Consider a system \$G(s) = \frac{1}{s}\$. It is marginally stable, as it has its only pole at \$s=0\$. However, if we apply a step input, the output is \$ tu(t) \$, which turns out to be unstable. But, the stability or instability of a system should not depend on the nature of the input. If it has a single pole at \$ s= 0\$, it should remain marginally stable, no matter what the input is.

So, how do we reconcile these conflicting notions?

  • 2
    \$\begingroup\$ It is marginally stable. If you apply an impulse, the response does not go to infinity. \$\endgroup\$
    – Chu
    Jan 16 '19 at 8:13
  • 1
    \$\begingroup\$ Marginal stability does not imply bibo stability, so there is nothing contradictory here \$\endgroup\$
    – AVK
    Jan 16 '19 at 16:36
  • \$\begingroup\$ @Chu, what if we apply a step \$\endgroup\$
    – SaJ
    Jan 17 '19 at 11:05
  • \$\begingroup\$ The input has nothing to do with the system's relative stability. Stability is a system property. In this case there's a pole at the origin, so it's on the boundary. \$\endgroup\$
    – Chu
    Jan 17 '19 at 15:28

As noted by AVK, there is more than one notion of stability. In terms of

  • internal stability, the integrator is marginally stable (though not asymptotically stable).

  • BIBO (bounded-input, bounded-output) stability, the integrator is unstable. Note that a bounded input such as a step produces an unbounded output, namely, a ramp.


I don`t hink that we can say that the "output turns out to be unstable". It is the whole system with feedback which is to be considered as stable or unstable (not its output). In the described case (H(s)=1/s), the system reacts upon a step input with a rising output, which corresponds with our expections and with the systems transfer function. Hence, it cannot be considered as unstable.

(However, we should realize that we speak about an ideal integrating system which cannot be realized in hardware).

  • \$\begingroup\$ But, the overall system is also unstable as the output increases without bound. Our expectation is that the output follows the step input as much as possible. But the output is tu(t), which will go to infinity as time increases. \$\endgroup\$
    – SaJ
    Jan 16 '19 at 10:40
  • \$\begingroup\$ No - for u(t)=0 the output remains (mathematically) at zero. But - as mentioned - it is a nonrealistic ideal model only. \$\endgroup\$
    – LvW
    Jan 16 '19 at 13:54

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