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How can analytically work out the gains knowing the required bandwidth and damping? The plant to be controlled is first order. The low pass filter is 2nd order in the feedback path.

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  • $\begingroup$ More information is required. What is the bandwidth and damping of the 2nd order LPF in relation to the required bandwidth and damping of the final closed loop system and the plant ? Please provide all information available with you. Why is PI controller required ? Why not just P ? $\endgroup$
    – AJN
    Commented Oct 3, 2022 at 15:18
  • $\begingroup$ Thanks for your reply. The bandwidth of the overall system should be 1000Hz and the damping isn't strictly specified but let's say 0.707. So bandwidth of the LPF i think also should be 1000Hz and bandwidth also something similar. Specs require a PI controller. $\endgroup$
    – Ben
    Commented Oct 3, 2022 at 15:29
  • $\begingroup$ So the details of the LPF are not available to you ? Are they fixed like the plant or can be tuned like the PI gains ? Please give all details about the problem. $\endgroup$
    – AJN
    Commented Oct 4, 2022 at 11:57

1 Answer 1

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The characteristic equation of the closed loop transfer function is the numerator of $$ 1 + \frac{A}{(s+B)}\frac{\omega_{LPF}^2}{(s^2+2\xi_{LPF}\omega_{LPF}+\omega_{LPF}^2)}\left(K_P + \frac{K_I}{s}\right) $$ which simplifies to $$ s \cdot (s+B) \cdot (s^2+2\xi_{LPF}\omega_{LPF}+\omega_{LPF}^2)+ A \cdot \omega_{LPF}^2 \cdot (K_P \cdot s + K_I) $$

By equating the coefficients of this equation to that of a desired system (whose dominant poles need to be near the magnitude $1000\cdot2\pi$), you might be able to work out the controller gains analytically.

Since there are 4 poles and only two tunable parameters, an analytic solution might not exist for any given arbitrary combination of 4 poles.

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  • $\begingroup$ Hi, Thank you for your reply. I want the closed loop system to be a 2nd order system. How do I equate this fourth order equation to that of 2nd order? $\endgroup$
    – Ben
    Commented Oct 11, 2022 at 13:03
  • $\begingroup$ AFAIK, you can't really reduce the order of a system. There is the concept of dominant poles 1, where one or two poles dictate most of the response of the system with other poles contributing less. IIRC, you can achieve that by placing the natural frequency of the third and fourth pole five or ten times (say $10,000\cdot 2 \pi$) that of the first two poles ($1000\cdot 2 \pi$). $\endgroup$
    – AJN
    Commented Oct 11, 2022 at 13:29
  • $\begingroup$ You need to equate two fourth order systems AFAIK. $\endgroup$
    – AJN
    Commented Oct 11, 2022 at 13:33

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