13 Hilbert Transform
13.1 Representation of the Potential Gradient in the Complex Plane
Let \(u\) and \(v\) denote the real and imaginary parts of a complex-valued function of the complex variable \(z = x + i y\), with \(f(x + iy) = u(x,y) + iv(x,y)\). The functions \(u\) and \(v\) are real-valued and differentiable at every point in an open subset \(\Omega \subset \mathbb{C}\). These functions map from \(\mathbb{R}^2\) to \(\mathbb{R}\).
A function \(f: \mathbb{C} \to \mathbb{C}\), \(f(z) = u + iv\), defined in a domain \(\Omega\), is called holomorphic if the real-valued functions \(u(x,y)\) and \(v(x,y)\) are continuously differentiable in \(\Omega\), and their partial derivatives satisfy the relations \[ \begin{align} u_x & = +v_y, \\ u_y & = -v_x, \end{align} \] at every point in \(\Omega\). These are the Cauchy-Riemann differential equations (CRD).
Consider a function \(f: \mathbb{C} \mapsto \mathbb{C}\). We want to demonstrate that this function is complex differentiable and holomorphic.
Definition of the derivative:
\[ f'(z) = \lim_{ h \to 0 } \frac{f(z + h) - f(z)}{h} \] What does this limit mean by thinking of the real and imaginary parts separately? Let’s replace \(z\) with \(x+iy\): \[ f'(x +iy) = \lim_{ h \to 0 } \frac{f(x + iy + h) - f(x + iy)}{h} \] Now \(h\) is also a complex number. Replace \(h\) by \(a + ib\) and take the limit as \(a\) and \(b\) together approach \(0\):
\[ f'(x +iy) = \lim_{ (a,b) \to (0,0) } \frac{f(x + iy + a + ib) - f(x + iy)}{a + ib} \] Now introduce a new function \(F: \mathbb{R}^{2} \mapsto \mathbb{C}\) that \(F(x,y) = f(x + iy)\). \(F\) is a function of two real variables, \(f\) is a function of a single complex variable.
\[ f'(x +iy) = \lim_{ (a,b) \to (0,0) } \frac{F(x + a, y + b) - F(x, y)}{a + ib} \] If that limit exists, then it is equal to the same limit when we just restrict to moving aong the real axis, i.e., when we approach \((0,0)\) just along the real axis. We set \(b=0\) and write \[ f'(x +iy) = \lim_{ a \to 0, b=0 } \frac{F(x + a, y + b) - F(x, y)}{a + ib} \] This is a restricted situation, but if the limit exists for \((a,b) \to (0,0)\) then this limit must also exist and approach to the same quantity. \[ f'(x +iy) = \lim_{ a \to 0 } \frac{F(x + a, y) - F(x, y)}{a} = \frac{\partial F}{\partial x}(x,y) \] Now we repeat the same story and instead of restricting to the real axis we restrict to the imaginary axis. \[ f'(x +iy) = \lim_{ a =\to= 0, b \to 0 } \frac{F(x + a, y + b) - F(x, y)}{a + ib} \] \[ f'(x +iy) = \lim_{ b \to 0 } \frac{F(x, y + b) - F(x, y)}{ib} = \frac{1}{i}\frac{\partial F}{\partial y}(x,y) \] If these limits are the same, then \[ \frac{\partial F}{\partial x}(x,y) = \frac{1}{i} \frac{\partial F}{\partial y}(x,y) \] Now introduce two new functions: \(u: \mathbb{R}^{2} \to \mathbb{R}\) and \(v: \mathbb{R}^{2} \to \mathbb{R}\), such that \[ F(x,y) = u(x,y) + i v(x,y) \] Now expand out the partial derivative \[ \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{1}{i} \frac{\partial F}{\partial y}(x,y) = \frac{1}{i}\left( \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} \right) = \left( \frac{1}{i}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \right) \] and equate the real and imaginary parts \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \frac{1}{i} \frac{\partial u}{\partial y} = i \frac{\partial v}{\partial x} \] which become the Cauchy-Riemann equations \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \] A function is called holomorphic when these equation are satisfied. The stronger condition however is that \(u\) and \(v\) are real differentiable functions and they satisfy the CRE.
13.2 Application to Gravitational Potential
Let \(V_y(y,z) := \partial_y V(y,z)\), and \(V_{yy} := \partial^2_{yy}V(y,z)\), etc., denote the partial derivatives of the gravitational potential in the respective coordinate directions.
