I have a task where I need to implement a filter circuit (active) for the following transfer function:


So, we have one pole and one zero. The bode plot is as below (dB/Hz): Bode plot of G(s)

However, I can't for the life of me seem to design a circuit to implement it. A simple RC filter doesn't have any zeroes, and everything else I've tried (such as the circuit on page 12 of this document) leaves me stuck with terms involving negative real values or \$-\frac{1}{s}\$ which don't map to a real component. (My reasoning is that resistors contain positive real components, inductors \$sL\$ and capacitors \$\frac{1}{sC}\$.

I'm not sure if I'm just solving this kind of equation incorrectly, or if I'm using "bad" topologies to approach this specific transfer function. How would you go about implementing it? Using op-amps is completely fine.

(For context, my circuit needs to reverse the effect of an unknown, "black box" filter. I got my gain function by figuring out the black box's transfer function and inverting it.)

  • \$\begingroup\$ A simple RC filter doesn't have any zeroes Hmm, so a high-pass section in an RC filter doesn't have any zeros? There are methods for synthesizing RC / RLC networks from transfer functions. You should study a book on the subject. \$\endgroup\$ Oct 16, 2019 at 10:34
  • \$\begingroup\$ Did you derive the TF from the bode plot? \$\endgroup\$
    – Andy aka
    Oct 16, 2019 at 11:06
  • \$\begingroup\$ @Andyaka Yes, using a calculated value for the pole and zero. \$\endgroup\$
    – Mack
    Oct 16, 2019 at 11:09
  • \$\begingroup\$ Are you sure about the pole in the right-half plane? \$\endgroup\$
    – Chu
    Oct 16, 2019 at 11:15
  • \$\begingroup\$ I had exactly the same question as Chu asked: the roots for the numerator and the denominator are positive: you have a right-half-plane zero (RHPZ) and a right-half-plane pole (RHPP). The RHPZ can be implemented in an all-pass filter but certainly not the RHPP which would lead to an instable system. Besides, if you factor 5447 and 961, the leading term is 5.66 indicating that this expression describes a system with gain at dc: cannot be implemented with simple \$RC\$ elements and an active circuit is needed. \$\endgroup\$ Oct 16, 2019 at 11:17

1 Answer 1


The best is to rewrite your transfer function in the proper low-entropy way, with a leading term. I have assumed the roots are all negative leading to left-half-plane pole and zero:

\$G(s)=\frac{s+5447}{s+961.3}=G_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ where:


\$\omega_z=5447\$ rd/s or 867 Hz

\$\omega_p=961.3\$ rd/s or 153 Hz

There are plenty of ways to implement such a filter. One way is to use an op-amp in an inverting configuration and have it followed by another inverting structure with a gain of -1:

enter image description here

You then calculate the values of the various components according to the below Mathcad sheet. These values can be determined using fast analytical techniques or FACTs.

enter image description here


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