Stability of an open loop controller

  • Thread starter icesalmon
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Given the following Controller equation Gol(s) and Plant equation Dol(s) for an open loop system the transfer function can be expressed as a ratio of polynomials where:
Gol(s) = b(s)/a(s)and Dol = c(s)/d(s).

For the open loop system the transfer function Tol = Gol(s)Dol(s) = b(s)c(s)/a(s)d(s), the roots of the characteristic equation (the denominator) of this transfer function cannot have any roots in the RHP.

What i'm confused about is that my notes say "An attempt to cancel unstable roots of a(s) of the plant by using c(s) of the controller will be useless. Although, cancelled, physically the unstable pole still remains. The slightest modelling uncertainty will cause the output to diverge"

This doesn't make sense to me as it seems to be saying that the mathematical modelling of the physical plant doesn't actually fully impact what happens physically. How is this possible? I would think that if you design a controller based off of an equation that counteracts the instability of the other controller, how can this same thing not happen when you physically build the thing?
 
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  • #2
anorlunda
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The slightest modelling uncertainty will cause the output to diverge"
I would think that if you design a controller based off of an equation that counteracts the instability of the other controller,
It is saying that in real life your counteraction can not be perfect, the modeling uncertainty becomes a fatal flaw.
 
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  • #3
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This makes more sense to me now, thank you!
 
  • #4
Baluncore
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A pole on the RHP is a vertical asymptote. It represents an exponentially increasing wave.
Placing a zero in the exact same position is more difficult than eliminating the pole from the RHP.
The only sure way to stabilise the system is to remove the pole from the RHP.
 

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