From: robert bristow-johnson on
On Apr 27, 10:49 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu>
wrote:
> Richard Owlett <rowl...(a)pcnetinc.com> wrote:
>
> (snip)
>
> > I was thinking in terms of Q = {resonant frequency}/{bandwidth}
> > Question may have become moot as Tim's answer triggered change of
> > how I defined my goals.
>
> It seems that there are different definitions for Q.
>
> Note, for example, the Wikipedia page Q_factor.
>
> Some of the difference comes from the difference between
> filters and oscillators, such that one is:
>
> Q = 2 pi * energy store / energy dissipated per cycle.

that is, i believe, the most general analog circuit definition. it
works for RLC series and RLC parallel circuits.

> The other is, as noted, (resonant frequency/(bandwidth),
> each in either cycles or radians.
>
> It seems that they are close for high Q, but not so close
> at lower Q.

there's a reason for this.

remember that even though the bilinear transform is not the only way
to map an analog filter to a digital filter, it can be help illustrate
the issue. the BLT maps the jw axis in the s-plane to the unit circle
in the z-plane just the "perfect" mapping of z=exp(s*T) does, except
that it maps a *different* point on the jw axis to some given point on
the unit circle. this means that every frequency response feature
that exists in the analog filter has a corresponding feature (of the
same dB gain) in the digital filter, except at a somewhat different
frequency. investigating this quantitatively leads one to what we
call "frequency warping" of the bilinear transform. every independent
frequency parameter of the analog design can be "fudged" so that,
after transforming to the digital filter using BLT, it ends up at what
the designer originally intended. i usually call that "pre-warping"
the frequency parameter.

for a 2nd-order LPF with Q of greater than 1/2 (if memory serves),
there is a peaking feature that gets more pronounced as the Q gets
higher. the BLT will not change the height in dB of that peaking
feature. similar for a BPF with the left and right "skirts" at a
constant level - the peak gain (measured linearly) is actually
proportional to Q. but, because of frequency warping of the BLT, the
analog bandwidth of the BPF is scrunched as the resonant frequency is
raised closer to Nyquist.

we often like to relate Q in an analog BPF to the *relative* bandwidth
as

1/Q = bw/f0

where bw and f0 (the resonant frequency) are measured in the same
units. i usually like to express relative bandwidth in terms of log
frequency (usually expressed in octaves) and we get a slightly
different relationship:

1/Q = 2*sinh(ln(2)/2 * BW)

where BW is in octaves. if BW is very small, the above gets close to

1/Q = ln(2) * BW

so it's pretty easy to see how BW in octaves gets related to the
relative bandwidth, bw/f0.

now, mapping analog bandwidth to digital bandwidth via BLT has a
little scrunching factor going on (because of frequency warping) and a
simple first-order approximation i have found for that scrunching
factor is

digital BW = sinc(f0/Nyquist) * (analog BW)

so i usually consider "Q" to be invariant between the corresponding
analog and digital frequency response (because the peak heights in the
frequency response remain unchanged), but i *do* consider the BW to be
changed. so the mapping between Q and BW in a digital filter is a
little bit adjusted from the analog:

1/Q = 2*sinh( ln(2)/2 * BW/sinc(f0/Nyquist) )

that's my spin on it.

r b-j
From: Magnum on
"cassiope" <fpm(a)u.washington.edu> wrote in message
news:a1905334-e7ac-4421-996b-c4ebdf636e7a(a)u31g2000yqb.googlegroups.com...
> Perhaps it would be useful if you would tell us how you define Q in
> the analog domain (there's more than one way).

ISTR that each way is a representation of the same formula.


From: Jerry Avins on
On 4/28/2010 3:44 AM, Magnum wrote:
> "cassiope"<fpm(a)u.washington.edu> wrote in message
> news:a1905334-e7ac-4421-996b-c4ebdf636e7a(a)u31g2000yqb.googlegroups.com...
>> Perhaps it would be useful if you would tell us how you define Q in
>> the analog domain (there's more than one way).
>
> ISTR that each way is a representation of the same formula.

Given high-enough Q. 3 is certainly high enough for folk music.

Jerry
--
"I view the progress of science as ... the slow erosion of the tendency
to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������
From: Mark on
On Apr 28, 5:37 am, Jerry Avins <j...(a)ieee.org> wrote:
> On 4/28/2010 3:44 AM, Magnum wrote:
>
> > "cassiope"<f...(a)u.washington.edu>  wrote in message
> >news:a1905334-e7ac-4421-996b-c4ebdf636e7a(a)u31g2000yqb.googlegroups.com....
> >> Perhaps it would be useful if you would tell us how you define Q in
> >> the analog domain (there's more than one way).
>
> > ISTR that each way is a representation of the same formula.
>
> Given high-enough Q. 3 is certainly high enough for folk music.
>
> Jerry
> --
> "I view the progress of science as ... the slow erosion of the tendency
>   to dichotomize." --Barbara Smuts, U. Mich.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

in the original post the op set the context of the question as a
"filter problem"

Often folks will say they need a high Q filter when they really mean
they need a high selectivity filter i.e. a filter with small
transition regions.

This is not really Q but more closely related to filter ORDER.

My guess is the op is asking about selectivity and order which is why
I tried to steer him to "taps".

Mark

From: Tim Wescott on
Magnum wrote:
> "cassiope" <fpm(a)u.washington.edu> wrote in message
> news:a1905334-e7ac-4421-996b-c4ebdf636e7a(a)u31g2000yqb.googlegroups.com...
>> Perhaps it would be useful if you would tell us how you define Q in
>> the analog domain (there's more than one way).
>
> ISTR that each way is a representation of the same formula.
>
>
They all converge to the same number pretty darn fast as the Q goes up,
but they don't necessarily match at low Q.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
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