MATLAB: Obtaining amplitude values from an FFT

amlitudefft

Hi there, I have made a single-sided spectrum using FFT of a time-domain signal. From that FFT, I would like to somehow get the amplitude at each of the frequencies from the signal.

If I am outputting an FFT that looks like the attached plot, which has about 4 distinct peaks, what should I do with this plot to get amplitude? I got it using

S = fft(eta)
P2 = abs(S/nFFT);

Where nFFT is the number of sampling points and eta is a time-domain signal.

Best Answer

To get the _correct) amplitudes for a one-sided fft, you need to change your code to:

S = fft(eta)/nFFT;
P2 = abs(S)*2;

Then plot ‘P2’.

The reason is that the total energy in the signal is conserved in the total fft, that, because it is symmetric, will have half the energy at the positive frequency and half at the negative. You have to multiply the amplitudes of the one-sided fft by 2 to plot the correct amplitudes.

The units of the Fourier-transformed amplitudes (using the code I changed) are the same units as the original signal.

For a full discussion, see the R2015a documentation for fft (link).

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MATLAB: How to make Narrow Bandpass filter

hello

seems I have already seen similar post somewhere... a buddy of yours ?

you did not mentionned the sampling frequency , but 24 kHz seemed correct

here my little contribution :


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% spectrogram dB scale
spectrogram_dB_scale = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% test data
Fs = 24000;
% t = 0:1/Fs:10-1/Fs;
fid = fopen('G4_12.dat','r')
[signal,samples] = fscanf(fid, '%12f');
fclose(fid);
% decimate
decim = 12;
signal = decimate(signal,decim);
Fs = Fs/decim;
samples = length(signal);
time =(1:samples)/Fs;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
n_max = fix(log2(samples));
NFFT = 2^(n_max-1);    % 2^n with n chosen so that NFFT is below length of signal (even after decimation) by factor > 2 so some averaging is doable
% NFFT = 2048/4;    % 2^n with n chosen so that NFFT is below length of signal (even after decimation) by factor > 2 so some averaging is doable
NOVERLAP = round(0.95*NFFT);
w = hanning(NFFT);     % Hanning window / Use the HANN function to get a Hanning window which has the first and last zero-weighted samples.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



% display : averaged FFT spectrum before filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sensor_spectrum, freq] = pwelch(signal,w,NOVERLAP,NFFT,Fs,'power');
sensor_spectrum_dB = 10*log10(sensor_spectrum);% convert to dB scale (ref = 1)

% sensor_spectrum_dB = sensor_spectrum_dB-min(sensor_spectrum_dB); % y axis shift so dB values are positive (min value = 0)
%% findpeaks
% df = Fs/NFFT;
% MAXW = 10*df;
% MPD = 20;
% MPP = 20;
% [peaks,loc,W,P] = findpeaks(sensor_spectrum_dB,'MinPeakDistance',MPD,'MaxPeakWidth',MAXW,'MinPeakProminence',MPP);
% freq_peaks = freq(loc);
freq_peaks = [159.0574, 359.3519, 459.4992, 709.8674];
peaks = interp1(freq,sensor_spectrum_dB,freq_peaks,'linear');
figure(1),plot(freq,sensor_spectrum_dB,'b',freq_peaks,peaks,'*r');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(' dB')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sg,fsg,tsg] = specgram(signal,NFFT,Fs,hanning(NFFT),NOVERLAP);  
% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg));     % NB : X=fft(x.*hanning(N))*4/N; % hanning only
% scalingf the dB range : 
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;
% plots spectrogram
figure(2);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title(['Spectrogram / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');
% keep only signal content extracted by bandpass filters 
fc = freq_peaks;      % Hz

bandwith = 8;      % Hz
f_low = fc-0.5*bandwith;
f_high = fc+0.5*bandwith;
N = 4;
signal_filtered = zeros(size(signal));
for ci = 1:length(fc)
    [b,a] = butter(N,2/Fs*[f_low(ci) f_high(ci)]);
    % add filtered signal to total final signal
    tmp = filtfilt(b,a,signal);
    signal_filtered = signal_filtered + tmp;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display : averaged FFT spectrum before / after notch filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sensor_spectrum_filtered, freq] = pwelch(signal_filtered,w,NOVERLAP,NFFT,Fs,'power');
sensor_spectrum_filtered_dB = 10*log10(sensor_spectrum_filtered);% convert to dB scale (ref = 1)
figure(4),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 filter','after filter');
xlabel('Frequency (Hz)');ylabel(' dB')
figure(5),plot(time,signal_filtered,'b');grid
title('Filtered signal');
xlabel('Time (s)');ylabel(' amplitude')

MATLAB: Frequency index to hertz

If the time-domain signal has a sampling frequency of 50 Hz and 100 data points, the one-sided frequency_domain plot will have a frequency vector going from 0 to 25 Hz with 51 points.

So in the frequency-domain plot, a peak at the 20th index would correspond to 25*20/51=9.8 Hz.

So 10 Hz is essentially correct. It is not exact because of the way the discrete Fourier transform is calculated and plotted. This is the sort of compromise that is inherent in all discrete signal processing.

Best Answer

Related Solutions

MATLAB: How to make Narrow Bandpass filter

MATLAB: Frequency index to hertz

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