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.
�����������������������������������������������������������������������
Reply by ●February 1, 20092009-02-01
On Jan 31, 9:21�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. �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.
> �����������������������������������������������������������������������- 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.
�����������������������������������������������������������������������
Reply by ●January 31, 20092009-01-31
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..
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.
�����������������������������������������������������������������������
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. �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.
Mark
Reply by HardySpicer●January 30, 20092009-01-30
On Jan 31, 6:31�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? �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. �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).
>
> 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�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? �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.
�����������������������������������������������������������������������