Hi again,
If we start to do the actual network analysis we find that we end
up with a voltage source:
VN*RNb/(RNa+RNb)
where
VN is any input voltage and
RNa is the 'top' resistor of the pot and
RNb is the 'bottom' resistor of the pot
and an impedance:
RNa||RNb+10k+1/sCN
for each input channel in the three capacitor circuit, and
for the single capacitor network we end up wtih the same
only the impedance is
RNa||RNb+10k+(1/sCN)||RNN
where
RNN is a bunch of resistors from the other two inputs lumped,
but then since we also have the requirement that C is large enough
to swamp the effects of frequency that means we can take the
limit as C --> +infinity, so guess what? We end up with the same
network:
RNa||RNb+10k
For both circuits!
(and of course the same equivalent input voltage).
That makes both circuits the same, given C is large.
But hey, you know what? Why dont we just let C approach
infinity to start with, without doing anything else, then look
at the two networks with these large C's in place...
With all C's approaching infinity, they become short circuits.
With all C's shorted out, both networks are:
EXACTLY THE SAME
The best way to see any differences when the C's are not large
(that is frequency affects operation) is to do a network analysis
of both circuits and then compare.
The simplest case for the three cap ciruit is to let all pots be set at
maximum, so that we end up with three networks of R+1/sC
in parallel, which is electrically equivalent to three caps in parallel
in series with three resistors in parallel (when they are all the same value).
This tells us that the total C from the three cap circuit is three times the
value of that of the one cap circuit, which means the frequency response
will be different.