Reply by Jon Harris October 22, 20042004-10-22
"canmc" <canmcbk@yahoo.com> wrote in message
news:a01b5002.0410220209.8b735d0@posting.google.com...
> "Jon Harris" <goldentully@hotmail.com> wrote in message
news:<2sj6n3F1l8fjeU1@uni-berlin.de>...
> > http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt
shows
> > how to compute peaking/notch filters (as well as some other common ones). > > String together 7 of those biquads in series and you should be in business. > > I have trouble with the formula in the case of BPF with constant skirt > gain in the above audio EQ cookbook. I don't understand why the peak > gain and the quality factor share the symbol Q. Could they be > different? Could anyone explain me? Thanks in advance.
I'm not an expert, but I _think_ it is just a coincidence. The higher the Q, the narrower the filter and hence the higher it peaks.
Reply by Jerry Avins October 22, 20042004-10-22
canmc wrote:

> "Jon Harris" <goldentully@hotmail.com> wrote in message news:<2sj6n3F1l8fjeU1@uni-berlin.de>... > >>http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt shows >>how to compute peaking/notch filters (as well as some other common ones). >>String together 7 of those biquads in series and you should be in business. > > > I have trouble with the formula in the case of BPF with constant skirt > gain in the above audio EQ cookbook. I don't understand why the peak > gain and the quality factor share the symbol Q. Could they be > different? Could anyone explain me? Thanks in advance.
Q determines the shape of the transfer function, including the peak gain. In LC and acoustic resonant circuits for which Q was originally defined, it is the ratio of the energy stored in the resonator to the energy dissipated in one cycle. The higher the ratio, the sharper and stronger the resonance. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by canmc October 22, 20042004-10-22
"Jon Harris" <goldentully@hotmail.com> wrote in message news:<2sj6n3F1l8fjeU1@uni-berlin.de>...
> http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt shows > how to compute peaking/notch filters (as well as some other common ones). > String together 7 of those biquads in series and you should be in business.
I have trouble with the formula in the case of BPF with constant skirt gain in the above audio EQ cookbook. I don't understand why the peak gain and the quality factor share the symbol Q. Could they be different? Could anyone explain me? Thanks in advance.
Reply by canmc October 16, 20042004-10-16
"Jon Harris" <goldentully@hotmail.com> wrote in message > 
> The way I would construct a peaking filter, and the way prescribed in the audio > EQ cookbook, would require a two-pole, two-zero structure: > http://www-ccrma.stanford.edu/~jos/filters/BiQuad_Section.html > > A peaking filter implemented like this would have unity gain at most > frequencies, but have a peak of the specified magntidue at the specified > frequency. Using this approach, you do have independence of adjustment of the > various sections, at least as I understand it. For example, if you had one > filter with a 3dB peak at 500 Hz and another filter with a 4dB peak at 2000 Hz, > the cascade/series combination of those is a filter with a ~3dB peak at 500 Hz > and a ~4dB peak at 2000 Hz. Of course, if the peaks are close enough or wide > enough, they will start to influence other nearby peaks. Does that help?
That sounds great. I'll try it. Thanks for your information.
Reply by canmc October 16, 20042004-10-16
davec@quantized.com (David Coffey) wrote in message news:<163e03dd.0410120752.1df4808e@posting.google.com>...
> canmcbk@yahoo.com (canmc) wrote in message news:<a01b5002.0410120109.640c8815@posting.google.com>... > > > > I tried implementing the filter with 5 peaks by cascading 5 two-pole > > filters (for simplicity, 2 notches are not mentioned here). Each of > > them is performed as shown in the link below: > > http://www-ccrma.stanford.edu/~jos/filters/Two_Pole.html > > > > According to this design, for each resonator, the peak frequency and > > 3dB-bandwidth are defined by the poles' angle and radius respectively, > > and the peak gain is controled by the numerator coefficient of the > > transfer function. For a resonator, it's ok. When cascading these > > resonators, I found that, in the magnitude response of the resulting > > system, the peak frequencies and the 3dB-bandwidths of the peaks are > > nearly the same as those of stand-alone resonators but the gains at > > peaks are changed. I also found out no way to individually adjust the > > gains at peak frequencies because each gain at a certain frequency > > will result in the same gain at all other ones. It's the problem. Can > > anyone help me? > > OK, just to clear things up, when you say 'cascading' do you mean > series connection i.e. taking the output from one section as the input > to the next? Because if this is the case, the gains will cascade through > the system. Say for instance the first section is specd. to have a gain of > 0.5 and the second 0.5, then the effective gain of the second section will > be 0.25. To put it another way, if you have an input source with a flat > frequency response form 50Hz - 5kHz(say), and you feed this into a reson > filter centred at 1kHz with a bandwidth of +/- 100Hz then take the output > of this and feed it into a second section centred at 2.5kHz you will get > very little out because the centre frequency of the second section is > operating low on the skirt of the first. > > As I previously pointed out, the system can be better specd. if > the sections are applied in parallel i.e run the 5 sections simultaneously > on the same input signal and sum the outputs. Also, the filter in the link > is fine but for a little extra computational effort you can place the zeros > at +/-1 on the unit circle to reject DC and frequencies approaching Nyquist. > This can also lead to an implementation more in line with the desired response. > This would be achieved by multiplying x[n-2] by -1 and including it in the > difference equation. > > The bottom line is that if the sections are fed in series, the gains > can't be controlled independantly because the gain of section 1 scales > the gain of 2,3... and 2 scales 3,4... and so on. > > Hope this helps, sorry if I'm missing something or misunderstanding your > terminology.
