I am trying to do the transfer function of this circuit and also trying to understand the frequency domain thing of the Laplace transform.

enter image description here

I want to do a transfer function of VA/V1, considering B to be ground.

After doing the node analysis and if my equations are correct, I have ended with this transfer function on the s domain.

enter image description here

A calculator online gave these answers for the cubic equation on the denominator

x1 = -915.52763
x2 = -42.23618 + i * 3304.67537
x3 = -42.23618 - i * 3304.67537

These are my problems:

  1. what do these numbers represent?
  2. the equation is on the denominator, what about the s100 on the nominator?
  3. how do I find the frequency performance of this circuit as the frequency changes?
  4. what does this function say about the circuit?
  • 3
    \$\begingroup\$ There is something strange in your transfer function (TF): 1) you have two energy-storing elements so \$D(s)\$ the denominator should be of order 2 not 3 2) for \$s = 0\$, in dc, your TF should simplify to a resistive divider involving \$R_2\$ and \$R_1\$. Finally, there is no zero meaning \$N(s)\$ should be 1. The correct result should be in the form of \$H(s)=H_0\frac{1}{1+b_1s+b_2s^2}\$ with \$H_0=\frac{R_2}{R_1+R_2}\$. Let me know if you want me to derive this TF as a formal response, it is easy with the fast analytical techniques or FACTs. \$\endgroup\$ Commented Jun 18, 2018 at 14:25
  • \$\begingroup\$ please do it... FACT? I will check the equations. \$\endgroup\$
    – Duck
    Commented Jun 18, 2018 at 14:33
  • \$\begingroup\$ The DC gain of your TF is zero, which is clearly wrong (should be 10/11 = 0.91) \$\endgroup\$
    – Chu
    Commented Jun 18, 2018 at 23:03

2 Answers 2


May I suggest that you start by simplifying the 2 resistors into one resistor. I'm suggesting this because it makes the math a lot easier and I think you have a mistake in your final formula. So, the voltage source reduces from 10 volts to 9.09 volts in series with a resistance of 90.90 ohms due to standard circuit theorums.

This then has a standard formula: -

$$H(s) = \dfrac{\omega_n^2}{s^2 +2\zeta\omega_n s+ \omega_n^2}$$ Where \$\omega_n = \dfrac{1}{\sqrt{LC}}\$

And \$\zeta= \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\$

The value for R is the newly recalculated value of 90.9 ohms. To offer a little more help, \$\omega_n\$ is the natural resonant frequency of the filter and \$\zeta\$ is called the damping ratio (also equal to the reciprocal of 2Q).

Now if you want to see what this looks like you can use this on-line calculator and plug-in the values: -

enter image description here

Resonance occurs at about 503 Hz and there is a peak in the response of about 11 dB. Circuit Q is about 3.5.

But don't forget that there is an overall attenuation (due to me simplying the two resistors into one resistor) so the formulas are multiplied by 0.909.

I haven't used your equations because I can see that there is an error in the final formula you derived.

  • \$\begingroup\$ Your answer is 100 light-years ahead of me. Let me take some time to digest it. \$\endgroup\$
    – Duck
    Commented Jun 18, 2018 at 17:07
  • \$\begingroup\$ B R I L L I A N T! The problem here is that I was not seeing the simplification, probably because my eye is not trained enough but I did two source transformations involving V1, R1 and R2 and saw that the whole thing could be replaced by a voltage source of 9,09V in series with a 90,90 resistor. BRILLIANT \$\endgroup\$
    – Duck
    Commented Jun 18, 2018 at 17:32
  • 1
    \$\begingroup\$ Well you didn't take as much time as I thought so kudos to you! \$\endgroup\$
    – Andy aka
    Commented Jun 18, 2018 at 17:33
  • \$\begingroup\$ Hi, if you don't mind I have 3 little questions about your answer. I have verified the equations and found this one for the gain: F(s) = 10^7(s^2 + 90.9s + 10^7). When I plug that on my graph app I get resonance at around 2kHz, not 503Hz as you. The second thing is this: what is the meaning of that peak? Is it the output bigger than the input? The third thing is this: we are doing the whole thing using R=90.90. Do I have to correct the values in 10%? I mean, if this is the R equivalent, isn't it equivalent? I am under the impression that you implied that. Thanks. \$\endgroup\$
    – Duck
    Commented Jun 22, 2018 at 12:59
  • \$\begingroup\$ I mean 10^7/(s^2 + 90.9s + 10^7) \$\endgroup\$
    – Duck
    Commented Jun 22, 2018 at 13:17

