It is cited in multiple places that the fact that a filter is causal (i.e. the impulse response is zero for t < 0) implies that the system function is analytical.

I couldn't find any proof of this, why is it the case?

  • $\begingroup$ You should be explicit with what you mean by "system function". (Might you mean "frequency response"?) And by what you mean by "analytical". (Might you mean something that has to do with the Hilbert Transform?) $\endgroup$ Oct 8, 2023 at 18:41
  • $\begingroup$ @robertbristow-johnson: The OP means analytic in the complex analysis sense of the word, which clearly has to do with the Hilbert transform. $\endgroup$
    – Matt L.
    Oct 8, 2023 at 18:48
  • $\begingroup$ How is $$ \frac{\mathrm{d}\Im m \{f(z)\}}{\mathrm{d}\Re e \{z\}} = \frac{\mathrm{d}\Re e \{f(z)\}}{\mathrm{d}\Im m \{z\}} $$ clearly related directly to the Hilbert Transform? $\endgroup$ Oct 8, 2023 at 19:06
  • $\begingroup$ I'm thinking that if $$ h(t) = 0 \qquad \forall t<0 \text{ and } h(t) \in \mathbb{R} $$ (what we mean by "causal impulse response") then $$ \Im m\{H(f)\} = -\mathscr{H} \Big\{\Re e\{ H(f)\} \Big\} \qquad \forall f \in \mathbb{R} $$ or that $H(f)$ is an "analytic signal". $\endgroup$ Oct 8, 2023 at 19:11
  • $\begingroup$ @robertbristow-johnson: I agree that the word "clearly" may have been an exaggeration. However, note the similarity between Cauchy's formula and the definition of the Hilbert transform. That's where it all comes from. $\endgroup$
    – Matt L.
    Oct 8, 2023 at 19:39

2 Answers 2


New answer:

I provide a new answer because I believe that this is a clearer and more direct way of explaining the relation between causality of the impulse response and analyticity of the corresponding transfer function. The old answer is not wrong but it relies on the Hilbert transform, which in hindsight appears to be a rather indirect approach. For the time being I will not delete the old answer, just in case anybody is interested.

I'll show that if a given transfer function $H(s)$ is analytic in the right half-plane, its corresponding impulse response $h(t)=\mathscr{L}^{-1}\{H(s)\}$ must be causal. The only additional assumption is that $\lim_{s\to\infty}H(s)=0$ for $\textrm{Re}\{s\}>0$ (i.e., in the right half-plane).

In order to prove that $h(t)=0$ for $t<0$ if $H(s)$ is analytic in the right half-plane, we consider the integral of $H(s)e^{st}$ over the closed contour $C$ which consists of the line segment on the imaginary axis from $s=-j\Omega$ to $s=j\Omega$ and of the semicircle in the right half-plane $s=\Omega e^{j\theta}$, with $\theta$ from $\pi/2$ to $-\pi/2$. Since $H(s)e^{st}$ is analytic inside $C$, we have from Cauchy's theorem


The integral $(1)$ can be split into the integral along the line segment on the imaginary axis and the integral along the semicircle $\Gamma$:

$$\oint_CH(s)e^{st}ds=\int_{-\Omega}^{\Omega}H(j\omega)e^{j\omega t}jd\omega+\int_{\Gamma}H(s)e^{st}ds\tag{2}$$

For $t<0$ and $\textrm{Re}\{s\}>0$ (i.e., in the right half-plane), the real part of the exponent of $e^{st}$ is negative, and since $H(s)$ tends to zero as $s\to\infty$ in the right-half plane, if follows from Jordan's lemma that the contribution of the integral along the semicircle $\Gamma$ tends to zero as $\Omega\to\infty$. With $(1)$ and $(2)$ we obtain

\begin{align} \lim_{\Omega\to\infty}\oint_CH(s)e^{st}ds &= \lim_{\Omega\to\infty}\left[\int_{-\Omega}^{\Omega}H(j\omega)e^{j\omega t}jd\omega+\int_{\Gamma}H(s)e^{st}ds\right] \\ &= j\int_{-\infty}^{\infty}H(j\omega)e^{j\omega t}d\omega \\ &= 2\pi j\, h(t) =0,\qquad t<0 \end{align}

This shows that under the given assumptions (analyticity of $H(s)$ in the right half-plane and $H(s)$ tending to zero as $s\to\infty$ in the right half-plane) the impulse response $h(t)$ is causal.