I'm agree with you about the inpossibility of the cascade configuration in independently controlling the gain at resonances. However, as I stated before, placing multiple resonators in parallel will result in some unexpected zeroes in the transfer function of the resulting system, causing some undesired notches in the resulting magnitude response. So both these configurations aren't good.
Reply by Jon Harris October 12, 20042004-10-12
"canmc" <canmcbk@yahoo.com> wrote in message
news:a01b5002.0410120109.640c8815@posting.google.com...
> "Jon Harris" <goldentully@hotmail.com> wrote in message
news:<2svubjF1qcaj5U1@uni-berlin.de>...
> > "canmc" <canmcbk@yahoo.com> wrote in message > > news:<2sj6n3F1l8fjeU1@uni-berlin.de>... > > > >
http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt
> > shows > > > > how to compute peaking/notch filters (as well as some other common
ones).
> > > > String together 7 of those biquads in series and you should be in
business.
> > > > > > > > "canmc" <canmcbk@yahoo.com> wrote in message > > > > news:a01b5002.0410051902.4cccb48e@posting.google.com... > > > > > Hi all. > > > > > I'm in trouble on designing this linear time-invariant digital > > > > > filter. It has five peaks and two notches in its magnitude frequency > > > > > response. Each peak/notch is specified by its frequency, 3dB-bandwidth > > > > > and magnitude. These parameters and of course, the sampling frequency, > > > > > can be chosen arbitrarily. Please give me any solution. Thanks in > > > > > advance. > > > > > Best regards, > > > > > Canmc > > > > > > Hi. > > > It's not simple as you think. Because it's impossible to control the > > > gain of each peak/notch in the desired magnitude response if we > > > combine some peak/notch filters in series. Moreover, it's not a good > > > idea to design a peak/notch filter using bilinear transform because > > > the nature of warping frequency will result in the difference between > > > the peak/notch frequency of an analog filter with that of a digital > > > one. > > > > Regarding your first issue, I guess I don't understand the problem well
enough
> > to know why this doesn't work. Regarding the second issue, the equations in
the
> > link I listed above already take into account the warping effect, so the > > frequencies are accurate. (However, the shapes of peaks are a little
"squished"
> > near the Nyquist frequency.) > > I tried implementing the filter with 5 peaks by cascading 5 two-pole > filters (for simplicity, 2 notches are not mentioned here). Each of > them is performed as shown in the link below: > http://www-ccrma.stanford.edu/~jos/filters/Two_Pole.html
The way I would construct a peaking filter, and the way prescribed in the audio EQ cookbook, would require a two-pole, two-zero structure: http://www-ccrma.stanford.edu/~jos/filters/BiQuad_Section.html A peaking filter implemented like this would have unity gain at most frequencies, but have a peak of the specified magntidue at the specified frequency. Using this approach, you do have independence of adjustment of the various sections, at least as I understand it. For example, if you had one filter with a 3dB peak at 500 Hz and another filter with a 4dB peak at 2000 Hz, the cascade/series combination of those is a filter with a ~3dB peak at 500 Hz and a ~4dB peak at 2000 Hz. Of course, if the peaks are close enough or wide enough, they will start to influence other nearby peaks. Does that help?
> According to this design, for each resonator, the peak frequency and > 3dB-bandwidth are defined by the poles' angle and radius respectively, > and the peak gain is controled by the numerator coefficient of the > transfer function. For a resonator, it's ok. When cascading these > resonators, I found that, in the magnitude response of the resulting > system, the peak frequencies and the 3dB-bandwidths of the peaks are > nearly the same as those of stand-alone resonators but the gains at > peaks are changed. I also found out no way to individually adjust the > gains at peak frequencies because each gain at a certain frequency > will result in the same gain at all other ones. It's the problem. Can > anyone help me?
