MATLAB: Fourier-Frequency analyse

fourierfrequency analyse

I need to do next: Draw a frequency image of following signal: t=0:0.0001:0.05; y=square(5*pi*50*t);

also it's required to draw real and imaginary part of frequency spectrum.

There is mine solution, so can anyone check it out

t=0:0.0001:0.05; y=square(5*pi*50*t); N=length(t); deltaf=1/0.05; f=(0:N-1)*deltaf;

disp('Efektivna vrijednost') Yef=sqrt(sum(y.*y)/N)

disp('Frekvencijska slika') F=(fft(y))*2/N;

figure, plot(f,F) axis([0 1200 -0.5 1])

figure, subplot(3,1,1)

plot(f,20*log10(real(F)/max(abs(F)))) axis([0 1200 -80 10])

subplot(3,1,2) plot(f,20*log10(imag(F)/max(abs(F)))) axis([0 1200 -80 10])

subplot(3,1,3) plot(f,20*log10(F/max(abs(F)))) axis([0 1200 -80 10])

Best Answer

Hey, It shows error here: y_freq = fftshift(fft(y))*dt;

because matrix dimensions doesn't agree here: f = -Nyq : df : Nyq;

I bypass that somehow, and I saw that our sinals in frequency spectrum looks exactly the same(I draw it without logarithm scale and for positive frequencies, thats why I multiply with 2 here F=(fft(y))*2/N;), but only difference is that ur values are 20 times smaller, for same frequency u get 20 times lower amplitudes. Why?

Related Solutions

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

MATLAB: Finding the dominant frequency of a time series data using fft matlab

The ‘frequency’ at 0 is the mean of your signal (or D-C offset). To eliminate it, subtract the mean before doing the fft.

This gives you the result you want:

Fs = 10; % sampling frequency 1 kHz
t =  [0,10,20,30,40,50,60,70,80,90]; % time scale
x = [10,120,130,120,120,100,123,456,78,89]; % time series
x = x - mean(x);                                            % <= ADDED LINE
plot(t,x), axis('tight'), grid('on'), title('Time series'), figure
nfft = 512; % next larger power of 2
y = fft(x,nfft); % Fast Fourier Transform
y = abs(y.^2); % raw power spectrum density
y = y(1:1+nfft/2); % half-spectrum
[v,k] = max(y); % find maximum
f_scale = (0:nfft/2)* Fs/nfft; % frequency scale
plot(f_scale, y),axis('tight'),grid('on'),title('Dominant Frequency')
fest = f_scale(k); % dominant frequency estimate
fprintf('Dominant freq.: true %f Hz, estimated %f Hznn\n', fest, fest)
fprintf('Frequency step (resolution) = %f Hznn\n', f_scale(2))

Best Answer

Related Solutions

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

MATLAB: Finding the dominant frequency of a time series data using fft matlab

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