I have a transfer function, for example:

$$H(s)= P\frac{10}{s(s+5)(s+9)}$$

See the figure below.

How do I make a linear RLC circuit from this transfer function, without using an op-amp, but a linear circuit from RLC passive components?

Compesates Responses

I have made a control circuit. I compensated the circuit so it has better settling time and good overshoot.

I usually make a simple circuit so it's easier for me to find the RLC circuit, but this one grew into a rather complex circuit and I find it confusing to make a circuit from the transfer function.

I'm trying to focus on a linear circuit. Not in the control domain; I only need to know how to go from transfer function to circuit.

I can figure out H(s) from an RLC circuit. Now I'm planning to make a method to reverse it from end to front.

Can somebody explain how to do this? I can't find any literature about it on websites.

I found out the transfer function can be separated to:

$$H(s) = 10\cdot\frac{1}{s}\cdot\frac{1}{s+5}\cdot\frac{1}{s+9}$$

Is this an answer of my problem, but it becomes an open-loop transfer function, not closed-loop. It is a closed loop but the circuit become open loop.

It becomes the same transfer function, but a different RLC circuit.

  • \$\begingroup\$ I think no. Its only the pole of it. No difficultty. \$\endgroup\$
    – adhitronic
    Feb 7, 2023 at 0:00
  • \$\begingroup\$ Right. I usually make a simple circuit so its kind easy for me to find the RLC circuit. But it growing into a rather complex circuit and i confusing to make a circuit from tf. \$\endgroup\$
    – adhitronic
    Feb 7, 2023 at 0:04
  • \$\begingroup\$ Im trying to focus in Linear Circuit. Not in control domain. Only need to know how to reverse from tf into circuit. \$\endgroup\$
    – adhitronic
    Feb 7, 2023 at 0:07
  • \$\begingroup\$ For simple explain. You may change the 3 into another one. For example 4. It is not an important for studying. Only a schems how it works. Even example in phase respons of tf. \$\endgroup\$
    – adhitronic
    Feb 7, 2023 at 0:22
  • \$\begingroup\$ So were the same. I have change to 9. Its for simplicities. Im trying to make a figure of the transfer function as simple as possible. Im already been confuse about those to. May because its stability in Phase response right. Lets say its imposible even can be realizing it. \$\endgroup\$
    – adhitronic
    Feb 7, 2023 at 0:42

1 Answer 1


I don't think it's possible. Recall from quadripoles (two port networks) theory the impedance parameters enter image description here

$$ \begin{pmatrix} V_1\\ V_2\\ \end{pmatrix} = \begin{pmatrix} z_{11} &z_{12}\\ z_{21} &z_{22}\\ \end{pmatrix} \cdot \begin{pmatrix} I_1\\ I_2\\ \end{pmatrix} $$

The transference is understood as \$V_2 / V_1\$ when \$I_2 = 0\$. That leaves us with \$H(s) = z_{21} / z_{11}\$ Now, you are asking for the transference

$$H(s)= P\frac{10}{s(s+5)(s+9)}$$

which has three poles and none zeros. But remember from impedance network synthesis theory that \$z_{11} (s)\$ and \$z_{21} (s)\$ must be positive real functions and therefore they are rational polynomial functions where the difference of poles and zeros must be equal or less than one.

So, the highest difference of poles or zeros in a transference function can't be more than two. With no zeros, you can have up to two poles, which is the case when \$z_{21}\$ has one pole and no zeros and \$z_{11}\$ has one zero and no poles.

To synthetize the transference you want, you will need at least two op-amps. I would use a topology for the two poles out of the origin (for example Sallen Key) in cascade with an integrator to add the pole in the origin.

If your transference had only two poles, that would be a simple RLC or RCRC second order circuit, which you can found the transference formula and design criteria in thousands of books and sites


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