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

I want to draw a frequency spectre of a square signal. Sampling is being performed at 10MHz, i want to see 4 polygons of that frequency

My current code is like this

close all;

%Define number of samples to take

fs = 100e3; %%% 100 kHz

f0 = 10e3; %Hz

%Define signal

t = 0:1/fs:1-1/fs;

dT = length(y);

t1=(0:dT-1)/f0/100;

dolzina=length(t);

y= zeros(1,dolzina);

y(1:fix(dolzina/2))=1;

y=[y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y];

signal = y;

figure(1)

plot(t1, signal);

%Take fourier transform

fftsignal = fft(y);

%apply fftshift to put it in the form

fftsignal = fftshift(fftsignal);

%Next, calculate the frequency axis, which is defined by the sampling rate

f = fs/2*linspace(0,30,fs);

fq = length(f)

%Since the signal is complex, we need to plot the magnitude to get it to

%look right, so we use abs (absolute value)

figure(2)

plot(f, abs(fftsignal));

i get an error, that i can't plot the transformed signal because of vectors not being the same length

can anyone help?

Answer 1

[Mathieu NOE]

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

[Mathieu NOE]

please remove following line otherwise you will get an error message

text(locs+.02,pks,num2str(freq(locs)))

[Iztok Grof]

Thank you very much for your answer

my bad there with stating that the sample frequency is 10MHz. I need to draw specters for 1kHz, 10kHz, 100Khz , 1Mhz and 10MHz thats why i made a mistake of giving the wrong info.

Tho the specifications are that the signal have lower amplitude of 0 and higher amplitude of 1. Thats why i used that code i posted

[Mathieu NOE]

hello again

regarding your comment : the signal have lower amplitude of 0 and higher amplitude of 1

is exactly what I suspected , so this is the simplest way to generate it (as in my code) :

signal = 0.5*(square(2*pi*f0*t)+1); % 0 / pos  square wave ( 0 / +1)

[Iztok Grof]

Thanks a lot mate :)