determine whether or not the following signals are periodic

clc
clear all
close all
t= -4 : 0.005 : 4;
a=1;
%FOR X(T)
subplot(2,1,1)
c1=2*pi;
c2=4*pi;
x2=a*cos(c1.*t);
x1=a*cos(c2.*t);
x=x1+x2
ac=xcorr(x,x);
[~,locs]=findpeaks(ac);
sol=mean(diff(locs)*0.1)
plot(t,x)
xlabel('time->')
ylabel('x(t)')
title('Plot For x(t)')
grid on
gtext('1841014009')
text(-1,-0.5, ['Period Of The Signal =', num2str(sol)])
%FOR Y(T)
subplot(2,1,2)
c1=2*pi;
c2=2;
y2=a*cos(c1.*t);
y1=a*cos(c2.*t);
y=y1+y2
ac=xcorr(y,y);
[~,locs]=findpeaks(ac);
sol=mean(diff(locs)*0.1)
plot(t,y)
xlabel('time->')
ylabel('y(t)')
title('Plot For y(t)')
grid on
gtext('1841014009')
text(-1,-0.5, ['Period Of The Signal =', num2str(sol)])

This is what i have done.

But i don't how to determine whether or not the following signals are periodic in MATLAB

[ytzhak goussha]

Hello,

the video in the attached link explains this well.

link

link2

In short, a function is pereodic if its fundemental periods (T0) is a rational number.

The T01 of cos(c1*t) = 2pi/c1 = 2pi/2pi = 1

The T02 of cos(c2*t) = 2pi/c2 = 2pi/4pi = 0.5

The T0 of x(t) is the product of the fundemental periods of its components i.e. T01/T02 = 1/0.5 = 2/1

This is a rational number meaning that x(t) is perodic with a period of 1*1 = 0.5/2 = 1

[Suraj Narayan]

Thank you.

Now I know where I'm doing wrong.

Thanks again, it's a great help.

[Suraj Narayan]

I have another doubt, can we use

r = rat(T01,T02)

to find whether it is a rational number or not.

[Walter Roberson]

NO.

rat() takes two parameters, the first of which is the number to convert, and the second is the tolerance. It cannot take two numbers with the intention of checking whether they are co-rational.

In IEEE 754 floating point, every representable finite number is rational. For example, pi() is approximated by 884279719003555 / 2^48 . You will not be able to use rat() to figure out whether a floating point number is rational.

[Suraj Narayan]

Thank again it's agreat help.

[Walter Roberson]

A signal f(t) is periodic if there is a time-independent delay time, dt, such that for every given time t, f(t) = f(t+dt) -- a time at which it is back at exactly the same state. Because f(t) = f(t+dt) then it also follows that f(t) = f(t+2*dt), f(t) = f(t+3*dt) and so on. The minimum interval dt for which the situation holds, is the "period" of the signal.

This definition can sometimes be more useful than the ratio of the periods. In particular for cos(f*t), there is an explicit formula:

>> expand(cos(f*t+dt))
cos(dt*f)*cos(f*t) - sin(dt*f)*sin(f*t)

And if we let dt*f be a multiple of 2*pi then cos(dt*f) would be 1 and sin(dt*f) would be 0, and the result would be the same as cos(f*t) no matter what t is, so this would be periodic.

It can be worth spending a few minutes to determine whether there is a smaller dt than dt = 2*pi/f .

For signals at two different frequencies, they will both be back at their original state at lcm() [Least Common Multiple] of the two dt values.

Answer 1

[Walter Roberson]

plot(t,x)

That should be plot(t,y) the second time.

sol=mean(diff(locs)*0.1)

why 0.1 ?

Finding the average difference between peaks in the correlation matrix does not tell you whether the signal is periodic.

Consider for example a square wave signal that pulses on for one unit, and each time the distance between pulses gets one unit further apart:

|^|_|^|__|^|___|^|____|^|_____|^|

The correlation measurement will have peaks that are not a constant distance apart, but despite not being constant distance, the mean of the distance is going to be some number. Like mean([1,3,6,10,15,21]) -> 9+1/3 . mean() of a series of values is always defined unless the values are empty or nan -- but the fact it exists does not tell you that the signal is periodic.

[Suraj Narayan]

Thank you

Yes, that should be plot(t,y).

Thanks for letting me know, And for why? 0.1 is wrong.

I'm a beginner to MATLAB, so there are a lot of mistakes.

As for where I got that idea, please follow the link