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Say there is an system with open-loop transfer function $$G(s)=\frac{1}{s-1}$$ The system is definitely unstable.

If I put an any nonlinear system in cascade with \$G(s)\$ & from the output, a negative unity feedback. Is it possible to make the system close-loop transfer function is stable? It is assumed that the higher order harmonics are filtered in the close loop path.

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  • \$\begingroup\$ Why do mention "any nonlinear system"? Is an additional linear feedback block not allowed? \$\endgroup\$ – LvW Mar 13 '15 at 14:35
  • \$\begingroup\$ May be. But my intention is not to make a liner system stable by using liner element, but using a nonlinear element. \$\endgroup\$ – Saprativ Saha Mar 13 '15 at 14:39
  • \$\begingroup\$ Well if the linearized small signal transfer function including the non-linear element is stable at every operating point along the non-linear element's characteristic then the overall system should be stable. Every practical control system in the world is evidence of this since there's no such thing as a perfectly linear element. \$\endgroup\$ – John D Mar 13 '15 at 15:04
  • \$\begingroup\$ Then please give me an example where a unstable system becomes stable using nonlinear element. @JohnD \$\endgroup\$ – Saprativ Saha Mar 13 '15 at 15:08
  • \$\begingroup\$ Conference: Applied Power Electronics Conference, APEC 2007 - Twenty Second Annual IEEE Source: IEEE Xplore ABSTRACT Digital controller provides capability to implement flexible control algorithm. The nonlinear PID function with power coefficients is proposed in this paper..... "Nonlinear PID in Digital Controlled Buck Converters" \$\endgroup\$ – John D Mar 13 '15 at 15:36
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If you connect a simple linear gain, K, in series with the 1/(s-1) block, giving a transfer function: K/(s-1), and then form a negative feedback closed loop system around this new block, the closed loop transfer function will be K/[s+(K-1)], which is stable if K>1.

Another simple block that is unstable in the open loop is an integrator, which has transfer function: 1/s. It's easy to see that this is unstable by considering a unit step input. This will result in a unit ramp output (integral of a constant, A, is ramp, At), and this goes to infinity for large t. However, closing the loop around the integrator (with or without a series gain, K) will give a stable closed loop with transfer function: K/(s+K), or 1/(s+1). As well as being stable, the closed loop system has unity steady state gain (or 'DC gain'), which is characteristic of closed loop systems that have a pure integrator in the forward path transfer function.

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  • \$\begingroup\$ Where is non-linear element? \$\endgroup\$ – Saprativ Saha Mar 14 '15 at 3:01
  • \$\begingroup\$ You could have a non-linear gain which is dependent on the instantaneous error signal magnitude, say, E(t). Perhaps: K=(1+E(t))^2. But why would the element need to be nonlinear in the first place? Normally we try to make things as linear as possible. \$\endgroup\$ – Chu Mar 14 '15 at 11:10
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We have a system with the response $$G(s)={1\over s-1},$$ and we'd like a system with an arbitary (say: stable) reponse $$H(s)=H_2(s)H_1(s)G(s).$$ We presume that we can add the blocks H1 and H2, in that order, into the loop, after G.

Now suppose that H2 alone is linear and would give a stable response. But first, H1 would undo the effects of G - it would be the inverse response to G. Depending on G, H1 might need to be nonlinear. The hand-wavey part is that I'm not sure whether the composition H(s) can be written as above if H1 is nonlinear. If it can, then there's your answer, and you have the steps needed to do it:

  1. Set H1 to be the inverse of G.

  2. Set H2 to the stable open loop response you want.

Of course all of this might be a bunch of granola, as I haven't dealt with s domain in ages :/

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