Due to the properties \[\nabla \times \mathbf{g} = \mathbf{0},\] and \[\Delta V = 0,\] the Cauchy-Riemann differential equations \[ V_{yz} - V_{zy} = 0 \qquad \text{and} \qquad V_{yy} + V_{zz} = 0 \] hold.
| CR equations | Gravity components |
|---|---|
| \(x+iy\) | \(z+iy\) |
| \(u+iv\) | \(V_y + i V_z\) |
| \(u_x\) | \(V_{yz}\) |
| \(u_y\) | \(V_{yy}\) |
| \(v_x\) | \(V_{zz}\) |
| \(v_y\) | \(V_{zy}\) |
| \(u_x = v_y\) | \(V_{yz} = V_{zy}\) |
| \(u_y = -v_x\) | \(V_{yy} = -V_{zz}\) |
We may therefore assume that \(V_y\) and \(V_z\) are the real and imaginary parts of a complex function \(G(s)\).
We say that \(G(s)\) is regular or holomorphic.
The functions \(V_y\) and \(V_z\) depend on \(y\) and \(z\), i.e., \(V_y = V_y(y, z)\) and \(V_z = V_z(y, z)\).
In the complex plane, the real coordinates \(y\) and \(z\) are combined into a complex coordinate \(s = z + iy\).
Thus, we obtain the regular function \(G : \mathbb{C} \to \mathbb{C}\) \[ G(s) = V_y(s) + i V_z(s). \]
13.2.1 Gravitational Acceleration in \(\mathbb{R}\)
With \(r^2 = (y - y')^2 + (z - z')^2\), the components of the gravitational acceleration are given by \[ V_y(y, z) = - 2 f \rho \iint_A \frac{y - y'}{r^2} \, dy' \, dz' \] and \[ V_z(y, z) = - 2 f \rho \iint_A \frac{z - z'}{r^2} \, dy' \, dz'. \]
13.2.2 Gravitational Acceleration in \(\mathbb{C}\)
The gravitational acceleration in the complex plane can be expressed as \[ G(s) = -2 f \rho i \iint_A \frac{1}{s - s'} \, dy' \, dz'. \]
Note that with \(s = z + iy\), \(s' = z' + iy'\) follows
\[ s - s' = (z - z') + i(y - y'). \] With complex conjugation we get \[ \overline{s - s'} = (z - z') - i(y - y'), \] such that \[ |s|^2 = s \cdot \overline{s}. \]
13.2.3 Computation of \(G\) in a Source-Free Region
Using the Cauchy Integral Formula, we find: Let \(G : \Omega \to \mathbb{C}\) be regular in a domain \(\Omega\), and let \(C : [a, b] \to \Omega \setminus \{z_0\}\) be a closed, positively oriented, piecewise \(C^1\)-continuous curve enclosing \(z_0 \in \Omega\). Then, \[ G(z_0) = \frac{1}{2 \pi i} \oint_C \frac{G(z)}{z - z_0} \, \mathrm{d}z. \] The values on the boundary completely determine those inside \(C\).
If \(z_0\) is located on \(C\), then the CIF reduces to \[ G(z_0) = \frac{1}{ \pi i} \oint_C \frac{G(z)}{z - z_0} \, \mathrm{d}z. \]
13.2.4 How to Choose the Integration Path?
By the Cauchy Integral Theorem, the integration path can be chosen arbitrarily. Since \(G(s) \to 0\) as \(y^2 + z^2 \to \infty\), \(z < 0\), the integration path is chosen to fully encompass the \(y\)-axis, returning along an arc in the upper half-plane, extending to infinity.
13.2.5 Derivation of the Hilbert Transform
We separate real and imaginary parts of \(G\) and obtain the Hilbert Transform Pair \[ V_y(y) = +\frac{1}{\pi} \int\limits_{-\infty}^\infty \frac{V_z(y')}{y' - y} \, dy' \] and \[ V_z(y) = -\frac{1}{\pi} \int\limits_{-\infty}^\infty \frac{V_y(y')}{y' - y} \, dy'. \]
Recognizing these as convolution integrals: \[ V_y(y) = -\frac{1}{\pi y} \ast V_z(y) = +\mathcal{H}\{V_z(y)\}, \] and \[ V_z(y) = +\frac{1}{\pi y} \ast V_y(y) = -\mathcal{H}\{V_y(y)\}. \]
In the frequency domain, convolution becomes multiplication.
