I would like to synthesize an arbitrary passive network based on a given transfer function (poles/zeros). The origin is the output impedance of a voltage regulator for which I want to find an equivalent circuit.

This is routinely done and I would like to automate this. The structure does not need to be the same as in the link; I am happy with an arbitrary passive network (ladder, series connections etc) that I can draw on a paper. In my case, I determine the output impedance by performing a sinusoidal (small signal) current sweep at the output and measure amplitude/phase of voltage.

$$ H(s) = \frac{v_o}{i_t} = \frac{8451 s^2 + 2.507\cdot10^{10} s + 6.126\cdot10^{10}}{s^2 + 8.681\cdot10^{10} s + 1.351\cdot10^{15}} $$

(Note that the example above is the pure regulator without the output caps which might be unrealistic; If this is an issue, an additional output cap can be added which generates an additional pole; my question is general and does not relate to these exact numbers)

The idea about Foster synthesis is to use partial fraction expansion to create a series network. Using MATLAB:

[r,p,k] = residue([8.4505e+003, 25.0724e+009, 61.2553e+009], [1.0000e+000 86.8070e+009 1.3513e+015])
r =
p =
k =

which is one resister (8.4505e+003) in series with two single pole networks. The single pole networks can be represented via parallel RC:

$$ \frac{r}{s-p} = \frac{\frac{1}{C}}{\frac{1}{RC}+s} $$

From here it can be seen that the resistors have to be negative which is of course not desired.

How can I easily make a passive network from that transfer function?

  • 2
    \$\begingroup\$ I'm just reading your writing and it's not clear to me, yet. What passive topology do you see as a way of achieving a 2nd order for both \$N_s\$ and for \$D_s\$? Could you please use the schematic editor and draw it out? Just to be clear, I mean. \$\endgroup\$
    – jonk
    Jun 4, 2022 at 1:20
  • 1
    \$\begingroup\$ Just breaking down your transfer function, I find:$$8.5405\:\text{k}\cdot\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}+288.83\:\text{m}\cdot\frac{2\zeta\,\omega_{_0}s}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}+45.331\:\mu\frac{\omega_{_0}^{\:2}}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}\\\\\text{where: }\omega_{_0}\approx 36.76\:\text{M}\frac{\text{rad}}{\text{s}}\text{ and }\zeta\approx 1180.73$$I may be a little curious about how one of those gain figures is to be passively achieved. \$\endgroup\$
    – jonk
    Jun 4, 2022 at 1:54
  • 1
    \$\begingroup\$ Show me a trial and error rlc example that comes close. I'm interested. \$\endgroup\$
    – jonk
    Jun 4, 2022 at 5:41
  • 1
    \$\begingroup\$ It seems @jonk's hints go unnoticed, sorry for spoiling. Passive "amplification" is, usually, not found in full biquads, only in very high Q filters, or some very clever, but unpractical circuits, and even then for a very limited bandwidth (in fact, the higher the Q, the narrower the BW), or the lower the Q the lower the "gain". And your high freq gain is flat and impossibly high (~100 GHz!!). Where does this tf come from? \$\endgroup\$ Jun 4, 2022 at 11:47
  • 1
    \$\begingroup\$ I think you want to synthesize a passive network that has a driving point impedance given by the Laplace expression. This is more complicated than you might think. You can check if it is possible at all by seeing if the expression is POSITIVE REAL. If it is not, a passive impedance cannot be synthesized, although active devices can be used. A simple process exists to realize any passive impedance, but uses ideal transformers. Realizing a general RLC impedance without transformers is more complicated. Are you doing this for simulation? \$\endgroup\$
    – user69795
    Jun 6, 2022 at 3:42

4 Answers 4


With Verbal Kint's contribution, I'll expand on thoughts behind my earlier comment(s).

