In signal processing, we define an analytic signal as a complex-valued signal which has no frequency components for $\omega<0$. It can be shown that the real part and the imaginary part of an analytic signal $x(t)=x_R(t)+jx_I(t)$ are related by the Hilbert transform:

\begin{align} x_I(t) & = \mathscr{H}\{x_R(t)\}\tag{1} \\ x_R(t) & = -\mathscr{H}\{x_I(t)\}\tag{2} \end{align}

In complex analysis, an analytic function (or holomorphic function) is a complex function of a complex variable that is infinitely differentiable, and its Taylor series about a given point converges to the function in some neighbourhood of that point.

The question is if there is some relationship between these two apparently unrelated meanings of the word "analytic", and if so, what is this relationship?

EDIT: I've just found a very similar question on the mathematics site, but the answers there are very concise without many details.


1 Answer 1


There is a relationship between these two concepts. Let the complex function $f(z)$ be analytic on and inside a simple closed curve $C$ in the complex plane. Then Cauchy's integral formula states that for any point $z$ inside $C$ we have

$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=2\pi j f(z)\tag{1}$$

For $z$ outside $C$, it follows from Cauchy's theorem that


Furthermore, for $z$ on the curve $C$ the following holds:

$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=\pi j f(z)\tag{3}$$

A proof of Eq. $(3)$ can be found in any text on complex analysis.

Let's now choose the curve $C$ as the real line from $-R$ to $R$ and the semi-circle $Re^{j\phi}$, $\phi\in[0,\pi]$, such that $C$ is closed. For $R\to\infty$, the curve $C$ "encloses" the upper half-plane. If $f(z)$ is analytic in the upper half-plane, and if it decays to zero sufficiently fast such that the contribution of the integral along the semi-circle with radius $R$ vanishes as $R\to\infty$, Eq. $(3)$ becomes

$$\int_{-\infty}^{\infty}\frac{f(\zeta)}{\zeta -x}d\zeta=\pi j f(x)\tag{4}$$

Note that the integral in $(4)$ is along the real line. With $z=x+jy$ and $f(z)=f_R(x,y)+jf_I(x,y)$, Eq. $(4)$ can be written as

$$\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)+jf_I(\zeta,0)}{\zeta -x}d\zeta=\pi j \big[f_R(x,0)+jf_I(x,0)\big]\tag{5}$$

Splitting Eq. $(5)$ into real and imaginary parts yields

\begin{align} f_I(x,0) & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)}{\zeta-x}d\zeta &&= \mathscr{H}\big\{f_R(x,0)\big\} \tag{6}\\ f_R(x,0) & = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_I(\zeta,0)}{\zeta-x}d\zeta &&= -\mathscr{H}\big\{f_I(x,0)\big\}\tag{7} \end{align}

where $\mathscr{H}\{\cdot\}$ denotes the Hilbert transform.

Equations $(6)$ and $(7)$ are identical to Eqs $(1)$ and $(2)$ in the question with $x_R(t)=f_R(t,0)$ and $x_I(t)=f_I(t,0)$. Consequently, the analytic function $f(z)$ equals an analytic signal on the real line.

Example 1:

Given the analytic signal $x(t)=e^{j\omega_0t}$, $\omega_0>0$, the corresponding analytic function is given by $f(z)=e^{j\omega_0z}$. Note that $f(z)$ satisfies all properties required for Eqs $(4)-(7)$ to be valid: it is analytic everywhere, and it decays rapidly for $|z|\to\infty$ and $\textrm{Im}\{z\}>0$ (i.e., in the upper half-plane).

Example 2:

We can also use functions with the required properties (analytic and decaying sufficiently fast in the upper half plane) to generate Hilbert transform pairs, or, equivalently, analytic signals. Let

$$f(z)=\frac{e^{j\omega_0 z}-1}{j\pi z},\qquad\omega_0>0\tag{8}$$

The function $f(z)$ is analytic everywhere (if we define $f(0)=\omega_0/\pi$), and it decays rapidly in the upper half plane $\textrm{Im}\{z\}>0$. Consequently, by evaluating $f(z)$ on the real line, we obtain an analytic signal and a Hilbert transform pair. For notational convenience, let's use $t$ to denote the real part of $z$. The corresponding analytic signal is given by

\begin{align} x(t) &= \frac{e^{j\omega_0 t}-1}{j\pi t} \\ & = \frac{\cos(\omega_0t)+j\sin(\omega_0t) - 1}{j\pi t} \\ & = \frac{\sin(\omega_0t)}{\pi t} + j\frac{1 - \cos(\omega_0t)}{\pi t} \tag{9} \end{align}

It has been shown in the answers to this question that the real and imaginary parts of $(9)$ indeed form a Hilbert transform pair. From the same answers, the Fourier transform of $x(t)$ can be easily derived:


showing (again) that $x(t)$ is an analytic signal.

