Periodic product of sinusoids

Question

(This is problem P-3.7 from the book 'Signal processing first')

Let $x(t) = 2\cos(\omega_1t)\cos(\omega_2t) = \cos([\omega_1 + \omega2]t)+\cos([\omega_2 - \omega_1]t)$

where $0 < \omega_1 < \omega_2$

Then what relation needs to hold for $\omega_1 + \omega2, \ \omega_2 - \omega1 ,\ \omega_1$ and $\omega_2$ such that $x(t)$ is periodic with period $T_0$, e.g $x(t) = x(t + T_0)$?

I know that $$ \gcd(\omega_1+\omega_2, \omega_2-\omega_1) = 2\pi/T_0$$ By the definition of the fundamental frequency. But I don't know if this yields any relationship for $\omega_1$ and $\omega_2$.

Answer 1

Consider $t = 0$ to start from and expand $x(t+T_0)$ and we get :

$$x(0) = cos(0) + cos(0) = 2$$

$$x(T_0) = cos((w_2+w_1)T_0) + cos((w_2-w_1)T_0) = 2$$

But that requires :

$$(w_2+w_1)T_0 = 2K\pi$$ $$(w_2-w_1)T_0 = 2L\pi$$

for some K, L integers.

Thus

$$w_2T_0 = (K+L)\pi$$ $$w_1T_0 = (K-L)\pi$$

And if we substitute these forms back into our original expressions :

$$x(t+T_0) = cos((w_2+w_1)t)cos((w_2+w_1)T_0)$$ $$ - sin((w_2+w_1)t)sin((w_2+w_1)T_0)$$ $$ + cos((w_2-w_1)t)cos((w_2-w_1)T_0)$$ $$ + sin((w_2-w_1)T_0)sin((w_2-w_1)t)$$

$$x(t+T_0) = cos((w_2+w_1)t)cos(2K\pi)$$ $$ - sin((w_2+w_1)t)sin(2K\pi)$$ $$ + cos((w_2-w_1)t)cos(2L\pi)$$ $$ + sin((w_2-w_1)T_0)sin(2L\pi)$$

The only terms we are left with are :

$$x(t+T_0) = cos((w_2+w_1)t) + cos((w_2-w_1)t) = x(t)$$

Hence the relation for the periods is :

$$w_2T_0 = (K+L)\pi$$ $$w_1T_0 = (K-L)\pi$$

for some K, L integers.