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Where did the following transfer function come from?

\$H(s) = -H_o\frac{\frac{\omega_o}{Q}s}{s^2 +\frac{\omega_o}{Q}s + \omega_o^2}\$

\$H_o\$ is the gain at resonant frequency.

I'm told it describes a "classical resonant circuit response" and so I guessed it might somehow be derived from the laplacian of a damped driven oscillator or something of the sort. It appears very generalized. I'm using it to build a multi negative feedback band pass filter.

The original equation appears in this derivation as well as the design:

http://eet.etec.wwu.edu/hardyc/project/docs/Filter%20Design/Multiple%20Feedback%20Bandpass.pdf

For instance, in the link I posted, the transfer function for the filter ends up being:

\$H(s) = - \frac{\frac{s}{R_1C}}{s^2 + \frac{2}{R_3C}s + \frac{1+R_1/R_2}{R_1R_2C^2}}\$

from which they deduce based on its similar form to the original equation I posted what \$\omega_0\$, \$Q\$ and other variables are.

Andrew
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  • It comes from a bandpass resonant filter and it is a **transfer function** and not a **"response"**. Other than that, it's unclear what you are asking here. – Andy aka Aug 20 '19 at 10:27
  • Did you follow the link I posted? It explains in the first couple sentences. Basically, they derive the transfer function for the filter and adjust it to polynomial form on the numerator and denominator and then infer what certain quantities are (such as Q of the filter) on the derived transfer function based off their place on this "generalized transfer function." I understand it's a transfer function - where does it come from and under what conditions does it apply? I will update my question to clarify – Andrew Aug 20 '19 at 10:32
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    Q) Where did it come from? A) Classical analysis of 2nd order systems. Do you understand pole zero diagrams, bode plots or s-plane analysis? – Andy aka Aug 20 '19 at 10:37
  • Right under the picture of the schematic they start showing you how to derive the function. This solves your 'where does it come from' question – Swedgin Aug 20 '19 at 10:38
  • As for 'under what conditions', IIRC the poles should be in the negative side of the s-plane to be stable – Swedgin Aug 20 '19 at 10:40
  • @Andyaka Okay, that is helpful. I only vaguely understand them. Do I need to understand pole zero diagrams and bode plots in order to grasp this equation? because everything in the derivation beyond that point is within my knowledge at this point. My presumption is that "s-plane" analysis is simply substituting \$s\$ for \$j\omega\$ and then working in Laplace space which is easier than lugging around imaginary numbers. I really wanted to justify that original equation to justify inferring what \$Q\$ and other variables were without getting too deep into the theory. Sorry for my ignorance – Andrew Aug 20 '19 at 10:50
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    [This answer](https://electronics.stackexchange.com/questions/275362/what-is-the-difference-between-frequency-response-and-transfer-function) might help (eventually LOL) – Andy aka Aug 20 '19 at 11:19
  • @Andyaka Yes, that's extremely helpful actually. Thanks! That visual is great – Andrew Aug 21 '19 at 05:22

2 Answers2

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There are many ways a transfer function can be written. The best and only way is the low-entropy form, a term forged by Dr. Middlebrook when working on Design-Oriented Analysis. The transfer function must be expressed in such a way you "see" where the poles and zeroes are located and how they affect the response.

In your expression, factor the numerator \$\frac{\omega_0s}{Q}\$ while in the denominator you factor \$\frac{\omega_0s}{Q}\$ also. Re-arrange the whole thing and you should get \$H(s)=-H_0\frac{1}{1+Q(\frac{\omega_0}{s}+\frac{s}{\omega_0})}\$ which is the correct form for expressing the transfer function of a second-order band-pass filter.

