Reply by Robert Orban February 1, 20092009-02-01
In article <d_WdnQ2KFd8i-
h_UnZ2dnUVZ_jqdnZ2d@giganews.com>, bantone@web.de 
says...
> > >>Hi, >> >>I had a hard time understanding the notion of "6dB per octave".
If I
>>have a single pole IIR low pass with the Fc = 1000 and Fs =
48000, for
>>example, >> >>a = [1.00000000000000 -0.87697646299276] >>b = 0.12302353700724 >> >>When doing freqz in Matlab, I was expecting >>-3db at 1000, >>-6dB at 2000, >>-9dB at 4000.... >>But what I have is >>-3@1000 >>-7@2000 >>-12.x@4000 >>-24.x@24000 >> >>Can anyone give my a hint what was wrong? Why it is -24dB at
24000Hz?
> >The digital frequency response is warped. >It doesn't look like a straight line going down >with 6db per octave (which is what you see >if you plot an analog response on a logarithmic scale - after >the rolloff at the beginning), >but it's a curve. >Remember that the distance from the pole is what determines >the amplitude. And you go from freq 0 to pi in a halfcircle in >the z-plane.
To add to the above post: You can approximate 6 dB/octave to any precision you like, but it takes more than one z-plane pole. At very least, you will need to add a zero, and you will get a much better approximation if you add more poles and zeros. Here are first, second and third-order approximations to your requirement. They are minimax magnitude approximations from 0 to 20 kHz: Zero # Real Imag. 1 -0.1698789 0.000000 Pole # Real Imag. 1 0.8720991 0.000000 MAXIMUM ERROR FROM 0.00 Hz TO 20000.00 Hz IS 0.1599127dB Zero # Real Imag. 1 -0.4590639 0.000000 2 -0.7892787E-01 0.000000 Pole # Real Imag. 1 -0.3661392 0.000000 2 0.8771473 0.000000 MAXIMUM ERROR FROM 0.00 Hz TO 20000.00 Hz IS 0.0042822dB Zero # Real Imag. 1 -0.6115832 0.000000 2 -0.2722988 0.000000 3 -0.4618174E-01 0.000000 Pole # Real Imag. 1 -0.5712548 0.000000 2 -0.1834612 0.000000 3 0.8773008 0.000000 MAXIMUM ERROR FROM 0.00 Hz TO 20000.00 Hz IS 0.0001147dB Error of the fifth-order approximation is basically down in the numerical noise: 1 -0.7513900 0.000000 2 -0.5236606 0.000000 3 -0.3078089 0.000000 4 -0.1286814 0.000000 5 -0.2164216E-01 0.000000 Pole # Real Imag. 1 -0.7370853 0.000000 2 -0.4922007 0.000000 3 -0.2562247 0.000000 4 -0.7020382E-01 0.000000 5 0.8773058 0.000000 MAXIMUM ERROR FROM 0.00 Hz TO 20000.00 Hz IS 0.0000004dB
Reply by Jerry Avins February 1, 20092009-02-01
makolber@yahoo.com wrote:

   ...

> OK, but using some number of coeff, i can create a perhaps complicated > response that after warping resembles a single pole -6 dB per octave > over a reasonable range of frequencies, no?
