Filter - Q vs Insertion Loss

[rfworks]

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 poles


Pi 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 poles
Q of pi network > Q of simple LC


Cascaded 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 poles
This cascaded network is a broadband (BB) network and hence QBB < Q simple LC < Q pi

Since the simple LC has 2 components, pi has 3 components, BB has 4 components, all non-ideal, resistive losses in LC < PI < BB

In summary,

                          Simple LC (2 pole)             PI (2 pole)                  BB (4 pole)
Q                           In between                         Max                           Min
Insertion Loss              Min                          In between                     Max


This 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.

[Benta]

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.

[T3sl4co1l]

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, Q cannot 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

[rfworks]

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?

[TimFox]

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.

[mawyatt]

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