Reply by robert bristow-johnson●July 1, 20082008-07-01
On Jun 30, 11:34�pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
wrote:
> robert bristow-johnson wrote:
> > On Jun 30, 3:16 pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
> > wrote:
>
> >>Greg Berchin wrote:
>
> >>>In the last audio EQ that I designed, the customers demanded boost/cut
> >>>symmetry in EQ filters, so I (re)defined Q accordingly.
>
> >>Exactly. The similar adjustment has to be applied for the shelving
> >>filters for tone control.
>
> > i guess i don't know what you mean, Vlad. �do you mean to adjust the
> > value of Q (when the shelf gain changes) so as to keep the steepest
> > possible monotonic slope of the shelf?
>
> > otherwise, what Q adjustment are you guys referring to?
>
> As noted by Greg, they want the boost and cut responses to be
> symmetrical. I.e. the cut response should look like a mirror image or
> the boost response.
i get that, and had since 1994. the cookbook peaking EQ does that,
but the Q or BW of that maybe needs a mapping to relate it to yours.
there are an infinite number of ways to accomplish this boost/cut
symmetry. my question is what basis did you use to chose one of those
ways.
> This applies to the EQ sections as well as to the
> tone control sections. With EQ sections, we have to adjust Q by gain,
> and with the shelving sections, we have to adjust the corner frequencies.
depends on how the corner frequencies are defined. since it's kinda
hard to define it as the 3 dB frequency when the boost/cut is less
than 3 dB, i've been defining it at the shelf-midpoint frequency
(since that exists for all boost and cut amounts). in the cookbook,
the boost and cut shelving filters are also symmetrical in the same
manner.
it's the same idea for the peaking EQ to define the bandedges at the
midpoint gain. those bandedges are consistently defined and exist for
any boost/cut gain.
r b-j
Reply by ●July 1, 20082008-07-01
On Jul 1, 6:57 am, "jungledmnc" <jungled...@gmail.com> wrote:
> Btw. I'm pretty newbie in this, what is exactly the bilinear transform in
> this case? I've read about the pole-zero method, but this does not seem too
> close. Just please give me a hint :-).
>
> And one more not very celver question : What is exactly the difference
> between filters like Chebyshev, biquad etc. I don't know if it is
> chebyshev, but it seems, that most of those filters are simply IIRs. So I
> would think, that the difference is basically the way how to compute filter
> coefficients to achieve certain frequency/phase responses. Am I correct?
>
> Thx.
http://en.wikipedia.org/wiki/Bilinear_transform
It's a method usually used for transforming an analog prototype filter
(Chebyshev, Butterworth, elliptic, Bessel, etc.) into a digital filter
that has a similar response. Google for these filter types and you can
learn all about them.
Jason
Reply by Greg Berchin●July 1, 20082008-07-01
On Jun 30, 7:20�pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> but that's the correct EE Q anyway for a notch. �it's what's in the
> cookbook (for notch, not for boost/cut peaking EQ).
Agreed, but non-intuitive for the customers, who were not EEs. Once
they saw for themselves just how narrow a 1/12 octave notch was
(compared with a 1/12 octave boost/cut EQ), they were sold.
Of course, then there was the problem of making 20 Hz 1/12 octave
notch filters work at 96 kHz sampling rate with fixed point
arithmetic.
Greg
Reply by jungledmnc●July 1, 20082008-07-01
Btw. I'm pretty newbie in this, what is exactly the bilinear transform in
this case? I've read about the pole-zero method, but this does not seem too
close. Just please give me a hint :-).
And one more not very celver question : What is exactly the difference
between filters like Chebyshev, biquad etc. I don't know if it is
chebyshev, but it seems, that most of those filters are simply IIRs. So I
would think, that the difference is basically the way how to compute filter
coefficients to achieve certain frequency/phase responses. Am I correct?
Thx.
Reply by Vladimir Vassilevsky●July 1, 20082008-07-01
robert bristow-johnson wrote:
> On Jun 30, 3:16 pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
> wrote:
>
>>Greg Berchin wrote:
>>
>>>In the last audio EQ that I designed, the customers demanded boost/cut
>>>symmetry in EQ filters, so I (re)defined Q accordingly.
>>
>>Exactly. The similar adjustment has to be applied for the shelving
>>filters for tone control.
>
>
> i guess i don't know what you mean, Vlad. do you mean to adjust the
> value of Q (when the shelf gain changes) so as to keep the steepest
> possible monotonic slope of the shelf?
>
> otherwise, what Q adjustment are you guys referring to?
As noted by Greg, they want the boost and cut responses to be
symmetrical. I.e. the cut response should look like a mirror image or
the boost response. This applies to the EQ sections as well as to the
tone control sections. With EQ sections, we have to adjust Q by gain,
and with the shelving sections, we have to adjust the corner frequencies.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com
Reply by robert bristow-johnson●June 30, 20082008-06-30
On Jun 30, 3:16�pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
wrote:
> Greg Berchin wrote:
> > On Jun 30, 1:59 pm, robert bristow-johnson <r...@audioimagination.com>
> > wrote:
>
> >>now, if you retain the same EE definition of Q for the peaking EQ (and
> >>the old analog EQs do that, e.g. pultec), then you find that if you
> >>leave Q constant, a cut of N dB is not the inverse filter of a boost
> >>of N dB. �the cut is a lot skinnier looking than the boost.
