MATLAB: Mean Amplitude of the signal

amplitudefftmean amplitudesignal

Hi, My problem is to calculate mean amplitude of the signal and plot it on the graph with signal course. Below you can see my source code:

      y % signal vector, where [3083 1]= size(y)
      Fs=100;
      L=length(y);
      NFFT = 2^nextpow2(L); 
      Y = fft(y,NFFT)/L;
      f = Fs/2*linspace(0,1,NFFT/2+1);
     n1=1/100; %1
     m1=L/100; %end
     [a b]=size(y);
     nov=(m1-n1)/(a-1);
     x=n1:nov:m1;
figure
subplot(2,1,1)
plot(f,2*abs(Y(1:NFFT/2+1)),color) % amplitude spectrum
subplot(2,1,2)
plot(x,y,'b') % course of the signal

Could you tell me how can I calculated mean amplitude based on this? Or maybe there exists another way to find it? When I used mean(abs(Y(1:NFFT/2+1))) and ploted it on the second subplot it was far away from signal…

Best Answer

Hi Karolina, No, it's proportional to the amplitude squared of the Fourier coefficients. You can get amplitude estimates like this:

t = 0:0.01:5-0.01;
x = cos(2*pi*4*t)+1/2*cos(2*pi*8*t)+randn(size(t));
xdft = fft(x);
xdft = (2/length(x))*xdft(1:length(x)/2+1);

Now, you have to be able to figure out that the 4-Hz component lives in DFT bin (21) and the 8-Hz component lives in DFT bin (41)

   abs(xdft(21))
   abs(xdft(41))

Give you the amplitude estimates for those frequencies.

   abs(xdft(16:51))

Gives you the amplitude estimates for 3-10 Hz.

Note that the spacing between DFT bins ist Fs/N where Fs is the sampling frequency and N is the length of the signal.

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MATLAB: Amplitude Error in amplitude Spectrum of FFT

I suspect that your frequency of interest is simply not falling on a DFT bin directly. Why do you think you need to use a power of two for the DFT?

For example, if I have data sampled at 1 kHz and I have 1000 points, then frequency increment in DFT bins is 1 Hz (the bins are 1 Hz apart) so a frequency of 100 Hz for example will fall directly on a bin.

   Fs = 1000;
   t = 0:1/Fs:1-1/Fs;
   x = 1.5*cos(2*pi*100*t);
   xdft = fft(x);
   freq = 0:Fs/length(x):Fs/2;
   xdft = xdft(1:length(x)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

See that the magnitude is estimated exactly. Now watch what happens when I pad to 1024:

   xdft = fft(x,1024);
   freq = 0:Fs/length(xdft):Fs/2;
   xdft = xdft(1:length(xdft)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

by padding to 1024, I have now made the frequency of interest fall between DFT bins and that makes my amplitude estimate inaccurate.

See https://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html

in the Signal Processing Toolbox documentation

MATLAB: Specter of a square signal(10Khz) at 10MHz frequency

hello

there is a much faster method to generate a square signal (using square )

see demo below for signal generation and fft analysis

NB your sampling freq is 100 kHz in the code, not 10 MHz as stated in the post

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



% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% dummy data
Fs = 100e3;     % sampling freq
f0 = 10e3;     % signal freq
duration = 1;   % signal duration (s)
samples = Fs*duration;
t = (0:samples-1)*1/Fs;
% signal = square(2*pi*f0*t); % neg / pos symetrical square wave ( -1 / +1)
signal = 0.5*(square(2*pi*f0*t)+1); % 0 / pos  square wave ( 0 / +1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 10000;    % 
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.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display : averaged FFT spectrum 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq,fft_spectrum] = myfft_peak(signal, Fs, NFFT, Overlap);
sensor_spectrum_dB = 20*log10(fft_spectrum);% convert to dB scale (ref = 1)
figure(1),plot(freq,sensor_spectrum_dB,'b');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(' dB')
text(locs+.02,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

Best Answer

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MATLAB: Specter of a square signal(10Khz) at 10MHz frequency

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