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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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – Andy aka Aug 20 at 10:27
  • \$\begingroup\$ 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 \$\endgroup\$ – Andrew Aug 20 at 10:32
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    \$\begingroup\$ 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? \$\endgroup\$ – Andy aka Aug 20 at 10:37
  • \$\begingroup\$ 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 \$\endgroup\$ – Swedgin Aug 20 at 10:38
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    \$\begingroup\$ This answer might help (eventually LOL) \$\endgroup\$ – Andy aka Aug 20 at 11:19
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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\$.

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  • \$\begingroup\$ Shouldn't Q be \$\frac{midfrequency}{bandwidth}\$ ? \$\endgroup\$ – Andrew Aug 21 at 5:18
  • \$\begingroup\$ This was extremely helpful by the way! Thank you for the fully worked out and explained example \$\endgroup\$ – Andrew Aug 21 at 5:37
  • \$\begingroup\$ Andrew - of course you are correct...Q=midfrequency/bandwidth. Thank you for editing... \$\endgroup\$ – LvW Aug 21 at 7:16
  • \$\begingroup\$ 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). \$\endgroup\$ – LvW Aug 21 at 7:42
  • \$\begingroup\$ 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}\$ ? \$\endgroup\$ – Andrew Aug 21 at 7:56
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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

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  • \$\begingroup\$ 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 \$\endgroup\$ – Andrew Aug 20 at 11:13
  • \$\begingroup\$ Good night then! \$\endgroup\$ – Verbal Kint Aug 20 at 11:17
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    \$\begingroup\$ 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... \$\endgroup\$ – LvW Aug 20 at 15:26
  • \$\begingroup\$ 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. \$\endgroup\$ – Verbal Kint Aug 20 at 15:33
  • \$\begingroup\$ 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 \$\endgroup\$ – Andrew Aug 21 at 5:13

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