>
> > In the last audio EQ that I designed, the customers demanded boost/cut
> > symmetry in EQ filters, so I (re)defined Q accordingly.
>
> Exactly. The similar adjustment has to be applied for the shelving
> filters for tone control.
i guess i don't know what you mean, Vlad. do you mean to adjust the
value of Q (when the shelf gain changes) so as to keep the steepest
possible monotonic slope of the shelf?
otherwise, what Q adjustment are you guys referring to?
r b-j
Reply by robert bristow-johnson●June 30, 20082008-06-30
On Jun 30, 4:51�pm, "jungledmnc" <jungled...@gmail.com> wrote:
> >i dunno what is wrong. �i am presuming that you are setting z=3De^(jw)
> >(where w=3D2*pi*f/Fs) and sticking that z into H(z) and then evaluating
> >the magnitude of the complex result, right? �there could be numerical
> >issues. �a while ago i posted a general magnitude frequency response
> >for a biquad (will work whether it's LPF or BPF or HPF or whatever
> >biquad).
>
> >http://groups.google.com/group/comp.dsp/msg/a1bc5b63ac56b686
>
> >maybe that will help.
>
> >> Thanks a lot.
>
> >FWIW
>
> >r b-j
>
> Well, used 64-bit and it got better. However I have a few more questions:
>
> 1) What are correct range of Q and f0 (I mean for all of the filter
> types)?2) Shelving filter seems a little bit weird, 'cos one must change Q
> according to frequency and even more - if you increase gain, the frequency
> seems to move to upwards. Is this correct? If so, how to compensate it, so
> when I set frequency, I can change Q and gain without problems?
in the cookbook, the shelving filters are odd-symmetric about the
shelf-midpoint frequency (ignoring frequency warping). the gain and
shelfslope are independent parameters, for the most part. as long as
we're not square rooting negative numbers, you can pair together any
gain with any other Q.
r b-j
Reply by robert bristow-johnson●June 30, 20082008-06-30
On Jun 30, 2:20�pm, Greg Berchin <gberc...@sentientscience.com> wrote:
> On Jun 30, 1:59�pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
> > now, if you retain the same EE definition of Q for the peaking EQ (and
> > the old analog EQs do that, e.g. pultec), then you find that if you
> > leave Q constant, a cut of N dB is not the inverse filter of a boost
> > of N dB. �the cut is a lot skinnier looking than the boost.
>
> In the last audio EQ that I designed, the customers demanded boost/cut
> symmetry in EQ filters, so I (re)defined Q accordingly.
that can be done in a few different ways. you can leave the Q 100%
the EE definition for boost and define it however it has to be for cut
(that is multiply by the gain of the cut). or you can leave it the EE
definition for cut and fudge it for boost (again by multiplying it by
the linear boost gain). or you can fudge it half as much for both
boost and cut (by multiplying it be sqrt(linear gain)). or distribute
the fudging anyone of a zillion other ways and still get symmetrical
boost and cut for the same setting of "Q". the cookbook uses the
equal fudge definition, not because audio practitioners like it (i
heard they don't), but because it is simple and consistent and i'll
just let the designer use their own Q to Q mapping.
>�For notch
> filters, though, I defined Q using the -3.01 dB points on either side
> of the center frequency. �The customers like that because it made the
> notches much narrower.
but that's the correct EE Q anyway for a notch. it's what's in the
cookbook (for notch, not for boost/cut peaking EQ).
r b-j
Reply by jungledmnc●June 30, 20082008-06-30
>i dunno what is wrong. i am presuming that you are setting z=3De^(jw)
>(where w=3D2*pi*f/Fs) and sticking that z into H(z) and then evaluating
>the magnitude of the complex result, right? there could be numerical
>issues. a while ago i posted a general magnitude frequency response
>for a biquad (will work whether it's LPF or BPF or HPF or whatever
>biquad).
>
>http://groups.google.com/group/comp.dsp/msg/a1bc5b63ac56b686
>
>maybe that will help.
>
>> Thanks a lot.
>
>FWIW
>
>r b-j
Well, used 64-bit and it got better. However I have a few more questions:
1) What are correct range of Q and f0 (I mean for all of the filter
types)?2) Shelving filter seems a little bit weird, 'cos one must change Q
according to frequency and even more - if you increase gain, the frequency
seems to move to upwards. Is this correct? If so, how to compensate it, so
when I set frequency, I can change Q and gain without problems?
Reply by Vladimir Vassilevsky●June 30, 20082008-06-30
Greg Berchin wrote:
> On Jun 30, 1:59 pm, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
>
>>now, if you retain the same EE definition of Q for the peaking EQ (and
>>the old analog EQs do that, e.g. pultec), then you find that if you
>>leave Q constant, a cut of N dB is not the inverse filter of a boost
>>of N dB. the cut is a lot skinnier looking than the boost.
>
> In the last audio EQ that I designed, the customers demanded boost/cut
> symmetry in EQ filters, so I (re)defined Q accordingly.
Exactly. The similar adjustment has to be applied for the shelving
filters for tone control.
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com