# Hilbert transform

Given an $h(t)$ that is nonsingular at $t=0$ and that is a causal response so that $h(t)=0$ for $t<0$, then its Fourier transform,

$H(\omega )=R(\omega )\pm iX(\omega ),$

(where $\omega =$ angular frequency) and $i={\sqrt {-1}}$ has the special property known as the Hilbert transform, expressed by

$X(\omega )={\frac {-1}{\pi \omega }}\star R(\omega )={\frac {-1}{\pi }}{\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {R(y)\;dy}{\omega -y}}$

$R(\omega )={\frac {-1}{\pi \omega }}\star X(\omega )={\frac {-1}{\pi }}{\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {X(y)\;dy}{\omega -y}}$

where $P.V.$ denotes the Cauchy principal value at discontinuities. If $H(\omega )$ vanishes for $\omega <0,$ its Fourier transform,

$h(t)+ix(t),$

has $h(t)$ and $x(t)$ forming a Hilbert transform pair. $h(t)$ and $x(t)$ have the same amplitude spectrum but differ in phase by $90^{\circ }$. $[h(t)+ix(t)]$ is called the analytic signal belonging to $h(t)$, and $x(t)$ is the quadrature signal corresponding to $h(t)$. Often used in complex trace analysis (q.v.). Named for David Hilbert (1862–1943), German mathematician.

## 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

$f(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)\;dw}{w-z}}.$

Here $z=x+iy$ is a complex variable, as is $w={\mbox{Re}}\;w+i{\mbox{Im}}\;w$ and $f(z)={\mbox{Re}}\;f+i{\mbox{Im}}\;f.$ the function $f$ is analytic meaning that the derivative of $f(z)$ with respect to $z$ exists (and similarly the derivative with respect to $w$ of $f(w)$ exists) for some region bounded by the closed curve $C$ called the contour of integration.

### Real and Imaginary Parts

All we have to do now is to substitute for $f(x)$ and $f(w)$ and equate the real and imaginary parts. When we do this, we obtain expressions for the imaginary and real parts of $f(z)$ :

${\mbox{Im}}\;f(z)=-{\frac {1}{2\pi }}\oint _{C}{\frac {{\mbox{Re}}\;f(w)\;dw}{w-z}}.$

${\mbox{Re}}\;f(z)={\frac {1}{2\pi }}\oint _{C}{\frac {{\mbox{Im}}\;f(w)\;dw}{w-z}}.$

Thus, for an analytic function $f(z)$, 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

"Figure 1: the contour C for a 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 $-R\rightarrow x-\epsilon$ and from $x+\epsilon \rightarrow R$, plus an integral over the smaller semicircle of radius $\epsilon$, which will tend to have a vanishing radius as $\epsilon \rightarrow 0$ plus an integral over the larger semicircle $C$, which will tend to have infinite radius, as $|R|\rightarrow \infty$. Because the point $z=x+iy$ 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 $f(z)$, we may write Cauchy's integral formula as the sum of three integrals

$0=\oint _{C}{\frac {f(w)\;dw}{w-z}}=\lim _{{|\epsilon |\rightarrow 0}}\lim _{{|R|\rightarrow \infty }}\left(\int _{{-R}}^{{x-\epsilon }}+\int _{{x+\epsilon }}^{{R}}\right)\left({\frac {f(w)\;dw}{w-z}}\right)+\int _{{|z|=\epsilon }}{\frac {f(w)\;dw}{w-z}}+\int _{{C_{1}}}{\frac {f(w)\;dw}{w-z}}$

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 $f(z)=o(1)$ as as $R\rightarrow \infty$. That is, we assume that $|f(z)|$ vanishes as $R\rightarrow \infty$. This makes the contribution over the larger semicircle vanish as $R\rightarrow \infty$.

If the contour $C$ may be expanded to infinity, with $z$ to be somewhere on the real axis, then we may replace the integral over $C$ with a Cauchy Principal value integral along the real axis. The Residue Theorem, when applied to a semicircular contour passing under $z$ yields $-\pi i{\mbox{Res}}(f(w),z)$ which called a half-residue yielding the special form of the Cauchy integral formula in terms of the Cauchy principal value

$\lim _{{|\epsilon |\rightarrow 0}}\lim _{{|R|\rightarrow \infty }}\left(\int _{{-R}}^{{x-\epsilon }}+\int _{{x+\epsilon }}^{{R}}\right)\left({\frac {f(w)\;dw}{w-z}}\right)=i\pi f(z)$ .

Substituting $f(z)={\mbox{Re}}\;f(z)+i{\mbox{Im}}\;f(z)$ and correspondingly $f(w)={\mbox{Re}}\;f(w)+i{\mbox{Im}}\;f(w)$ we obtain the equivalent two expressions for the real and imaginary parts of $f(z)$ in terms of Cauchy principal value integrals

${\mbox{Im}}\;f(z)=-{\frac {1}{\pi }}{\mbox{P.V.}}\int {\frac {{\mbox{Re}}\;f(w)\;dw}{w-z}}$

and

${\mbox{Re}}\;f(z)={\frac {1}{\pi }}{\mbox{P.V.}}\int {\frac {{\mbox{Im}}\;f(w)\;dw}{w-z}}$ .

