FFT plot accuracy & Scaling

Currently struggling on knowing whether my fft plot is entirely accurate. Trying to also achieve the correct scaling as well. I have looked at a lot of posts regarding this but still not achieving the same sort of scale. I am trying to achieve this:

But I am getting this:

Fs = 5000/157
data = x
L = length(data)
Y = fft(data)/L
f = Fs/2*linspace(0,1,L/2+1);
plot(f,2*abs(Y(1:L/2+1)))
xlim([0.08 0.78])
set(gca, 'YScale', 'log')
title('Duffing')
xlabel('f (Hz)')
ylabel('Amplitude')

Any help would be much appreciated

[VBBV]

%if true
Y = fft(data/L);

Try above

[VBBV]

Also reduce Fs

%f true
Fs = 5000/1570

[Bledar Ferati]

Yeah I pretty much get the exact same graph

[VBBV]

Same as you wanted as in first graph or second ?

[Bledar Ferati]

Not much changed sorry it still looks like the second graph.

[Mathieu NOE]

hi

FYI, this is my fft / averaging / scaling function - please use it as you like :

  function  [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
    %FFT peak spectrum of signal  (example sinus amplitude 1   = 0 dB after fft).
    %   signal - input signal, 
    %   Fs - Sampling frequency (Hz).
    %   nfft - FFT window size
    %   Overlap - buffer overlap % (between 0 and 0.95)

    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;
    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

[dpb]

" it still looks like the second graph."

Where you divide by L is immaterial so that is/was a do-nothing operation.

The change of Fs = 5000/1570 for Fs = 5000/157 will have changed Fmax by factor/divisor of 10 which seems misguided since your peak frequencies seem to match well.

Your magnitude seems to match pretty well, too; that the fine structure isn't the same is undoubtedly owing to difference in the time series data from which you started vis a vis the reference figure.

Dunno there's much anybody here can tell you without a lot more background information.

[Bledar Ferati]

Yeah thats what I was thinking as well, in terms of accuracy its pretty much there anyways. Thank you all for the help!

[VBBV]

%true
L = max(data)
f = (Fs/2)*linspace(0,1,L)
plot(f(1:L:end), 2*abs(Y(1:L:end)))

Use max value of data instead of length