Notes from: "A Practical Method of Designing RC Active Filters"

I want to start out by reflecting on the TR-50 paper by R. P. Sallen & E. L. Key, dated 6 May 1954 (I'm not aware of a link on the web to it at this time.) This earlier edition is still controlled by the US Air Force but has been unclassified and has a more material in it than the abridged version that was published the next year, in 1955.

The authors focus' is on active networks, with active gain stages using vacuum tubes, and therefore only give a small nod towards passive networks (as a basis upon which to build these active network filters.) But they do provide some handy results to apply in parsing 2nd order transforms of the form (where \$a_i\$ and \$b_i\$ are all real, positive constants such that \$a_i\ge 0\$ and \$b_i\gt 0\$:

$$G_s = \frac{N_s}{D_s}=\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0}$$

Given that, they focus on \$D_s\$ and provide that if \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$d=\frac{b_1}{\sqrt{b_2 \,b_0}}\$ (\$d=2\zeta\$ in modern usage), then \$D_s\$ can be factored as:


and where the zeros of \$D_s\$ (poles, when placed in the denominator) line on a circle with radius \$\omega_{_0}\$ with the real part at \$-\frac12 d\,\omega_{_0}=\zeta\,\omega_{_0}\$ in the under-damped case when \$\zeta\le 1\$. (The over-damped case has all the zeroes directly located on the negative real axis.)

The shape is determined solely by \$d\$ (or \$\zeta\$ or \$Q\$), while \$\omega_0\$ determines the positions of the zeroes in the frequency domain and \$b_0\$ is merely a relative amplitude value.


Verbal Kint prefers (as I see it) the use of \$Q\$ over \$\zeta\$ or \$d\$. The relationship is: \$Q=\frac1{2\zeta}=\frac1{d}\$.

Which is preferred depends on where your brain is at, I think. I don't think of it as an always \$Q\$ or always \$\zeta\$ kind of thing.

When dealing with under-damped situations, I tend to think more in terms of \$Q\$.

When dealing with wide bandpass situations (over-damped, for sure) then I tend to think more in terms of \$\zeta\$.

And when it's just a matter of mathematics I tend to stick closely to \$\zeta\$ as I prefer that over fractions of the form \$\frac1{Q}\$, which is more of a pain when writing out equations.

Breaking Down Your Transfer function

In your case, \$a_2=8.4505\times 10^3\$, \$a_1=25.0724\times 10^9\$, \$a_0=61.2553\times 10^9\$, \$b_2=1\$, \$b_1=86.807\times 10^9\$, and \$b_0=1.3513\times 10^{15}\$. Then it is just a matter of following the above recipe:

$$\begin{align*} &\text{Numerator} &&\text{Denominator} \\\\ \omega_{_0}^{\:'}&=\sqrt{\frac{a_0}{a_2}\vphantom{\frac{b_0}{b_2}}}\approx 2.692\times 10^3 & \omega_{_0}&=\sqrt{\frac{b_0}{b_2}}\approx 36.76\times 10^6 \\\\ d^{\,'}=2\zeta^{\,'}&=\frac{a_1}{\sqrt{a_2 \,a_0\vphantom{b_2 \,b_0}}}\approx 1102.0 &d=2\zeta&=\frac{b_1}{\sqrt{b_2 \,b_0}}\approx 2361.5 \end{align*}$$

This results in the following:

$$\begin{align*} G_s&=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2+d^{\,'}\left(\frac{s}{\omega_{_0}^{\:'}}\right)+1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \end{align*}$$

Before pursuing this as Verbal Kint did (I'll get to that in a later section), I'd like to follow through with a different process that leads to an interesting, separate result:

$$\begin{align*} G_s&=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2+d^{\,'}\left(\frac{s}{\omega_{_0}^{\:'}}\right)+1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{d^{\,'}\left(\frac{s}{\omega_{_0}^{\:'}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{\omega_{_0}}{\omega_{_0}^{\:'}}\right)^2\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{\left(\frac{\omega_{_0}}{\omega_{_0}^{\:'}}\right)\left(\frac{d^{\,'}}{d}\right)d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\frac{a_2}{b_2}\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{a_1}{b_1}\frac{d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{a_0}{b_0}\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1} \end{align*}$$

The last result above is what's important. It converts what appears to be an abstract and somewhat impenetrable pair of 2nd order polynomials into the key ideas required to understand their combined behaviors. (The first term is a high-pass, the middle term is a band-pass, and the final term is a low-pass and the gains for each are separated out, now.)

