4th order Butterworth filter Q

[AndrewT]

I'll rephrase that to hopefully help with understanding.

A 1st order Butterworth is simply an RC filter but two conditions must be met. The source impedance must be zero and the load impedance must be infinite. The Q is 1/sqrt(2) when the loadings conditions are met and the R&C are perfect devices.

A 2nd order Butterworth is an RLC (passive) filter or an active two pole filter. Again the source and loading conditions must be met. The Q=1/sqrt(2)

A 3rd order Butterworth is a 1st order + a 2nd order. The final Q is still 1/sqrt(2)
Since the RC 1st order has a Q=1/sqrt(2), then the 2nd order must have a Q=1.
Again all the loading conditions of the two filters must be met.

If you read the filter parameters of the 1st order Butterworth and compare them to the first line (stage 1) of the third order filter you will see they are identical and similarly stage 1 of the 5th order is identical. & the 7th & the 9th etc.

 

[speaker dave]

Hi,


I don't have the DCX2496 but I'm using a computer based DSP. I can cascade parametric filters but I don't think I'm getting the correct slopes. I was using the Q-factors mentioned earlier.

1st Butterworth = 0.5Q
2nd Butterworth = 0.71Q
3rd Butterworth = 0.5Q + 1Q

In your setup you are using a 2nd order with Q of .5 for both your 1st and 3rd order filters. This needs to be a first order instead.

Look at the slopes. You want to see 6 then 12 then 18 dB per Octave. Instead it looks like 12, 12 and 24.

Search through the DSP software or manual for "real pole" or "1st order". As mentioned by Andrew a single real pole has no Q. A section with Q of .5 is really two poles (second order) so your slopes are off. This also explains why the first two curves are similar: they are both 2nd order with slightly different Q, so only different at the rolloff corner.

Does anybody know why the phase rolls up??

David S.

 

[AndrewT]

An RC filter inherently has a Butterworth characteristic if properly loaded. i.e. Q=1/sqrt(2)

If one adds (cascades) two filters then the Q of each are multiplied to give the effective Q of the cascaded pair.

If two Butterworth 2pole filters are cascaded then the two Qs are 1/sqrt(2).
and the resulting Q is 1 / [sqrt(2)*sqrt(2)] = 1 / [2] = 0.5 This is still a 4pole filter, but it is no longer Butterworth.

Linkwitz Reilly uses precisely this cascade for their LR4 crossover.

Look at the Qi in the table.
Assume a value of 0.71 for the dashes (-) and multiply the Qi for each filter group. All the multiplicands are 1/sqrt(2) ~0.71, i.e. Butterworth.

 

[AndrewT]

Does anybody know why the phase rolls up?

the phase difference of a single pole filter at the -3dB roll off frequency is 45degrees.
Cascade two identical single pole filters (each properly terminated) and the total phase difference at F-3dB frequency is the sum of 45+45degrees i.e. 90degrees.

If the frequency is taken to the limit then the phase difference becomes 90degrees. Cascading two of these at the frequency limit results in 180degree phase difference, i.e. exactly out of phase.

The rule is add the phase differences for each filter at the frequency being investigated.
Take 3 different filters with turn over frequencies that could result in a 60degree & 45degree and 1.5degree phase differences at the specific frequency being considered. The total phase difference at the output of the 3 different cascaded filters is 60+45+1.5=106.5degrees.

 

[speaker dave]

Andrew, I know you are very knowledgable on this subject. Why does the phase curve, of a low pass filter, roll up?

David S.

 

[boris81]

Thank you, JohnPM, Speaker Dave and AndrewT!

My understanding of filter math is very limited. What I'm getting from this discussion is that odd orders Butterworth can not be described by the Q-factor alone.

I might be able to compile a biquad implementation but I don't know how to compute the coefficients a0,a1,a2, b0,b1,b2 for a Butterworth LP and HP.

 

[speaker dave]

In your setup you are using a 2nd order with Q of .5 for both your 1st and 3rd order filters. This needs to be a first order instead.

David S.

 

[AndrewT]

Why does the phase curve, of a low pass filter, roll up?

lead or lag in current phase relative to voltage phase.

