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What does "6db per octave" mean in the digital world?

Started by Unknown January 29, 2009
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?

Thanks a lot.
rayaudiolabs@gmail.com wrote:
 
> 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
6dB is a factor of 10**0.6 or about 4 in power, which is a factor of 2 in voltage. For a low pass filter, that means as you double the frequency the voltage goes down by a factor of two. A single RL or RC filter well past the cutoff frequency, does that. Hopefully others will give you a better answer for the digital case, but otherwise it means a filter with similar properties to a single (pole) RL or RC filter. -- glen
rayaudiolabs@gmail.com wrote:
> 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....
?? Isn't that -3 dB/octave?
> But what I have is > -3@1000
The filter is down 3 dB at turnover.
> -7@2000
I would have expected closer to -6
> -12.x@4000
2 octaves @ -6 dB/octave = -12 dB. Where's the mystery?
> -24.x@24000
24000 is fs/2. The relation to an analog filter has broken down.
> Can anyone give my a hint what was wrong? Why it is -24dB at 24000Hz?
Can you draw Bode plots starting from asymptotic approximations? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
>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. gr Bjoern
>Thanks a lot. >
<rayaudiolabs@gmail.com> wrote:

>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,
For Fc << Fs/2 you should see a 6 dB per octave slope for the octaves immediately above Fc. But as you approach any significant fraction of Fc/2 the response will deviate from 6 dB/octave. This is because a sampled-time one-pole IIR does not, and cannot, exactly emulate its continuous-time counterpart. This is a case of "what you see is what you get". Steve
On Jan 30, 3:36&#2013266080;pm, "banton" <bant...@web.de> wrote:
> >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 &#2013266080;-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. &#2013266080;And you go from freq 0 to pi in a halfcircle in > the z-plane. &#2013266080; > > &#2013266080;gr > Bjoern > > >Thanks a lot.
It isn't if you sample high enough though. Hardy
>> 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
On 30 Jan, 02:40, rayaudiol...@gmail.com wrote:
> Hi, > > I had a hard time understanding the notion of "6dB per octave".
The term is historical and originates from the days when analog filters were designed from table look-ups and rule-of-thumb formulas, and built from components with 20% tolerance on the values. It means that you should see a straightis line with slope 6 dB per octave when the filter response is plotted in a log-log Bode plot.
> If I > have a single pole IIR low pass with the Fc = 1000 and Fs = 48000,
Here you no longer deal with an analog filter, but a discrete-time filter. You might have designed this filter from a '6dB per octave' analog prototype, but in this case the frequency response is warped so a term like '6dB per octave' needs not make any sense in discrete time domain. Rune
On Jan 30, 4:04&#2013266080;am, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 30 Jan, 02:40, rayaudiol...@gmail.com wrote: > > > Hi, > > > I had a hard time understanding the notion of "6dB per octave". > > The term is historical and originates from the days when analog > filters were designed from table look-ups and rule-of-thumb formulas, > and built from components with 20% tolerance on the values. > > It means that you should see a straightis line with slope 6 dB > per octave when the filter response is plotted in a log-log > Bode plot. > > > If I > > have a single pole IIR low pass with the Fc = 1000 and Fs = 48000, > > Here you no longer deal with an analog filter, but a discrete-time > filter. You might have designed this filter from a '6dB per octave' > analog prototype, but in this case the frequency response is warped > so a term like '6dB per octave' needs not make any sense in discrete > time domain. > > Rune
That's the key to the issue that the OP pointed out; transforming an analog IIR filter to a discrete-time one forces you to map the entire imaginary axis in the s-plane (i.e. the frequency response of the filter over all frequencies from -infinity to +infinity) to the unit circle in the z-plane (which has a finite circumference). The mapping between the two planes that is often used, the bilinear transform, is pretty accurate near zero frequency, but as you get closer to fs/2 the approximation by necessity breaks down, as you're sort of "cramming" the tails of the s-plane's imaginary axis into the portion of the circle near fs/2. Jason
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. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;