Instantaneous Frequency

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• Author: Diego Inácio
• GitHub: github.com/diegoinacio
• Notebook: instantaneous_frequency_FM.ipynb

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Analytical approach to continuous Instantaneous Frequency and Frequency Modulation.

%matplotlib inline
import matplotlib.pyplot as plt
from IPython.display import Audio
import numpy as np

from _utils import *

fs = 44100   # sampling rate 44.1 kHz
T = 3        # duration in seconds
t = np.linspace(0, T, fs*T)

Instantaneous frequency

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A modulated signal can be expressed as: $$ \large x(t)=a(t)\cos(\phi(t)) $$ where:

• The instantaneous amplitude or envelope is given by $\large a(t)$;
• The instantaneous phase is given by $\large \phi(t)$;
• The instantaneous angular frequency is given by $\large \omega(t)=\frac{d}{dt}\phi(t)$;
• The instantaneous ordinary frequency is given by $\large f(t)=\frac{1}{2\pi}\frac{d}{dt}\phi(t)$.

Examples

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Modulation using instantaneous frequency

Linear frequency modulation

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Given a modulation frequency, defined by:

$$ \large f(t)=f_a+\frac{(f_b-f_a)t}{T} $$

where $f(t)$ is linearly interpolated from $f_a$ to $f_b$.

fa = 880      # frequency A
fb = 100      # frequency B

# Linear interpolation from fa to fb
ft = fa + (fb - fa)*t/T

audiovis(ft)

audio mono

The modulated signal whithout instantaneous frequency is:

$$ \large x(t)=sin(2\pi f(t)t) $$

In this example the carrier frequency is zero ($\phi_c=0$).

xt = np.sin(2*np.pi*ft*t)
Audio(xt, rate=fs)

spectrogram(xt, flim=[0, 1000])

audio mono

If we considere:

$$ \large \phi(t)=\int\omega(t)dt=2\pi\int f(t)dt $$

the modulated signal by the instantaneous frequency could be expressed by:

$$ \large x_m(t)=sin(\phi(t))=sin\left(2\pi\left[f_at+\frac{(f_b-f_a)t^2}{2T}\right]\right) $$

xm = np.sin(2*np.pi*(fa*t + (fb - fa)*t**2/(2*T)))
Audio(xm, rate=fs)

spectrogram(xm, flim=[0, 1000])

audio mono

Exponencial frequency modulation

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$$ \large f(t)=f_a+\frac{f_b-f_a}{1+e^{-k\left(\frac{t}{T}-\frac{1}{2}\right)}} $$

$f(t)$ is a mudulation frequency which interpoles from $f_a$ to $f_b$ exponentially. It's inspired by the sigmoid function where $k$ is the slope of the middle point.

fa = 880      # frequency A
fb = 100      # frequency B
k = 12        # slope of the middle point

e_k = np.exp(-k*(t/T - 1/2))

# Exponential interpolation
ft = fa + (fb - fa)/(1 + e_k)

audiovis(ft)

audio mono

xt = np.sin(2*np.pi*ft*t)
Audio(xt, rate=fs)

spectrogram(xt, flim=[0, 1200])

audio mono

The modulated signal by the instantaneous frequency could be expressed by:

$$ \large x_m(t)=sin(\phi(t))=sin\left(2\pi\left[f_at-\frac{(f_b-f_a)T}{k}\left[\ln(e^{-k\left(\frac{t}{T}-\frac{1}{2}\right)}) - \ln(e^{-k\left(\frac{t}{T}-\frac{1}{2}\right)}+1)\right]\right]\right) $$

xm = np.sin(2*np.pi*(fa*t - (fb - fa)*T/k*(np.log(e_k) - np.log(e_k + 1))))
Audio(xm, rate=fs)

spectrogram(xm, flim=[0, 1000])

audio mono

Sinusoidal frequency modulation

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$$ \large f(t)=f_a+\frac{f_b-f_a}{2}[sin(2\pi f_m t) + 1] $$

$f(t)$ is a mudulation frequency which oscillates from $f_a$ to $f_b$ given a sinusoidal function where $f_m$ is the ordinary froquency.

fa = 110      # frequency A
fb = 440      # frequency B
fm = 1        # oscillator's ordinary frequency

# frequency oscillator
ft = fa + (fb - fa)/2*(np.sin(2*np.pi*fm*t) + 1)

summary(ft)

       min:        110.0000
  1st Quar:        158.3288
    median:        275.0000
      mean:        275.0000
  3rd Quar:        391.6712
       max:        440.0000
     sigma:        116.6722

audiovis(ft)

audio mono

xt = np.sin(2*np.pi*ft*t)
Audio(xt, rate=fs)

spectrogram(xt, flim=[0, 3000])

audio mono

The modulated signal by the instantaneous frequency could be expressed by:

$$ \large x_m(t)=sin(\phi(t))=sin\left(2\pi\left[f_at+\frac{f_b-f_a}{2}\left[-\frac{cos(2\pi f_m t)}{2\pi f_m}+t\right]\right]\right) $$

xm = np.sin(2*np.pi*(fa*t + (fb - fa)/2*(-np.cos(2*np.pi*fm*t)/(2*np.pi*fm) + t)))
Audio(xm, rate=fs)

spectrogram(xm, flim=[0, 660])

audio mono