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Normalize graph chebyshev filter

조회 수: 2(최근 30일)
Juan Chehin
Juan Chehin 2021년 8월 23일
댓글: Mathieu NOE 2021년 9월 2일
Hello, I have this code that corresponds to an analog chebyshev low pass filter, which must have a cutoff frequency of 250 Hz at 0.7, when graphing with freqs I get a signal that does not cut at 0.7. ¿How do I normalize the filter to match the point (250,0.7)?
n = 3; Rp = 0.5;
Wn = [250 2000];
[b a]=cheby1(n,Rp,Wn,'s')
[h,w]=freqs(b,a)
plot(w,abs(h))
  댓글 수: 2
Star Strider
Star Strider 2021년 8월 23일
Chebyshev filter passbands and stopbands are defined differently than for other filters. The stopband is the frequency at which the filter characteristic equals the passband ripple amplitude, not the -6 dB point as with other filter types.

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Mathieu NOE
Mathieu NOE 2021년 8월 23일
hello
I tried an iterative correction method . I plot the frf of the filter in dB and 0.7 then correspond to the - 3 dB cutoff amplitude.
Got a good result within 8 iterations :
Code :
n = 3; Rp = 0.5;
Wn = [250 2000];
[b a]=cheby1(n,Rp,Wn,'s') ;
w = logspace(2,4,300); % Frequency vector
[h,ww]=freqs(b,a,w) ;
FRF_dB = 20*log10(abs(h));
FRF_dB2 = FRF_dB;
semilogx(w,FRF_dB);
for ci = 1:8 % loops
% find the - 3dB crossing points
[f1,s0_pos,f2,s0_neg] = crossing_V7(FRF_dB2,w,-3);
cor_coeff1 = Wn(1)/f1;
cor_coeff2 = Wn(2)/f2;
% 2nd iteration
W = [W(1)*cor_coeff1 W(2)*cor_coeff2];
[b a]=cheby1(n,Rp,W,'s') ;
[h,ww]=freqs(b,a,w) ;
FRF_dB2 = 20*log10(abs(h));
end
% check again the - 3dB crossing points
[f11,s0_pos,f22,s0_neg] = crossing_V7(FRF_dB2,w,-3);
semilogx(w,FRF_dB,'b',w,FRF_dB2,'r',f11,s0_pos,'dr',f22,s0_neg,'dr');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [t0_pos,s0_pos,t0_neg,s0_neg] = crossing_V7(S,t,level,imeth)
% [ind,t0,s0,t0close,s0close] = crossing_V6(S,t,level,imeth,slope_sign) % older format
% CROSSING find the crossings of a given level of a signal
% ind = CROSSING(S) returns an index vector ind, the signal
% S crosses zero at ind or at between ind and ind+1
% [ind,t0] = CROSSING(S,t) additionally returns a time
% vector t0 of the zero crossings of the signal S. The crossing
% times are linearly interpolated between the given times t
% [ind,t0] = CROSSING(S,t,level) returns the crossings of the
% given level instead of the zero crossings
% ind = CROSSING(S,[],level) as above but without time interpolation
% [ind,t0] = CROSSING(S,t,level,par) allows additional parameters
% par = {'none'|'linear'}.
% With interpolation turned off (par = 'none') this function always
% returns the value left of the zero (the data point thats nearest
% to the zero AND smaller than the zero crossing).
%
% [ind,t0,s0] = ... also returns the data vector corresponding to
% the t0 values.
%
% check the number of input arguments
error(nargchk(1,4,nargin));
% check the time vector input for consistency
if nargin < 2 | isempty(t)
% if no time vector is given, use the index vector as time
t = 1:length(S);
elseif length(t) ~= length(S)
% if S and t are not of the same length, throw an error
error('t and S must be of identical length!');
end
% check the level input
if nargin < 3
% set standard value 0, if level is not given
level = 0;
end
% check interpolation method input
if nargin < 4
imeth = 'linear';
end
% make row vectors
t = t(:)';
S = S(:)';
% always search for zeros. So if we want the crossing of
% any other threshold value "level", we subtract it from
% the values and search for zeros.
S = S - level;
% first look for exact zeros
ind0 = find( S == 0 );
% then look for zero crossings between data points
S1 = S(1:end-1) .* S(2:end);
ind1 = find( S1 < 0 );
% bring exact zeros and "in-between" zeros together
ind = sort([ind0 ind1]);
% and pick the associated time values
t0 = t(ind);
s0 = S(ind);
if ~isempty(ind)
if strcmp(imeth,'linear')
% linear interpolation of crossing
for ii=1:length(t0)
%if abs(S(ind(ii))) > eps(S(ind(ii))) % MATLAB V7 et +
if abs(S(ind(ii))) > eps*abs(S(ind(ii))) % MATLAB V6 et - EPS * ABS(X)
% interpolate only when data point is not already zero
NUM = (t(ind(ii)+1) - t(ind(ii)));
DEN = (S(ind(ii)+1) - S(ind(ii)));
slope = NUM / DEN;
slope_sign(ii) = sign(slope);
t0(ii) = t0(ii) - S(ind(ii)) * slope;
s0(ii) = level;
end
end
end
% extract the positive slope crossing points
ind_pos = find(sign(slope_sign)>0);
t0_pos = t0(ind_pos);
s0_pos = s0(ind_pos);
% extract the negative slope crossing points
ind_neg = find(sign(slope_sign)<0);
t0_neg = t0(ind_neg);
s0_neg = s0(ind_neg);
else
% empty output
ind_pos = [];
t0_pos = [];
s0_pos = [];
% extract the negative slope crossing points
ind_neg = [];
t0_neg = [];
s0_neg = [];
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% % Addition:
% % Some people like to get the data points closest to the zero crossing,
% % so we return these as well
% [CC,II] = min(abs([S(ind-1) ; S(ind) ; S(ind+1)]),[],1);
% ind2 = ind + (II-2); %update indices
%
% t0close = t(ind2);
% s0close = S(ind2);
end
  댓글 수: 7
Mathieu NOE
Mathieu NOE 2021년 9월 2일
thanks !!

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