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What is the best way to design a circuit based on the following transfer function: (0.0364*s)/(0.0002*s + 1) ? I am trying to design the derivative stage of a PIDF controller using op-amps

EDIT: This is part of a bigger problem where i have to design a PIDF controller to achieve the minimum possible settling time for the plant function: $$\ G(s)= \frac{100*20*100}{(s+10)(s+20)(s+100)}$$

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  • $\begingroup$ Welcome to Engineering! This looks like a "homework question" (notice the quotation marks). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$
    – Wasabi ♦
    Commented Feb 23, 2017 at 22:32

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To my knowledge the simplest way to accomplish the transfer function...

$$ G(s) = \frac{E_o(s)}{E_i(s)} = \frac{0.0364s}{0.0002s + 1} = \frac{KTs}{Ts +1} $$

...is the following high-pass filter circuit: High pass filter with opamp gain

The transfer function of this circuit is:

$$ \frac{E_o(s)}{E_i(s)} = \frac{R_2+R_3}{R_3}\frac{R_1 C s}{R_1 C s + 1} $$

So let $R_1C = T = 0.0002$ and $\frac{R_2+R_3}{R_3} = K = 0.0364/0.0002$. Then solve for $R_1$, $R_2$, $R_3$ and $C$. You'll have some wiggle room as to what values you pick, because you have two equations and four parameters. There are multiple solutions and you will have to decide what combination of values work best for your application.

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  • $\begingroup$ Is there any way to modify the above circuit so i can get negative output? Or do i need to connect it to an inverter circuit? $\endgroup$
    – Shivalnu
    Commented Feb 23, 2017 at 22:12
  • $\begingroup$ I'm not entirely sure what you mean. This is just a low-pass filter. If you give it a constant negative input it will eventually approach a constant negative output. Similarly for a constant positive input. Generally it depends on what you supply your op-amp with. If you give it +12 V and -12 V with respect to ground then it will be able to output both positive and negative voltages up to the supply voltages. $\endgroup$
    – ConjuringFrictionForces
    Commented Feb 23, 2017 at 23:16
  • $\begingroup$ Btw, I haven't included the supply terminals in the diagram above, just the input and output terminals. $\endgroup$
    – ConjuringFrictionForces
    Commented Feb 23, 2017 at 23:17

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