[sdiy] Voltage divider with RC filter - help with the basics, please

Tom Wiltshire tom at electricdruid.net
Wed Jul 24 20:40:00 CEST 2013

On 24 Jul 2013, at 09:58, "cheater00 ." <cheater00 at gmail.com> wrote:

> Hi Russell,
> On Tue, Jul 23, 2013 at 4:27 PM, Russell McClellan
> <russell.mcclellan at gmail.com> wrote:
>> Hey Tom,
>> Richie already gave a complete and accurate response, but maybe you'd
>> be interested in a standard, general technique for this sort of
>> passive network.
>> First, I convert everything to "impedances" - this means treated the
>> capacitor as an "impeder" with impedance 1/sC.  The units of impedance
>> are ohms, and resistors translate directly - their impedance is R.
>> Don't worry about what the "s" "means" other than that it depends on
>> frequency (it's a fancy math thing called a "laplace variable").  The
>> cool thing about impedances is that they behave and combine exactly
>> like resistors - you can use the 1/R_parallel = 1/R_1 + 1/R_2 rule for
>> parallel impedances and you can use addition to find serial impedance.
>> Using these two tools you can quickly build up your whole network
>> into a single equation - this equation is called the "Transfer
>> Function" of your system.  Interpreting this transfer function is
>> where your lack of formal training may get you into trouble - but, if
>> you've correctly made a low-pass filter, your function should look
>> like x/(1 + ys) where x and y are combinations of R and C values.  If
>> you can't use algebra to turn your equation into that form, it isn't a
>> lowpass filter.  As richie mentioned, if it's something like (1 +
>> xs)/(1+ys), it could be a shelf filter or an allpass filter (i.e., a
>> phasor).  In the lowpass case above, the cutoff is just 1/(y*2Pi).  Of
>> course there's deeper reasons why these equations work the way they
>> do, but that's probably out of scope for an e-mail exchange.
>> The reason I prefer to use this slightly pedantic method to analyze
>> even relatively simple networks of capacitors and resistors is I don't
>> have very much intuition.  This method wouldn't be used directly by
>> experienced folks except in relatively complicated situations.  For
>> me, it can be hard to guess whether your network even is a low-pass
>> filter just by looking at it.  Calculating the "cut-off" doesn't mean
>> much if you don't even know whether it's a low-or-high pass!
> That's the most amazing, down-to-earth description of obtaining
> transfer functions I've seen. I feel I finally stand a chance of
> understanding the process! Thanks!
> D.

I concur! Thanks.


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