 

[speaker dave]

lead or lag in current phase relative to voltage phase.

Phase of a low pass filter should roll down in the stop band. It should have negative phase. Its either something wierd from a digital implimentation or the plotting routine is wrong.

David S.

 

[boris81]

Phase of a low pass filter should roll down in the stop band. It should have negative phase. Its either something wierd from a digital implimentation or the plotting routine is wrong.
David S.

It might be a bug in HOLMImpulse. I will install ARTA when I get a chance and compare.
http://www.diyaudio.com/forums/multi-way/145662-holmimpulse-measurements-practice-2.html#post1854986


I think I found the formulas for 1st order LP and HP biquad filters.
From Cascade form IIR Filters | Comp.DSP | DSPRelated.com

H(z) = (b0 + b1*z^-1)/(a0 + a1*z^-1)

1st order LPF: H(s) = 1/(s+1)

b0 = sin(w0)
b1 = sin(w0)
a0 = cos(w0) + sin(w0) + 1
a1 = sin(w0) - cos(w0) - 1

1st order HPF: H(s) = s/(s+1)

b0 = cos(w0) + 1
b1 = -(cos(w0) + 1)
a0 = cos(w0) + sin(w0) + 1
a1 = sin(w0) - cos(w0) - 1

Now I figure out the hard part of how to implement it in the DSP and compile it.

 

[JohnPM]

Phase of a low pass filter should roll down in the stop band. It should have negative phase. Its either something wierd from a digital implementation or the plotting routine is wrong.

If that refers to the Holm plot it will be due to the zero time set incorrectly for the impulse response (peak is occurring earlier than it should on the time axis so the response has in effect a negative time delay and a correspondingly increasing phase lead as frequency rises).

 

[boris81]

i got it

I got it. Thank you all, I would've never figured it out on my own.

Fs = 300Hz
green = Butterworth 1st order
red = Butterworth 2nd order
blue = Butterworth 3rd order

 

[boris81]

To implement any order any common type filter follow the look-up guide at the end of this datasheet.
http://focus.ti.com/lit/ml/sloa088/sloa088.pdf

This was explained earlier but I will repeat it. For Butterworth:
1st order -> 1st order real pole filter (Q is not needed in the biquad implementation)
2nd order -> 2nd order with Q = 0.707
3rd order -> 1st + 2nd with Q = 1.
4th order -> 2nd with Q = 0.54 + 2nd with Q = 1.31
and so on

To compute the biquad coefficients follow the famous Audio-EQ-Cookbook for 2nd order and these formulas for 1st order
Cascade form IIR Filters | Comp.DSP | DSPRelated.com

Keep in mind that different biquad implementations scale the coefficients to speed up processing so you might have to divide by a0,2,-1 or some other scaling factor.

 

[AndrewT]

when combining mixed Q values, it is better to place the higher Q after the lower Q.
This avoids the peak in the frequency response needlessly using up headroom.

 

[neyurt]

Hey guys,

I have been reading this thread and got an idea how to implement Linkwitz-Riley Filters in APO Equalizer for active crossovers.

Is my cascading filters correct to produce these curves.I'm trying to create them using BW filters:

Can someone verify if this is correct? 1000hz is used in these examples.

#L-R2 12db/octave
Filter: ON HPQ Fc 1000 Hz Q 0.5


#LR-4 24db/octave
#Cascading 2 2nd-Order BW Filters
Filter: ON HPQ Fc 1000 Hz Q 0.7071
Filter: ON HPQ Fc 1000 Hz Q 0.7071



#LR-8 48db/octave
#Cascading 2 4th-order Butterworth
#4th-order BW is: 4th order -> 2nd with Q = 0.541 + 2nd with Q = 1.306
Filter: ON HPQ Fc 1000 Hz Q 0.541
Filter: ON HPQ Fc 1000 Hz Q 1.306
Filter: ON HPQ Fc 1000 Hz Q 0.541
Filter: ON HPQ Fc 1000 Hz Q 1.306


Q-values reference: http://csserver.evansville.edu/~ric...ynthesis/resources/handouts/ButterQValues.pdf