Finding the fundamental frequency of a periodic signal

Question

Suppose we have the signal $$x(t) = e^{j\omega_1 t} + e^{j\omega_2 t} + e^{j\omega_3 t},$$ where all the frequencies are rationally related (that is, the ratio of any pair of frequencies is a rational number).

• How do I prove the fact that the fundamental frequency of the expression above is given by $\omega_0 = \gcd(\omega_1, \omega_2, \omega_3)$?

  My attempt is as follows: assume that the signal has a fundamental period $T_0$ which is related to the fundamental frequency by $\omega_0 = 2\pi/T_0$. Now for the signal to be periodic with this period as the fundamental period the following equations must be satisfied $$\omega_1 T_0 = 2 \pi l$$ $$\omega_2 T_0 = 2\pi m$$ $$\omega_3 T_0 = 2\pi n,$$ where $l, m, n \in \mathbb{Z}$ and their values are such that $T_0$ is minimized.

  • Where do I go from here?
  • How do I show that the fundamental frequency $\omega_0 = 2\pi/T_0$ is the greatest common divisor of $\omega_1$, $\omega_2$, and $\omega_3$?

Answer 1

1) You are almost there if you have found an $ l,m,n$ and a $T_0$ that fit your equations. Let $p$ be the largest common factor of $ l,m,n$. If $p=1$ you are done, otherwise divide $ l,m,n$ and $T_0$ by $p$ and now you are done.

2) From your own equations:

$$ \omega_1 = \omega_0 \cdot l $$ $$ \omega_2 = \omega_0 \cdot m $$ $$ \omega_3 = \omega_0 \cdot n $$

You already know that $l,m,n$ don't have a common factor greater than one.

Hope this helps.

Ced

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Followup:

Taking it from the top. You are supposing that you know the frequency ratios:

$$ \frac{\omega_1}{\omega_2} = \frac{a}{b} $$ $$ \frac{\omega_2}{\omega_3} = \frac{c}{d} $$

So $a,b,c,d$ are known. Therefore:

$$ \frac{l}{m} = \frac{a}{b} $$ $$ \frac{m}{n} = \frac{c}{d} $$

Cross multiply:

$$ lb = ma $$ $$ md = nc $$

Let $ m = bc $ and therefore $ l = ac $ and $ n = bd $

You now have an initial $l,m,n$ set that you can possibly reduce as stated above. Then the fundamental frequency, $ \omega_0 $, can be solved for using any of the three values.

$$ \omega_0\ = \omega_1 / l = \omega_2 / m = \omega_3 / n $$

Finally, $T_0 = 2\pi / \omega_0 $ gives you the period.

Quite Easily Done

Note, it is possible that none of $l,m,n$ are equal to one.

Answer 2

you show that $$x(t+T_0) = x(t)$$ for all $-\infty<t<+\infty$ and that $T_0$ is the smallest such period that this is the case.