Why and what should I do if a signal ampiltude decreased so much after it passed a low-frequency and a high-frequency butterpass filter

Hello,

I was trying to remove the noise of a heartsound, then I found in a reference that I should use a butterpass filter. But when I apply it to the signal the ampiltude decreased from 0.1 to 0.000001. I was wondering whether it is correct or how can I amplify the signal？

[y,fs] = audioread(total_path);
y = y(:,1);
dt = 1/fs;
total_t = length(y)*dt;
t = (0:dt:(total_t)-dt)';
fc=40; % the cut-off frequency for low-pass filter
filter_parameters.N=10; % order of the filter
filter_parameters.type='butter';%apply type
% of filter ('Butterworth')
filter_parameters.lphp='low'; % Set the filter to low-pass
[b,a]=make_digital_filter(fc,fs,filter_parameters);% make filter parameters
D1=filter(b,a,y);% apply the low-pass filter
N_IR=fs; % set length, in samples, of impulse
         % rewponse to calculate
[h,th,H,fH]=i_f_response(b,a,fs,N_IR);%calculate the impulse and frequency response
figure
plot(th,h);% plot the impulse response
title('impulse response');
xlabel('time(s)');
figure
plot(fH,abs(H));%plot the amplitude response
title('amplitude response');
xlabel('frequency(Hz)');
ylabel('gain');

% Apply the high-pass filter
filter_parameters.lphp='high'; % Set the filter to high-pass
fc=190; % the cut-off frequency for high-pass filter
filter_parameters.N=4; % order of the filter
[b,a]=make_digital_filter(fc,fs,filter_parameters);% make new filter parameters
D2=filter(b,a,D1);
figure
plot (t,D2,'c');

Answer 1

[Star Strider]

The filtering operation removed much of the signal energy.

To amplify it, just multiply it by a constant:

D2 = D2*10;

Choose whatever value works best in your application.

.

[Jingyi Yuan]

Thanks!

[Star Strider]

As always, my pleasure!

[Jingyi Yuan]

Sorry it's me again, after I applying another signal (also heart sound) the low pass filter returns NaN, can I ask what might happened?

[Star Strider]

There are two possibilities, the first being that there are NaN values in the data, and the second that the filter (if different from the first one that worked) is unstable. (Transfer function realisations can produce unstable filters, so second-order-section implementation is always preferable.) I suggest using the bandpass function (and its friends) with the name-value pair 'InpulseResponse','iir' for best results. Those functions design a very efficient eliptic filter.

I cannot determine the specific cause otherwise.

[Jingyi Yuan]

Thank you so much for your reply, yes, I think it is the problem of the butterpass filter. The NaN result has been changed after I decrease the order of butterpass filter. Thanks again!

[Star Strider]

As always, my pleasure!

I still suggest using the second-order-section implementation. See Limitations for details.