The below Mathcad sheet shows that responses are identical. In your circuit, determine the denominator \$D(s)\$ using the fast analytical circuit techniques (FACTs) from which you obtain \$Q\$ and \$\omega_0\$. Factor \$N(s)\$ to match the formula I gave and your leading term represents the peaking gain \$H_0\$ (20 dB in the given example).

enter image description here

Verbal Kint
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  • alright I've got to go to sleep now but I will run over your math tomorrow and give you feedback or accept your answer. Thanks for taking your time to explain! Just didn't want you to feel ignored for the extended silence in the meantime – Andrew Aug 20 '19 at 11:13
  • Good night then! – Verbal Kint Aug 20 '19 at 11:17
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    Verbal Kint...you have mentioned the "correct form" for a transfer function. My question: Do you consider the the first equation (generalized form) as given in the problem description as not "correct"? I think, each form is "correct" as long as we make no mathematical error during rewriting... – LvW Aug 20 '19 at 15:26
  • Hello LvW, of course, the other forms are perfectly valid but they do not yield any insight in the transfer function. I believe transfer functions in which the terms are ordered in a meaningful way are easier to use for meeting design goals. – Verbal Kint Aug 20 '19 at 15:33
  • okay so this and LvW's comment as well as Andy aka's link to a visual of the "s-plane" really all fit together to help me understand this. My electronics class (a one semester race through analog and digital) didn't cover Laplace space. I think I might still have a couple questions, I just need to process this a bit more – Andrew Aug 21 '19 at 05:13
  • Hi Verbal Kint, I know what you mean, but I must disagree. Example: Bandpass with series LC and grounded R. Calculation result: H(s)=R/(R+sL+1/sC) . Without knowing the expressins for the key parameters (Q, wp) it is not so easy to rewrite this equation in the form which you propose as "meaningful" and "easier to use for meeting design goals". For practical reasons, it is much simpler to create a classical second-order polynom in the denominator.....and it is obvious that we only have to multiply numerator and denominator with sLC. More than that, it is easy to verify the role of wp and Qp. – LvW Aug 21 '19 at 07:36
  • Allo LvW, in your expression, factor \$R\$ and you have \$H(s)=\frac{1}{1+s\frac{sL}{R}+\frac{1}{sRC}}\$. Then via a simple manipulation, you can rewrite your expression as \$H(s)=\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}\$ with \$Q=\sqrt{\frac{L}{C}}\frac{1}{R}\$ and \$\omega_0=\frac{1}{\sqrt{LC}}\$. I don't see a conflict with what I previously wrote. Just a matter of rearranging things I guess. Tschuess! – Verbal Kint Aug 21 '19 at 08:59
  • Hi Verbal Klint....you simply misunderstood my comment. I did NOT dispute that a manipulation is possible to arrive at your proposed form. And, of course, there is no "conflict". HOWEVER: I think, for a newcomer it is not so easy to see what he has to do to arrive at the expression as proposed by you. What is the multiplication factor? You speak about a "simple" manipuation"...but it is not so simple to find the corresponding factor. How would YOU find this factor without knowing the Q expression in advance? Therefore, I think that creating a 2nd-order polynom is much simpler. – LvW Aug 21 '19 at 09:32
  • Ok, I see what you mean. Well, if you have you denominator starting by \$1+...\$ and you know the formalized form \$1+Q(...\$ you want to match then finding \$Q\$ and \$\omega_0\$ is easy. Again, people can start with the simplest approach they know and step-by-step learn how to write things in a more advanced manner. That is at least what I would recommend. – Verbal Kint Aug 21 '19 at 10:49
  • That's the thing ... I came across FACTs because of verbal here and it perked my interest. I have spent the last couple of weeks on my hol reading C.Basso "linear circuit transfer function" where the concept is captured elegantly and I have to agree, it is the "correct way". It definitely isn't the only way but the benefits definitely deserve the title. The only downside is breaking old habits and I think I have 2nd order systems grok'ed... Higher (ie LISN with output EMI) are... Interesting and I haven't grasped that yet, doesn't mean I won't carry on as the benefits are clear –  Aug 24 '19 at 11:51
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Andrew - let us start with a simple RLC-bandpass (series L-C and a grounded resistor R). I think, such an example will lead us to an anser for your problem. The transfer function is simple to find (s=jw):

\$H(s)=\frac{sRC}{1+sRC+s²LC}=\frac{\frac{sR}{L}}{\frac{1}{LC}+s\frac{R}{L}+s²}\$

An interpretation of this complex equation will (probably) answer your question.

