Contradiction in stability of derivative system with doublet impulse response.

Question

A linear time-invariant (LTI) system is BIBO stable iff the region of convergence (ROC) of its Laplace transform includes the entire imaginary axis. The Laplace transform of the doublet is S with ROC = all S-plane. Then it includes the imaginary axis. ِDoublet is the impulse response of a derivative system. But we know that a derivative system is unstable. How can I explain this contradiction?

Answer 1

The condition you state for BIBO stability is not complete. A necessary condition for BIBO stability is that the transfer function has region of convergence that includes the right half plane, and it be uniformly bounded on the imaginary axis.

The exact condition for BIBO stability is that the impulse response must be in $L^1$ ($\;L^1(-\infty,\infty)$ for non-causal systems and $L^1(0,\infty)$ for causal ones) and may include delta functions of the form $\sum_{k=0}^\infty \alpha_k \delta(t-t_k)$ provided the sequence $\left\{ \alpha_k \right\}$ is absolutely summable. Impulse responses of in this class have Laplace transform that is analytic in the open right half plane, and uniformly bounded on the imaginary axis.

The differentiator (of any order) does not have a transfer function that is bounded on the imaginary axis, and therefore violates this necessary condition.