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I'm looking for the transmittance of this filter.

schematic

simulate this circuit – Schematic created using CircuitLab

I need help on how to deal with a third order transfer function. I suppose there is a way to have a product of second order form and a first order one, but what is the method to do it?

Here is the transfer function:

$$H(s)=\frac{(RL_2C_1s^2)}{(RL_2C_1s^2+RL_1C_1s^2+L_1L_2C_1s^3+L_2s+R}$$

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    \$\begingroup\$ Why can't you just calculate it or use a simulator? \$\endgroup\$ Commented Sep 23 at 19:16
  • \$\begingroup\$ Why do you want to split the 3rd-order function into a 2nd-order and a 1st-order function? Do you have problems to alculate the transfer function? \$\endgroup\$ Commented Sep 24 at 7:09
  • \$\begingroup\$ Can you tell me the meaning of "transmittance"? \$\endgroup\$ Commented Sep 24 at 13:34
  • \$\begingroup\$ A transfer function is the ratio between an input and an output. You haven't indicated the output in your schematic. It is likely the voltage across \$L_2\$ but such a detail should not be left out. \$\endgroup\$ Commented Sep 24 at 13:38
  • \$\begingroup\$ @LvW I'm trying to express the transfer function in a standard form where the denominator is factored into a second-order and a first-order term, so that I can identify ω0 and the quality factor Q, but I'm not sure how to proceed or if it's even possible \$\endgroup\$ Commented Sep 24 at 18:38

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To determine this transfer function, I will use the fast analytical circuits techniques or FACTs, as described in my last book on the subject. The principle consists of identifying the resistance \$R\$ driving the energy-storing elements - the \$C\$ and the \$L\$ - to form the time constants of the circuit defined as \$\tau=RC\$ and \$\tau=\frac{L}{R}\$. With these time constants obtained without writing a single line of algebra, it becomes possible to determine the transfer function in a low-entropy form, with well-ordered coefficients.

enter image description here

The difficulty then comes from the factorization of the denominator is one wants to cascade a 2nd-order expression followed by a pole, as in this 3rd-order circuit. You need to compare the times constants and, depending on their values, group certain of them to write the denominator. This document from CoPEC describes the process, it starts with the low-\$Q\$ approximation, page 51.

I used Mathcad for this exercise and I have obtained an expression which is well factored but the leading term does not match the gain at resonance yet:

enter image description here

However, you can see the good matching between the reference brute-force expression and my expression featuring a 2nd-order polynomial followed by an inverted pole.

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  • \$\begingroup\$ Thanks for your reply seems to be the best way to solve my problem ! do you mean that the factorisation is only possible if the time constants are well separated ? which order ? and if not, does it means that factorisation is not always possible ? \$\endgroup\$ Commented Oct 5 at 10:33
  • \$\begingroup\$ Please have a look at the document I refer to in my answer. When the time constants are well separated, the factorization is simple, e.g. 3 cascaded poles in a 3rd-order circuit. But if the time constants are spread differently, when two are close enough for instance, you can group them into a 2nd-order polynomial. \$\endgroup\$ Commented Oct 5 at 17:59
  • \$\begingroup\$ Thanks a lot it's clear and I'm going to read your document. \$\endgroup\$ Commented Oct 6 at 8:14
  • \$\begingroup\$ I don't understand the Tau23 : from my understanding, I suppose that we look for the resistance which drives the reactance 3 with the reactance 2 in HF state is that right ? in this case what do we consider the reactance 1 ? and why the result is L3/rs ? (what is Rs) ? For the Tau123, I also don't understand the result \$\endgroup\$ Commented Oct 12 at 14:27
  • \$\begingroup\$ The comments section is a bit small to detail the process : ) but yes, tau23 means 2 is in HF and you look at \$R\$ driving 3, 1 being left in its reference state, e.g. dc. I use \$R_{INF}\$ and \$R_s\$ to avoid indeterminacies so that I can later simplify. So when a configuration returns 0 ohms or infinity (open circuit), I respectively consider \$R_s\$ and \$R_{INF}\$ instead. You can have a look at my APEC 2016 seminar on FACTs or consider the book I referenced in my answer. \$\endgroup\$ Commented Oct 12 at 19:27

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