General implementation of Butterworth filter questions

Started by Randall August 29, 2010
Having used up 5 days already, cutting and pasting the internet source code
for Butterworth filters, only to have the audio spectrum analyzer show a
terrible mess, I have to drop back to first principles in order to get
something to work correctly.

1. I have the general Butterworth normalized polynomials for order-n (or
can generate them)  See http://en.wikipedia.org/wiki/Butterworth_filter

2. I am told to substitute for s, the value c*(z-1)/(z+1) for the bilateral
transform where c = cotangent(wc*T/2) where T is the sample time.  See
http://www.apicsllc.com/apics/Sr_3/Sr_3.htm.

This will give a rational polynomial equation in z, with the coefficients
for the difference equation.

But I am having problems getting things to work out correctly.
For wc = 1/2, T = 1 sec, and order 3, I get

B_n(3) = 1 / (s^3 + 2*s^2 + 2*s + 1)
c = cot(1/4) = 3.9131736..
s = 3.9131736*(z-1)/(z+1) and thence

       99.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423
H(z)   ---------------------------------------------------
                   z^3 + 3*z^2 + 3*z + 1

Which is no where close to octave's answer for the command
[b,a]=butter(3,1/2)
b = 1/6 1/2 1/2 1/6
a = 1 0 1/3 0

(we notice that sum(b) = sum(a) = 4/3

What steps am I missing to take H(z) above to match octave's answer? Or am
I misunderstanding octave's answer and misapplying it?  I am trying to
automate the answers to H(z) for any order n and cut off frequency wc.





Randall wrote:
> Having used up 5 days already, cutting and pasting the internet source code > for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly.
Nobody gives up good stuff for free. Butterworth filter design is simple: either do it yourself or pay money for someone doing it for you. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
On Aug 30, 3:07=A0am, "Randall" 
wrote:
> Having used up 5 days already, cutting and pasting the internet source co=
de
> for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly. > > 1. I have the general Butterworth normalized polynomials for order-n (or > can generate them) =A0Seehttp://en.wikipedia.org/wiki/Butterworth_filter > > 2. I am told to
Those aren't even halfway close to first principles: You have only dug up stuff you don't understand and accepted orders to do stuff beyond your abilities. 'First principles' would be to 1) Find a book on the subject 2) Read it 3) Contemplate it 4) Test the recipes and algorithms 5) Iterate items 1)-4) as many times as necessary The question you ask is like riding a bicycle: Requires some effort and pain to learn, but ridiculously easy once you know how. Rune
Randall  wrote:

>c = cot(1/4) = 3.9131736..
I must admire you for using a slide rule in this day and age. cot(1/4) = 3.916317364645940... Steve
>Having used up 5 days already, cutting and pasting the internet source
code
>for Butterworth filters, only to have the audio spectrum analyzer show a >terrible mess, I have to drop back to first principles in order to get >something to work correctly. > >1. I have the general Butterworth normalized polynomials for order-n (or >can generate them) See http://en.wikipedia.org/wiki/Butterworth_filter > >2. I am told to substitute for s, the value c*(z-1)/(z+1) for the
bilateral
>transform where c = cotangent(wc*T/2) where T is the sample time. See >http://www.apicsllc.com/apics/Sr_3/Sr_3.htm. > >This will give a rational polynomial equation in z, with the coefficients >for the difference equation. > >But I am having problems getting things to work out correctly. >For wc = 1/2, T = 1 sec, and order 3, I get > >B_n(3) = 1 / (s^3 + 2*s^2 + 2*s + 1) >c = cot(1/4) = 3.9131736.. >s = 3.9131736*(z-1)/(z+1) and thence > > 99.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423 >H(z) --------------------------------------------------- > z^3 + 3*z^2 + 3*z + 1 > >Which is no where close to octave's answer for the command >[b,a]=butter(3,1/2) >b = 1/6 1/2 1/2 1/6 >a = 1 0 1/3 0 > >(we notice that sum(b) = sum(a) = 4/3 > >What steps am I missing to take H(z) above to match octave's answer? Or
am
>I misunderstanding octave's answer and misapplying it? I am trying to >automate the answers to H(z) for any order n and cut off frequency wc.
The answer to your problem is easy as pi. :-) Steve
On Aug 29, 9:07=A0pm, "Randall" 
wrote:
> Having used up 5 days already, cutting and pasting the internet source co=
de
> for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly. > > 1. I have the general Butterworth normalized polynomials for order-n (or > can generate them) =A0Seehttp://en.wikipedia.org/wiki/Butterworth_filter > > 2. I am told to substitute for s, the value c*(z-1)/(z+1) for the bilater=
al
> transform where c =3D cotangent(wc*T/2) where T is the sample time. =A0Se=
ehttp://www.apicsllc.com/apics/Sr_3/Sr_3.htm.
> > This will give a rational polynomial equation in z, with the coefficients > for the difference equation. > > But I am having problems getting things to work out correctly. > For wc =3D 1/2, T =3D 1 sec, and order 3, I get > > B_n(3) =3D 1 / (s^3 + 2*s^2 + 2*s + 1) > c =3D cot(1/4) =3D 3.9131736.. > s =3D 3.9131736*(z-1)/(z+1) and thence > > =A0 =A0 =A0 =A099.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423 > H(z) =A0 --------------------------------------------------- > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0z^3 + 3*z^2 + 3*z + 1 > > Which is no where close to octave's answer for the command > [b,a]=3Dbutter(3,1/2) > b =3D 1/6 1/2 1/2 1/6 > a =3D 1 0 1/3 0 > > (we notice that sum(b) =3D sum(a) =3D 4/3 > > What steps am I missing to take H(z) above to match octave's answer? Or a=
m
> I misunderstanding octave's answer and misapplying it? =A0I am trying to > automate the answers to H(z) for any order n and cut off frequency wc.
See if this helps: http://www.claysturner.com/dsp/Butterworth%20Filter%20Formulae.pdf Clay
On 8/29/2010 6:07 PM, Randall wrote:


