MATLAB: How to choose Spectrogram parameter

dspfftfrequencyMATLABsignal processingspectrogramstft

I have a signal with 3Hz frequency and runs from 0:10 second. The signal is zeroed everywhere except from 2 to 4 and 7 to 8 second in the signal as per the image below. I tried to to get the spectrogram but it didn't give correct representation and accurate one. When i surf the spectrogram i can see the 3 signal but the time is shifted. How to choose the correct parameters for spectrogram ? My code:

t=0:1/50:10;
x=sin(2*pi*t*3);
x[1:100]=0;
x[200:350]=0;
x[400:501]=0;
[s,ff,tt,p]=spectrogram(x,50,25,2048,50);
surf(tt,ff,(p),'edgecolor','none'); axis tight; view(0,90);

Thank you

Best Answer

Where you set it to zero, you should be using parentheses and not brackets. not really sure what you were plotting since that syntax should just throw an error.

This should create what you would expect to see, two pulses at around 3 Hz

t=0:1/50:10;
x=sin(2*pi*t*3);
x(1:100)=0;
x(200:350)=0;
x(400:501)=0;
[s,ff,tt,p]=spectrogram(x,50,25,2048,50);
surf(tt,ff,(p),'edgecolor','none'); axis tight; view(0,90);

Related Solutions

MATLAB: Can I set a range for spectrogram Analysis

Hi Ivy,

There are several issues with the code that you have posted. Each of these issues falls into one of two broad categories:

Issues with how the code generates the source signal x(t), and
Issues with how the code attempts to analyze and visualize the signal using the spectrogram function.

The most important issue is related to category #2. The sampling rate Fs is 1000 samples per second, while the number of samples in each FFT is 1024 samples. As a result, the frequency resolution of the spectrogram is nearly 1 hertz:

dF = Fs/NFFT = 1000/1024 = 0.9766 hertz

The frequency of the source signal is 50+rand hertz. Since the rand function generates a uniformly distributed random value between 0 and 1, the frequency of the signal will vary between 50 and 51 hertz, which is a range of only 1 hertz. This range is only just slightly larger than the resolution of the spectrogram, so that most of the time the spectrogram will not be able to detect it.

So to address this issue, you can either increase NFFT to a larger value, or increase the range of frequencies in the source signal.

As I mentioned, there are other issues in the code, but the frequency resolution is probably the most important one.

HTH.

Rick

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

hello Anna

look at my fft function at the end of my code

clc
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% data
data = importdata('SFC5S_nov25_ST_1_1.txt');
dt = 1e-6; % 1 micro seconds
signal = data(:,1);
samples = length(signal);
Fs = 1/dt; % sampling frequency (Hz)
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 1;
if decim>1
    signal = decimate(signal,decim);
    Fs = Fs/decim;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 512;    % 
OVERLAP = 0.95;
% spectrogram dB scale
spectrogram_dB_scale = 80;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums 
% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq, sensor_spectrum] = myfft_peak(signal,Fs,NFFT,OVERLAP);
% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);
% apply A weigthing if needed

if option_w == 1
    pondA_dB = pondA_function(freq);
    sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
    my_ylabel = ('Amplitude (dB (A))');
else
    my_ylabel = ('Amplitude (dB (L))');
end
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(my_ylabel);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sg,fsg,tsg] = specgram(signal,NFFT,Fs,hanning(NFFT),floor(NFFT*OVERLAP));  
% 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
% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(fsg);
    sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
    my_title = ('Spectrogram (dB (A))');
else
    my_title = ('Spectrogram (dB (L))');
end
% saturation of the dB range : 
% saturation_dB = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
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([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');
function pondA_dB = pondA_function(f)
	% dB (A) weighting curve
	n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
	r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
	pondA = n./r;
	pondA_dB = 20*log10(pondA(:));
end
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: Can I set a range for spectrogram Analysis

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

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