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Why during demodulation, the demodulated signal might double its frequency? A minimal example:

from scipy.signal import hilbert, periodogram
import numpy as np
import matplotlib.pyplot as plt

fs = 20000 t = np.arange(0, 5, 1 / fs) sig = np.cos(t * 2 * np.pi * 7) carrier = np.cos(t * 2 * np.pi * 1000) measured = sig * carrier

f, pxx = periodogram(np.abs(hilbert(measured)), fs, detrend=False) plt.plot(f, pxx) plt.xlim([-0.2, 30]) plt.grid() plt.show()

enter image description here

Gideon Genadi Kogan
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1 Answers1

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Assume that we have a signal $$x(t)=\cos(\omega t)$$ with a carrier $$c(t) = \cos(\omega_c t).$$ Than, for the amplitude modulated signal, $$y(t)=x(t)c(t),$$ the analytic signal is $$A(y(t))=x(t)e^{-j\omega_c t}.$$ Here, $x(t)$ can be extracted by $$\left|A(y(t))\right|=\left|x(t)\right|$$.

Now, for our case, following this derivation: $$\left|x(t)\right|=\left|\cos(\omega t)\right|=\frac{2}{\pi }+\frac{4}{\pi } \sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2m \omega t)$$

Here, for $m=1$, the frequency is double the original frequency.

Other derivation could be based on $$\left|\cos(\omega t)\right|=\cos(\omega t)\cdot\text{sgn}\left[\cos(\omega t)\right]$$

Gideon Genadi Kogan
  • 1,156
  • 5
  • 17