Bandstop starts at 2.7 dB instead of 0 dB. Why? Demo Code from a Matlab Book

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Wolf Zimmermann
Wolf Zimmermann 2020-7-29
评论： Shae Morgan 2020-8-14
hello guys, i have this demo code from a matlab lecture which first makes a LP-butterworth Filter and then makes a LP->BS Transformation. I don't get why the plot in the Bandstop starts around 2.7 dB or something and not at 0 dB. in this book is also a LP->BP Transformation with the formulas i need to change in the code and this is waht i need. but also in this transformation, the BP starts at its cutoff frequencies and raises to 150 dB for example (with my values). And i really don't get, why and on which parametres it depends, that the "limit" is not at 0 dB.
This is the Code for the Notch Filter which is working, but also boosts frequencies until the cutoff frequencies are reached:
% Auxillary computation for analog BW lowpass
% dsp_iirdes_1 * mw * 2017
%% Analog Butterworth LP
dD = .1; dS = .2;
fD = 2.2; fS = 4.8; % in kH
epsilon = sqrt(1/(1-dD)^2 - 1);
lambda = sqrt(1/dS^2 - 1);
N = ceil((log(lambda/epsilon)/log(fS/fD)));
f3dB = fD/epsilon^(1/N);
%% Poles in s domaine (normalized)
ps = []; a = 1i*pi/(2*N);
for k=1:N
ps = [ps exp(a*(2*k+N-1))];
end
%% Bilinear transform (normalized)
Omega = pi/5; % target for 3dB normalized radian frequency
alpha = tan(Omega/2);
pz = (1+alpha*ps)./(1-alpha*ps);
%% Filter coefficients
a = poly(pz); b = poly([-1 -1 -1]);
%% Pol-zero diagram, H(1)=1
% IIR LP filter to IIR BS filter
% dsp_iirdsg_2.m * mw * 2017
%% Prototype low-pass (Butterworth)
bLP = [3 10 10 3];
aLP = [1 -1.76004 1.18289 -0.27806];
wcLP = .2*pi; % 3dB corner frequency (design)
pLP = roots(aLP); zLP = roots(bLP); % poles and zeros
fvtool(bLP*sum(aLP)/sum(bLP),aLP) % filter viewer
%% LP-BS-Transform
wL = .4*pi; wU = .5*pi;
alpha = cos((wU+wL)/2)/cos((wU-wL)/2);
beta = tan((wU-wL)/2)*tan(wcLP/2);
for k = 1:length(zLP)
m = 2*(k-1) + 1;
P = 2*alpha*(1-pLP(k))/((1-beta)*pLP(k)-beta-1);
Q = ((1+beta)*pLP(k) + beta - 1)/((1-beta)*pLP(k)-beta-1);
p5(m) = -P/2 + sqrt(P^2-4*Q)/2;
p5(m+1) = -P/2 - sqrt(P^2-4*Q)/2;
P = 2*alpha*(1-zLP(k))/((1-beta)*zLP(k)-beta-1);
Q = ((1+beta)*zLP(k) + beta - 1)/((1-beta)*zLP(k)-beta-1);
z5(m) = -P/2 + sqrt(P^2-4*Q)/2;
z5(m+1) = -P/2 - sqrt(P^2-4*Q)/2;
end
a5 = poly(p5); b5 = poly(z5);
fvtool(b5,a5) % filter viewer

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回答（1 个）

Shae Morgan
Shae Morgan 2020-8-10
编辑：Shae Morgan 2020-8-10
If a filter is a perfect rectangular filter, then filtering begins at the cutoff frequency and goes to the desired attenuation/gain.
In other filters, the cutoff frequency is commonly defined as the "half-power" point in the function. That is to say, when the function attenuates by half of the initial power (~3 dB = 10*log10(1/2)) that is where the cutoff frequency is.
This is my guess as to why there is a cuttoff around 3 dB. As you increase the filter attenuation rate (filter slope) then that 3 dB down point will get closer and closer to your desired cutoff frequency

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