# Hilbert transform

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.

## Contents

## 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 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

Formally we may write:

indicating that 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 Integral 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.

Here we consider the contour to be an integration along the real axis. Because the point is a singularity, this integral can only be made sense of as a Cauchy Principal Value evaluation, where real axis contribution is sum of the limits from the left plus the limits from the right on the real axis. This is represented by the small semi-circular portion of the contour passing under the pole. The contour is closed over the larger semicircular arc which is of radius See Figure 1. The resulting integrals are equivalent to the formal expressions above

Here the *const.* is a possible contribution from the larger semi-circular part of that is at distance from
the origin. The factor of that appeared on the original Cauchy integral equation no longer appears,
as the value of the Cauchy principal value integral is equivalent to a a *half residue* evaluation at the pole at .

We now restrict the evaluation to functions as as . That is, we assume that as . This makes the contribution over the larger semicircle 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*. Note that the minus sign has been absorbed into the denominator.

Thus, we have a definition of a forward Hilbert transform 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 integrand. This is done to yield the convolution form

The proper form of the inverse Hilbert transform, according to this derivation, is

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^{[1]}

### 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, owing to 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 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,

As we are considering to be a complex variable, 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

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.

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

When we close in the lower half-plane of the integration contour is clock-wise, 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 contination of these two integrations. This function is valid for all except possibly

### Final result

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.

## References

- ↑ Titchmarsh, E (1948), Introduction to the theory of Fourier integrals (2nd ed.), Oxford University: Clarendon Press (published 1986), ISBN 978-0-8284-0324-5.