help needed for Butterworth crossover response

Hi all,

I implemented a butterworth crossover with the matlab (code given below).

The code is working perfectly for a 4th order butterworth crossover(i.e.
24 dB/Octave). Th  added response for the lowpass and high pass filters of
the butterworth crossover gives 3 dB increase at the corner frequency which
is expected.

But the same is not obtained when i use 2nd and 3rd order butterworth
filters. Instead of having a 3 dB increase at the cutoff i am getting all
pass response for 3rd order and a dip at cutoff for 2nd order butterworth
filter.

Can anybody help me with this? your help is very much appreciated.

See the code below for your reference.

%Butterworth crossover

filterorder=4;
cutoff=1000;
fs=44100;

    [b,a] = butter(filterorder,cutoff/fs,'low');
    h1=freqz(b,a,44100);
    
    semilogx(10*log(abs(h1)),'r');
    hold
    [b,a] = butter(filterorder,cutoff/fs,'high');
    h2=freqz(b,a,44100);
    
    semilogx(10*log(abs(h2)),'b');
    h=h1+h2;
    semilogx(10*log(abs(h)),'g');
    grid  



Regards
Gangadhar  

gangadhar.m wrote:
> Hi all,
> 
> I implemented a butterworth crossover with the matlab (code given below).
> 
> The code is working perfectly for a 4th order butterworth crossover(i.e.
> 24 dB/Octave). Th  added response for the lowpass and high pass filters of
> the butterworth crossover gives 3 dB increase at the corner frequency which
> is expected.
> 
> But the same is not obtained when i use 2nd and 3rd order butterworth
> filters. Instead of having a 3 dB increase at the cutoff i am getting all
> pass response for 3rd order and a dip at cutoff for 2nd order butterworth
> filter.

The phase shift at the cutoff frequency depends on the order of the 
filters. You should combine the filter outputs so they add in phase.

> Can anybody help me with this?   your help is very much appreciated.

I like this phrase. Anybody can certainly help you with this depending 
on how much is the real appreciation.


Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

Hi,

As you said, the phase shift is playing a major role. Can you please tell
me how should I proceed now to achieve 3 dB increase at the cut off
frequency for all butterworth crossovers(2nd,3rd and 4th orders)
irrsepective of change in the order. Actually same kind of behaviour is
observed in implememntation of LR Crossovers also (i.e i got correct flat
all pass response for LR 4 Crossover and a big dip at cutoff in case of LR
2 Crossover).

Is there any way for the phase adjustment while combining the filter
responses?

I really appreciate your help in this.

Thanks and regards
Gangadhar


>
>
>gangadhar.m wrote:
>> Hi all,
>> 
>> I implemented a butterworth crossover with the matlab (code given
below).
>> 
>> The code is working perfectly for a 4th order butterworth
crossover(i.e.
>> 24 dB/Octave). Th  added response for the lowpass and high pass filters
of
>> the butterworth crossover gives 3 dB increase at the corner frequency
which
>> is expected.
>> 
>> But the same is not obtained when i use 2nd and 3rd order butterworth
>> filters. Instead of having a 3 dB increase at the cutoff i am getting
all
>> pass response for 3rd order and a dip at cutoff for 2nd order
butterworth
>> filter.
>
>The phase shift at the cutoff frequency depends on the order of the 
>filters. You should combine the filter outputs so they add in phase.
>
>> Can anybody help me with this?   your help is very much appreciated.
>
>I like this phrase. Anybody can certainly help you with this depending 
>on how much is the real appreciation.
>
>
>Vladimir Vassilevsky
>
>DSP and Mixed Signal Design Consultant
>
>http://www.abvolt.com
>

gangadhar.m wrote:
> Hi,
> 
>    ... Can you please tell
> me how should I proceed now to achieve 3 dB increase at the cut off
> frequency for all butterworth crossovers(2nd,3rd and 4th orders)
> irrsepective of change in the order.   ...

Butterworth is Butterworth. If you change it in any way, it's not 
Butterworth any more. Try Linkwitz-Reilly.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Dear Jerry,

As i mentioned earlier i get the same kind of response in case of Linkwitz
Riley filters also..... In net, i find papers where they have flat all pass
response for Linkwitz Riley crossovers(LR-2, LR-3  and LR-4 filters)

but for me, all pass response is observed only for the LR-4 crossover and
big dig at cutoff for the other LR Crossovers. Similar cases are observed
for Butterworth crossovers also.

Since we are having butterworth crossovers available, i assume that all
butterworth crossovers shoud be able to exhibit 3dB increase at the cutoff
irrsepective of their order. How can I achieve that?

Please help me if my understanding is not correct.

Regards
Gangadhar
>gangadhar.m wrote:
>> Hi,
>> 
>>    ... Can you please tell
>> me how should I proceed now to achieve 3 dB increase at the cut off
>> frequency for all butterworth crossovers(2nd,3rd and 4th orders)
>> irrsepective of change in the order.   ...
>
>Butterworth is Butterworth. If you change it in any way, it's not 
>Butterworth any more. Try Linkwitz-Reilly.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
>

I got the answer. I have to use Polarity inversion also as specified in the
document availabe at the link
www.lake.com.au/documentation/Lake%20Contour%20Classical%20Crossover%20Notes.pdf


The document says that

"When using 24 dB and 48 dB L-R crossovers, there is no need to reverse
the polarity of adjacent crossover channels. However, the 12 dB and 36 dB
L-R crossovers will require polarity reversals.  In order to provide the
desired constant magnitude summation, the polarity of an output channel
must be inverted."