The easiest way to determine this transfer function is to use the fast analytical circuits techniques or FACTs. In these techniques, you consider the circuit in different conditions to determine its time constants. This will lead you to a so-called low-entropy format in which you should see gain, poles and zeros if any.

With your circuit, you have two energy-storing elements: a capacitor \$C_1\$ and an inductor \$L_2\$. As they have independent state variables, this is a 2nd-order circuit. As such, the denominator \$D(s)\$ must obey the following expression: \$D(s)=1+b_1s+b_2s^2\$. From this expression, you can rework it and put it under the canonical form \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$.

We start with \$s=0\$. Redraw your circuit in which the inductor is replaced by a short circuit (its impedance is \$sL_2\$ which reduces to 0) while the capacitor is open-circuited (its impedance is \$\frac{1}{sC_1}\$ and is infinite at dc). The circuit looks likes this:

enter image description here

In dc, the gain \$H_0\$ is a simple resistive divider: \$H_0=\frac{R_2}{R_1+R_2}\$. Now, reduce the excitation source to 0 V: replace \$V_{in}\$ by a short circuit and "look" through the connecting terminals of \$L_2\$ and \$C_1\$ while the second element is put in its dc state (open-circuited cap and shorted inductor). In the first case, for \$\tau_1\$ you "see" \$R_1||R_2\$ which implies a time constant equal to \$\tau_1=C_1(R_1||R_2)\$. Do the same for \$\tau_2\$ involving \$L_2\$ and you see an infinite resistance leading to a 0-s time constant. Now, set the capacitor in its high-frequency state (a short circuit) and "see" the resistance offered by \$L_2\$'s terminals in this mode: \$\tau_{12}=\frac{L_2}{R_1||R_2}\$. You can now calculate \$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}\$ and rewrite it into the canonical form given below:

enter image description here

Now, you can determine the same expression using the classical brute-force approach with a Thévenin-equivalent model. Nothing wrong here but a) it is likely to be a complicated exercise b) you may make mistakes - I would : ) and c) you will have to inject more energy to get the canonical form I gave. With this form, you have the resonant frequency \f_0\$ set to 503.3 Hz, a dc gain of -0.83 dB and a peaking of 10.8 dB. This is what the below shot confirms:

enter image description here

The FACTs have lead you there straight away to the compact form you need without writing a single line of algebra, just going through simple sketches that you inspect. If you made a mistake, just fix the guilty drawing and update the time constant. If you are solving transfer functions whether you have passive or active sources, the FACTs cannot be beaten!

  • \$\begingroup\$ it will take me a while to digest what you have said. I will comment here when I do. Your answer is amazing!!!!!!!!!!!! \$\endgroup\$
    – Duck
    Commented Jun 18, 2018 at 21:15
  • \$\begingroup\$ you lost me here... "Do the same for τ2 involving L2 and you see an infinite resistance leading to a 0-s time constant." What is exactly tau? \$\endgroup\$
    – Duck
    Commented Jun 18, 2018 at 22:04
  • \$\begingroup\$ \$\tau\$ is the time constant associated with the energy-storing element: \$RC\$ for a capacitor and \$\frac{L}{R}\$ for an inductor. The term \$R\$ is what you need to determine by inspection, when "looking" through the component connections. Have a look at this PPT for an introduction to the FACTs: cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ Commented Jun 19, 2018 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.