Note that it also now becomes more obvious why \$a_i\$ can be zero while \$b_i\$ cannot be.

As you can see, the whole thing breaks into a nicely expressed standard form.

And this is where I got the gain factors for my earlier comment: \$\frac{a_2}{b_2}\approx 8.5405\:\text{k}\$, \$\frac{a_1}{b_1}\approx 288.83\:\text{m}\$ and \$\frac{a_0}{b_0}\approx 45.331\:\mu\$ (this last one is Verbal Kint's \$G_0\$.)

You can feel free to replace \$d\$ with either \$2\zeta\$ or \$\frac1{Q}\$. Doing that and multiplying everything by \$\frac{\omega_{_0}^{\:2}}{\omega_{_0}^{\:2}}\$ yields:

$$\begin{align*} G_s&=\frac{a_2}{b_2}\cdot\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}+\frac{a_1}{b_1}\cdot\frac{2\zeta\,\omega_{_0}s}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}+\frac{a_0}{b_0}\cdot\frac{\omega_{_0}^{\:2}}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}} \end{align*}$$

Which is the other standardized form and the one I indicated in comments, earlier.

A Short Note

More commonly, a transfer function only includes one of the above three terms and is either entirely a low-pass, entirely a high-pass, or entirely a band-pass transfer function.

But either of these simpler passive networks will result in a transfer function that includes all three terms:


simulate this circuit – Schematic created using CircuitLab

Other arrangements with different elements can be readily produced. But the above are, in my mind anyway, two of the simplest ones using ideal components that can get you there.


Let's return to (I promised to get back to Verbal Kint's answer):

$$\begin{align*} G_s&= \frac{a_0}{b_0} \cdot \left[ \frac{ \left( \frac{s}{\omega_{_0}^{\:'}} \right)^2+d^{\,'}\left( \frac{s}{\omega_{_0}^{\:'}} \right) + 1 }{ \left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1 } \right] \end{align*}$$

Verbal Kint specified that \$G_0=\frac{a_0}{b_0}\$ and focused his attention upon on the roots. In the over-damped case (your numerator and denominator are both way over-damped) the solutions are:

$$\begin{align*}\text{zeros}\left\{\begin{array}{l}s_1&=-\omega_{_0}\left(\zeta-\sqrt{\zeta^2-1}\right)&\text{and}& \omega_{_\text{L}}=-s_1\\s_2&=-\omega_{_0}\left(\zeta+\sqrt{\zeta^2-1}\right)&\text{and}& \omega_{_\text{H}}=-s_2 \end{array}\right.\end{align*}$$

In the over-damped case, both \$\omega_{_\text{H}}\$ and \$\omega_{_\text{L}}\$ are real-valued. So, the transfer function can also be written out as:

$$\begin{align*} G_s&=G_0\cdot\left[\frac{\left(1+\frac{s}{\omega_{_\text{L}}^{\:'}}\right)\left(1+\frac{s}{\omega_{_\text{H}}^{\:'}}\right)}{\left(1+\frac{s}{\omega_{_\text{L}}}\right)\left(1+\frac{s}{\omega_{_\text{H}}}\right)}\right] \end{align*}$$