Thank you all
Gangadhar
>
>Dear Jerry,
>
>As i mentioned earlier i get the same kind of response in case of
Linkwitz
>Riley filters also..... In net, i find papers where they have flat all
pass
>response for Linkwitz Riley crossovers(LR-2, LR-3  and LR-4 filters)
>
>but for me, all pass response is observed only for the LR-4 crossover
and
>big dig at cutoff for the other LR Crossovers. Similar cases are
observed
>for Butterworth crossovers also.
>
>Since we are having butterworth crossovers available, i assume that all
>butterworth crossovers shoud be able to exhibit 3dB increase at the
cutoff
>irrsepective of their order. How can I achieve that?
>
>Please help me if my understanding is not correct.
>
>Regards
>Gangadhar
>>gangadhar.m wrote:
>>> Hi,
>>> 
>>>    ... Can you please tell
>>> me how should I proceed now to achieve 3 dB increase at the cut off
>>> frequency for all butterworth crossovers(2nd,3rd and 4th orders)
>>> irrsepective of change in the order.   ...
>>
>>Butterworth is Butterworth. If you change it in any way, it's not 
>>Butterworth any more. Try Linkwitz-Reilly.
>>
>>Jerry
>>-- 
>>Engineering is the art of making what you want from things you can get.
>>¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
>>
>

gangadhar.m wrote:
> Dear Jerry,
> 
> As i mentioned earlier i get the same kind of response in case of Linkwitz
> Riley filters also..... In net, i find papers where they have flat all pass
> response for Linkwitz Riley crossovers(LR-2, LR-3  and LR-4 filters)
> 
> but for me, all pass response is observed only for the LR-4 crossover and
> big dig at cutoff for the other LR Crossovers. Similar cases are observed
> for Butterworth crossovers also.
> 
> Since we are having butterworth crossovers available, i assume that all
> butterworth crossovers shoud be able to exhibit 3dB increase at the cutoff
> irrsepective of their order. How can I achieve that?
> 
> Please help me if my understanding is not correct.

A Butterworth low-pass filter is one whose frequency-response 
derivatives are all zero at w = 0. Band-pass and high-pass Butterworth 
filters can be derived from it by the usual transformations or derived 
directly by setting all derivatives to zero at w = infinity (high-pass) 
or at the center frequency (band-pass). A consequence is that the 
transfer function of a Butterworth low-pass prototype of order n is

                      H(s) = 1/sqrt(1 + (w/w_c)^2n)

That's it, unalterably. If that response isn't suitable, then 
Butterworth isn't suitable. This response can only be approximated by a 
digital filter, and the approximation is usually crude by the standards 
of those who listen with their calculators instead of with their ears.

There is another degree of freedom with crossovers. It is not written in 
stone that the corner frequencies of the high-pass and low-pass sections 
must be equal.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins wrote:


> A Butterworth low-pass filter is one whose frequency-response 
> derivatives are all zero at w = 0. Band-pass and high-pass Butterworth 
> filters can be derived from it by the usual transformations or derived 
> directly by setting all derivatives to zero at w = infinity (high-pass) 
> or at the center frequency (band-pass). A consequence is that the 
> transfer function of a Butterworth low-pass prototype of order n is
> 
>                      H(s) = 1/sqrt(1 + (w/w_c)^2n)
> 
> That's it, unalterably. If that response isn't suitable, then 
> Butterworth isn't suitable. This response can only be approximated by a 
> digital filter, and the approximation is usually crude by the standards 
> of those who listen with their calculators instead of with their ears.

Actually, if a filter is specified by the frequency response, going 
digital improves it. Warping provides for the better roloff then in the 
analog case.


> There is another degree of freedom with crossovers.

There are many degrees of freedom if you have nothing more important to do.


Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

Vladimir Vassilevsky wrote:

    ...

> Actually, if a filter is specified by the frequency response, going 
> digital improves it. Warping provides for the better roloff then in the 
> analog case.

Warping sharpens the band edge. Whether that's an improvement depends on 
circumstance. A filter with sharper-than-Butterworth transition isn't a 
Butterworth filter.

   ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

In article <nNydnXeVrvc0IwDbnZ2dnUVZ_gCdnZ2d@giganews.com>, 
gangadhar.m@jasmin-infotech.com says...
>
>
>I got the answer. I have to use Polarity inversion also as specified in the
>document availabe at the link
>www.lake.com.au/documentation/Lake%20Contour%20Classical%20Crossover%
20Notes.pdf
>
>
>The document says that
>
>"When using 24 dB and 48 dB L-R crossovers, there is no need to reverse
>the polarity of adjacent crossover channels. However, the 12 dB and 36 dB
>L-R crossovers will require polarity reversals.  In order to provide the
>desired constant magnitude summation, the polarity of an output channel
>must be inverted."

To add a bit of info: Odd-order Butterworth highpass and lowpass pairs all 
have the allpass summation property regardless of the polarity of the 
summation. However, the order of the resulting allpass is different depending 
on polarity. If you want to have the summation be an allpass with the minimum 
number of poles, you need to invert polarity for 3rd-order (18 dB/oct), 7th 
order, 11th order... For example, the allpass sum of a 3rd-order Butterworth 
has two poles if the summation is in uninverted polarity and one pole if the 
sum has one input with inverted polarity (view with monospaced font):

                                   3              2         
          1                       s              s  - s + 1 
_____________________ + _____________________ = ____________
  3      2                3      2                2         
 s  + 2+s  + 2+s + 1     s  + 2+s  + 2+s + 1     s  + s + 1 
                                   3               
    
          1                       s              1 - s 
_____________________ - _____________________ = _______
  3      2                3      2               s + 1 
 s  + 2+s  + 2+s + 1     s  + 2+s  + 2+s + 1


Polarity for 5th order, 9th order... should not be inverted,

Finally, the allpass summation property is preserved if the Butterworth filter 
is transformed into an IIR digital filter via the binlear transform.