fs = 16e3;
%range = [50 8000];
CF1=linspace(50, 8000, 32) -50;
CF2=linspace(50, 8000, 32) +50;
Can't use the first and last elements of CF1 and CF2 respectively. CF1(1) = 0, which is not a valid input for CutOffFrequency1. This issue could be dealt with by making the first filter a low pass. CF2(end) is greater than the Nyquist frequecy.
for ii = 1:numel(CF1)-2
bpfilt{ii} = designfilt( ...
'bandpassfir', ...
'FilterOrder',20, ...
'CutoffFrequency1',CF1(ii+1), ...
'CutoffFrequency2',CF2(ii+1), ...
'SampleRate',fs);
[h{ii},f] = freqz(bpfilt{ii}.Coefficients,1,4*8192,fs);
end
Frequency responsee of each filter
figure
subplot(211)
plot(f,db(abs([h{:}])));
subplot(212);
plot(f,180/pi*(angle([h{:}])));
Frequency response of cascaded filters in series; this doesn't look like very useful.
Edit: In fact, the values on the magnitude plot on are so small that they are probably rounding error and are effectively zero (in absolute value). Are you sure you want the filters cascaded?
figure
subplot(211)
plot(f,db(abs(prod([h{:}],2))));
subplot(212);
plot(f,180/pi*(angle(prod([h{:}],2))));
Unfortunately, digitalFilter objects can't be mulitplied.
In theory, the coefficients of each filter can be combined as follows into a single filter
b = 1;
for ii = 1:numel(bpfilt)
b = conv(b,bpfilt{ii}.Coefficients);
end
Compare the frequency response of the cascade to the cascade of the frequency responses
hcascade = freqz(b,1,f,fs);
figure
subplot(211)
hold on
plot(f,db(abs(prod([h{:}],2))));
plot(f,db(abs(hcascade)));
subplot(212);
hold on
plot(f,180/pi*(angle(prod([h{:}],2))));
plot(f,180/pi*angle(hcascade))
In practice, that doesn't look like such a good idea.
If going to pursue this path, it's probabably best to use a loop and filter the input in stages. That will work, in theory, for forward filtering. Not sure if that's appropriate for forward/backward filtering.