Computing the values I find:

$$\begin{align*} \omega_{_\text{L}}^{\:'}&\approx 2.44314 &f_{_\text{L}}^{\:'}&\approx 388.83\:\text{mHz} \\ \omega_{_\text{H}}^{\:'}&\approx 2.96697\times 10^6& f_{_\text{H}}^{\:'}&\approx 472.208\:\text{kHz} \\ \omega_{_\text{L}}&\approx 15.5667\times 10^3 &f_{_\text{L}}&\approx 2.47752\:\text{kHz} \\ \omega_{_\text{H}}&\approx 86.807\times 10^9 &f_{_\text{H}}&\approx 13.816\:\text{GHz} \end{align*}$$

Note that Verbal Kint asked you if \$\omega_{_\text{H}}\approx 13.816\:\text{GHz}\$ is useful as a pole. It's probably not as its factor is very, very close to 1 for any frequency most of us encounter. (Of course, you can't just drop it out without also then being forced to modify the numerator in some reasonable way, as well. You don't get to have numerator that is a higher order than the denominator.)

Relationships between \$\omega_{_\text{H}}\$, \$\omega_{_\text{L}}\$, and \$\zeta\$

For the over-damped situation, it turns out that \$\omega_{_\text{H}}\$ and \$\omega_{_\text{L}}\$ relate directly to \$\zeta\$:

$$\zeta=\frac12\frac{\omega_{_\text{L}}+\omega_{_\text{H}}}{\sqrt{\omega_{_\text{L}}\cdot \omega_{_\text{H}}}}=\frac12\frac{\omega_{_\text{L}}+\omega_{_\text{H}}}{\omega_{_0}}$$

It's pretty easy to see that as \$\omega_{_\text{L}}\to \omega_{_\text{H}}\$ then \$\zeta\to 1\$. But for \$\omega_{_\text{H}}\gg \omega_{_\text{L}}\$, then \$\zeta\to \frac12\sqrt{\omega_{_\text{H}}}\$.

It follows that wide bandwidth 2nd order filters tend to have large values for \$\zeta\$ (and therefore very small values for \$Q\$.) This fact helps in designing such filters. If \$\zeta\$ is large, then the design will tend towards using separate low-pass and high-pass filters, as separate sections that are concatenated. However, if \$\zeta\$ is small (\$Q\$ is larger), then the design will tend towards a unified bandpass design, instead.

There are a couple of interesting terms used to describe the fractional bandwidth of a system. One is \$B_{_\text{F}}=\frac{\omega_{_\text{H}}-\omega_{_\text{L}}}{\omega_{_0}}\$. The other is \$B_{_\text{F}}^{\:'} =2\frac{\omega_{_\text{H}}-\omega_{_\text{L}}}{\omega_{_\text{H}}+\omega_{_\text{L}}}\$. These are related to each other in this way: \$B_{_\text{F}}=\zeta\cdot B_{_\text{F}}^{\:'}\$. It turns out that \$0 \le B_{_\text{F}}^{\:'}\le 2\$ but that \$B_{_\text{F}}\to \sqrt{\frac{\omega_{_\text{H}}}{\omega_{_\text{L}}}}\$ for large ratios of \$\frac{\omega_{_\text{H}}}{\omega_{_\text{L}}}\$.

Just be aware that fractional bandwidth doesn't have a single definition in the literature.


I'm still wondering what passive network you imagine duplicating your original transfer function. (The comment expression you gave certainly doesn't cut it.) Though perhaps you had also simplified your expression?

I remain curious and interested. Draw out a schematic and provide part values. Not another equation. I'd like to see the topology and details and I'll work out the transfer equation on my own.

But I suspect that by hacking things around, you did in fact find functions that were usefully close to the original and felt you had succeeded. (Not strictly as a matter of mathematics but as a practical matter.) If so, all you likely did is, by trial and error, find a usefully simplified expression.

I think Verbal Kint was trying to move you towards a similar place but through a rigorous mental approach rather than trial and error.

Added Notes Per Comments from OP

If I understand correctly, this applies to voltage transfer functions.