1.) We know that the denominator of a transfer function is identical to the "characteristic polynom" of a second-order linear system. Hence we try to find the roots of this polynom (identical to the poles of the transfer function):

\$\frac{1}{LC}+\frac{R}{L}s+s²=0\$

with \$s_{1,2}=-R/2L \pm \sqrt{(R/2L)^2-1/LC}\$

Because we are interested in frequencies (knowing s=a+jw) we rewrite the result:

\$s_{1,2}=-R/2L \pm j\sqrt{1/LC-(R/2L)^2}\$

Hence we have a complex pole pair \$s_1=Re+j(Im)\$ and \$s_2=Re-j(Im)\$ with

real part \$a_o=-R/2L\$ and imag. part \$w_o=\sqrt{1/LC-(R/2L)^2}\$.

(\$a_o\$ and \$w_o\$ are the "eigenvalues" of the system; note that here \$w_o\$ is NOT the resonant frequency).

This complex number describes the position of the pole pair in the s-plane.

2.) It is interesting to find the magnitude of the vector from ther origin to the pole and it is easy to show that \$\sqrt{a_o^2+w_o^2}=1/\sqrt{LC}\$.

This frequency is called "pole frequency": \$w_p=1/\sqrt{LC}\$.

Interpretation: Going back to the original transfer function (example) , we see that at \$w=w_p\$ the denominator has its MINIMUM (the real parts add to zero) and the transfer function \$H(s)\$ has its MAXIMUM. Hence, for a bandpass, the pole frequency \$w_p\$ is identical to the midfrequency \$w_p=w_m\$.

3.) Now we want to analyze and describe the position of the pole. For this purpose, we use the angle phi between the real axis and the vector \$w_p\$. And we define a quality factor \$Q_p\$ - knowing that for \$a_o=0\$ (pole on the imag. axis) we have no damping with \$R=0\$. The circuit will oscillate and we allocate an infinite \$Q_p\$ value to this case. This allows us to define: \$Q_p=\frac{1}{2\cos{\phi}}\$ with \$\cos{\phi}=|a_o|/w_p\$.

Interpretation: Again going back to the original function (example) we see that in this case \$Q_p=\frac{1}{2\cos{\phi}}=w_p/2|a_o|\$ and \$w_p/Q_p=2|a_o|=R/L\$

This expression \$R/L=w_p/Q_p\$ appears in denominator as a factor for the linear s-term as well as in the numerator.

4.) Knowing that each second-order bandpass has the same form (active or passive, excluding the gain \$A_{max}\$ at \$w=w_p\$) we can generalize these results and find the general second-order bandpass function:

\$H(s)=A_{max}(\frac{s(w_p/Q_p)}{w_p^2+s(w_p/Q_p)+s^2})\$

Comment 1: Remember, for a band pass we have midfrequncy \$w_m=\$pole frequency \$w_p\$ (don`t mix this with \$w_o\$ as given in your task description).

Comment 2: The definition of the pole Q (\$Q_p\$) as given above has the following advantage: For a second-order bandpass the pole Q is identical to the "classical" quality factor \$Q_p=Q=3dB\$-\$\frac{bandwidth}{midfrequency}\$. Example: For \$\phi=0\$ we have a double real pole with \$\frac{1}{2\cos{\phi}}=Q_p=Q=0.5\$.

Comment 3: It is rather easy to design each bandpass (active or passive) because it is only necessary to compare term-by-term the general transfer function with the actual transfer function (using the corresponding parts values of the circuit). This comparison gives the necessary design equations for \$A_{max}\$, \$Q\$ and \$w_p\$.

