Hilbert transform

In mathematics and signal processing, the Hilbert transform is a specific singular integral that takes a function, u(t) of a real variable and produces another function of a real variable H(u)(t). The Hilbert transform is given by the Cauchy principal value of the convolution with the function ${\displaystyle 1/(\pi t)}$ (see § Definition). The Hilbert transform has a particularly simple representation in the frequency domain: It imparts a phase shift of ±90° (π/2 radians) to every frequency component of a function, the sign of the shift depending on the sign of the frequency (see § Relationship with the Fourier transform). The Hilbert transform is important in signal processing, where it is a component of the analytic representation of a real-valued signal u(t). The Hilbert transform was first introduced by David Hilbert in this setting, to solve a special case of the Riemann–Hilbert problem for analytic functions.

The Hilbert transform of u can be thought of as the convolution of u(t) with the function h(t) = ⁠1/πt⁠, known as the Cauchy kernel. Because 1/t is not integrable across t = 0, the integral defining the convolution does not always converge. Instead, the Hilbert transform is defined using the Cauchy principal value (denoted here by p.v.). Explicitly, the Hilbert transform of a function (or signal) u(t) is given by

$${\displaystyle \operatorname {H} (u)(t)={\frac {1}{\pi }}\,\operatorname {p.v.} \int _{-\infty }^{+\infty }{\frac {u(\tau )}{t-\tau }}\,\mathrm {d} \tau ,}$$

provided this integral exists as a principal value. This is precisely the convolution of u with the tempered distribution p.v. ⁠1/πt⁠. Alternatively, by changing variables, the principal-value integral can be written explicitly as

$${\displaystyle \operatorname {H} (u)(t)={\frac {2}{\pi }}\,\lim _{\varepsilon \to 0}\int _{\varepsilon }^{\infty }{\frac {u(t-\tau )-u(t+\tau )}{2\tau }}\,\mathrm {d} \tau .}$$

When the Hilbert transform is applied twice in succession to a function u, the result is

$${\displaystyle \operatorname {H} {\bigl (}\operatorname {H} (u){\bigr )}(t)=-u(t),}$$

provided the integrals defining both iterations converge in a suitable sense. In particular, the inverse transform is ${\displaystyle -\operatorname {H} }$. This fact can most easily be seen by considering the effect of the Hilbert transform on the Fourier transform of u(t) (see § Relationship with the Fourier transform below).