No, it applies to all transfer functions. In voltage-in/voltage out cases the units are \$\frac{V}{V}\$, which can be treated as unitless. But in cases where the input units are current, for example, and the output is voltage then the gain value will carry units (Ohms, for example.)

However, my example is about the output impedance, described via transfer function H(s) (apologize if this is not the right nomenclature) which I want to find an equivalent circuit. I am not sure if active filter design relates to this.

Again, you can have a gain with units, such as Ohms. For example, suppose the following case:


simulate this circuit

The transfer function is:

$$\begin{align*} G_s&=A\cdot\left[\frac{1}{\left(1+\frac{s}{\omega_{_\text{L}}}\right)\left(1+\frac{s}{\omega_{_\text{H}}}\right)}\right] \end{align*}$$

In this case, \$A=\frac{R_1\,R_4}{R_1 + R_2 + R_3 + R_4}\$ and the units are definitely \$\Omega\$.

I am not sure if active filter design relates to this.

The mathematics approach is similar. I'm not sure there's a problem. But I suppose you can identify one?

Does this apply to an output impedance as well?

Yes. As indicated above.

I don't mind the exact structure of the equivalent passive circuit, I just want to be able to draw it. Could be a ladder, shunt networks or similar.

Then perhaps you should refer to Verbal Kint's reference here.

I don't want to provide, nor do I have the time or inclination to do so, a general approach for arbitrary order. I'm not sure one exists as I imagine the set of possible answers has often more than one item in it. I don't believe such a proof exists, but if you can prove that there is always one and only one possible answer in the set, I may yet give it a hack. But I'll need to see the mathematical proof before I bother.

  • \$\begingroup\$ Hi @jonk, if I understand correctly, this applies to voltage transfer functions. Of course gain >1 will need active gain (or some sort of resonance). However, my example is about the output impedance, described via transfer function H(s) (apologize if this is not the right nomenclature) which I want to find an equivalent circuit. I am not sure if active filter design relates to this. Does this apply to an output impedance as well? I stick with my belief that this is possible: signalintegrityjournal.com/articles/… \$\endgroup\$
    – divB
    Jun 6, 2022 at 22:02
  • \$\begingroup\$ My problem is very similar to that with the exception (a) Possibly more complex transfer function (=more poles/zeroes due to C/ESR/ESL) (b) I don't mind the exact structure of the equivalent passive circuit, I just want to be able to draw it. Could be a ladder, shunt networks or similar. \$\endgroup\$
    – divB
    Jun 6, 2022 at 22:04
  • \$\begingroup\$ @divB The gain in front of these equations isn't necessarily unitless. It can have Ohms, for example, or Siemens. Etc. The gain can carry units. The rest cannot. \$\endgroup\$
    – jonk
    Jun 7, 2022 at 3:28
  • \$\begingroup\$ @divB If you find disagreeable axioms or a single case where I fail to apply appropriate logic in reaching sound conclusions feel free to point them out. But this is pretty much cut-and-dried stuff, bounced around by dozens if not hundreds of minds from almost 100 years ago, and I'm drawing from the better of those who applied themselves. I may be misrepresenting them. If I am, that is my mistake and not theirs. Also, please read my comment to this question. There, I point out the units are Ohms for the gain part of the transfer equation. \$\endgroup\$
    – jonk
    Jun 7, 2022 at 3:54