Hi;

I wrote these notes for a class I took on filters this spring. If you 
find any errors, pls let me know:

butterworth based IIR digital filter design:

http://12000.org/my_notes/IIR_digital_filter_design/index.htm

I have list of butter lowpass normalized poly's here for up to N=20

http://12000.org/my_notes/list_of_normalized_low_pass_butterworth/index.htm

For butter analog filter design

http://12000.org/my_notes/butterworth_analog_filter_design/index.htm

hth

--Nasser



On 8/29/2010 6:07 PM, Randall wrote:


Hi;

I wrote these notes for a class I took on filters this spring. If you 
find any errors, pls let me know:

butterworth based IIR digital filter design:

http://12000.org/my_notes/IIR_digital_filter_design/index.htm

I have list of butter lowpass normalized poly's here for up to N=20

http://12000.org/my_notes/list_of_normalized_low_pass_butterworth/index.htm

For butter analog filter design

http://12000.org/my_notes/butterworth_analog_filter_design/index.htm

hth

--Nasser



On Aug 29, 9:07=A0pm, "Randall" 
wrote:
> Having used up 5 days already, cutting and pasting the internet source co=
de
> for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly. > > 1. I have the general Butterworth normalized polynomials for order-n (or > can generate them) =A0Seehttp://en.wikipedia.org/wiki/Butterworth_filter > > 2. I am told to substitute for s, the value c*(z-1)/(z+1) for the bilater=
al
> transform where c =3D cotangent(wc*T/2) where T is the sample time. =A0Se=
ehttp://www.apicsllc.com/apics/Sr_3/Sr_3.htm.
> > This will give a rational polynomial equation in z, with the coefficients > for the difference equation. > > But I am having problems getting things to work out correctly. > For wc =3D 1/2, T =3D 1 sec, and order 3, I get > > B_n(3) =3D 1 / (s^3 + 2*s^2 + 2*s + 1) > c =3D cot(1/4) =3D 3.9131736.. > s =3D 3.9131736*(z-1)/(z+1) and thence > > =A0 =A0 =A0 =A099.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423 > H(z) =A0 --------------------------------------------------- > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0z^3 + 3*z^2 + 3*z + 1 > > Which is no where close to octave's answer for the command > [b,a]=3Dbutter(3,1/2) > b =3D 1/6 1/2 1/2 1/6 > a =3D 1 0 1/3 0 > > (we notice that sum(b) =3D sum(a) =3D 4/3 > > What steps am I missing to take H(z) above to match octave's answer? Or a=
m
> I misunderstanding octave's answer and misapplying it? =A0I am trying to > automate the answers to H(z) for any order n and cut off frequency wc.
See if this helps: http://www.claysturner.com/dsp/Butterworth%20Filter%20Formulae.pdf Clay
>Having used up 5 days already, cutting and pasting the internet source
code
>for Butterworth filters, only to have the audio spectrum analyzer show a >terrible mess, I have to drop back to first principles in order to get >something to work correctly. > >1. I have the general Butterworth normalized polynomials for order-n (or >can generate them) See http://en.wikipedia.org/wiki/Butterworth_filter > >2. I am told to substitute for s, the value c*(z-1)/(z+1) for the
bilateral
>transform where c = cotangent(wc*T/2) where T is the sample time. See >http://www.apicsllc.com/apics/Sr_3/Sr_3.htm. > >This will give a rational polynomial equation in z, with the coefficients >for the difference equation. > >But I am having problems getting things to work out correctly. >For wc = 1/2, T = 1 sec, and order 3, I get > >B_n(3) = 1 / (s^3 + 2*s^2 + 2*s + 1) >c = cot(1/4) = 3.9131736.. >s = 3.9131736*(z-1)/(z+1) and thence > > 99.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423 >H(z) --------------------------------------------------- > z^3 + 3*z^2 + 3*z + 1 > >Which is no where close to octave's answer for the command >[b,a]=butter(3,1/2) >b = 1/6 1/2 1/2 1/6 >a = 1 0 1/3 0 > >(we notice that sum(b) = sum(a) = 4/3 > >What steps am I missing to take H(z) above to match octave's answer? Or
am
>I misunderstanding octave's answer and misapplying it? I am trying to >automate the answers to H(z) for any order n and cut off frequency wc.
The answer to your problem is easy as pi. :-) Steve
Randall  wrote:

