FFT of ON OFF signal

clc,close all,clear all

codn=70;

% fc=6e+3;

fs=36000;

bode=1000;

code=round(rand(1,codn));

code_len=round(1/bode/(1/fs));

for i=1:codn

x((i-1)*code_len+1:code_len*i)=code(i);

end

figure(1)

plot(x)

axis([0 2520 -0.5 1.5])

y=abs(fft(x))

figure(2)

title('orignal baseband signal')

xlabel('time');

ylabel('amplitude')

df=-fs/2:fs/length(y):fs/2-1; (PROBLEM IN THIS FREQUENCY RANGE )

plot(df,y)

This is my code which i have write to take FFT of ON OFF signal but issue is that i am unable to find exact frequency spectru(df=-fs/2 : fs/length(y) : fs/2-1) of ON OFF signal

Answer 1

[David Goodmanson]

Hi imran,

you appear to be pretty close on this. The code below uses ftshift to put f = 0 at the middle of the freq array, and uses a freq array with the same span as yours but shifted appropriately.

Another modoification concerns dc offset in the time domain. The pulse train has an average value of approximately 1/2 (not exactly 1/2 due to the random nature of the pulses) which leads to a large uninteresting spike at f = 0 that dominates the freq domain data. Subtracting 1/2 off of the data before the fft gives a reasonable peak at f = 0.

Also the fft is divided by the number of points N to give the corrrect amplitude in the freq domain.

codn=70;
%  fc=6e+3;
fs=36000;
bode=1000;
code=round(rand(1,codn));
code_len=round(1/bode/(1/fs))
 for ii=1:codn   
     x((ii-1)*code_len+1:code_len*ii)=code(ii);
 end
 
 x2 = x-(1/2);    % get rid of most of the dc peak

% set up time and frequency arrays 
fs = 36000;
N = length(x);
delt = 1/fs;
delf = fs/N;
tvec = (1:N)*delt;
fvec = (-N/2:N/2-1)*delf;    % shifted frequency array
figure(1)
plot(tvec,x2)
ylim([-1 1.5])
y = fftshift(fft(x2)/N);    
figure(2)
plot(fvec,abs(y))