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Let's say we have a control system that ends up having two poles at the origin. I've simulated this and it appears as though this takes the system into instability.

I just wanted to confirm, does having two repeated poles at the origin make a system unstable? I know that having a pole at the origin results in a marginally stable system, but I am not sure about multiple repeated poles at the origin.

If so, why is the case (that multiple poles at the origin result in an unstable system?

Edit: For clarification, the system model I am talking about is as follows:

$$ G_P = \frac{1}{ks^2}$$ Where \$k\$ is a constant positive value (\$k>0\$)

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  • \$\begingroup\$ If you have a zero also at the origin then that cancels a pole so, to be sure about this you should be more direct in stating the TF. \$\endgroup\$
    – Andy aka
    Nov 14, 2022 at 18:05
  • \$\begingroup\$ Hi Andy, what I meant was a system transfer function with only two poles at the origin. I.e. \$ G_P = \frac{1}{ks^2}\$ \$\endgroup\$ Nov 14, 2022 at 18:11

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The system \$H(s) = \frac{1}{s^2} \$ is open loop unstable. Take the inverse Laplace transform to find the impulse response

$$h(t) = \mathscr{L}^{-1} \bigg\{ \frac{1}{s^2} \bigg\} =t$$

and it shows that \$h(t) \to \infty \ \text{for} \ t\to \infty\$. Hence, the system is unbounded for any input.

If only one pole is at the origin, \$H(s) = \frac{1}{s}\$, the system is open loop stable for an impulse-input, but unstable for a step input.

$$h(t) = \mathscr{L}^{-1} \bigg\{ \frac{1}{s} \bigg\} =1$$ $$s(t) = \mathscr{L}^{-1} \bigg\{ \underbrace{\frac{1}{s}}_{\text{system}} \cdot \underbrace{\frac{1}{s}}_{\text{step}} \bigg\} =t \ \ \ \text{(step response)}$$

However, a system with multiple poles in the origin is not necessarily closed loop unstable. In the case with two poles in the origin we have

$$H_\text{cl} (s) = \frac{1}{1+H(s)} = \frac{1}{1+\frac{1}{s^2}} =\frac{s^2}{s^2+1} $$

The step response shows a BIBO-stable system

$$s(t) = \mathscr{L}^{-1} \bigg\{ \frac{s^2}{s^2+1} \cdot \frac{1}{s} \bigg\} = \cos(t).$$

So even one pole in the origin makes the system open loop unstable. But one or two poles in the origin doesn't make the system closed loop unstable. In the case of three poles in the origin, however, the system is both open and closed loop unstable:

$$G_\text{cl}(s) = \frac{1}{1+\frac{1}{s^3}} = \frac{s^3}{s^3+1} $$ $$\mathscr{L}^{-1} \bigg\{ \frac{s^3}{s^3+1} \cdot \frac{1}{s} \bigg\} =\frac{1}{3}e^{-t} + \frac{2}{3}e^{\frac{1}{2}t}\cos \bigg(\frac{\sqrt{3}t}{2} \bigg) \to \infty \ \text{for} \ t\to \infty. $$

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2 poles alone isn't unstable -- it can't reach 180 ° phase (only asymptotically approach it). But any additional pole can make the phase exceed 180 °, although stability will depend on the (DC) gain and the 3rd pole (e.g. does the total phase reach 180 ° before gain falls below 0 dB).

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    \$\begingroup\$ The pole is at zero i.e. an integrator, so the phase shift is independent of frequency. Or put another way, the asymptote is everywhere. \$\endgroup\$ Nov 14, 2022 at 18:37
  • \$\begingroup\$ Phase shift for a free integrator is -90 exactly. For two, -180 exactly. \$\endgroup\$
    – RussellH
    Nov 14, 2022 at 19:28
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This system is unstable in any sense. To verify this, one can consider the corresponding state space system $$ \left\{\begin{array}{lll} \dot x_1&=&x_2\\ \dot x_2&=&u\\ y&=&x_1/k. \end{array} \right. $$ The solution this system for some initial conditions \$x_1(0)=x_1^0\$, \$x_2(0)=x_2^0\$ and input \$u(t)\$ is $$ x_2(t)= \int_0^t u(\tau)\,d\tau +x_2^0 $$ $$ x_1(t)= \int_0^t x_2(s)\,ds +x_1^0=\int_0^t \left( \int_0^s u(\tau)\,d\tau +x_2^0 \right)\,ds +x_1^0= \int_0^t \int_0^s u(\tau)\,d\tau\,ds +x_2^0 t +x_1^0 $$ The term \$\int_0^t \int_0^s u(\tau)\,d\tau\,ds\$ means that for a bounded input the system can give an unbounded output, that is, our system is not bibo-stable. The term \$x_2^0 t\$ means that any arbitrarily small deviation \$x_2^0\ne 0\$ of the state from the equilibrium leads to an unbounded growth of \$x_1(t)\$, which implies the absence of marginal stability.

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Separated poles on the imaginary axis result in marginal stability.

2 or more coincident poles on the imaginary axis result in instability.

The origin is on the imaginary axis and so 2 coincident poles at the origin (2 integrators) results in instability.

You can visualise this firstly by considering a single pole at the origin (a single integrator). If a practical impulse (one with a short pulse width) is input to an integrator then the integrator's output will ramp up and then stay at a constant output level after the practical impulse has ended and the integrator's input has returned to zero. Any increase in the integrator's input will send its output to +infinity and any decrease in the integrator's input will send its output to -infinity. The output is finely balanced and the integrator is said to be marginally stable.

Now if we connect a second integrator to the output of the first integrator so we have two coincident poles at the origin (2 integrators or 1/s^2) and apply a practical impulse to the first integrator then its output will hold steady at some finite output (as described above) and this constant output from the first integrator will drive the second integrator's output to infinity. Hence with two series integrators, represented by 2 coincident poles at the origin, we have instability.

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