>c = cot(1/4) = 3.9131736..
I must admire you for using a slide rule in this day and age. cot(1/4) = 3.916317364645940... Steve
On Aug 30, 3:07=A0am, "Randall" 
wrote:
> Having used up 5 days already, cutting and pasting the internet source co=
de
> for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly. > > 1. I have the general Butterworth normalized polynomials for order-n (or > can generate them) =A0Seehttp://en.wikipedia.org/wiki/Butterworth_filter > > 2. I am told to
Those aren't even halfway close to first principles: You have only dug up stuff you don't understand and accepted orders to do stuff beyond your abilities. 'First principles' would be to 1) Find a book on the subject 2) Read it 3) Contemplate it 4) Test the recipes and algorithms 5) Iterate items 1)-4) as many times as necessary The question you ask is like riding a bicycle: Requires some effort and pain to learn, but ridiculously easy once you know how. Rune

Randall wrote:
> Having used up 5 days already, cutting and pasting the internet source code > for Butterworth filters, only to have the audio spectrum analyzer show a > terrible mess, I have to drop back to first principles in order to get > something to work correctly.
Nobody gives up good stuff for free. Butterworth filter design is simple: either do it yourself or pay money for someone doing it for you. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Having used up 5 days already, cutting and pasting the internet source code
for Butterworth filters, only to have the audio spectrum analyzer show a
terrible mess, I have to drop back to first principles in order to get
something to work correctly.

1. I have the general Butterworth normalized polynomials for order-n (or
can generate them)  See http://en.wikipedia.org/wiki/Butterworth_filter

2. I am told to substitute for s, the value c*(z-1)/(z+1) for the bilateral
transform where c = cotangent(wc*T/2) where T is the sample time.  See
http://www.apicsllc.com/apics/Sr_3/Sr_3.htm.

This will give a rational polynomial equation in z, with the coefficients
for the difference equation.

But I am having problems getting things to work out correctly.
For wc = 1/2, T = 1 sec, and order 3, I get

B_n(3) = 1 / (s^3 + 2*s^2 + 2*s + 1)
c = cot(1/4) = 3.9131736..
s = 3.9131736*(z-1)/(z+1) and thence

       99.5744*z^3 - 200.04249*z^2 + 144.6923*z - 36.22423
H(z)   ---------------------------------------------------
                   z^3 + 3*z^2 + 3*z + 1

Which is no where close to octave's answer for the command
[b,a]=butter(3,1/2)
b = 1/6 1/2 1/2 1/6
a = 1 0 1/3 0

(we notice that sum(b) = sum(a) = 4/3

What steps am I missing to take H(z) above to match octave's answer? Or am
I misunderstanding octave's answer and misapplying it?  I am trying to
automate the answers to H(z) for any order n and cut off frequency wc.