[sdiy] Talking of Q

David G. Dixon dixon at mail.ubc.ca
Fri Jun 17 19:07:51 CEST 2011


> I've been trying to work through the maths of a few 
> filter-related equations recently.
> 
> With regards to calculating Q...
> 
> Can anyone provide some Q range/factor amounts for some real 
> world filter designs?
> 
> For example, the ARP2500 filter is claimed to have a Q range 
> of 500+ - is this pretty standard?
> 
> Just trying to put the bits I'm trying to understand into a 
> context that I understand... if that makes sense...

AFAIAC, there is no shortcut to understanding this stuff which bypasses
slogging through the math(s).

Here's my approach to understanding filter behaviour:

1. Derive the filter transfer equation(s) by solving current balances around
the nodes where the filter output signal voltages are manifested (generally
at the "business end" of the filter capacitors), making sure to incorporate
resonance gain behaviour in the calculations.  This is basic circuit
analysis, and there is no way around it.  For ease of derivation, denote all
sRC terms simply as "S".

2. Carry out all algebraic operations within the transfer term.  For
example, if there is a term such as (S+1)^3, then replace it with S^3 + 3S^2
+ 3S + 1.

3. Replace all "S" terms with "jw" and carry out all indicated algebraic
operations on the imaginary number.  Remember: j^2 = -1, j^3 = -j, and j^4 =
1.

4. Gather all the terms with "j" (the "imaginary" parts) and all the terms
without "j" (the "real" parts) into separate equations in both numerator and
demoninator.

5. Calculate the magnitude of the transfer function (in other words, the
actual voltage out over the voltage in) by taking the square root of the sum
of the squares of the real and imaginary parts in both numerator and
demoninator.

Here's an example.  The second stage of a four-pole COTA filter with a
single inverting resonance gain stage between the fourth and first stages
has the following transfer function:

V_2/V_0 = (S+1)^2/[(S+1)^4 + G]

where G is the resonance feedback gain.  That's the result of step 1.  Now,
step 2, the algebra:

V_2/V_0 = [S^2 + 2S + 1]/[S^4 + 4S^3 + 6S^2 + 4S + 1 + G]

Now, step 3, replacing S with jw and carrying out the algebra on j:

V_2/V_0 = [(jw)^2 + 2jw + 1]/[(jw)^4 + 4(jw)^3 + 6(jw)^2 + 4jw + 1 + G] =
[-w^2 + 2jw + 1]/[w^4 - 4jw^3 - 6w^2 + 4jw + 1 + G]

Now, step 4, gathering numerator and demoninator into real and imaginary
parts:

V_2/V_0 = [(1 - w^2) + j(2w)]/[(w^4 - 6w^2 + 1 + G) + j(4w - 4w^3)]

And finally, step 5, calculating the amplitude by taking the square roots of
the sum of squares of the real and imaginary parts, top and bottom:

{V_2/V_0} = [(1 - w^2)^2 + (2w)^2]^(1/2)/[(w^4 - 6w^2 + 1 + G)^2 + (4w -
4w^3)^2]^(1/2)

Plot this last function in Excel as a function of w.  You will find that it
goes to infinity (i.e., the demoninator goes to zero) at the cutoff
frequency (w = 1) at a gain of G = 4.  This is exactly where the COTA filter
begins to oscillate.  This fact is, of course, easy to see from the result
of step 4, since, at w = 1, the denominator becomes:

(w^4 - 6w^2 + 1 + G) + j(4w - 4w^3) = (1 - 6 + 1 + G) + j(4 - 4) = G - 4

If there's a simpler way to understand filter behaviour, then I haven't
found it.




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