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
what is the roll-off factor and cutoff-slope of Butterworth filter?
Started by ●November 18, 2004
Reply by ●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 ●November 29, 20042004-11-29
Country_Chisel wrote:> "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. �����������������������������������������������������������������������
