I am trying to better understand the properties of improper systems $H(s) = \frac{b(s)}{a(s)}$, for which the order of the numerator $b(s)$ is greater than the order of the denominator $a(s)$ (in the broader context of continuous-time linear time-invariant systems). I have trouble reconciling the application of region of convergence (ROC) rules and my intuition.

Let us take the simplest example that is the pure differentiator, i.e., $H(s)=s$.

Application of ROC rules

  • Causality. The system has a pole at $s=\infty$. Thus, the ROC is not a right-sided plane. So the system is not causal.

  • BIBO stability. The ROC includes the imaginary axis $s=j\omega$. The system is BIBO stable.

My intuition

  • Causality. The differentiator can be expressed as the limit of a forward difference or as the limit of a backward difference. So depending on how we define the differentiator, it can be a causal system or an anti-causal system.

  • BIBO stability. I have two intuitions here.

    (1) The derivative of a bounded input like the step function can be unbounded as it gives the Dirac function. So the system is not BIBO stable.

    (2) The impulse response of the differentiator is $h(t) = \delta'(t)$, i.e., the derivative of the Dirac function. This is not an absolutely integrable function as $\int_{-\infty}^{\infty} |\delta'(t)| dt = \infty$. See https://en.wikipedia.org/wiki/Unit_doublet. So the system is not BIBO stable.

How can I reconcile the ROC analysis and my intuition? Where am I wrong?


Causality. A LTI system with impulse response $h(t)$ is said to be causal, that is, it has the property that the value of the output at time $t_0$ depends on the values of the input and output for all $t$ up to time $t_0$ but no further, i.e., only for $t \leq t_0$, if and only if $h(t)=0$ for all $t < 0$.

BIBO stability. A LTI system with impulse response $h(t)$ is said to be BIBO stable, that is, it has the property that its output $y$ is bounded whenever its input $u$ is bounded, if and only if $$\int_{-\infty}^{\infty} h(t) dt \triangleq \| h \|_1 < \infty $$ in which case we can bound the peak of the output by $$ \| y \|_\infty \leq \| h \|_1 \| u \|_\infty $$ where $\|u\|_\infty \triangleq \mathrm{sup}_{t\in\mathbb{R}} u(t)$.

  • $\begingroup$ How deep do you need to go? Normally in real-world engineering, such systems are either virtual (i.e., they're a result of block diagram manipulation) and their wacky effects are canceled out by other blocks so the aggregate system is proper, or they're a sign that you're taking a bad approach and you need to discard that path of inquiry. In neither case do you have to worry about the fine mathematical points. $\endgroup$
    – TimWescott
    Commented Feb 8, 2020 at 15:43
  • 1
    $\begingroup$ Hi @TimWescott. Thank you for your comment. I do understand the limited implications in real-world applications. Nevertheless, I am interested in better understanding this mathematical curiosity from a theoretical viewpoint. $\endgroup$
    – Marca85
    Commented Feb 8, 2020 at 16:18
  • 1
    $\begingroup$ I think your intuition is correct, but it really depends on how causality and BIBO stability are defined. (Although I think you're on safer ground with the BIBO argument). For me, the more important aspect of such a system is that if it's used in the real world, you immediately run into the problem that noise tends to be very wideband, and any too-enthusiastic lead-lag approximation of a differentiator saturates itself with noise. $\endgroup$
    – TimWescott
    Commented Feb 8, 2020 at 18:40
  • $\begingroup$ Thanks again @TimWescott. I will edit my post to add the definitions that I have in mind (for sake of completeness). $\endgroup$
    – Marca85
    Commented Feb 8, 2020 at 18:53
  • 1
    $\begingroup$ Strictly speaking one cannot implement a pure differentiator in the analog domain -- everything has delay if you pay attention to high enough frequencies. $\endgroup$
    – TimWescott
    Commented Feb 8, 2020 at 20:58

1 Answer 1


Note that the imaginary axis is not inside the ROC because $|H(j\omega)|\to\infty$ for $\omega\to\infty$. So the system is definitely unstable, regardless of causality, which confirms your intuition.

Concerning causality, the ROC is neither a right half-plane nor a left half-plane (both because of the pole at $s\to\infty$), which is consistent with a (generalized) function concentrated at $t=0$.


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.