[sdiy] Passive filters and impedances

[Richie Burnett]

Do you actually *need* to know the transfer functions Tom, or can you just 
simulate in Spice?  They are complicated and messy to work out because of 
the series & parallel combinations of complex impedances, although you're 
doing it right starting at the output side and combining the impedances 
until you get back to the input.  The final equation reduces down to the 
potential divider equation with R1 as the top impedance in the divider, and 
the series/parallel combination of all the other complex impedances in the 
network as the bottom impedance in the potential divider.  This is the sort 
of problem that EE students are made to solve because it tests their maths 
skills but never ever use in practice!!!

In general we don't usually build filters like this because the lack of 
buffering causes loading effects that move the filter poles.  However, if 
you just want to make a quick and dirty 2nd or 3rd order filter without 
op-amp buffers, a good way to reduce the un-wanted interactions is to make 
the impedance of each RC filter stage about 10 times greater than the 
impedance of the preceding stage.  It doesn't stop the interaction, but it 
reduces it considerably.  The downside is that the output impedance of the 
filter will probably be quite high, so you'll likely need a buffer to follow 
it at least.

It might be okay at something like a cheap PWM reconstruction filter for 
generating a slowly varying control signal.  But for most other practical 
filtering applications I can think of you'd probably want something that's a 
bit more optimal.

-Richie,


-----Original Message----- 
From: Tom Wiltshire
Sent: Saturday, October 31, 2015 4:24 PM
To: synthdiy diy
Subject: [sdiy] Passive filters and impedances

Hi All,

I'm back to learning about impedances again. I've been looking at the 
following situations:

http://electricdruid.net/images/PassiveFilters.png

This shows a first order, second order, and third order passive filter, each 
feeding a destination which has its own impedance Rimp. Rimp might be 
infinity, in which case we could ignore it, or it might not.

For the first order case, I've been able to derive the transfer function by 
thinking of it as a potential divider with impedances. Ignoring Rimp for a 
moment, we have:

Vout / Vin = (1 / sC) / (R1 + (1/sC))

This works fine. Including Rimp doesn't make it much more complicated, 
except that we replace  1/sC with (1/sC || Rimp)

However, I can't make the second order (or by extension third order) case 
work the same way. It still looks like a potential divider, with the output 
across C2, but the total impedance is a bit more involved. I think we have 
to work backwards in from the end, so we have the output across C2. We have 
R2 in series with C2, which is then in parallel with C1, and the result of 
that is in series with R1. So the total impedance is something like:

Ztotal = R1+ (1/sC1 || (R2 + 1/sC2)) (ignoring Rimp again)

So my transfer function is:

Vout / Vin = 1/sC2  /  (R1+ (1/sC1 || (R2 + 1/sC2)))

Is this correct? Am I on the right lines with this "potential divider" 
understanding of the situation?

Thanks,
Tom


_______________________________________________
Synth-diy mailing list
Synth-diy at dropmix.xs4all.nl
http://dropmix.xs4all.nl/mailman/listinfo/synth-diy


-----
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2014.0.4842 / Virus Database: 4447/10922 - Release Date: 10/31/15