Paralled caps – sort of. Help calculating

[lion]

Something I've been wondering about.
I understand parallel capacitors sums up to the total of the cap values – ie 10n//10n = 20nF.
But what happens if you insert at low(er) value series resistor on one of the caps?

In a simple low pass filter as an example with 47k series resistance and a 1n cap to ground.
Fc = 3,388 Hz

Add a parallel 2n2 cap to the first one
1n//2n2 = 3n2 > Fc = 1,059 Hz

Now insert a series resistor of say 10k between the 2n2 and ground – the HF roll off is obviously reduced, but I would like to know what is really going on, and how one can calculate it?
As a hint to my thinking and interest it seem to me there's more than a reduction of the HF bleed happening with the insertion of low value resistors. Same action with a guitar tone control, most of the rotation there's an increase in roll off, but near the end (where the series res is low) something else is happening too.

Any help much appreciated.

[merlinb]

But what happens if you insert at low(er) value series resistor on one of the caps?

Things get more complicated. Basically you have created two filters operating side by side. One is an ordinary low-pass, and the other is a low-pass shelving filter (a filter that doesn't keep on cutting forever, but levels off). The overall frequency response then depends on which one starts doing the cutting first...

[lion]

Thanks for the reply.
Interesting. I would like to explore it further. I wonder if there's calculators/simulators (that doesn't take a master degrees to operate) that could be used for seeing the results of different combinations of standard lowpass and shelvings - with different RC values.

[merlinb]

Thanks for the reply.
Interesting. I would like to explore it further. I wonder if there's calculators/simulators (that doesn't take a master degrees to operate) that could be used for seeing the results of different combinations of standard lowpass and shelvings - with different RC values.

I reommend Tina TI. It does everything, but is easier to use than PSpice:
http://www.ti.com/tool/tina-ti

For a simple RC filter the cut-off frequency is:
f = 1 / (2 * pi * R *C)

For a low-pass shelving filter the "pole" (the frequency where the response start falling) is:
f = 1 / (2 * pi * [R1+R2] *C)
The resonse then falls at 6dB/octave until you reach the "zero" (the frequency where it levels off again), which is:
f = 1 / (2 * pi * R2 *C)
Where R2 is the second resistor (the one directly in series with the cap).

But as I said, when you combine the two filters in one, it gets more complicated.

[ashcat_lt]

For a low-pass shelving filter the "pole" (the frequency where the response start falling) is:
f = 1 / (2 * pi * [R1+R2] *C)
The resonse then falls at 6dB/octave until you reach the "zero" (the frequency where it levels off again), which is:
f = 1 / (2 * pi * R2 *C)
Where R2 is the second resistor (the one directly in series with the cap).

That right there is the simple, clear, concise answer that I was searching for all last week!  Nobody, anywhere that I could find, gave that basic breakdown.  I simmed it and hacked at it and got it wrong anyway.  This helps much, thanks!

[lion]

Merlin, thanks a lot for the details and the sim link.
I've downloaded TINA and will give it a try. Might come running back for help though.