Problems with Butterworth f. when order>=10

Hello everybody.

Thank you always for your help.
I have a question regarding Butterworth filters with order >.

I was out of the office, so couldn't use MATLAB. Instead I used Octave. I tried to do the following:

>[b a]=butter(10,0.015);
>y=filter(b,a,x);

for a original signal x.(a sin signal)

Now when ploting I can see this gives me a filtered signal (although a little strange in some values, still good

But when I tried the following

>[b a]=butter(10,0.01);
>y=filter(b,a,x);

the result is not a filtered sinusoid at all, but a signal that goes to infinite!

Now I am at lost why I can not construct butterworth filters of order 10 with a normalized cutoff of less than 0.015. Is it a theoretical constraint? is it a bug in octave? (havent tried it in Matlab)

can anyone share some insight in this?

Kansai Start a New Thread

Hi kansai,
Could you attach you butterworth coefficients calculated out of
octave, someone may check your results on matlab.

B.R.
Wayne

On 8/7/09, a...@yahoo.co.jp wrote:
> Hello everybody.
>
> Thank you always for your help.
> I have a question regarding Butterworth filters with order >.
>
> I was out of the office, so couldn't use MATLAB. Instead I used Octave. I
> tried to do the following:
>
>>[b a]=butter(10,0.015);
>>y=filter(b,a,x);
>
> for a original signal x.(a sin signal)
>
> Now when ploting I can see this gives me a filtered signal (although a
> little strange in some values, still good
>
> But when I tried the following
>
>>[b a]=butter(10,0.01);
>>y=filter(b,a,x);
>
> the result is not a filtered sinusoid at all, but a signal that goes to
> infinite!
>
> Now I am at lost why I can not construct butterworth filters of order 10
> with a normalized cutoff of less than 0.015. Is it a theoretical constraint?
> is it a bug in octave? (havent tried it in Matlab)
>
> can anyone share some insight in this?
>
> Kansai Start a New Thread

HI Wayne

Thanks for the reply

The coefficients are:

1) in the first case
[b a]=butter(10.0.015)

b=4.5446e-017
4.5446e-016
2.0451e-015
5.4535e-015
9.5436e-015
1.1452e-014
9.5436e-015
5.4535e-015
2.0451e-015
4.5446e-016
4.5446e-017

a 1.0000
-9.69876
42.33414
-109.51313
185.93260
-216.48723
175.06136
-97.08115
35.33393
-7.62163
0.73988

for this case after
Y=filter(b,a,x) I obtain a curve that resembles a sin curve to some degree
however for

b) second case

[b a]=butter(10, 0.01)

b
8.2784e-019
8.2784e-018
3.7253e-017
9.9340e-017
1.7385e-016
2.0862e-016
1.7385e-016
9.9340e-017
3.7253e-017
8.2784e-018
8.2784e-019

a 1.0000
-9.79918
43.21271
-112.93005
193.68525
-227.79613
186.06003
-104.21312
38.30719
-8.34475
0.81805

for this case after
Y=filter(b,a,x) I obtain a curve that doesnt resemble a sin curve at all

I will appreciate some insight into why this happen.
Thank you

Kansai

Wayne wrote:
Hi kansai,
Could you attach you butterworth coefficients calculated out of
octave, someone may check your results on matlab.

B.R.
Wayne

On 8/7/09, a...@yahoo.co.jp wrote:
> Hello everybody.
>
> Thank you always for your help.
> I have a question regarding Butterworth filters with order >.
>
> I was out of the office, so couldn't use MATLAB. Instead I used Octave. I
> tried to do the following:
>
>>[b a]=butter(10,0.015);
>>y=filter(b,a,x);
>
> for a original signal x.(a sin signal)
>
> Now when ploting I can see this gives me a filtered signal (although a
> little strange in some values, still good
>
> But when I tried the following
>
>>[b a]=butter(10,0.01);
>>y=filter(b,a,x);
>
> the result is not a filtered sinusoid at all, but a signal that goes to
> infinite!
>
> Now I am at lost why I can not construct butterworth filters of order 10
> with a normalized cutoff of less than 0.015. Is it a theoretical constraint?
> is it a bug in octave? (havent tried it in Matlab)
>
> can anyone share some insight in this?
>
> Kansai Start a New Thread