Note that \[ \mathcal{H}(\mathcal{H}(u))(t) = -u(t), \] which follows from the effect of the Hilbert transform on the Fourier transform of \(u(t)\) (see next section).
13.2.6 What is the Fourier Spectrum of \(-\frac{1}{\pi y}\)?
\[ -\frac{1}{\pi} \int\limits_{-\infty}^\infty \frac{e^{-i k y}}{y} \, dy = i \, \mathrm{sign}(k). \]
Derivation using the Residue Theorem in Complex Analysis: Replace \(\frac{1}{y}\) with \(\lim_{a \to 0} \frac{y}{y^2 + a^2}\), \(y = u + iv\), apply the Residue Theorem, and integrate around poles \(+ia\) and \(-ia\).
The goal is to find \[ \mathcal F\{ -\frac{1}{\pi y}, k\} = -\frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{-i k y}}{y}\, \dd y. \] We replace \(\frac{1}{y}\) with \(\frac{y}{y^2 + a^2}\) and evaluate the integral in the limit \[ \lim_{a \to 0} \int_{-\infty}^\infty \frac{y}{y^2 + a^2} e^{-i k y}\, \dd y. \] The integrand is a complex function of \(y = u + iv\), hence \[ e^{-iky}=e^{-iku}e^{+kv} \] The roots of the polynomial in the denominator are at \(y=\pm ia\).
We use the residual theorem for a simple pole \[ \text{Res} f(y) = \lim_{y \to y_{\text{pole}}}(y - y_{\text{pole}}) f(y) = \frac{1}{2 \pi i} \oint f(y) \, \dd y. \] Since the path of integration is arbitrary, we choose a path such that \[ f(y) \to 0 \begin{cases} k<0 \text{ and } v \to +\infty \\ k<0 \text{ and } v \to -\infty \end{cases} \]
This reduces the integration to the real \(y\)-axis from \(-\infty\) to \(+\infty\).
For \(k<0\) we enclose the pole at \(y=-ia\) and integrate clock-wise (sign!) \[ \int_{-\infty}^\infty \frac{y}{y^2 + a^2} e^{-i k y}\, \dd y = -\oint \frac{y}{y^2 + a^2} e^{-i k y}\, \dd y = -2 \pi i \text{Res} f(y). \]
\[ \begin{align} -2 \pi i \text{Res}f(y) & = -2 \pi i \lim_{y \to -ia} (y+ia)f(y) \\ & = -2 \pi i \lim_{y \to -ia} (y+ia) \frac{y e^{-i k y}}{(y-ia)(y+ia)} \\ & = -2 \pi i \frac{-ia e^{+i^2 k a}}{-ia -ia} \\ & = -\pi i e^{-ka} \end{align} \] Now \[ \lim_{a\to 0} (- \pi i e^{-ka}) = -\pi i, \] which leads to the final result \[ -\frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{-i k y}}{y}\, \dd y = \begin{Bmatrix} +i \qif k>0 \\ -i \qif k<0 \end{Bmatrix} = i \, \text{sign}(k). \]
Thus, the spectrum of \(-\frac{1}{\pi y}\) is \(i \, \mathrm{sign}(k)\)!
For negative and positive wave numbers \(k\), the spectrum of \(V_y\) (or \(V_z\)) is multiplied by \(-i\) and \(+i\), respectively.
The Hilbert transform is a multiplier operator. The multiplier is \(\Phi_H(k) = -i \, \mathrm{sign}(k)\), where \(\mathrm{sign}\) is the signum function.
By Euler’s formula, \[ \Phi_H(k) = \begin{cases} +i = e^{+i\pi/2} & \qfor k < 0 \\ 0 & \qfor k = 0 \\ -i = e^{-i\pi/2} & \qfor k > 0 \end{cases} \]
The Hilbert transform has the effect of shifting the phase of the negative wavenumber components of the signal by +90\(\textdegree\) and the phase of the positive wavenumber components by -90\(\textdegree\).
When applied twice, the phase of the negative and positive wavenumber components are shifted by +180\(\textdegree\) and -180\(\textdegree\), resp., which are equivalent amounts. The signal is negated, because \[ (\Phi_H(k))^2 = e^{\pm i \pi} = -1 \qfor k \ne 0. \]