)$ and $\mathbf{e}_1=(1)\in\mathbb{C}^1$.

The preimages of the orbits of $\mathbf{e}_1$ generated by any generator $g\in\mathfrak{su}(n)$ of $\mathrm{SU}(n)$ under $H$ are therefore $$\Re\left( \exp(igt)\mathbf{e}_1 \right)=\frac{1}{2}\left( \exp(igt)+\exp(-igt) \right)\mathbf{e}_1$$ where we used $g^\dagger = g$. This suggests that $H$ should map $$H:\frac{1}{2}\left( \exp(igt)+\exp(-igt) \right)\mathbf{e}_1 \mapsto \left(1,\exp(igt)\mathbf{e}_1\right)$$ And here appears a problem. The map is supposed to work for all generators $g$, so also specifically for $-g$. If we plug in $-g$, the left hand side remains unchanged, but the right hand side is mapped to its complex conjugate. Hence, the map is not well defined. It is this very problem that makes using the magnitude of $\omega$ mandatory in the original Hilbert transform map. The generators in the Lie algebra form a vector space, and there is no canonical way to map all generators to a half space.

We have to equip the Lie algebra with an additional structural element, and that is a continuous map $$|\cdot|:\mathfrak{su}(n)\to\mathfrak{su}(n)$$ with the property $|g|\in \{g,-g \}$ and $|g|=|-g|$ for all $g$.

With this we have the generalisation of the Hilbert transform to unitary groups, given by the map $$H:\frac{1}{2}\left( \exp(igt)+\exp(-igt) \right)\mathbf{e}_1 \mapsto \left(1,\exp(i|g|t)\mathbf{e}_1\right)$$

The geometric interpretation of this transform is that a vector valued signal is decomposed into a real non-negative amplitude signal and a rotation in a complex vector space. The rotation is so the corresponding point on the Lie manifold traces a path whose time-oriented tangent is always in the same half-space, generalising the concept of positive frequency.

This does not exactly answer your original question though. You can however embed a lower dimensional signal into $\mathbb{R}^n$ and then use this generalisation to find an $n$-dimensional Hilbert transform with any number of generalised components that you desire.

$\endgroup$
  • $\begingroup$ @LaurentDuval, you're welcome. I actually enjoy to talk a bit about the more mathematically abstract aspects of my work. May I ask what the motivation or inspiration for this question has been? $\endgroup$ – Jazzmaniac May 12 '16 at 20:46
  • $\begingroup$ So manifold I cannot summarize: redundancy, Hilbert tranform craze, and tetrahedral sensors google.com.ar/patents/WO2002068996A8?cl=en. Be happy to discuss that if interested $\endgroup$ – Laurent Duval May 12 '16 at 21:01
1
$\begingroup$

In this paper: "On the p-norm of the truncated Hilbert transform" by McLean and Elliot they say that the Hilbert transform is the only bounded integral operator on $L^p(\mathbb{R})$ that commutes with both translation and dilations.

This paper by King, "Operator basis for analytic signal construction", might be of interest as it talks about the above property and also gives an overview of a "fractional Hilbert transform".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.