Old answer:

A causal impulse response $h(t)$ satisfies


where $u(t)$ is the unit step function. With


the Fourier transform of Eq. $(1)$ is

\begin{align} H(j\omega)&=\frac{1}{2\pi}H(j\omega)\star\left(\pi\delta(t)+\frac{1}{j\omega}\right)\\ &=\frac{1}{2}H(j\omega)-\frac{j}{2}H(j\omega)\star \frac{1}{\pi\omega}\tag{3} \end{align}

where $H(j\omega)$ is the Fourier transform of $h(t)$ (i.e., the frequency response) and $\star$ denotes convolution. Noting that convolution with $1/(\pi\omega)$ corresponds to the Hilbert transform, we can write Eq. $(3)$ as


with $\mathscr{H}\{\cdot\}$ denoting the Hilbert transform. Splitting $(4)$ into real and imaginary parts, we arrive at

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

where $H_R(\omega)$ and $H_I(\omega)$ are the real and imaginary parts of $H(j\omega)$, respectively.

Since the real and imaginary parts of $H(j\omega)$ are related by the Hilbert transform, the function $H(j\omega)$ can be analytically extended from the imaginary axis to the right half-plane, which results in the transfer function $H(s)$ being analytic for $\textrm{Re}\{s\}>0$, assuming that $H(s)$ decays sufficiently fast for $|s|\to\infty$ and $\textrm{Re}\{s\}>0$. For more mathematical details see ref 1 and especially ref 2, and the second part of this answer.

  • 1
    $\begingroup$ Still, Matt, I have something to learn here. How is the meaning of Analytic signal (in the sense we learn in Electrical Engineering) related to the meaning of complex Analytic function (in the sense we learn taking a Complex Variables course from the Mathematics department)? I had always thunked that the two different disciplines meant different things. $\endgroup$ Oct 8, 2023 at 19:43
  • 4
    $\begingroup$ @robertbristow-johnson: That's a very good question, and for a long time I've wanted to ask (and answer) this here. I hope I'll have time to do that some time soon. The two things are not different at all. Analytic signals are analytic (in the complex analysis sense of the word) in the upper half plane (if we make $t$ a complex variable). $\endgroup$
    – Matt L.
    Oct 8, 2023 at 20:44
  • $\begingroup$ (Following story with anticipation. Eating popcorn.) $\endgroup$ Oct 9, 2023 at 3:49
  • $\begingroup$ Me too! Looking forward to it $\endgroup$ Oct 15, 2023 at 13:57
  • 1
    $\begingroup$ @DanBoschen: You can find the question (and my answer) here. $\endgroup$
    – Matt L.
    Oct 17, 2023 at 8:04

As it happens, you need not go far from the concept of disperion (Kramers Kronig) relations (that caused you to discover the fact that you couldn't find any proof of) and the definition of the (signal processing) filter.

You can read in the Wiki article on KK relations, section Related proof from the time domain, that

  • A causal impulse response can be expressed as the sum of an even function and an odd function,
  • The even and odd parts of a time domain waveform correspond to the real and imaginary parts of its Fourier integral,
  • Multiplication by the sign function in the time domain corresponds to the Hilbert transform (i.e. convolution by the Hilbert kernel (1/πω) in the frequency domain,
    -- all these items illustrated with the picteresque plots of what can easily be interpreted as filter transfer function representations in time and frequency domains.

While not a rigorous math, all this gives an intuitive but still convincing proof of both validity of KK relations and their intriguing connection with (signal processing) filter causality.

If you are interested in profound and charming mathematics behind the connection between the differential equations of mathematical physics and causality, you can read, for example, a book Evolutionary Equations, Picard's Theorem for Partial Differential Equations, and Applications by Seifert, Trostoff, and Waurick, published by Birkhäuser and available in open access.

Never go far from the ship -- Epictetus.


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