I would first rewrite the expression in a low-entropy form in which the leading term appears. I assume it is in ohms if you modeled an output impedance. Then, I would check the quality factor and, if low enough, the low-\$Q\$ approximation should hold well. Let's see how it works with a quick Mathcad sheet:

enter image description here

Once the equation is reformatted, you can check the quality factor of the numerator and the denominator and confirm it is extremely low. That way, you can factor both expressions as cascaded poles and zeroes. Actually, one pole is stratospheric so it can be neglected without affecting the transfer function (TF). And if you check the magnitude and phase plots for the various TFs, they perfectly superimpose:

enter image description here

With this last expression, one needs to find a structure satisfying the given time constants. Now, the question you may ask yourself is regarding how closely do you need to match the exact expression. If you deal with a voltage regulator, it is unlikely that you want to observe phenomenon beyond 1 MHz unlike this is a very special case? In the simplified expression, we see a low-frequency zero which makes the impedance take off at \$\approx\$ 0.4 Hz with a +1 slope then a zero makes it flat at 2.5 kHz and, finally, a second zero restarts a +1 slope but at almost 500 kHz. Unless you have an extremely-high crossover frequency, wouldn't a simple pole-zero combination for this output impedance model be good enough?


The following method will allow you to turn any transfer function of the form...

$$H(s) = \frac {N_0 + N_1 s + N_2 s^2 + N_3 s^3 ...} {D_0 + D_1 s + D_2 s^2 + D_3 s^3 ...}$$

Into a circuit consisting of resistors, capacitors, and op-amps. We are using op-amps, so it's not totally passive. That makes sense because the transfer function may indeed have gain even at DC. Also this method only works if the roots of the denominator are real.

  1. First find the roots R0...RN of the denominator so that we can rewrite H(s) as...

$$H(s) = \frac {N_0 + N_1 s + N_2 s^2 + N_3 s^3 ...} {(s-R_0) + (s-R_1) + (s-R_2) + (s-R_3) ...}$$

  1. Use partial fraction expansion to rewrite the H(s) as...

$$H(s) = C_0\frac {N_0 + N_1 s + N_2 s^2 ...} {(s-R_0)} + C_1\frac {N_0 + N_1 s + N_2 s^2 ...} {(s-R_1)} + C_2\frac {N_0 + N_1 s + N_2 s^2 ...} {(s-R_2)}$$

  1. Next take the inverse Laplace transform of each fraction. Note that...

$$L^{-1} [\frac {C} {(s-R)}]=Ce^{-Rt}$$

And that multiplying s^X in the Laplace domain is equivalent to taking the derivative X times in the time domain. Therefore...

$$L^{-1} [C \frac {s^X} {(s-R)}]=(-R)^XCe^{-Rt}$$

so the entire inverse laplace transform is going to have the form...

$$H(t) = C_0 N_0 e^{-R_0t} + C_0 N_1 (-R_0)e^{-R_0t}+ C_0 N_2 (-R_0)^2e^{-R_0t} ... + C_1 N_0 e^{-R_1t} + C_1 N_1 (-R_1)e^{-R_1t}+ C_1 N_2 (-R_1)^2e^{-R_1t} ... + C_2 N_0 e^{-R_2t} + C_2 N_1 (-R_2)e^{-R_2t}+ C_2 N_2 (-R_2)^2e^{-R_2t} ... $$

Note that this is just a sum of decaying exponentials...

Each exponential term can be modeled as a resistor + capacitor and a gain factor. And then turned into the following op-amp circuit.


simulate this circuit – Schematic created using CircuitLab

I leave it to you to figure out the resistor and capacitor vlues to come up with the proper time constant and gain on each RC network. Note that filters with negative gain (from the odd powers of s) connect to the negative input of the final op-amp, and the filters with positive gain (from even powers of s) connect to the positive terminal.


Use the Brune method: https://dspace.mit.edu/handle/1721.1/10661. This method works by removing just enough of the real part of the impedance to create some poles on the jw axis. Then the poles are realized by a coupled inductor and capacitor network. This is done for each complex pole pair. It is difficult to physically make exact coupled inductors, but if this is just for simulation, it is easy. Active impedances can be synthesized by using negative resistors in the same procedure.
The first thing you would want to do is to frequency scale your expression so that you won’t run into numerical precision problems with your calculations. Brune’s paper is fine, but you can also read about it in textbooks on electrical network synthesis.


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