0
$\begingroup$
import numpy as np
import matplotlib.pyplot as plt

j = 1j
pi = np.pi


interval = 0.5
fs = 1024
Ts = 1/fs
N = np.uint(interval*fs)
t = np.linspace(0, interval - Ts, N)
freq = 2*pi/N * np.arange(N)
f0 = 16 

# x(t) = sin(2pi * f0 * t)
# a[n] = x(n*Ts) = sin(2pi * f0 * n * Ts)
n = np.arange(N)
a = np.sin(2*pi*f0*Ts*n) 



# visualisation de a
# on ajoute a droite la valeur de gauche pour la periodicite
plt.subplot(311)
plt.plot( t, a)

# calcul de A
A = 1/N * np.fft.fft(a)

# visualisation de A
# on ajoute a droite la valeur de gauche pour la periodicite

plt.subplot(312)
plt.plot(freq, np.real(A))

plt.ylabel("Real Part")

plt.subplot(313)
plt.plot(freq, np.imag(A))


plt.ylabel("Imaginary Part")

plt.show()

So I have a signal $$x(t) = \sin(2\pi \cdot f_0 \cdot t),\ t = 0\rightarrow0.5s$$

I wish to sample the signal at $f_s = 1024\ \mathrm{H_z} \longrightarrow T_s = 1/1024$

So I make $$a[n] = x(n\cdot T_s) \longrightarrow a[n] = \sin(2\pi\cdot f_0\cdot T_s\cdot n)$$

But clearly, this makes the discrete signal have $f = f_0\cdot T_s$.

So when I FFT $a[n]$, the 2 frequency spike is at the wrong location. You can see that the $\sin$ spike is at $2\pi\cdot f_0\cdot T_s$, instead of $2\pi\cdot f_0$

enter image description here

I don't want to make $$a[n] = \sin(2\pi\cdot f_0\cdot n)$$ because then I only get the values of $x(t)$ at positions $t = 2\pi\cdot f_0\cdot n$

Which step did I forget (or did wrong)? How to sample the $x(t)$ so that the FFT is correct? Thank you.

$\endgroup$
  • $\begingroup$ The relative frequency in discrete time can be defined as $f=F/F_s$ which you got it right with your equation $f=f_0 * T_s$. If you perform FFT on $a[n]$ you may see the spectral content concentrated at $f$. $\endgroup$ – jomegaA Feb 24 at 11:13
  • $\begingroup$ The FFT of $\sin$ whose spectrum might resemble like a spike must be at $f=f_0*T_s$, if you plotted using relative frequency instead of $n$ $\endgroup$ – jomegaA Feb 24 at 11:16
  • $\begingroup$ so what should I add so that the graph has 2 spike at 2*pi*16 ? (the original frequency in continuous time) $\endgroup$ – Duke Le Feb 24 at 11:44
  • $\begingroup$ You may use fftshift (python equivalent I have no idea)... $\endgroup$ – jomegaA Feb 24 at 13:02
0
$\begingroup$
close all;
clear all;

interval        = 0.5;
f0              = 16; 
t               = 0:1/2048:interval-(1/2048)
x_t             = sin(2*pi*f0*t);

figure(1);
subplot(2, 1, 1);
plot(t, x_t);
axis([0 interval])
hold on;

fs              = 1024;
N               = (interval * fs);
sample_index    = [0:N-1];
x_n             = sin(2*pi*(f0/fs)*sample_index);

n               = 0:1/fs:interval-(1/fs)

stem(n, x_n);


X               = fftshift(fft(x_n, N));

df              = fs / N;
sampleIndex     = -N/2:N/2-1;
relative_f      = sampleIndex * df;

subplot(2, 1, 2);
stem(relative_f, abs(X));
xlabel('f (Hz)'); ylabel('|X(k)|');
axis([min(relative_f) max(relative_f)])

enter image description here

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.