Analytical signal and Hilbert Transform

For a real time series x(t), its analytic signal x(t) is defined as

x(t) = x(t) – iH[x(t)]

Let us consider an example of a monochromatic signal 𝑥(𝑡) = 5 sin(10𝑡 + 3).

Figure 1

Now, let us consider a more complex function

x(t) = 1*sin(2𝜋10𝑡 + 0.3) + 2*sin(2𝜋20𝑡 + 0.2) + 3*sin(2𝜋30𝑡 + 0.4).

Figure 2

We can clearly observe that the Hilbert transform estimates the instantaneous frequency of a signal for monocomponent signals only.

Matlab Codes:

clear; close all; clc

fs = 1e4;

t = 0:1/fs:1;

f1=10;

f2=20;

f3=30;

x=1*sin(2*pi*f1*t+0.3)+2*sin(2*pi*f2*t+0.2)+3*sin(2*pi*f3*t+0.4);

%x=5*sin(2*pi*10*t+3);

y = hilbert(x);

figure(1)

plot(t,real(y),t,imag(y))

%xlim([0.01 0.03])

legend('real','imaginary')

title('Hilbert Function')

figure(2)

subplot(3,1,1)

env=abs(y);

plot(t,x)

xlabel('Time')

title('Envelope')

hold on

plot(t,env)

legend('original','envelope')

subplot(312)

instph=fs/(2*pi)*unwrap(angle(y));

plot(t,instph)

xlabel('Time')

ylabel('Phase (in rad)')

grid on

title('Instantaneous Phase')

subplot(313)

instfreq = fs/(2*pi)*diff(unwrap(angle(y)));

plot(t(2:end),instfreq)

xlabel('Time')

ylabel('Hz')

grid on

title('Instantaneous Frequency')

Application of Hilbert Transform:

Figure 3

Aliasing and Nyquist Condition

Let us consider a simple sinusoidal signal

x(t) = 5sin(10t + 3)

This is a signal with an amplitude of 5, the angular frequency of 10 and phase of 3. The Nyquist condition demands that the angular frequency, mod(𝜔) ≤ pi/delta t.

For the above figure 1, we took 𝜔 to be 10 and pi/delta t is 99.5, so the Nyquist condition satisfies and we do not see the aliasing. Here the sampling rate, Δ𝑡 is 0.0316.

Let us see what happens when we take the pi/delta t to be less than 𝜔.

In this case, we took the sampling rate, Δ𝑡 to be 0.3307. Here the pi/delta t is 9.5000 which is less than 𝜔. Hence the Nyquist condition is not satisfied and there is a clear aliasing. The sinusoidal signal does not appear to be sinusoidal.

Figures