Given an that is nonsingular at and that is a causal response so that for , then its Fourier transform,
(where angular frequency) and has the special property known as the Hilbert transform, expressed by
where denotes the Cauchy principal value at discontinuities. If vanishes for its Fourier transform,
has and forming a Hilbert transform pair. and have the same amplitude spectrum but differ in phase by . is called the analytic signal belonging to , and is the quadrature signal corresponding to . Often used in complex trace analysis (q.v.). Named for David Hilbert (1862–1943), German mathematician.
- 1 A Derivation of the Hilbert Transform
- 2 References
A Derivation of the Hilbert Transform
There are varied notations and varied sign conventions depending on choices that may be made, but this has no effect on the usage of the Hilbert transform, as long as consistency is maintained.
The Cauchy Integral Formula
Our derivation here begins with the Cauchy Integral Formula, which is a standard result from the theory of functions of a complex variable
Here is a complex variable, as is and the function is a complex-valued function of . The fuction is analytic meaning that the derivative of with respect to exists (and similarly the derivative with respect to of exists) for some region bounded by the closed curve called the contour of integration.
Real and Imaginary Parts
All we have to do now is to substitute for and and equate the real and imaginary parts. When we do this, we obtain expressions for the imaginary and real parts of :
Thus, for an analytic function , there is a relationship between its real and imaginary parts. These formal expressions provide a context for the Hilbert transform.
Integration contour and the Cauchy Principal Value integral
By the Cauchy's Theorem we may deform an integration contour in the complex plane any way we want as long as we do not try to cross a pole, branch point, or other singularity, which are the places where the integrand fails to be analytic. See Figure 1.
Here we consider the integration contour to consist of a portion along the real axis from and from , plus an integral over the smaller semicircle of radius , which will tend to have a vanishing radius as plus an integral over the larger semicircle , which will tend to have infinite radius, as . Because the point is a singularity, the portion of the integration contour along the real axis can only be made sense of as a Cauchy Principal Value evaluation.
Returning to the complex-valued function , we may write Cauchy's integral formula as the sum of three integrals
The 0 results by Cauchy's theorem from the fact that there is no pole inside the full contour.
We now restrict the evaluation to functions as as . That is, we assume that vanishes more quickly than as . Because the arclength of the larger semicircle grows as a function of this restriction guarantees that the contribution of the integral over the larger semicircle vanishes as .
If the contour may be expanded to infinity, with to be somewhere on the real axis, then we may replace the integral over with a Cauchy Principal value integral along the real axis. The Residue Theorem, when applied to a semicircular contour passing under yields which called a half-residue yielding the special form of the Cauchy integral formula in terms of the Cauchy principal value
Substituting and correspondingly we obtain the equivalent two expressions for the real and imaginary parts of in terms of Cauchy principal value integrals
Defining the Hilbert transform
The first expression defines the forward Hilbert transform, which we write substituting and for and as
Here, the sign is naturally positive on the integral. The minus sign as been absorbed into the integrand by switching the order of and in the denominator of the original integrand. This is done to yield the convolution form
The proper form of the inverse Hilbert transform, is given by the second Cauchy principal value integral above as
It is common that the same integral definition is used for both forward and inverse transforms. The double application of the Hilbert transform yields the negative of the original expression
The Fourier representation of the Hilbert Transform
If we begin with the convolution form of the Hilbert transform
we know from the results on convolution that convolution in the time domain is multiplication in the frequency domain, such that we may write the Hilbert transform formally as
Here, denotes the forward Fourier transform.
We now will calculate the Fourier transform of .
We will consider the contour integral
with a being a closed counter-clockwise path of integration that includes the origin. By the residue theorem, this integral is easily evaluated as
Choice of contour
As in the previous discussion, we deform the contour of integration to create a real-valued integral along the real axis. This integral makes sense as a Cauchy Principal Value integral. That is, we avoid the singularity in the integrand at . In addition to this contribution, there is a a half-residue evaluation on the semi-circular arc due to the pole at by performing a limiting process on either side of the singularity.
The third part is an integral over the contour the larger semi-circle, but this vanishes by a result known as known as Jordan's Lemma
from which we may write,
Here we are considering to be a complex variable, such that . There are constraints on the values of such that this integral may exist. We note that the exponential part of the integrand may be rewritten as
where and .
Here the factor of
represents an exponential decay if the exponent is negative, and an exponential growth if the sign of the exponent is positive. Because we are deforming the contour so as to send it to infinity, we must have a decaying integrand. Otherwise, the integrand will blow up, and the integral will not exist. If the integrand does not blow up in a given half plane of then the integral represents a function that is analytic in that half plane.
Hence, when for the case when the integration contour closes in the upper-half-plane of we must have and when when the contour closes in the lower-half-plane of we must have for the integral to exist.
When we close in the lower half-plane of the integration contour is clockwise, causing each of the residue evaluations to be negative and yielding
Hence, the Fourier transform of is
The appearance of the function may seem a bit mysterious. Using this function is the only way to combine the result from both integrations into a unified expression. Each integration defines a result that is only good in the respective half-plane where the respective integral exists. The function is the analytic continuation of these two integrations. This function is valid for all except possibly
Hence, for this choice of Fourier transform exponent sign conventions
This result is without the minus sign. If we were to define the forward and inverse Fourier transform with the opposite exponent sign convention, then the result would have a minus sign.
- Titchmarsh, E (1948), Introduction to the theory of Fourier integrals (2nd ed.), Oxford University: Clarendon Press (published 1986), ISBN 978-0-8284-0324-5.