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I'm a bit confused regarding the stability of a differentiator and an integrator. When an integrator is added to the system it decreases the stability as it adds a pole at origin and bends the root locus of the system transfer function towards right of the plane but adding a differentiator increases the stability as it adds a zero at origin, which bends the root locus of the system transfer function to the left. When considering the transfer function of a unity feedback system as \$G(s)={1/s}\$, the root locus tends towards the right of the plane..

Hence my question is: What is the stability of the system with transfer function \${1/s}\$?

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  • \$\begingroup\$ What exactly is your question? \$\endgroup\$ – uint128_t Dec 30 '15 at 17:30
  • \$\begingroup\$ what is the stability of the system with transfer functin {1/s} \$\endgroup\$ – Manasa Harini Dec 30 '15 at 17:32
  • \$\begingroup\$ It cannot be unstable because there is no feedback stuff to make it unstable. \$\endgroup\$ – Andy aka Dec 30 '15 at 18:34
  • \$\begingroup\$ It's BIBO unstable, since we can find a bounded input (eg a unit step) for which the output is unbounded (in this case, a ramp) \$\endgroup\$ – Chu Dec 30 '15 at 19:05
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When an integrator is added to the system it decreases the stability ... but adding a differentiator increases the stability

This is not always true since it depends on the system you are adding the integartor (or differentiator) to. To analyze closed loop stability you need to include both the open loop system and the compensator.

The stability margins of a simple integrator by itself are:

gain margin: infinite

phase margin: 90 degrees

You can determine this by looking at the Bode Plot for the integrator.

This basically means that if you close the (negative) feedback loop on the integrator you can indefinitely increase the gain and not create an unstable system. But if you somehow shift phase in the loop by 90 degrees, the closed loop will become unstable.

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For a linear system, stability is usually defined in terms of the "bounded input/bounded output" (BIBO) criteria. For a pure integrator, the system is not stable, since any constant input (except for 0) creates an unbounded output.

Another way to look at it, is that a unit impulse response should have a bounded L1-norm. The impulse response of an integrator is the unit Heaviside function (0 when t < 0, and 1 when t >= 0): $$ r(t) = \int_{-\infty}^t\delta_0(\tau)d\tau= H(t) = \begin{cases}0,\qquad t < 0\\ 1, \qquad t \ge 0\end{cases} $$ But then $$ \| r \|_1 = \int_{-\infty}^\infty |r(t)| dt = \int_{-\infty}^\infty H(t)dt = \int_0^\infty dt = \lim_{T \to \infty}\int_0^T dt = \lim_{T\to\infty} T = \infty, $$ So the L1-norm is not bounded.

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