# Overall Q of a fourth order LPF and higher order LPF cascaded

If I end up cascading low pass filters to derive higher order filters, how should i calculate the Q of the overall filter? For example if i have the following transfer function $$H(s)=\frac{1}{(s^2 +s \frac{1}{Q1}+1)\cdot(s^2 +s \frac{1}{Q2}+1) }$$ Then what should be the Q of the LPF filter? I think the Q of the last stage should dominate the overall Q so should be Q2, am i correct?

In general if i was to cascade more stages then what should be the overall response?

• When asking for the Q of the "overal filter" you should have a corresponding definition in mind . Do you? No - that is not possible because there is not such a definition.
– LvW
May 19, 2020 at 19:13

Each 4th order filter is itself composed of a pair of 2nd order sections.

Typically, one of those sections has a fairly high Q, the other has a low Q (overdamped). Thus, the rising response of the higher Q section below resonance cancels the early roll-off of the lower Q section; above resonance, both filters are rolling off, with a relatively sharp corner at resonance.

An 8th order filter would NOT be composed of two 4th order filters but four 2nd order filters. Typically two would be high-Q (underdamped) to different extents, while two would be overdamped - each matching its under-damped counterpart. (As pointed out in a comment, there are other realisations especially in passive L-C filters : this answer is biased towards active filters such as Sallen & Key. Even in the FDNR implementations I've seen, it is possible to distinguish sections with the same spread of Q values).

Exactly where the resonances and Qs are placed determines the filter alignment; choosing these is commonly done by following polynomials such as Bessel, Butterworth, Chebyshev or Cauer polynomials according to the compromise you want between phase response, passband flatness, or steepness of attenuation.

See for example the 2nd, 4th, 6th and 8th order polynomials under "Normalized Butterworth polynomials" in the above page, where for example the 4th order polynomial is given as:

(s^2+0.7654s+1)(s^2+1.8478s+1)

Characterising high order filters is done in terms of these polynomials, rather than a specific overall Q value.

• Ok, so if i casscade two RC second order low pass filter, with the same Q then would that filter work? SInce both the filters provide the same amount of damping?
– RAN
May 19, 2020 at 17:22
• It would work but for most purposes it would eb far from an optimal filter.
– user16324
May 19, 2020 at 17:23
• Right understood, thank you.
– RAN
May 19, 2020 at 17:24
• @Brian Drummond..."An 8th order filter would NOT be composed of two 4th order filters but four 2nd order filters". It should be added that this is ONE possible method only (cascading 2nd-order sections). Higher-order active filters - as 6th or 8th order - in most cases, are derived from passive ladder structures and realized using two classical techniques: Active-L simulation or FDNR technique (applying the BRUTON transformation).
– LvW
May 19, 2020 at 19:11

Q, or quality factor applies only to 2nd order filters and Q is equal to the transfer function gain at the natural resonant frequency of that 2nd order filter. Trying to apply some variation of Q factor to a 4th order system is missing the point.

• Does that mean that Q of fourth order filter does not exist?
– RAN
May 19, 2020 at 17:02
• I can’t possibly say whether someone hasn’t erroneously contrived some figure of merit for higher order filters based on individual q factors. If they have, then I believe that they are also missing the point of what QF is. May 19, 2020 at 17:05