I am trying to find the transfer function for the following second order model, but running into difficulty as I can't manipulate it into the standard form required for me to extract time constants.

The circuit is as follows:

Battery model

So far I have worked from this equation for the impedance:

$${\frac{\left( {\it R1}+{\frac {1}{s{\it C1}}} \right) \left( {\it R2}+{ \frac {1}{s{\it C2}}} \right)}{{\it R1}+{\frac {1}{s{\it C1}}}+{\it R2}+{\frac {1}{s{\it C2}}}}}+R3$$

and factored out the frequency independent gain: $$k={\frac {{\it R1}\,{\it R2}}{{\it R1}+{\it R2}}}+{\it R3}$$

to arrive at somewhere near what I think is standard form, but missing an \$\omega_{o}\$ term. I wanted to keep the factor of \$s^2\$ in the denominator unity.

I have this: $$k.{\frac{{s}^{2}+{\frac { \left( {\it R1}\,{\it C1}+{\it R2}\,{\it C2}+{\it R3} \,{\it C2}+{\it R3}\,{\it C1} \right) s}{ \left( {\it R1}\,{\it R2}+{ \it R3}\,{\it R1}+{\it R3}\,{\it R2} \right) {\it C2}\,{\it C1}}}+{ \frac {1}{ \left( {\it R1}\,{\it R2}+{\it R3}\,{\it R1}+{\it R3}\,{ \it R2} \right) {\it C2}\,{\it C1}}}}{{s}^{2}+{\frac { \left( {\it C2}+{\it C1} \right) s}{ \left( {\it R1}+ {\it R2} \right) {\it C2}\,{\it C1}}}}}$$

and I'm stuck. I have spent a day manipulating it in various ways, not taking the factor k out, but can't get it into a recognisable form. Intuitively I see it is a mixture of high-pass and band pass.

I wonder if part of my problem is that I am assuming an input current of \$I_{b}\$ for the calculation of transfer impedance \$Z={\frac {V_{b}} {I_{b}} }\$, when the circuit must be considered open circuit for transfer function calculation? Any help appreciated.

  • \$\begingroup\$ Transfer impedance or just impedance into the terminals? Also, this is a first order impedance - there will be no 2nd order terms in this by inspection. \$\endgroup\$
    – Andy aka
    Sep 17 '13 at 10:00
  • \$\begingroup\$ Just impedance looking into the terminals. I want to identify the impedance spectrum of a battery by measuring its voltage response to a current stimulus, hence the reference to a transfer function. I thought it was second order as I see two corner frequencies in the bode plot, and there are two energy storage elements? \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 10:08
  • \$\begingroup\$ R1 and C1 together with R2 and C2 can be equated to \$R_x\$ in series with \$C_y\$ i.e. just a simple 1st order filter. \$\endgroup\$
    – Andy aka
    Sep 17 '13 at 10:17
  • \$\begingroup\$ I still can't see it - if I omit R3 and go from there I still always end up with an s^2 term in the denominator, hence two poles? \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 10:59
  • \$\begingroup\$ @ZetaSeeker, there are two poles and two zeros. I'm writing an answer now. \$\endgroup\$ Sep 17 '13 at 11:00

First,the proper term is driving point impedance since it is the ratio of the voltage across and current through a one-port.

Now, since the impedance goes to infinity as the frequency goes to zero, the factor you've pulled out is the high frequency asymptotic impedance but I don't think it's useful in this case to do that.

I think a transparent form for the driving point impedance is:

\$Z_{eq} = R_3 + \dfrac{1}{s(C_1 + C_2)}\dfrac{s^2(R_1R_2C_1C_2) + s(R_1C_1 + R_2C_2) + 1}{s(R_1 + R_2)C_1||C_2 + 1} \$

There are clearly two poles; one at \$s = 0 \$ and one at \$s =-\dfrac{1}{(R_1 + R_2)C_1||C_2}\$

The numerator is 2nd order so there are two zeros. You can factor the numerator to find the zeros (of the numerator) at \$s = -\dfrac{1}{R_1C_1} \$ and \$s = -\dfrac{1}{R_2C_2} \$

As \$s \rightarrow \infty \$, the 2nd term approaches

\$\dfrac{R_1R_2}{R_1 + R_2} = R_1||R_2 \$

The two parallel RC networks have an equivalent impedance given by:

\$Z = (R_1 + \dfrac{1}{sC_1})||(R_2 + \dfrac{1}{sC_2}) = \dfrac{R_1R_2 + \frac{R_1}{sC_2} + \frac{R_2}{sC_1} + \frac{1}{s^2C_1C_2}}{R_1 + R_2 + \frac{1}{sC_1} + \frac{1}{sC_2}}\$

