Limited band white noise block simulink

Hi, i am using limited band white noise block on simulink. I adjust the two parameters in it - noise power and correlation time, but i cannot figure out how to find the range of frequencies of the limited white noise. Ideal white noise has all the frequencies but here it is limited. Any help will be highly appreciated!

Answer 1

[Mathieu NOE]

hello

as the doc says , the Band-Limited White Noise block produces output at a specific sample rate, which is related to the correlation time of the noise.

if we use the default settings with tc = 0.1 , we can observe that this duration corresponds to the stairs length (zero hold duration)

now the spectrum of this signal will be limited , with a higher limit frequency corresponding to the inverse of this time

here a time reduced to 1e-3 s , the signal is sampled at Fs = 10 kHz

we can clearly see the drop at f = 1 kHz (and multiple of it)

in conclusion, this signal can be viewed as a good approximation of a "true" white noise only between 0 and a fraction of f_upper_limit = 1/ correlation time , let's say f_upper_limit = 0.25/correlation time, so here it would be 0 to 250 Hz.

If you want to further reduce the bandwith of the signal , simply add a bandpass filter

[Mathieu NOE]

fyi, this is the m code to generate the above plots

% simulink signals 
signal = out.simout;
time= out.tout;

[samples,channels] = size(signal);
dt = mean(diff(time));
Fs = 1/dt;


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 1000;    % for maximal resolution take NFFT = signal length (but then no averaging possible)
OVERLAP = 0.75; % overlap will be used only if NFFT is shorter than signal length


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : time domain plot
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure(1),
plot(time,signal,'b');grid on
title(['Time plot  / Fs = ' num2str(Fs) ' Hz ']);
xlabel('Time (s)');ylabel('Amplitude');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[freq, spectrum_raw] = myfft_peak(signal,Fs,NFFT,OVERLAP);
ind = freq>0;
freq = freq(ind);
spectrum_raw = spectrum_raw(ind);

figure(2),semilogx(freq,20*log10(spectrum_raw),'b');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('Amplitude (dB)');


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

[Devyani Varshney]

@Mathieu NOE How are you able to generate the signal with tc=1e-3 but sampled at 10 kHz? the simulink just logs the signal at tc instead of 10 kHZ.

[Mathieu NOE]

hello @Devyani Varshney

yes usually the sampling rate is equal to tc and therefore the spectrum is almost flal as we would only be able to visualize the fft up to the F_nyquist = 0.5/tc - which correspond to the red box in my previous example

now, just FYI, if you want to force the acquisition to a faster rate (10 kHz here) you need to select the (fixed) sampling rate in both the solver parameter window

and also in the "to workspace" block

NB that tc is still = 10-3 in the parameter window of the white noise generator block :

hope it helps !