Oh dear.
The "name dropping" just won't stop, will it? Now Riley is on the list.
Anyway, it looks interesting. I don't have the time to look at it now, but will come back when possible.
Just for your reference: the "Linkwitz-Riley" filter has nothing to do with cascading first-order filters. It's somewhat more sophisticated.
You seem to have missed this passage:
And there are even other approximations aimed at some other design problems (e.g.: the Linkwitz–Riley —or simply Linkwitz— used in some audio crossovers to ensure that the combined audio outputs of both a low-pass and a high-pass filter is flat. Incidentally, your filter formed by two 1st-order cascaded filters is the low-pass half of a 2nd-order Linkwitz filter).
OP seems to be applying the terminology promptly and consistently, and the above information has not been disputed or corrected so there would be no reason to expect otherwise. I don't get what you're blaming them of.
SilverSolder: for higher order filters, in the active filter case, the poles can simply be placed as needed for each pair, and standard filter tables give the coefficient for each pair. That coefficient gives, whatever it gives -- damping factor, imaginary component, Q factor, etc. These are all equivalent (one defines the other), but have different values so which one is applicable must be clear. In any case, follow the scaling formulas given with the table (or, use a calculator that automates all this for you).
The one catch is, if you do an odd-order filter, you can put the single real pole* out in front as a passive RC. This is in turn loaded by the input impedance of the filter connected to it, so all its values change as well. Obviously, this can be avoided at the expense of one additional op-amp (follower), if so desired. Easy enough to demonstrate in your simulation -- you'll find that by adding the real pole out front, you can make the complex pole just a bit sharper (peakier) while maintaining flatness (Butterworth-ish transfer function); which is, I forget how much exactly, a sqrt(3) or something?
*Poles fall on a semicircle, so evenly dividing that arc by an odd number leaves exactly one on the real axis (midpoint of the arc). Even order has all complex pole pairs.
For additional poles beyond this, of course you'll get other roots and powers. There's a certain pattern or symmetry to them, though it's hard to grasp (and, rightfully so, it's ultimately about factoring polynomials).
For the passive network, very hand-wavingly: when the network can be bisected, the impedance ratio between that section (i.e., Zo = sqrt(L/C)) and its surroundings, defines a system Q factor, and thus a damping ratio or what have you. The Q factor of each section, is the same as the Q factor required for each pole pair in the active filter.
This is a hand-waving explanation, because it's hard to bisect a general ladder network, and for the most part, all values interact with each other, so that it doesn't make sense to consider a given section in isolation, you most likely have to work with the whole.
It's easiest seen, I think, in terms of a network of coupled resonators: when the Q factor is high and the coupling is small, the bandwidth and peaking due to any given resonator, or pair thereof, is at its simplest: the system Q goes as 1/(coupling factor). Simply put, as long as the component Q is large enough to ignore (X >> R), then it will be damped by the system (source and/or load resistance), and thus we can design some desired amount of damping (filter sharpness).
There used to be a web calculator that did exactly this, showing the coupling factors for a N-order matrix of resonators, of given filter type -- which I thought was interesting as it's simultaneously not that useful (it leaves a ton of work, going from coupling factors to a real design; and at the frequencies where you'll be relying on resonators, that means mechanical design at that!), and yet also one of the most basic, yet abstract, demonstrations of how filters are formulated -- a set of resonators and couplings.
Like, this kind of explains it, it works with coupling factors -- but notice the response is skewed a bit, because the coupling isn't fixed, but frequency-dependent (coupling cap!), and so the asymptotes are asymmetrical. When using the same coupling factors with transformers (coupled inductors), the response is symmetrical. (Or with resistors, but then you get all the insertion loss.)
https://www.ivarc.org.uk/uploads/1/2/3/8/12380834/2._qandkmethod.pdfthere are a number of other calculators (online right now) too, mostly working with the same form as well.
I don't know that this... really helps, intuitively speaking? I'd say it was formative, for me, simultaneously not really understanding the form of such a thing ("what the heck is this matrix, I just want L's and C's..!?"), while also reducing it to its simplest form: a series of coupling factors, symmetrical between ports, and proportioned according to the filter type.
Or that this is even wanted, as the question [so far] concerns active filters -- which are simpler, for all the reasons the above has made obvious.
Irregardless, to wrap this up; you can transform such a design, into an LPF (or any other, or vice versa), and so also understand that an LPF is also a chain of coupled resonators -- heavily skewed (to a full passband for one asymptote!) by its construction (alternating L/C ladder), and it's just a lot harder to tease out what counts as a resonator, because everything is tightly coupled, which is also why the response is so hard to tune by adjusting components one at a time.
Or, put very simply, tuning single components is a poor way to do it because the cutoff frequency goes as 1/sqrt(L*C) and varying one random inductor at a time, affects Fc in part, but also the Q of everything around it, as well as itself. So, yeah.
Or alternately, going the direct route: simply solving for the transfer function in terms of component values, we find the value of each and every component shows up in the coefficient for basically every power of ω. So not only is it hard to go back and solve for any L or C value in the first place, but that's on top of solving the damn polynomial (which is nontrivial). So we simply go to the filter tables -- prepared for us by those wizened in the arcanum of filters
and computers -- and select what L and C we need based on the prototype. And finally, tweak around that to arrive at standard commercial values. (Fortunately, passive filters aren't too unstable with respect to component tolerances; the Sallen-Key type is a bit more critical however.)
Tim
Home
Help
Search
Login
Register
