How to find frequency of given signal?

fs=2400;
fo=50;
t=0:100;
xp=sin(2*pi*(49.98/fs)*t)+sin(2*pi*(50.01/fs)*t)+sin(2*pi*(49.99/fs)*t)+sin(2*pi*(50.01/fs)*t);

How to find frequency and rate of change of frequency?

[Star Strider]

Reference — Find period cycle and frequency for signal

Answer 1

[Walter Roberson]

The frequency is the reciprocal of the time needed for all of the components to first complete an integer number of cycles simultaneously:

format long g
fs = 2400

fs = 
        2400

time_for_one_cycle = lcm(lcm(lcm(lcm(4998, 5001), 4999), 5001), fs*100)

time_for_one_cycle = 
         1.66599993336e+15

frequency = 1/time_for_one_cycle

frequency = 
       6.0024012004802e-16

[49.98/fs, 50.01/fs, 49.99/fs, 50.01/fs] * time_for_one_cycle

ans = 1×4

1.0e+00 * 34694448612222 34715273611389 34701390278611 34715273611389

That last line shows that at that time, the different frequencies have indeed completed an integer number of cycles.

[Shanmuka]

Can we plot the correlation of the below two signals output in the same graph?

1)xp=sin(2*pi*(50/fs)*t)

2)xp=sin(2*pi*(49.98/fs)*t)+sin(2*pi*(50.01/fs)*t)+sin(2*pi*(49.99/fs)*t)+sin(2*pi*(50.01/fs)*t)

[Walter Roberson]

Yes, you can plot the correlation and the two signals in the same plot.

fs=2400;
fo=50;
t=0:100;
xp1=sin(2*pi*(50/fs)*t);
xp2=sin(2*pi*(49.98/fs)*t)+sin(2*pi*(50.01/fs)*t)+sin(2*pi*(49.99/fs)*t)+sin(2*pi*(50.01/fs)*t);
[r,lags] = xcorr(xp1,xp2);
plot(t, xp1, t, xp2);
hold on
stem(lags, r);
hold off
legend({'50', '50-ish', 'xcorr'});

You have to look hard to see the plots of the signals.

[Walter Roberson]

Also note that the units for the xcorr is "lag" (number of signal points) and the units for the other two lines is "seconds".

[Paul]

@Walter Roberson

I got a different answer (Reference)

fs = sym(2400);
syms t real
%fo=50;
%t=0:100;
T = fs./[49.98 50.01 49.99 50.01];
xp(t) = sum(sin(2*pi./T*t));

Ratios of T1/Tj

ratios = simplifyFraction(T(1)./T(2:end))

ratios = 

They are rational, so xp(t) is periodic

Get the LCM of the denominators

[~,den] = numden(ratios);
K = lcm(den);

The fundamental period of xp(t) is (i.e., smallest value of T that satisfies x(t) = x(t + T) )

Txp = K*T(1)

Txp = 

240000

Verify that xp(t) has period Txp

simplify(xp(t) - xp(t + Txp))

ans = 

0

Your solution is a period in that it satisfies the defintion of a periodic signal, but its not the fundamental period

time_for_one_cycle = lcm([4998, 5001, 4999, 5001,fs*100])

time_for_one_cycle = 

1665999933360000

simplify(xp(t) - xp(t + time_for_one_cycle))

ans = 

0

Answer 2

[Image Analyst]

What do you mean by frequency? That signal has several frequencies in it, not just one. From Fourier theory, any signal can be thought of as a sum of all frequencies. In general you can find out the relative power in all frequencies in a signal by using fft or pwelch

[Walter Roberson]

When frequency is considered to be the number of complete cycles per second, then you need to figure out the length of a complete cycle. Which I do in my Answer... it's pretty long. And the answer is different than the average number of pairs of zero crossings per second.