\$ = \dfrac{s^2R_1R_2C_1C_2 + s(R_1C_1 + R_2C_2) + 1}{s^2(R_1 + R_2)C_1C_2 +s(C_1 + C_2)}\$

\$ = \dfrac{1}{s(C_1+C_2)}\dfrac{s^2R_1R_2C_1C_2 + s(R_1C_1 + R_2C_2) + 1}{s(R_1 + R_2)C_1||C_2 +1} \$

  • \$\begingroup\$ Thank you. I've corrected the title. I thought extracting the k factor was premature myself, and now I see it confirmed. My mistake lay in trying to match a biquadratic standard form for the denominator with unity s^2 + (w_o/Q) s + w_o^2. I could understand the w_o^2 term disappearing because of the pole at d.c., but then I couldn't understand why the (w_o/Q) term didn't also disappear in that case. While I don't fully understand that yet, your answer has enabled me to work out the time constants I need and I can carry on with the rest of the work - thanks. \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 12:54
  • \$\begingroup\$ Alfred, I'm having problems with understanding how there are two poles and how therefore this is a 2nd order system. I'm also not 100% sure about the formula you got for (R1, C1) & (R2, C2) in parallel. It's probably me being a bit dumb. \$\endgroup\$
    – Andy aka
    Sep 17 '13 at 20:32
  • \$\begingroup\$ @Andyaka, for the two RC networks in parallel, at low enough frequencies, the circuit looks like two parallel capacitors so there is a pole at zero frequency and there is a 1st order roll off of impedance with frequency. At high enough frequency, the circuit looks like two resistors in parallel so the impedance approaches a constant. However, there are clearly two zeros. If there weren't another pole, the impedance at high frequencies would have a 1st order increase with frequency so there must be another pole. I'll edit my answer to show how I got the formula. \$\endgroup\$ Sep 18 '13 at 0:47

What's the problem? I haven't checked the detail of your algebra but lets do a sanity check. At dc, the impedance of the network is infinite. So put s=0 into your equation and the denominator becomes zero, so you have infinite impedance.

At high frequencies, the impedance must be R3 plus the parallel combination of R1 and R2. So let the \$s^2\$ terms dominate in your equation and the result is k. Looking good.

On the back of an envelope I just got something which can be written as :-


... which looks just like your expession.

It might though be better to write it as :-


since being a passive network, the roots must be real. You can of course plug the coefficients of your numerator into the quadratic formula to solve for \$\omega_1\$ and \$\omega_2\$ but I'll leave that to you!

  • \$\begingroup\$ Thank you for your response, unfortunately as a new member I can not vote this as useful yet, but I will revisit when I am able to. It is useful and correct but Alfred Centauri has answered more explicitly so he gets the vote. Thanks. \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 12:07

Yep, you've run into a brick wall I'm afraid. It is not possible to get a general analytic reverse Laplace transfer function for anything more than a first-order (or resistive-loaded first order) RC circuit. Not series RC, not butterworth, nothing. This will not simplify unless you make some elements equal to each other, e.g. R1=R2 and C1=C2. Then, for a second-order circuit you will be able to find a reverse laplace transform with any textbook list of standard transforms.

You will have to go another route than Laplace transforms if you want to solve a generic form of your circuit.

If you go third-order, even equalling resistors and capacitors will not yield you a standard form.

Also, a great way of learning to manipulate these kinds of equations is to use Mathematica, Matlab or some other maths package to help you fiddle with simplifying your equations.

  • \$\begingroup\$ I'm not sure I fully understand - by reverse do you mean to take the inverse Laplace transform? Or do you mean I can't obtain the general solution this way? So I shouldn't be so lazy with my s terms and express it in terms of jw instead? Thanks, I'll try that. Funny you mention it, this was the problem that finally persuaded me to get to grips with Maple yesterday - much quicker and less error-prone when fiddling with equations! \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 10:15
  • \$\begingroup\$ Ah, I see the problem here. When you're transforming your \$j\omega\$ equation (this is called a time-domain equation) into an equation with \$s\$, the latter equation is called the Laplace transform of your equation. \$s\$ is called the Laplace invariant. The equation now exists in the Laplace domain. When you then try to backtransform it into an equation with \$j\omega\$ this is called the inverse Laplace transformation. See also en.wikipedia.org/wiki/… or in any textbook on the matter. \$\endgroup\$
    – user36129
    Sep 17 '13 at 11:04
  • \$\begingroup\$ Thanks, I think the problem was less about what domain I was working in and more about me chasing the standard form. I am happy to have the expression in s (or jw) because I already have the dft of my input signal, and I will do an inverse dft of the output response I get in the s domain. I think the method is valid, if a bit cheeky to a mathematician... \$\endgroup\$
    – ZetaSeeker
    Sep 17 '13 at 13:03

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