Reply by Jerry Avins October 12, 20042004-10-12
David Coffey wrote:

  ...

> OK, just to clear things up, when you say 'cascading' do you mean > series connection i.e. taking the output from one section as the input > to the next? Because if this is the case, the gains will cascade through > the system. Say for instance the first section is specd. to have a gain of > 0.5 and the second 0.5, then the effective gain of the second section will > be 0.25. To put it another way, if you have an input source with a flat > frequency response form 50Hz - 5kHz(say), and you feed this into a reson > filter centred at 1kHz with a bandwidth of +/- 100Hz then take the output > of this and feed it into a second section centred at 2.5kHz you will get > very little out because the centre frequency of the second section is > operating low on the skirt of the first.
The phase shifts through parallel filters are important too. A notch will develop at some frequency where the amplitudes are nearly equal and the phases are shifted oppositely nearly 90 degrees. The filters best suited for this application are probably recursive, but convolutional FIRs will be easier to design. It's the simplest choice if the processing power is available, Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by David Coffey October 12, 20042004-10-12
canmcbk@yahoo.com (canmc) wrote in message news:<a01b5002.0410120109.640c8815@posting.google.com>...
> "Jon Harris" <goldentully@hotmail.com> wrote in message news:<2svubjF1qcaj5U1@uni-berlin.de>... > > "canmc" <canmcbk@yahoo.com> wrote in message > > news:a01b5002.0410090055.7e3c675@posting.google.com... > > > "Jon Harris" <goldentully@hotmail.com> wrote in message > news:<2sj6n3F1l8fjeU1@uni-berlin.de>... > > > > http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt > shows > > > > how to compute peaking/notch filters (as well as some other common ones). > > > > String together 7 of those biquads in series and you should be in business. > > > > > > > > "canmc" <canmcbk@yahoo.com> wrote in message > > > > news:a01b5002.0410051902.4cccb48e@posting.google.com... > > > > > Hi all. > > > > > I'm in trouble on designing this linear time-invariant digital > > > > > filter. It has five peaks and two notches in its magnitude frequency > > > > > response. Each peak/notch is specified by its frequency, 3dB-bandwidth > > > > > and magnitude. These parameters and of course, the sampling frequency, > > > > > can be chosen arbitrarily. Please give me any solution. Thanks in > > > > > advance. > > > > > Best regards, > > > > > Canmc > > > > > > Hi. > > > It's not simple as you think. Because it's impossible to control the > > > gain of each peak/notch in the desired magnitude response if we > > > combine some peak/notch filters in series. Moreover, it's not a good > > > idea to design a peak/notch filter using bilinear transform because > > > the nature of warping frequency will result in the difference between > > > the peak/notch frequency of an analog filter with that of a digital > > > one. > > > > Regarding your first issue, I guess I don't understand the problem well enough > > to know why this doesn't work. Regarding the second issue, the equations in the > > link I listed above already take into account the warping effect, so the > > frequencies are accurate. (However, the shapes of peaks are a little "squished" > > near the Nyquist frequency.) > > I tried implementing the filter with 5 peaks by cascading 5 two-pole > filters (for simplicity, 2 notches are not mentioned here). Each of > them is performed as shown in the link below: > http://www-ccrma.stanford.edu/~jos/filters/Two_Pole.html > > According to this design, for each resonator, the peak frequency and > 3dB-bandwidth are defined by the poles' angle and radius respectively, > and the peak gain is controled by the numerator coefficient of the > transfer function. For a resonator, it's ok. When cascading these > resonators, I found that, in the magnitude response of the resulting > system, the peak frequencies and the 3dB-bandwidths of the peaks are > nearly the same as those of stand-alone resonators but the gains at > peaks are changed. I also found out no way to individually adjust the > gains at peak frequencies because each gain at a certain frequency > will result in the same gain at all other ones. It's the problem. Can > anyone help me?
OK, just to clear things up, when you say 'cascading' do you mean series connection i.e. taking the output from one section as the input to the next? Because if this is the case, the gains will cascade through the system. Say for instance the first section is specd. to have a gain of 0.5 and the second 0.5, then the effective gain of the second section will be 0.25. To put it another way, if you have an input source with a flat frequency response form 50Hz - 5kHz(say), and you feed this into a reson filter centred at 1kHz with a bandwidth of +/- 100Hz then take the output of this and feed it into a second section centred at 2.5kHz you will get very little out because the centre frequency of the second section is operating low on the skirt of the first. As I previously pointed out, the system can be better specd. if the sections are applied in parallel i.e run the 5 sections simultaneously on the same input signal and sum the outputs. Also, the filter in the link is fine but for a little extra computational effort you can place the zeros at +/-1 on the unit circle to reject DC and frequencies approaching Nyquist. This can also lead to an implementation more in line with the desired response. This would be achieved by multiplying x[n-2] by -1 and including it in the difference equation. The bottom line is that if the sections are fed in series, the gains can't be controlled independantly because the gain of section 1 scales the gain of 2,3... and 2 scales 3,4... and so on. Hope this helps, sorry if I'm missing something or misunderstanding your terminology.
