Filter - Q vs Insertion Loss - Page 1
Forum section: electronics / beginners

author: rfworks
author_rank: Newbie
timestamp: February 15, 2022, 06:43:13 pm
content: Hi,Is there any relation between Q of a filter and its insertion loss? Intuitively, the more poles you have in a filter, faster the roll off hence higher Q. With more poles, comes more components (which have resistive losses) and hence higher the insertion loss.However, I tried to analyze a few different topologies.Simple LC (low pass):H(s) = 1/[1+L*C*s^2] ---> 2 polesPi network (direction from input to output is -> cap C1 in shunt arm, ind L in series arm, cap C2 in shunt arm):H(s) = 1/[1+L*C2*s^2] ---> 2 polesQ of pi network > Q of simple LCCascaded network (direction from input to output is -> ind L1 in series arm, cap C1 in shunt arm followed by cap C2 in series arm, ind L2 in shunt arm):H(s) = [L2*C2*s^2]/[L1*L2*C1*C2*s^4 + (L1*C1 + L1*C2 + L2*C2)*s^2 + 1] ----> 4 polesThis cascaded network is a broadband (BB) network and hence QBB < Q simple LC < Q piSince the simple LC has 2 components, pi has 3 components, BB has 4 components, all non-ideal, resistive losses in LC < PI < BBIn summary,Simple LC (2 pole) PI (2 pole) BB (4 pole)Q In between Max MinInsertion Loss Min In between MaxThis summary contradicts the intuitive statement I made initially in the thread: Higher Q means more poles, more poles means more components and hence higher the insertion loss.Can anyone help resolve this discrepancy and let me know which analysis is corrrect?Thanks in advance.

author: Benta
author_rank: Super Contributor
timestamp: February 15, 2022, 07:55:19 pm
content: Where your assumption fails is in stating that higher Q means more poles.That's not true. If you need a sharper roll-off or resonance point, you may need more poles/zeros. But that's got nothing to do with Q.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 15, 2022, 08:00:08 pm
content: Something's missing: no source or load impedance has been defined, I think? So the Q is infinite in all cases!In practice, there are, well, about as many Q's as there are impedances you can rub together -- you don't have to take just the intrinsic component Q -- but there are two in particular that are of significant note.Component Q: the Q of the two-terminal element itself. Real inductors and capacitors have some amount of resistance, and this defines the Q of the component itself.By example, notice that the Q of a resonator (as a two-terminal LC component) goes to zero at resonance: its reactance cancels out, presenting just resistance, and so the Q is zero by definition. This might seem at odds with intuition (surely a resonator has some Q defined by the quality of its resonance?), but serves to illustrate we must be careful with our definitions.And for that case, when we do have a 2-terminal component with some poles in its impedance: we might wish to use another definition of Q, like the bandwidth of a resonant peak or trough. Thus we disregard the terminal impedance, understanding that the component has some internal structure that isn't so well defined by instantaneous impedance alone.Alternately, we might craft a basic model of the component, around a frequency of interest, and consider the Qs of those representative elements. This can result in some quite peculiar values (consider the microscopic C and ludicrous L of a quartz crystal equivalent circuit!), but gives a correct answer within the scope of that model (i.e., the, whatever xxH inductor, plus 100Ω-ish ESR, gives the, whatever 100k-odd Q factor we find so useful).The other important definition is system Q. Say we take a 1uH 1Ω inductor and connect it in series between two 50Ω ports (source and load). The series circuit has 101 ohms total resistance and 1uH inductance, giving a -3dB point of ~16MHz. At the corner frequency, X_L = R, so the system Q is 1 at this point. At passband frequencies, Q << 1, and stopband, Q >> 1 (rising eventually towards component Q). Note that, in the pass and transition bands, component Q affects insertion loss and sharpness, and system Q has a hard ceiling of component Q (you can't connect any (positive) resistors around a component and get a higher Q than its internal resistance dictates!).So, I think as your transfer functions are lacking any source or load resistance, Qcannot be defined-- there also cannot be any well-defined power transfer in such a system, at least, finite nonzero amounts for all [finite nonzero] frequencies (I think?).Tim

author: rfworks
author_rank: Newbie
timestamp: February 15, 2022, 08:09:42 pm
content: If we have more poles in the system, the roll off will be sharper and narrower the bandwidth.Since Q is inversely proportional to bandwidth, doesn't the above statement hold true?

author: TimFox
author_rank: Super Contributor
timestamp: February 15, 2022, 08:15:34 pm
content: As stated above, a band-pass filter is not properly characterized by Q.In general, with a single R-L-C resonant circuit, the loss is determined by the ratio of the loaded and unloaded Q of the resonance.Therefore, you want to use high-Q components (L and C), even though the bandwidth is determined by the loaded Q when the source and load resistances are included.

author: mawyatt
author_rank: Super Contributor
timestamp: February 15, 2022, 09:04:45 pm
content: Time Domain interpretations of "Q" are also interesting. Like ratio of energy stored/dissipated within a waveform cycle and pulse response damping (under-over-critical). In general, I've never particularly cared for the use of "Q" to define something, seems to loose a term and has too many interpretations.Best