Andrew
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LvW
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  • Shouldn't Q be \$\frac{midfrequency}{bandwidth}\$ ? – Andrew Aug 21 '19 at 05:18
  • This was extremely helpful by the way! Thank you for the fully worked out and explained example – Andrew Aug 21 '19 at 05:37
  • Andrew - of course you are correct...Q=midfrequency/bandwidth. Thank you for editing... – LvW Aug 21 '19 at 07:16
  • Andrew, in order to evaluate the information as given in Andy aka`s referenced paper it is helpful to know that the damping coeff. (greek symbol "theta") is related to the "pole quality factor Qp" by the following relation: theta=1/(2Qp). – LvW Aug 21 '19 at 07:42
  • Yes, thank you, I was able to figure that out. Is there a reason \$Q_p\$ isn't simply the reciprocal of the damping coefficient? Why is the factor of \$\frac{1}{2}\$ necessary? Why must \$\zeta=\frac{1}{2}\cdot\frac{1}{Q}\$ ? – Andrew Aug 21 '19 at 07:56
  • I guess I'm not understanding the significance of Comment #2. I understand that a \$\phi = 0\$ would give us a Q of 0.5 which would give us two poles lying on the Real axis at \$-\pi\$. I guess I don't understand why you chose to define Q as \$\frac{1}{2}\cdot\frac{1}{\cos{\phi}}\$ . Why did the \$\frac{1}{2}\$ need to be included in the definition of Q? Thanks in advance – Andrew Aug 21 '19 at 08:25
  • Andrew...I did not define the pole-Q (Qp) - it was defined long time ago by someone else. It was defined following the parameter definitions for the differential equation of a damped oscillating systems (damping coeficient "theta"). Hence, theta=cos(phi) and Qp=1/(2*theta). Only in this case is the circuits Q of a band pass (midfrequency/bandwidth) numerically identical to the pole-Q (Qp) which appears in the transfer function. That is the back ground and the main advantage of this (arbritrary) definition. – LvW Aug 21 '19 at 09:23
  • Okay, so two questions: 1) for \$\phi = 0\$ wouldn't we get a double pole at \$\omega = 0\$ ? What would that mean? Resonant frequency is zero Hz? A low pass filter? 2) how did we initially define \$s\$ as \$=j\omega\$ which would appear that s ONLY has an imaginary part but when we solve the polynomial suddenly has has a real part (\$a_o\$) AND imaginary part (\$\omega_o\$) ? Thank you again for your help – Andrew Aug 22 '19 at 23:54
  • Andrew...(1) yes, a real double pole for a bandpass (Q=0.5). Midfrequency still as definded (wm=wp=real part of the (real) double pole); a lowpass could have the same denominator, however, the zeros(numerator) determine the BP character.; (2) We set s= sigma+jw for pole (and zero) analysis only. This is a nice and clear method to compare different filter types. However, complex frequencies do not exist in practice - so for analyzing the real magnitude and phase response we set s=jw. – LvW Aug 23 '19 at 09:28
  • Got it. Thank you so much!!! – Andrew Aug 23 '19 at 10:38
  • You are welcome.....system theory of filters is a very interesting area....for example: At least 30 different topologies for a 2nd-order bandpass (passiv, activ). Under IDEAL conditions all structures show the same behaviour...however, under real conditions (tolerances, non-idealities, neglections,...) all are different. And it is a challenging task for a engineer to select a proper circuit for a particular task. – LvW Aug 23 '19 at 14:19
  • hmm, I didn't realize that. Very fascinating. Obviously I'm a total noobie at this and unintentionally jumped into the deep end with this active MFB filter. Probably should have started with a passive filter but seemed too easy to me... I was never taught s-plane analysis although I'm acquainted with Laplace transforms from learning Diff EQs. Just a physicist fumbling around with electronics and learning in the process... I'll be around to post more questions on this forum. Hope you'll be around to answer! Thanks again! – Andrew Aug 24 '19 at 02:09