Reply by canmc October 12, 20042004-10-12
"Jon Harris" <goldentully@hotmail.com> wrote in message news:<2svubjF1qcaj5U1@uni-berlin.de>...
> "canmc" <canmcbk@yahoo.com> wrote in message > news:a01b5002.0410090055.7e3c675@posting.google.com... > > "Jon Harris" <goldentully@hotmail.com> wrote in message > news:<2sj6n3F1l8fjeU1@uni-berlin.de>... > > > http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt > shows > > > how to compute peaking/notch filters (as well as some other common ones). > > > String together 7 of those biquads in series and you should be in business. > > > > > > "canmc" <canmcbk@yahoo.com> wrote in message > > > news:a01b5002.0410051902.4cccb48e@posting.google.com... > > > > Hi all. > > > > I'm in trouble on designing this linear time-invariant digital > > > > filter. It has five peaks and two notches in its magnitude frequency > > > > response. Each peak/notch is specified by its frequency, 3dB-bandwidth > > > > and magnitude. These parameters and of course, the sampling frequency, > > > > can be chosen arbitrarily. Please give me any solution. Thanks in > > > > advance. > > > > Best regards, > > > > Canmc > > > > Hi. > > It's not simple as you think. Because it's impossible to control the > > gain of each peak/notch in the desired magnitude response if we > > combine some peak/notch filters in series. Moreover, it's not a good > > idea to design a peak/notch filter using bilinear transform because > > the nature of warping frequency will result in the difference between > > the peak/notch frequency of an analog filter with that of a digital > > one. > > Regarding your first issue, I guess I don't understand the problem well enough > to know why this doesn't work. Regarding the second issue, the equations in the > link I listed above already take into account the warping effect, so the > frequencies are accurate. (However, the shapes of peaks are a little "squished" > near the Nyquist frequency.)
I tried implementing the filter with 5 peaks by cascading 5 two-pole filters (for simplicity, 2 notches are not mentioned here). Each of them is performed as shown in the link below: http://www-ccrma.stanford.edu/~jos/filters/Two_Pole.html According to this design, for each resonator, the peak frequency and 3dB-bandwidth are defined by the poles' angle and radius respectively, and the peak gain is controled by the numerator coefficient of the transfer function. For a resonator, it's ok. When cascading these resonators, I found that, in the magnitude response of the resulting system, the peak frequencies and the 3dB-bandwidths of the peaks are nearly the same as those of stand-alone resonators but the gains at peaks are changed. I also found out no way to individually adjust the gains at peak frequencies because each gain at a certain frequency will result in the same gain at all other ones. It's the problem. Can anyone help me?
Reply by Jon Harris October 11, 20042004-10-11
"canmc" <canmcbk@yahoo.com> wrote in message
news:a01b5002.0410090055.7e3c675@posting.google.com...
> "Jon Harris" <goldentully@hotmail.com> wrote in message
news:<2sj6n3F1l8fjeU1@uni-berlin.de>...
> > http://www.harmony-central.com/Computer/Programming/Audio-EQ-Cookbook.txt
shows
> > how to compute peaking/notch filters (as well as some other common ones). > > String together 7 of those biquads in series and you should be in business. > > > > "canmc" <canmcbk@yahoo.com> wrote in message > > news:a01b5002.0410051902.4cccb48e@posting.google.com... > > > Hi all. > > > I'm in trouble on designing this linear time-invariant digital > > > filter. It has five peaks and two notches in its magnitude frequency > > > response. Each peak/notch is specified by its frequency, 3dB-bandwidth > > > and magnitude. These parameters and of course, the sampling frequency, > > > can be chosen arbitrarily. Please give me any solution. Thanks in > > > advance. > > > Best regards, > > > Canmc > > Hi. > It's not simple as you think. Because it's impossible to control the > gain of each peak/notch in the desired magnitude response if we > combine some peak/notch filters in series. Moreover, it's not a good > idea to design a peak/notch filter using bilinear transform because > the nature of warping frequency will result in the difference between > the peak/notch frequency of an analog filter with that of a digital > one.
Regarding your first issue, I guess I don't understand the problem well enough to know why this doesn't work. Regarding the second issue, the equations in the link I listed above already take into account the warping effect, so the frequencies are accurate. (However, the shapes of peaks are a little "squished" near the Nyquist frequency.)