Analytic signal

Remarks

With the Fourier transform $\hat f(\xi) = \int_\R f(x) e^{-2\pi i x \xi}\, dx$, one has $\widehat{Hf}(\xi) = -i \sgn(\xi) \hat f(\xi)$, so $$\widehat{f_a}(\xi) = \big(1 + \sgn(\xi)\big) \hat f(\xi),$$ which is $2 \hat f(\xi)$ for $\xi > 0$ and $0$ for $\xi < 0$. Since $f$ is real, $f = \operatorname{Re} f_a$, and $f_a$ is determined by $f$. One writes $f_a(t) = A(t) e^{i\phi(t)}$ with $A = \abs{f_a}$ the instantaneous amplitude and $\phi$ the instantaneous phase.

Example: $f(x) = \frac{1}{1 + x^2}$. Its Hilbert transform is $\frac{x}{1 + x^2}$, so $f_a(x) = \frac{1 + ix}{1 + x^2} = \frac{1}{1 - ix} = \frac{i}{x + i}$. The function $z \mapsto \frac{i}{z + i}$ is holomorphic on the upper half-plane, as the definition predicts.

Non-example: the clause "real-valued" matters. For the complex function $g = \frac{i}{x + i}$ one has $Hg = -i g$, so $g + i Hg = 2g \ne g$, and the real part of $g + iHg$ is $2 \operatorname{Re} g$, not $\operatorname{Re} g$.