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I am trying to understand the concept of Q factor for higher order circuit. Consider the circuit as shown in the figure enter image description here

I calculated Zin using Matlab and the transfer function looks as follows: $$Zin(s)=\frac{{\left(c_1 \,c_2 \,c_3 \,l_1 \,l_2 \,r_1 \,r_2 \right)}\,s^5 +{\left(c_2 \,c_3 \,l_1 \,l_2 \,r_1 +c_1 \,c_3 \,l_1 \,l_2 \,r_1 +c_1 \,c_2 \,l_1 \,l_2 \,r_1 \right)}\,s^4 +{\left(c_2 \,c_3 \,l_2 \,r_1 \,r_2 +c_2 \,c_3 \,l_1 \,r_1 \,r_2 +c_1 \,c_3 \,l_1 \,r_1 \,r_2 \right)}\,s^3 +{\left(c_3 \,l_2 \,r_1 +c_2 \,l_2 \,r_1 +c_2 \,l_1 \,r_1 +c_1 \,l_1 \,r_1 \right)}\,s^2 +{\left(c_3 \,r_1 \,r_2 \right)}\,s+r_1 }{{\left(c_1 \,c_2 \,c_3 \,l_1 \,l_2 \,r_2 \right)}\,s^5 +{\left(c_2 \,c_3 \,l_1 \,l_2 +c_1 \,c_3 \,l_1 \,l_2 +c_1 \,c_2 \,l_1 \,l_2 +c_1 \,c_2 \,c_3 \,l_2 \,r_1 \,r_2 \right)}\,s^4 +{\left(c_2 \,c_3 \,l_2 \,r_2 +c_2 \,c_3 \,l_2 \,r_1 +c_2 \,c_3 \,l_1 \,r_2 +c_1 \,c_3 \,l_2 \,r_1 +c_1 \,c_3 \,l_1 \,r_2 +c_1 \,c_2 \,l_2 \,r_1 \right)}\,s^3 +{\left(c_3 \,l_2 +c_2 \,l_2 +c_2 \,l_1 +c_2 \,c_3 \,r_1 \,r_2 +c_1 \,l_1 +c_1 \,c_3 \,r_1 \,r_2 \right)}\,s^2 +{\left(c_3 \,r_2 +c_2 \,r_1 +c_1 \,r_1 \right)}\,s+1}$$

which is a fifth order transfer function. Now if i substitute the values of components and convert the denominator to second order sections in the form of $$s^2 + 2\delta \omega\cdot s + \omega^2$$ then i can find the Q factor of each section. The overall Q will be simply the average of the Q factors, in this case I will have two Q factors.

Is my analysis correct?

RAN
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    Q-factor for a higher-than-second-order filter is a myth and irrelevant. Of course, if you have some document or weblink that finds some use for it then please leave a link. – Andy aka Oct 04 '21 at 11:48
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    Why do you think that "the overall Q will be simply the average of the Q factors,"? I think, this is an assertion without any technical background. – LvW Oct 04 '21 at 12:24
  • @Andyaka I agree, hence I am breaking the network into second order sections. Regardless, the circuit which I have posted must have a Q factor correct? If yes, then how to determine it? – RAN Oct 04 '21 at 13:05
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    Q-factor is a 2nd-order phenomena @RAN - additionally if you have multiple 2nd order filters of the same type and shape that are cascaded, some people will attribute the resulting (and apparent) higher Q by multiplying the transfer functions but not averaging the individual Q-factors. – Andy aka Oct 04 '21 at 13:08
  • @Andyaka Ok, but is the method of determining q factor of a higher order network by breaking it into second order sections, finding the Q factor of each sections, then multiplying all Qs correct? – RAN Oct 04 '21 at 13:15
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    There is no valid mechanism for establishing a Q-factor for higher order networks. Sometimes (in some people's opinion) multiplying the transfer function can be used to determine an overall Q-factor but, it's a flawed route and has no true meaning. – Andy aka Oct 04 '21 at 13:39
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    RAN-as mentioned already, the Q-factor (correct: Pole-Q) is a quantity which describes the position of a SECOND-ORDER pole in the s-plane. Hence, such a parameter does not and cannot exist for higher.order circuits. – LvW Oct 04 '21 at 14:41
  • @LvW Here is how i understand the concept: If a system is comprised of 'n' second order sections then there should exist 'n' second order poles with a Q factor determined for each section. The overall Q of the system must exist, although I agree to Andy that it is very difficult to determine. But the Q factor must exist even for higher order networks. – RAN Oct 04 '21 at 15:07
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    Yes - for a sequence of n second-order stages, we have n second-order pole pairs - and, of course, n different Q-values (describing each pole pair). However, what is the reason for you to expect that "the overall Q of the system must exist"? With the same logic you could require that an "overall" pole frequency must exist. – LvW Oct 04 '21 at 15:47
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    @RAN What you are asking is equivalent to: "What is the discrimant of this 4th order polynomial \$P(x) =x^4+5x^3+9x^2+7x+2 \$?" It doesn't makes sense to talk about a discriminant for any polynomial above or below 2nd order, just like with the Q-factor. Sure you can split the polynomial up into two second order polynomials and find the individual discriminants, but obviously there is no "overall discrimant". The same holds for the Q-factor. It doesn't make sense to talk about an "overall Q". – Carl Oct 04 '21 at 18:35
  • Well, assuming that the network can be decomposed to series (or parallel) R,L,C components, then the Q factor can be found out. Can it be then said that the Q factor of the network is equal to the Q factor of the RLC circuit? – RAN Oct 07 '21 at 06:55
  • @Carl, I agree and understand your reply. Thank you – RAN Oct 07 '21 at 06:56
  • @Andyaka, if Q can be defined as the ratio or stored energy to power loss, ie, reactive power or energy over active power, why is the resulting value for a higher order circuit "nonsensical"? – jrive Feb 26 '23 at 17:34
  • What "resulting value" do you refer to @jrive? – Andy aka Feb 26 '23 at 19:53
  • @Andy --the ratio of stored energy to power loss per cycle – jrive Mar 21 '23 at 21:45
  • It applies only to a 2nd order system. – Andy aka Mar 21 '23 at 22:36
  • @Andy, why would this ratio of energy stored to power loss per cycle only apply to 2nd order systems? – jrive Jun 01 '23 at 20:23
  • Look up the derivation of damping factor. It's a second order thing that Q is directly related to. – Andy aka Jun 01 '23 at 22:56

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