In a completely analogous manner we can show the relation between the causality of a time domain function and the analyticity of the corresponding transfer function. If $h(t)$ is the impulse response of a causal LTI system, i.e., if $h(t)=0$ for $t<0$ is satisfied, it is well-known that the real and imaginary parts of its Fourier transform (the frequency response) are related by the Hilbert transform. Let $H(j\omega)$ be the Fourier transform of $h(t)$ with real and imaginary parts $H_R(\omega)$ and $H_I(\omega)$, respectively:

$$\mathscr{F}\big\{h(t)\big\} = H(j\omega) = H_R(\omega) + jH_I(\omega) $$

Then the following Hilbert transform relations hold:

\begin{align} H_R(\omega) &= \mathscr{H}\big\{H_I(\omega)\big\}\tag{11} \\ H_I(\omega) &= -\mathscr{H}\big\{H_R(\omega)\big\}\tag{12} \end{align}

Now let's evoke Eq. $(3)$, but this time we choose the curve $C$ as the straight line connecting the points $z=jR$ and $z=-jR$ on the imaginary axis, and the semi-circle $Re^{j\phi}$, $\phi\in[-\pi/2,\pi/2]$, connecting those two points via the right half-plane. For a stable system, the transfer function $H(s)$ will decay sufficiently fast in the right half-plane, such that the contribution to the total integral $(3)$ of the integral over the semi-circle vanishes for $R\to\infty$, and we're left with an integral over the imaginary axis:

\begin{align} \lim_{R\to\infty}\oint_C\frac{H(\zeta)}{\zeta-s}d\zeta &= \int_{j\infty}^{-j\infty}\frac{H(\zeta)}{\zeta-s}d\zeta \\ & = \int_{-\infty}^{\infty}\frac{H(j\Omega)}{s-j\Omega}jd\Omega\tag{13} \end{align}

For $s$ on $C$, i.e., on the imaginary axis, we have $s=j\omega$ and, using $(13)$, Eq. $(3)$ becomes

$$\int_{-\infty}^{\infty}\frac{H(j\Omega)}{\omega-\Omega}d\Omega = \pi j H(j\omega)\tag{14}$$

Splitting $(14)$ into real and imaginary parts yields

\begin{align} H_R(\omega) &= \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{H_I(\Omega)}{\omega-\Omega}d\Omega \\ H_I(\omega) &= -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{H_R(\Omega)}{\omega-\Omega}d\Omega \end{align}

which are just the Hilbert transform relations $(11)$ and $(12)$. I.e., the Hilbert transform relations are equivalent to Eq. $(14)$, which is a consequence of Eq. $(3)$ IF $H(s)$ is analytic in the right half-plane and if it decays sufficiently fast in the right half-plane. In sum, causality (and stability) of an LTI system cause its transfer function $H(s)$ to be analytic in the right half-plane. This is equivalent to $H(s)$ satisfying the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis.

Depending on the field, the Hilbert transform relations $(11)$ and $(12)$ are also referred to as Kramers-Kronig relations or as dispersion relations (e.g., here and here).

Example 3:

Given the causal and stable impulse response


we know from the above that the corresponding transfer function


must be analytic and decaying in the right half-plane, both of which is straightforward to verify. Furthermore, $H(s)$ must satisfy the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis. Splitting $H(j\omega)$ into real and imaginary parts gives


which is a well-known Hilbert transform pair (see this answer or this table).

  • $\begingroup$ Thank you Matt. I'll read through this later. When I think of "analytic function" I think of the Cauchy equation: $$ \frac{\partial \Re e \{f(z)\}}{\partial \Im m\{z\}}=\frac{\partial \Im m \{f(z)\}}{\partial \Re e\{z\}}$$ and vise versa $\endgroup$ Oct 10, 2023 at 18:09
  • $\begingroup$ @robertbristow-johnson: The Cauchy-Riemann equations are of course satisfied in the region where the function is analytic, but if the function also decays in a half-plane then the Hilbert transform relations are satisfied on the real (or imaginary) axis. $\endgroup$
    – Matt L.
    Oct 10, 2023 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.