CODEWITHSUNDEEP
pythonnumpydft
clearseplex
clearseplex

Reputation: 729

Discrete Fourier Transform Unexpected Peak

I am implementing a DFT and transformed a sine wave (f = 440 Hz) as test. However there is a peak in the transformed signal where I don't expect one (see figure below). Can someone see why it is there?

My code:

import numpy as np
import matplotlib.pyplot as plt

t = 0.1
freq = 440
Fs = 2000
T = 1/Fs
N = int(Fs * t)

# signal information                 
omega = 2 * np.pi * freq       # angular frequency of the sine wave
t_vec = np.arange(N) * T     # time vector for plotting
y = np.sin(omega * t_vec)  # sine input signal

def DFT(y, N):
    dft = []
    angle = (-1 * 1j * 2 * np.pi) / (N)
    for k in range(int(N)):
        X = 0
        for n in range(int(N)):
            X += y[n] * np.exp(angle * k * n)
        dft.append(X)
    return dft

freq_vec = Fs * np.arange((N))/N

dft_signal = np.abs(DFT(y = y, N = N))

#Plots
fig, (ax1, ax2) = plt.subplots(2)
ax1.plot(t_vec, y, label='$y = numbers')
ax1.set(ylabel='Amplitude', xlabel='Time [s]', title = 'Input Signal')
ax2.plot(freq_vec, dft_signal, linewidth=5)
ax2.set(yscale='log', xscale='log', ylabel='Amplitude', xlabel='Frequency [Hz]', title = 'DFT of Input Signal')
plt.tight_layout() 
plt.show()

enter image description here

Upvotes: 0

Views: 214

Answers (1)

Romain F
Romain F

Reputation: 407

Your code seems good to me. What you see are simply the negative frequencies since the Fourier Transform of a sine function is made of a peak in the positive frequencies and one peak in the negative. Don't plot it in xlog scale and it will appear more obvious! Then it is just a matter of fftshift to put the zero frequency in the middle or at the beginning of the array (like your implementation).

Remember that if your frequency range is Fs your DFT only goes from -Fs/2 to +Fs/2

Upvotes: 1

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