Obtaining 2 highest peaks of spectrum in dB

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Ashlyn McCann 2021-3-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/769642-obtaining-2-highest-peaks-of-spectrum-in-db

评论： Mathieu NOE 2021-3-12

Pr= 20*(10^-6); %reference sound pressure level
n=length(y);
npoints= 2^nextpow2(n);
Ss = abs(fft(y)); % gets the spectrum, y is in Pa
SS = Ss/2
f = (0:n-1)*(fs/n); %gets x axis in frequencies
plot(f,SS)
[maxpks, maxloc]=findpeaks(SS(1:40000),'MinPeakProminence',5,'MinPeakDistance',50);
figure(1)
plot(f,(SS))
hold on
plot(f(maxloc), maxpks, 'r*')
sortpks=sort(maxpks);
[j,h]=size(sortpks);
h1=(sortpks(j)) %obtain highest peak
h2=(sortpks(j-1)) %obtain second highest peak
Pr= 20*(10^-6); %reference sound pressure level
H1 = (10*log10(h1.^2/Pr^2)) %convert to dB This value is double what it should be
H2 = (10*log10(h2.^2/Pr^2)) %convert to dB This value is just not picking up the value in dB of highest peak
H1H2 = (H1 - H2)

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Ashlyn McCann 2021-3-11

I can't seem to pick off the two highest values and correctly convert them to dB

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Answer 1

Mathieu NOE 2021-3-11

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/769642-obtaining-2-highest-peaks-of-spectrum-in-db#answer_645142

hello again !

you should probably use findpeaks once the spectrum is in dB (my own preference)

otherwise , you have seen in your other post how to use it with sorting options

plot regression line through first 5 peaks - MATLAB Answers - MATLAB Central (mathworks.com)

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Mathieu NOE 2021-3-11

在 MATLAB Online 中打开

hello

according to your graph, your focus seems to me to be limited to the lower frequency range, so IMHO I would in first place decimate the data by a factor 3 to 5 to reduce the amount of data processing and speed up the fft

next I would do averaging - what you don't do , to get a smoother output (to the price of a reduced frequency resolution) - pwelch can do it , and also use some overlap (50 to 75% is my default setting)

FYI this is my typical FFT code for audio signal analysis :

clc
clearvars
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% dummy data
Fs = 10000;
samples = 20000;
t = (0:samples-1)'*1/Fs;
signal = sin(2*pi*50*t)+2*sin(2*pi*440*t)+1e-2*randn(samples,1); % two sine + some noise
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 5;
if decim>1
    signal = decimate(signal,decim);
    Fs = Fs/decim;
    samples = samples/decim;
    
end
t = (0:samples-1)*1/Fs;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = Fs;    % so delta f = 1 Hz
Overlap = 0.75;
w = hanning(NFFT);     % Hanning window / Use the HANN function to get a Hanning window which has the first and last zero-weighted samples.
%% notch filter section %%%%%%
% H(s) = (s^2 + 1) / (s^2 + s/Q + 1)
fc = 50;       % notch freq
wc = 2*pi*fc;
Q = 5;         % adjust Q factor for wider (low Q) / narrower (high Q) notch
                % at f = fc the filter has gain = 0
w0 = 2*pi*fc/Fs;
alpha = sin(w0)/(2*Q);
            b0 =   1;
            b1 =  -2*cos(w0);
            b2 =   1;
            a0 =   1 + alpha;
            a1 =  -2*cos(w0);
            a2 =   1 - alpha;
            
% analog notch (for info)
num1=[1/wc^2 0 1];
den1=[1/wc^2 1/(wc*Q) 1];
% digital notch (for info)
num1z=[b0 b1 b2];
den1z=[a0 a1 a2];
freq = (fc-1:0.01:fc+1);
[g1,p1] = bode(num1,den1,2*pi*freq);
g1db = 20*log10(g1+1e-6);
[gd1,pd1] = dbode(num1z,den1z,1/Fs,2*pi*freq);
gd1db = 20*log10(gd1+1e-6);
figure(1);
plot(freq,g1db,'b',freq,gd1db,'+r');
title(' Notch: H(s) = (s^2 + 1) / (s^2 + s/Q + 1)');
legend('analog','digital');
xlabel('Fréquence (Hz)');
ylabel(' dB')
 
% now let's filter the signal 
signal_filtered = filtfilt(num1z,den1z,signal);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display : averaged FFT spectrum before / after notch filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq,fft_spectrum] = myfft_peak(signal, Fs, NFFT, Overlap); %
sensor_spectrum_dB = 20*log10(fft_spectrum);% convert to dB scale (ref = 1)
% demo findpeaks
[pks,locs]= findpeaks(sensor_spectrum_dB,'SortStr','descend','NPeaks',2);
[freq,fft_spectrum_filtered] = myfft_peak(signal_filtered, Fs, NFFT, Overlap);
sensor_spectrum_filtered_dB = 20*log10(fft_spectrum_filtered);% convert to dB scale (ref = 1)
figure(2),plot(freq,sensor_spectrum_dB,'b',freq,sensor_spectrum_filtered_dB,'r');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
legend('before notch filter','after notch filter');
xlabel('Frequency (Hz)');ylabel(' dB')
text(freq(locs)+1,pks,num2str(freq(locs)))
function  [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
% FFT peak spectrum of signal  (example sinus amplitude 1   = 0 dB after fft).
% Linear averaging
%   signal - input signal, 
%   Fs - Sampling frequency (Hz).
%   nfft - FFT window size
%   Overlap - buffer overlap % (between 0 and 0.95)
signal = signal(:);
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
    s_tmp = zeros(nfft,1);
    s_tmp((1:samples)) = signal;
    signal = s_tmp;
    samples = nfft;
end
% window : hanning
window = hanning(nfft);
window = window(:);
%    compute fft with overlap 
 offset = fix((1-Overlap)*nfft);
 spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
%     % for info is equivalent to : 
%     noverlap = Overlap*nfft;
%     spectnum = fix((samples-noverlap)/(nfft-noverlap));	% Number of windows
    % main loop
    fft_spectrum = 0;
    for i=1:spectnum
        start = (i-1)*offset;
        sw = signal((1+start):(start+nfft)).*window;
        fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft);     % X=fft(x.*hanning(N))*4/N; % hanning only 
    end
    fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling
% one sidded fft spectrum  % Select first half 
    if rem(nfft,2)    % nfft odd
        select = (1:(nfft+1)/2)';
    else
        select = (1:nfft/2+1)';
    end
fft_spectrum = fft_spectrum(select);
freq_vector = (select - 1)*Fs/nfft;
end

Ashlyn McCann 2021-3-12

Oh that actually answers my question, I added that line to my code after doing the fft of my data. Thank you!

Mathieu NOE 2021-3-12

OK

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Obtaining 2 highest peaks of spectrum in dB

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