How analyze PMU data in the form of sinusoidal waveform?

Question

Jan Ali 2021-3-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/786959-how-analyze-pmu-data-in-the-form-of-sinusoidal-waveform

评论： Jan Ali 2021-3-31

Dear experts.

I need your help with PMU dataset analysis. I am wondering how to plot sinusoidal waveform if we have PMU data then apply time-frequency analysis method such as Fourier Transform, Wavelet, etc.

I appreciate any tips, guide and help.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Mathieu NOE 2021-3-29

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/786959-how-analyze-pmu-data-in-the-form-of-sinusoidal-waveform#answer_661714

simple_demo_wav.m

hello

I am not sure what your data represents, but FYI, this is a code to do spectral analysis (averaged fft and spectrogram)

(attached)

6 个评论
显示 4更早的评论隐藏 4更早的评论

Mathieu NOE 2021-3-30

在 MATLAB Online 中打开

hello again

so I finally almost understood what's in the data file - except the last three columns....

see comments in the beginning on my code - I am not 100% that your data are to be analysed with FFT or other spectral analysis tools , because simply there is not much "dynamic" behaviour to be seen .

clc
clearvars
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% data
% data has 19 columns 
% first  columns are date and time : day / month / year / hour / minutes /  seconds / milliseconds
% those data will not be displayed
% sampling time  = 20 milliseconds (see constant increment in column 7 )
% data to be retrieved start at column 8 (to 19) : first line below as
% example : 
% 7	8	2019	16	35	0	47	230,51	-120,56	230,31	-120,62	231,13	-120,68	49,98	49,98	49,99	-3,29	-3,25	-3,29
% so finally there are 12 columns of interest - looks like a three phase voltage data record : 
% Voltage phase 1 / angle phase 1 / Voltage phase 2 / angle phase 2 / Voltage phase 3 / angle phase 3
% / Frequency phase 1 / Frequency phase 2 / Frequency phase 3 / the last 3 columns are ... what ?
% % NB : 
% 1/ all 3 Voltage are very steady => no real need to do any frequency analysis , it's only a DC value
% 2/ same regarding Frequency phase 1 / 2 / 3 , very steady , it's only a DC value
% 3/ all 3 voltage phases evolution is very similar - what are we supposed
% to do on this ? I am not aware that doing a fft on a phase plot is
% meaningfull...
% 4/ don't know what are the last 3 columns ??
dt = 20e-3;
Fs = 1/dt;
data = importdata('data.xls');
signal = data(:,8:19); % remove date ant hour/min/seconds data
[samples,channels] = size(signal);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 256;    % 
OVERLAP = 0.75;
% 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;
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 1;
if decim>1
    for ck = 1:channels
    newsignal(:,ck) = decimate(signal(:,ck),decim);
    Fs = Fs/decim;
    end
   signal = newsignal;
end
samples = length(signal);
time = (0:samples-1)*1/Fs;
%%%%%% legend structure %%%%%%%%
for ck = 1:channels
    leg_str{ck} = ['Channel ' num2str(ck) ];
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : time domain plot
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1),plot(time,signal);grid on
title(['Time plot  / Fs = ' num2str(Fs) ' Hz ']);
xlabel('Time (s)');ylabel('Amplitude');
legend(leg_str);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : 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(2),plot(freq,sensor_spectrum_dB);grid on
df = freq(2)-freq(1); % frequency resolution 
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(df,3) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);
legend(leg_str);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 3 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for ck = 1:channels
    [sg,fsg,tsg] = specgram(signal(:,ck),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+ck);
    imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
    axis('xy');colorbar('vert');grid on
    df = fsg(2)-fsg(1); % freq resolution 
    title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(df,3) ' Hz / Channel : ' num2str(ck)]);
    xlabel('Time (s)');ylabel('Frequency (Hz)');
end
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 percentage of overlap % (between 0 and 0.95)
[samples,channels] = size(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
    s_tmp = zeros(nfft,channels);
    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*ones(1,channels));
        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

Mathieu NOE 2021-3-31

hello

I see there is a lot of publications about PMU on the internet

maybe you should take contact with some of the authors , they may have the raw data you need

all the best

Jan Ali 2021-3-31

Many thanks Mathieu for your kind help. I really appreciate it.

I will try to my best. It is my thesis assignment, however the time is very constrained.

请先登录，再进行评论。

How analyze PMU data in the form of sinusoidal waveform?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论
显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

How analyze PMU data in the form of sinusoidal waveform?

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

6 个评论 显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

6 个评论
显示 4更早的评论隐藏 4更早的评论