### Defining the Hilbert transform

The first expression defines the forward Hilbert transform, which we write substituting $t$ and $\tau$ for $z$ and $w$ as

${{\mathcal H}}(f(t))\equiv {\frac {1}{\pi }}{\mbox{P.V.}}\int _{{\infty }}^{{\infty }}{\frac {f(\tau )\;d\tau }{t-\tau }}.$

Here, the sign is naturally positive on the integral. The minus sign as been absorbed into the integrand by switching the order of $t$ and $\tau$ in the denominator of the integrand. This is done to yield the convolution form

${{\mathcal H}}(f(t))={\frac {1}{\pi t}}\star f(t).$

The proper form of the inverse Hilbert transform, is given by the second Cauchy principal value integral above as

$f(t)\equiv -{\frac {1}{\pi }}{\mbox{P.V.}}\int _{{\infty }}^{{\infty }}{\frac {{{\mathcal H}}(f(\tau ))\;d\tau }{t-\tau }}.$

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]

$-f(t)={{\mathcal H}}[{{\mathcal H}}(f(t))].$

### The Fourier representation of the Hilbert Transform

If we begin with the convolution form of the Hilbert transform

${{\mathcal H}}(f(t))={\frac {1}{\pi t}}\star f(t)$

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

${{\mathcal H}}(f(t))={\frac {1}{2\pi }}\int _{{-\infty }}^{{\infty }}{{\mathcal F}}\left({\frac {1}{\pi t}}\right){\hat {f}}(\omega )e^{{-i\omega t}}d\omega$

Here, ${{\mathcal F}}$ denotes the forward Fourier transform.

We now will calculate the Fourier transform of $1/\pi t$.

${{\mathcal F}}\left({\frac {1}{\pi t}}\right)=\int _{{-\infty }}^{{\infty }}{\frac {1}{\pi t}}e^{{i\omega t}}\;dt$

We will consider the contour integral

$I=\int _{{C}}{\frac {e^{{i\omega t}}}{t}}\;dt$

with a $C$ being a closed counter-clockwise path of integration that includes the origin. By the residue theorem, this integral is easily evaluated as

$I=2\pi i.$

"Figure 2: contour for t<0 and positive frequency"

### 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 $t=0$. In addition to this contribution, there is a a half-residue evaluation on the semi-circular arc due to the pole at $t=0.$ 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

$2\pi i={\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {e^{{i\omega t}}}{t}}\;dt+\pi i$

from which we may write,

$i={\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {e^{{i\omega t}}}{\pi t}}\;dt.$

Here we are considering $t$ to be a complex variable, such that $t=a+ib$. There are constraints on the values of $\omega$ such that this integral may exist. We note that the exponential part of the integrand may be rewritten as

$e^{{i\omega t}}=e^{{i|\omega |\operatorname{sgn}(\omega )(a+ib)}}=e^{{i\omega a}}\cdot e^{{-|\omega |\operatorname{sgn}(\omega )b}},$

where $a={\mbox{Re}}\;t$ and $b={\mbox{Im}}\;t$.

Here the factor of

$e^{{-|\omega |\operatorname{sgn}(\omega )b}}$

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 $t$ then the integral represents a function that is analytic in that half plane.

Hence, when ${\mbox{Im}}\;t<0,$ for the case when the integration contour closes in the upper-half-plane of $t,$ we must have $\operatorname{sgn}(\omega )=+1$ and when ${\mbox{Im}}\;t>0,$ when the contour closes in the lower-half-plane of $t,$ we must have $\operatorname{sgn}(\omega )=-1$ for the integral to exist.

"Figure 3: contour for t>0 and negative frequency"

When we close in the lower half-plane of $t$ the integration contour is clockwise, causing each of the residue evaluations to be negative and yielding

$-2\pi i={\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {e^{{i\omega t}}}{t}}\;dt-\pi i.$
$-i={\mbox{P.V.}}\int _{{-\infty }}^{{\infty }}{\frac {e^{{i\omega t}}}{\pi t}}\;dt$

Hence, the Fourier transform of $1/\pi t$ is

${{\mathcal F}}\left({\frac {1}{\pi t}}\right)=-i\operatorname{sgn}(\omega ).$

The appearance of the function $-i\operatorname{sgn}(\omega )$ 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 $i\operatorname{sgn}(\omega )$ is the analytic continuation of these two integrations. This function is valid for all $\omega$ except possibly $\omega =0.$

### Final result

Hence, for this choice of Fourier transform exponent sign conventions

${{\mathcal H}}(f(t))={\frac {1}{2\pi }}\int _{{-\infty }}^{{\infty }}i\operatorname{sgn}(\omega ){\hat {f}}(\omega )e^{{-i\omega t}}\;d\omega .$

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

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