# How to find the delta cycles (change in the number of cycles) in two sine signals with nearly identical frequencies?

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I have these two signals generated from the following code;

clc

clear all

close all

%%

f = 4000;

t = 0:1e-5:50e-3;

sig1 = sin(2*pi*(f)*t);

sig2 = sin(2*pi*(f+50)*t);

noiseAmplitude = 0.05;

sig1_with_noise = sig1 + noiseAmplitude * randn(1, length(sig1));

sig2_with_noise = sig2 + noiseAmplitude * randn(1, length(sig2));

%%

figure

plot(t,sig1_with_noise)

hold on

plot(t,sig2_with_noise)

ylim([-1.1 1.1]);

These are two sinusoids (with a small noise) with nearly identical frequencies. Their frequencies differ by 50 Hz as you can see in the above code. But since these are not perfectly identical, one of the waveforms overtakes the other as a function of time, meaning the number of cycles over time from one signal will be greater than the other one. I want to find the delta_cycles Vs time plot keeping the sig_1 waveform as the reference. Currently, I can just tell by looking that the delta_cycles will be ~ 2.5 cycles over the time duration that showed in the top figure. But I want to plot the delta_cycles Vs time. Should I just take the difference? Or should I use the derivative to see the change? Thanks in advance.

Plot:

##### 3 Comments

Matt Gaidica
on 17 Dec 2020

### Accepted Answer

VBBV
on 17 Dec 2020

%ue

plot(t,(sig1_with_noise-sig2_with_noise)./(2*pi))

As you said, you can take the difference between two signals and divide by 2*pi to get the delta number of cycles and plot it.

##### 4 Comments

VBBV
on 17 Dec 2020

Edited: VBBV
on 17 Dec 2020

### More Answers (1)

David Goodmanson
on 18 Dec 2020

Edited: David Goodmanson
on 19 Dec 2020

Hi Jay,

Using the hilbert transform on the signal gives the so-called analytic signal. The transform creates an imaginary part and adds it to the original signal (the real part) to gilve a complex signal of the form

const*exp(2*pi*i*f*t)

as in fig(1).

[NOTE that the Matlab definition of the hilbert transform is NONstandard. The actual hilbert transform gives just the red waveform in fig 1. Mathworks adds in the original signal (blue) as well, to create the full analytic signal ]

Finding the angle of the analytic signal, and using unwrap, shows an angle that increases linearly with with t as in fig(2). As you can see the higher freq signal has the steeper slope. Taking the ratio of the two analytic signals gives the relative phase as in fig(3). Since the difference frequency is 50 Hz and the total time is 50 msec, you would expect the accumulated phase difference to be

2*pi*50*50e-3 = 15.7 radians

which is what happens.

The code works with f = 4000, but I dropped f to 1000 to make things easier to see.

f = 1000;

t = 0:1e-5:50e-3;

sig1 = sin(2*pi*(f)*t);

sig2 = sin(2*pi*(f+50)*t);

noiseAmplitude = 0.0;

sig1_with_noise = sig1 + noiseAmplitude * randn(1, length(sig1));

sig2_with_noise = sig2 + noiseAmplitude * randn(1, length(sig2));

s1ana = hilbert(sig1_with_noise); % analytic signal

s2ana = hilbert(sig2_with_noise); % analytic signal

figure(1)

plot(t,real(s1ana),t,imag(s1ana))

xlim([0 10e-3])

grid on

angle1 = unwrap(angle(s1ana));

angle2 = unwrap(angle(s2ana));

figure(2)

plot(t,angle1,t,angle2)

legend(['sigl'; 'sig2'],'location','east')

anglerel = unwrap(angle(s2ana./s1ana));

figure(3)

plot(t,anglerel)

##### 0 Comments

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