The more complex you make the filter, the closer you can bring your approximate straight-line response to fs/2. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by February 1, 20092009-02-01
On Jan 31, 9:21&#2013266080;pm, Jerry Avins <j...@ieee.org> wrote:
> makol...@yahoo.com wrote: > > On Jan 30, 6:13 pm, Jerry Avins <j...@ieee.org> wrote: > >> makol...@yahoo.com wrote: > >>> x-no-archive: > >>>> Ok, I see what you mean. > >>>> But still, you can't avoid the warping and get a straight line > >>>> by raising the sampling rate. > >>>> It's just that you get a bigger portion of the response to be > >>>> less curved. &#2013266080;At the end of the response approaching pi you will > >>>> again get strong deviation from the analog response. > >>>> Of course it's wrong what I said in my message before; > >>>> the responses do not look the same at different Fs. &#2013266080; > >>>> The shape of a filter with cutoff at 0.5 pi will > >>>> allways look the same, but 0.5 pi isn't the same cutoff freq > >>>> if you change the sampling rate (I am stating the obvious, but > >>>> that made me write my (wrong) statement). > >>> I thought there was a technique to adjust the coeffefficents to pre- > >>> warp the response so that the response remains closer to "ideal" over > >>> a wider frequency range... &#2013266080; But of course, as you say, when you > >>> actually get to Fs/2, it's all over. > >> Prewarping can move a single critical (read important) frequency to > >> where it is needed, but it cannot straighten a curved response. > > >> Jerry > > > OK thank you I didn't know that.. > > > But why would that be so? &#2013266080;If you can choose coefficients to create > > any arbitray shape (within reason) then why could you not choose a set > > of coef that create a response that is curved in some crazy way such > > that AFTER warping was a nice -6 dB per octave shape over a limited > > range anyway... I > > > I'm not saying you are wrong, I'm just trying to understand why.. > > You can create nearly any arbitrary shape, but not with just a single pole. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;- Hide quoted text - > > - Show quoted text -
OK, but using some number of coeff, i can create a perhaps complicated response that after warping resembles a single pole -6 dB per octave over a reasonable range of frequencies, no? Mark
Reply by Jerry Avins January 31, 20092009-01-31
makolber@yahoo.com wrote:
> On Jan 30, 6:13 pm, Jerry Avins <j...@ieee.org> wrote: >> makol...@yahoo.com wrote: >>> x-no-archive: >>>> Ok, I see what you mean. >>>> But still, you can't avoid the warping and get a straight line >>>> by raising the sampling rate. >>>> It's just that you get a bigger portion of the response to be >>>> less curved. At the end of the response approaching pi you will >>>> again get strong deviation from the analog response. >>>> Of course it's wrong what I said in my message before; >>>> the responses do not look the same at different Fs. >>>> The shape of a filter with cutoff at 0.5 pi will >>>> allways look the same, but 0.5 pi isn't the same cutoff freq >>>> if you change the sampling rate (I am stating the obvious, but >>>> that made me write my (wrong) statement). >>> I thought there was a technique to adjust the coeffefficents to pre- >>> warp the response so that the response remains closer to "ideal" over >>> a wider frequency range... But of course, as you say, when you >>> actually get to Fs/2, it's all over. >> Prewarping can move a single critical (read important) frequency to >> where it is needed, but it cannot straighten a curved response. >> >> Jerry >> > > OK thank you I didn't know that.. > > But why would that be so? If you can choose coefficients to create > any arbitray shape (within reason) then why could you not choose a set > of coef that create a response that is curved in some crazy way such > that AFTER warping was a nice -6 dB per octave shape over a limited > range anyway... I > > I'm not saying you are wrong, I'm just trying to understand why..
You can create nearly any arbitrary shape, but not with just a single pole. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by January 31, 20092009-01-31
On Jan 30, 6:13&#2013266080;pm, Jerry Avins <j...@ieee.org> wrote:
> makol...@yahoo.com wrote: > > x-no-archive: > >> Ok, I see what you mean. > >> But still, you can't avoid the warping and get a straight line > >> by raising the sampling rate. > >> It's just that you get a bigger portion of the response to be > >> less curved. &#2013266080;At the end of the response approaching pi you will > >> again get strong deviation from the analog response. > > >> Of course it's wrong what I said in my message before; > >> the responses do not look the same at different Fs. &#2013266080; > >> The shape of a filter with cutoff at 0.5 pi will > >> allways look the same, but 0.5 pi isn't the same cutoff freq > >> if you change the sampling rate (I am stating the obvious, but > >> that made me write my (wrong) statement). > > > I thought there was a technique to adjust the coeffefficents to pre- > > warp the response so that the response remains closer to "ideal" over > > a wider frequency range... &#2013266080; But of course, as you say, when you > > actually get to Fs/2, it's all over. > > Prewarping can move a single critical (read important) frequency to > where it is needed, but it cannot straighten a curved response. > > Jerry >
OK thank you I didn't know that.. But why would that be so? If you can choose coefficients to create any arbitray shape (within reason) then why could you not choose a set of coef that create a response that is curved in some crazy way such that AFTER warping was a nice -6 dB per octave shape over a limited range anyway... I I'm not saying you are wrong, I'm just trying to understand why.. Mark
Reply by Jerry Avins January 30, 20092009-01-30
makolber@yahoo.com wrote:
> x-no-archive: >> Ok, I see what you mean. >> But still, you can't avoid the warping and get a straight line >> by raising the sampling rate. >> It's just that you get a bigger portion of the response to be >> less curved. At the end of the response approaching pi you will >> again get strong deviation from the analog response. >> >> Of course it's wrong what I said in my message before; >> the responses do not look the same at different Fs. >> The shape of a filter with cutoff at 0.5 pi will >> allways look the same, but 0.5 pi isn't the same cutoff freq >> if you change the sampling rate (I am stating the obvious, but >> that made me write my (wrong) statement). >> >> > > I thought there was a technique to adjust the coeffefficents to pre- > warp the response so that the response remains closer to "ideal" over > a wider frequency range... But of course, as you say, when you > actually get to Fs/2, it's all over.
