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So while answering how to design a PI controller for a first order time delayed system (Question Here )

Here is the closed loop equation to a control system:

$$ G_C(s) = \frac{\frac{K}{T}(1-sT)(s)} { s^3 + (\frac{1}{T} + a - KK_p)s^2 + (\frac{a}{T} + \frac{KK_P}{T} +K_I)s+\frac{KK_I}{T}} $$

Question: How do you deal with normalizing the numerator in your closed loop transfer function when the filter is unstable? (Pole on RH of plane)

Typically you introduce a filter before your controller that does:

$$ \frac{1} {\frac{K}{T} (1-sT)(s)} $$

to normalize the numerator

But the filter itself is unstable because of the term:

$$ \frac{1}{(1-sT)}$$ is unstable for a step response which would create an issue realizing the system at all.

One way I've thought about dealing with this is multiplying it by its complex conjugate $$ \frac{(1+sT)} {(1+sT)}$$

but im not really sure about the merits of it.

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  • $\begingroup$ Good question. Controls have never been my strong suit, but are you sure that you would want to add a filter to the loop with that response? Since it's a closed-loop system, adding a filter $F(s)$ to the feedback arm doesn't just multiply the closed-loop transfer function by $F(s)$. Also, I'm not sure what multiplication by the complex conjugate would do; the pole is still there in the right-half plane. $\endgroup$ – Jason R Apr 20 '12 at 13:28
  • $\begingroup$ the complex conjugate is a time delay. $\endgroup$ – CyberMen Apr 23 '12 at 12:55
  • $\begingroup$ I'm still not sure what you mean. $\frac{1+sT}{1+sT} = 1$, not a time delay. And if you introduce that filter inside the feedback loop, it doesn't just multiply the closed-loop transfer function (because of the feedback). If you were trying just to cancel the zero, you would want it outside of the loop. As you noted, however, such a filter would be unstable. It's possible that this is just par for the course with PI control; excessive delay in the loop causes instability due to the integrator. Note that if the delay is small in the original system, $e^{-sT} \approx 1$, and could be neglected. $\endgroup$ – Jason R Apr 25 '12 at 2:08
  • $\begingroup$ @JasonR I was thinking about reformulating the equation by using the complex conjugate to write a more suitable circuit. $\endgroup$ – CyberMen Apr 25 '12 at 12:46
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    $\begingroup$ Why do you want to normalise the numerator? $\endgroup$ – lxop Mar 19 '13 at 23:11
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In general, in order to stabilise a system, however complex, if you have the transfer function $G(s)$, you introduce a feedback loop with a new function $F(s)$.

Write the closed loop transfer function for the new system with the added $F(s)$, and then find $F(s)$ in order for the new system to be stable. This is like the first exercise in any control book in order to offer an example of stabilising a system via negative Feedback.

Check the book from Ogata on control engineering for reference.

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