# What are the properties of continuous-time improper systems?

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

• 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 8 at 15:43
• 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 8 at 16:18
• 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 8 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 8 at 18:53
• 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 8 at 20:58

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 $$t=0$$.