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If there are multiple poles at origin, that system is unstable. So a type 2 system will be have s^2 in denominator. So this should be unstable right? Correct me if I am wrong.

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  • \$\begingroup\$ There are many transfer functions that have s^2 in the denominator but that circuit isn't unstable. \$\endgroup\$ – Andy aka Sep 19 at 6:57
  • \$\begingroup\$ depends upon the gain and the phase, perhaps provided by an explicit "compensator" \$\endgroup\$ – analogsystemsrf Sep 19 at 7:13
  • \$\begingroup\$ Are you referring to open-loop poles or closed-loop poles? \$\endgroup\$ – Chu Sep 19 at 9:07
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In general, a system is stable, if all real parts of the poles are negative.

It is well described here


Is a type 2 system is always unstable?

No, the order of a system says nothing about its stability.

If there are multiple poles at origin, that system is unstable.

This is a marginal case. It can be either unstable or slightly stable, depending on how many poles there are at the origin.

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The characteristic equation is the equation which is the denominator of the transfer function. If the characteristic equation contains an S^2 term then the system is a second order system. The values of the poles of the system can be found by equating the characteristic equation to zero and solving for S. This will give you the roots of the characteristic equation.

If the poles (roots of the characteristic equation) are all negative then the system will be stable. If the characteristic equation has at least one positive root then the system will be unstable. If the roots of the characteristic equation contain an imaginary part (a +jb) then the real part of the roots must be negative for the system to be stable.

To summarise, for the system to be stable, all the poles (roots of the characteristic equation) must lie on the lefthand side of the S plane.

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