```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

```
```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;
```
```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
```
```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;
```
```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

```
```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;
```
```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
```
```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
```
```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
```
```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;
```