Second order driving point impedance of RC network

Question

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:

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.

Answer 1

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} \$

Answer 2

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 :-

$$Z=k.\frac{s^2+2\zeta\omega_ns+\omega_n^2}{s(s+a)}$$

... which looks just like your expession.

It might though be better to write it as :-

$$Z=k.\frac{(s+\omega_1)(s+\omega_2)}{s(s+\omega_3)}$$

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!

Answer 3

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.