> "lucy" <losemind@yahoo.com> wrote in message
> news:cnjhpg$18e$1@news.Stanford.EDU...
>
>>Hi all,
>>
>>I am doing this in digital domain... I have used the Butterwork lowpass
>>filter grabbed from Internet:
>>
>> % X and Y matrices with ranges normalised to +/- 0.5
>> x = (ones(rows,1) * [1:cols] - (fix(cols/2)+1))/cols;
>> y = ([1:rows]' * ones(1,cols) - (fix(rows/2)+1))/rows;
>>
>> radius = sqrt(x.^2 + y.^2);
>>
>> f = 1 ./ (1.0 + (radius ./ cutoff).^(2*n));
>>
>>
>>Then "f" is the frequency domain representation of the filter. n is the
>>order of the filter; cutoff is the normolized cutoff frequency between [0,
>>0.5]...
>>
>>Now I want to compute the roll-off factora nd cutoff-slope of this n-order
>>Butterworth filter...
>>
>>How do I get those? Can anybody give me some pointers?
>>
>>Thanks a lot
>>
>>
>
>
> The asymptotic roll-off of an nth order is just 20n dB/decade or 6n
> dB/octave.
I don't think it's obvious enough that n above is the filter order. It
isn't obvious at all (to me) that this asymptotic rolloff applies to
filters of any type, not just Butterworth, or that it applies exactly to
analog filters, approximately to digital filters with cutoffs well below
the sample rate, and not at all to digital filters with cutoffs as high
as fs/4. Impulse-invariant and bilinear designs are different from one
another. As far as I'm concerned, there's no such thing as a digital
Butterworth filter. There are digital filters with Butterworth analog
prototypes which, in a less august and professional forum than comp.dsp,
might be called "more-or-less Butterworth maybe" filters.
Jerry
--
Engineering is the art of making what you want from things you can get.
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Reply by Country_Chiel●November 28, 20042004-11-28
"lucy" <losemind@yahoo.com> wrote in message
news:cnjhpg$18e$1@news.Stanford.EDU...
> Hi all,
>
> I am doing this in digital domain... I have used the Butterwork lowpass
> filter grabbed from Internet:
>
> % X and Y matrices with ranges normalised to +/- 0.5
> x = (ones(rows,1) * [1:cols] - (fix(cols/2)+1))/cols;
> y = ([1:rows]' * ones(1,cols) - (fix(rows/2)+1))/rows;
>
> radius = sqrt(x.^2 + y.^2);
>
> f = 1 ./ (1.0 + (radius ./ cutoff).^(2*n));
>
>
> Then "f" is the frequency domain representation of the filter. n is the
> order of the filter; cutoff is the normolized cutoff frequency between [0,
> 0.5]...
>
> Now I want to compute the roll-off factora nd cutoff-slope of this n-order
> Butterworth filter...
>
> How do I get those? Can anybody give me some pointers?
>
> Thanks a lot
>
>
The asymptotic roll-off of an nth order is just 20n dB/decade or 6n
dB/octave.
Country Chiel
Reply by lucy●November 18, 20042004-11-18
Hi all,
I am doing this in digital domain... I have used the Butterwork lowpass
filter grabbed from Internet:
% X and Y matrices with ranges normalised to +/- 0.5
x = (ones(rows,1) * [1:cols] - (fix(cols/2)+1))/cols;
y = ([1:rows]' * ones(1,cols) - (fix(rows/2)+1))/rows;
radius = sqrt(x.^2 + y.^2);
f = 1 ./ (1.0 + (radius ./ cutoff).^(2*n));
Then "f" is the frequency domain representation of the filter. n is the
order of the filter; cutoff is the normolized cutoff frequency between [0,
0.5]...
Now I want to compute the roll-off factora nd cutoff-slope of this n-order
Butterworth filter...
How do I get those? Can anybody give me some pointers?
Thanks a lot