Better rough equation for RC bandpass filter ? - Page 1
Forum section: electronics / projects

author: MathWizard
author_rank: Super Contributor
timestamp: February 16, 2022, 01:35:27 pm
content: Looking up RC bandpass filters, I found lots of places just using the eqn's for a separate LP and HP, but that's not very accurate at all is it ? For example R1,2=1k, C1=32n C2=159n would make a 5kHz LP and 1kHz HP, but together, both those freq's are at about the same mag as center Freq of the actual output.So what's a better approximate eqn?I tried to work out the transfer func. for a 4 element RC BP, and I ended up with a quadratic of s , over another quadratic of s. Even if that's right, I haven't learned the next steps yet.

author: ferdieCX
author_rank: Regular Contributor
timestamp: February 16, 2022, 02:00:52 pm
content: Hi, you will find an article about it inhttps://worldradiohistory.com/Wireless_World_Magazine.htmDownload the October 1969 issue and read the page 462

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2022, 02:30:07 pm
content: Right, they'll load each other, though you'll get close to no interaction if you set R2 >> R1.(Assuming a ladder network with series R1, shunt C1, series C2, shunt R2.)Note that Q < 1, so you can't expect a very narrow band, with respect to the center frequency (i.e. Fc/BW < 1). So the "independent poles" approximation isn't too far off, in general -- for the general case where those Fc's are very different. Of course with more error when they're closer together, and more still when the impedances are similar.All of which you should find in the transfer function, hidden cryptically as it is in that expression. Let's see... it should be of overall form s / (s^2 + w_0^2), with the zero at the origin (s) saying gain rises from DC to some point, then a pole cancels it out (one of the denominator factors) and that's your flat passband, then another pole sets the HF roll-off. You say there's two zeroes; that must be related to the loading effect, I think? I'd have to think about it a bit to figure how that should be... As a hand-wave, though, consider the case for C2 >> C1: in this case, for most (mid and high) frequencies, C2 is effectively a short and R2 acts in parallel with C1, thus acting in parallel further with R1 -- you have a resistor divider of R1 into R2, with a lowpass response given by R1 || R2 (i.e. the Thevenin resistance at C1), and C1. This represents the case for Fc(HP) << Fc(LP), a fairly useless condition, and/or Z2 << Z1, also fairly useless (strong loading effect). I'm not sure exactly what the consequence of this all is -- for practical values -- but it looks like it'll have a pole-zero cancellation effect, where for marginal values of these ratios (i.e. LP and HP too close, or significant loading effect), you get an asymmetrical passband.Tempted to run through the equation myself and see; it's not too terrible a problem to work out. Only a pair of quadratics after all, as you say!Also to see the other case, HP into LP (series C1, shunt R1, series R2, shunt C2), and the difference between them if any.As for practical filter specs -- -3dB is where |H| = sqrt(2)/2), take the magnitude, equate, and solve for s -- there will be four solutions, as a quadratic of quadratics (i.e., it's s^4, but no s^3 or s terms, so really a quadratic in s^2), because anything that works for positive frequencies also works for negative.Oh, uh, this isn't evenquitethe complete picture, because it should be -3dB with respect to passband peak, not absolute -- so you should solve for insertion loss first (d|H|/ds = 0, find maximum), and then use -3dB of that to solve for the above -- yeah, it can get pretty messy, unfortunately. If you find one or both of the above assumptions are handy (wide freq or impedance ratios), then you can apply those and some simplification should drop out (for one, insertion loss should be close to 0dB under those assumptions).Aside, I usually work these in Fourier (complex, s --> jw), and I don't actually remember offhand if magnitude is anything special in Laplace domain? Taking the magnitude of a complex rational polynomial of course is rather a pain. Maybe I should use s more often...Tim

author: ferdieCX
author_rank: Regular Contributor
timestamp: February 16, 2022, 03:24:11 pm
content: Some time ago, I derived the equations for one of the possible configurationsSorry, the comments are in Spanish

author: MathWizard
author_rank: Super Contributor
timestamp: February 16, 2022, 05:44:01 pm
content: I tried making an audio amp, w/ BP input using the Rin of the amp as R2, then when saw the sim didn't match I got of into this. At least the simulator makes it easy to tune.Ok I started from the Z-param's and wrote out the gain that way. If I had a higher order of s in the denominator than numerator, then I could have tried partial fractions, I can't remember for equal or positive-ratio orders.No worries, I need to learn more about all that anyways and Fourier transforms for multi-frequency signals.And so it's more akin to making some waves, fit in some box, and have the desired amplitude. at the edges or some output hole?. So there's reflections going on ? Or is that really at higher frequencies where real energy gets bounced back, or how does that apply ? Or maybe actual only apparent power gets reflected, and prefect resistors don't have that and real power doesn't either, only apparent power ?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: February 16, 2022, 05:59:14 pm
content: I think all roads lead back to Zevrev-https://archive.org/details/HandbookOfFilterSynthesis/page/n25/mode/2up

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2022, 07:17:17 pm
content: Wow, Z params? No need for RF port analysis here, plain AC analysis will do! Will be the same result in the end; but, I would expect that method to take longer to work out?And yeah, for the same reason, not really any need to worry about waves and ports. In a transmission line context, RC filters aren't great, as the insertion loss is substantial and the impedance isn't very flat due to the vector sum of resistor and capacitor. (They can be handy when the signal isn't very important, like for EMI filtering of low bandwidth signals, or when crudely (AC) terminating short transmission lines.)Tim

author: mawyatt
author_rank: Super Contributor
timestamp: February 16, 2022, 07:29:34 pm
content: Yep, Zverev is the Bible of all filter designsEdit: BTW about 40 years ago (when Apple II and PC came out) we developed a numerical method of doing Laplace Transforms and Inverse Laplace Transforms with results in Time and Frequency Domains. The method was based upon the Central Limit Theorem. This undertaking was done because we had some non-linear S domain functions that involved roots of S and exponentials of S and the standard look up tables had none of these type S functions.Wish I could remember where that numerical method program (in Basic) ended up, likely on some old 360KB floppiesBest,