Prewarping can move a single critical (read important) frequency to where it is needed, but it cannot straighten a curved response. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by January 30, 20092009-01-30
x-no-archive:
> > Ok, I see what you mean. > But still, you can't avoid the warping and get a straight line > by raising the sampling rate. > It's just that you get a bigger portion of the response to be > less curved. &#2013266080;At the end of the response approaching pi you will > again get strong deviation from the analog response. > > Of course it's wrong what I said in my message before; > the responses do not look the same at different Fs. &#2013266080; > The shape of a filter with cutoff at 0.5 pi will > allways look the same, but 0.5 pi isn't the same cutoff freq > if you change the sampling rate (I am stating the obvious, but > that made me write my (wrong) statement). > >
I thought there was a technique to adjust the coeffefficents to pre- warp the response so that the response remains closer to "ideal" over a wider frequency range... But of course, as you say, when you actually get to Fs/2, it's all over. Mark
Reply by HardySpicer January 30, 20092009-01-30
On Jan 31, 6:31&#2013266080;am, "banton" <bant...@web.de> wrote:
> >banton wrote: > >>>> The digital frequency response is warped. > >>>> It doesn't look like a straight line going down > >>>> with 6db per octave (which is what you see > >>>> if you plot an analog response on a logarithmic scale - after > >>>> the rolloff at the beginning), > >>>> but it's a curve. > >>>> Remember that the distance from the pole is what determines > >>>> the amplitude. =A0And you go from freq 0 to pi in a halfcircle in > >>>> the z-plane. =A0 > > >>>> gr > >>>> Bjoern > > >>>>> Thanks a lot. > >>> It isn't if you sample high enough though. > > >>> Hardy > > >> What? &#2013266080;The frequency response shape of a digital 1 pole filter looks > >> the same, regardless of the sampling frequency. > > >Not so. Warping depends on f/fs. > > >Jerry > >-- > >Engineering is the art of making what you want from things you can get. > > > > Ok, I see what you mean. > But still, you can't avoid the warping and get a straight line > by raising the sampling rate. > It's just that you get a bigger portion of the response to be > less curved. &#2013266080;At the end of the response approaching pi you will > again get strong deviation from the analog response. > > Of course it's wrong what I said in my message before; > the responses do not look the same at different Fs. &#2013266080; > The shape of a filter with cutoff at 0.5 pi will > allways look the same, but 0.5 pi isn't the same cutoff freq > if you change the sampling rate (I am stating the obvious, but > that made me write my (wrong) statement). > > thanks, > Bjoern
That's right but who cares? You always have a finite bandwidth you operate over. Hardy
Reply by HardySpicer January 30, 20092009-01-30
On Jan 30, 10:02&#2013266080;pm, "banton" <bant...@web.de> wrote:
> >> The digital frequency response is warped. > >> It doesn't look like a straight line going down > >> with 6db per octave (which is what you see > >> if you plot an analog response on a logarithmic scale - after > >> the rolloff at the beginning), > >> but it's a curve. > >> Remember that the distance from the pole is what determines > >> the amplitude. =A0And you go from freq 0 to pi in a halfcircle in > >> the z-plane. =A0 > > >> gr > >> Bjoern > > >> >Thanks a lot. > > >It isn't if you sample high enough though. > > >Hardy > > What? &#2013266080;The frequency response shape of a digital 1 pole filter looks > the same, regardless of the sampling frequency. > > gr. > Bjoern
The higher you sample relative to the cut-off of your filter, the nearer it is to analogue. So it will have the right slope. Hardy
Reply by Jerry Avins January 30, 20092009-01-30
banton wrote:

   ...

> Ok, I see what you mean. > But still, you can't avoid the warping and get a straight line > by raising the sampling rate. > It's just that you get a bigger portion of the response to be > less curved. At the end of the response approaching pi you will > again get strong deviation from the analog response. > > Of course it's wrong what I said in my message before; > the responses do not look the same at different Fs. > The shape of a filter with cutoff at 0.5 pi will > allways look the same, but 0.5 pi isn't the same cutoff freq > if you change the sampling rate (I am stating the obvious, but > that made me write my (wrong) statement).
You see it now. I'm glad I helped you to think it out. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;