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So I recently worked with a problem where I was given a bode plot where magnitude of the transfer function (the ratio between the output and input voltage) was plotted against frequency of the system. From there, I needed to construct a circuit only from this.

I didn't quite know where to start. First, I thought that I could split it into different regions, for the constant, increasing and decreasing parts. From there I thought that I may deduce what the transfer function has to look like. But still it doesn't simplify the problem for me that much since it's hard from the algebraic expressions directly know what type of circuitry it has to correspond to.For instance this bode diagram:

enter image description here

I know this question is quite broad, but I'd wonder if there's any good strategy for reconstructing a circuit given the frequency response? Maybe there's someone out there who does this for a job, and I'd like to know more about your thought process when solving such problems. I understand that it also takes alot of experience, but for a newbie, what's the general way to do this?

Tanamas
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  • You mean just a RCL circuit or are e.g. transistors allowed? – Qmechanic Sep 29 '22 at 10:58
  • if you provide Bode diagram, it would be possible to have a better idea of your problem. Maybe a low-order system is enough, without going into the whole theory of systems and identification techniques – basics Sep 29 '22 at 11:02
  • @Qmechanic Yes, RCL including op-amps. – Tanamas Sep 29 '22 at 12:59
  • @Farcher Is there a way to move the question without having to delete it? – Tanamas Sep 29 '22 at 12:59
  • @basics Sure, I'll add one I think would be reasonable for me to being able to solve. – Tanamas Sep 29 '22 at 13:00
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    [What is migration and how does it work?](https://meta.stackexchange.com/questions/10249/what-is-migration-and-how-does-it-work) but why not just cut and paste your post? – Farcher Sep 29 '22 at 13:05

3 Answers3

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This is the Bode plot of a second-order system with subcritical damping \$\xi < 1\$ and natural frequency \$\omega_n = 1.\$, that can be represented in time domain as

\$\ddot{x}(t) + 2\xi\omega_n \dot{x}(t) + \omega_n^2 x(t) = \dfrac{f(t)}{m}\$

and in frequency domain as

\$\hat{x}(\omega) = \dfrac{1}{m} \dfrac{1}{\omega_n^2 - \omega^2 + i 2\xi \omega_n \omega} \hat{f}(\omega) = G(\omega) \hat{f}(\omega)\$.

You can find the values of \$m\$ and \$\xi\$ parameters looking at the plot.

Gain and value of \$m\$. Looking at the value of the transfer function in $\omega = 0$ (steady response), you can find the value of the parameters \$m\$

\$G(0) = \dfrac{1}{m \omega_n^2}\$.

Looking at the plot and using the value of \$|G(\omega)|_{dB} = 6 \, dB\$, recalling that \$\omega_n = 1.0\$, using the definition of decibel (\$A_{dB} = 20 \log_{10}A\$) you can find that

\$\dfrac{1}{m} = \omega_n^2 G(0) = G(0) = 10^{6/20} \approx 2\$\$\qquad \rightarrow \qquad m = 0.5\$.

Damping coefficient \$\xi\$. Then you can loop over a set of values of \$\xi \in [0,1]\$ to find the best value of \$\xi\$, the one that gives you the most similar plots. I tried, and it seems to me that

\$\xi = 0.4\$

is a good choice.

basics
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For a more general answer, consider: Padé approximant

I haven't written an algorithm for this before, but the existence more or less makes constructing a circuit and fitting curves straightforward, which I've done many times [but not so often that it's worth writing the algorithm, heh].

As for proof of applicability, this is sufficient for transfer functions of linear circuits because every RLC network has a corresponding rational frequency response, i.e., circuits are as expressive as ratios of polynomials are.

Tim Williams
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  • Thank you for the link! I looked into it and I do understand it, however, I'm still concered about how we from the transfer function (ratios of polynomials) can see how a circuit has to be constructed? I hope I made myself clear enough :) – Tanamas Sep 30 '22 at 17:58
  • @Tanamas To construct a ladder network for example, either subtract {R, L, C}, then invert and subtract again, and so on. Each inversion represents the combination of parallel branches, and each subtraction represents placing either a series (subtracted) or shunt (subtracted when inverted) element. Which element(s) are chosen at each stage depend on which best fits the curve. This is most obviously applicable to a 1-port impedance, but similar methods work for transfer functions. Note that, as a transfer function (as in your curve), one cannot solve for impedance for example. – Tim Williams Sep 30 '22 at 20:01
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This is an area of control theory called system identification which is the technique of deriving the transfer function of the system given the bode plot.

That is the bode plot of a second order system evident by the -12 dB/octave roll-off. An octave is a doubling in frequency and looking at the gain drop between say, ω = 4 rad/sec and ω = 8 rad/sec, a -12 dB drop is evident. A roll-off of -12 dB/octave implies a double pole and the double pole occurs at ω = 1 rad/sec. This is evident because of the peaking.

There is also a dc gain, k of 6 dB = 2.

The general form of the second order transfer function is -

$$ \frac{k\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}. $$

Looking at the standard tables I see that for an amount of peaking equal to 8 db, the corresponding value of ζ (the damping ratio) is equal to 0.2.

I have already said that ωn = 1 rad/sec and k = 2.

Therefore the transfer function for that bode plot is -

$$ \frac{2}{s^2 + 0.4s + 1}. $$

There is a pair of complex conjugate poles.

The circuit corresponding to that transfer function could be any number of open loop or closed loop systems built from any number of circuit configurations.

To give an example...

Consider an RLC series circuit which is a second order open loop system with the damping provided by the R term.

The transfer function of a series RLC circuit is given below -

Transfer function

Comparing this transfer function to the generalised 2nd order form given above and the transfer function of the bode plot we get -

Equations

If we let C = 1 F then for an Wn = 1 rad/s we get L = 1 H

Looking at the s term in the transfer functions and comparing coefficients we get -

Equation

and a value for R = 0.4 ohms.

Looking at the resulting circuit and simulation we can see that there is a peaking of 8 dB (2.5 times) at the resonant peak, the same amount as in the original Bode plot.

Circuit

This circuit would require an amplifier with a gain of 6 dB (2 times) before it to provide for k being equal to 2 in the transfer function.

This isn't really a practical circuit with those very large values for L & C but hopefully it illustrates the idea well enough.