5

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?

Definitions

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 $\|u\|_\infty \triangleq \mathrm{sup}_{t\in\mathbb{R}} u(t)$.

Marca85
  • 51
  • 2
  • 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. – TimWescott Feb 08 '20 at 15:43
  • 1
    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. – Marca85 Feb 08 '20 at 16:18
  • 1
    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. – TimWescott Feb 08 '20 at 18:40
  • Thanks again @TimWescott. I will edit my post to add the definitions that I have in mind (for sake of completeness). – Marca85 Feb 08 '20 at 18:53
  • except using analog means, i dunno how one implements a differentiator without delay. representing it with a forward difference is acausal. representing it with a backward difference ($1-z^{-1}$) introduces a half-sample delay. – robert bristow-johnson Feb 08 '20 at 19:21
  • 1
    Strictly speaking one cannot implement a pure differentiator in the analog domain -- *everything* has delay if you pay attention to high enough frequencies. – TimWescott Feb 08 '20 at 20:58

1 Answers1

0

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 $t=0$.

Matt L.
  • 78,282
  • 5
  • 66
  • 149