Is $x(t) = \sin(3t) + \cos\left({2\over3}t\right) + \cos(\pi t)$ periodic?

Question

I have the function $x(t) = \sin(3t) + \cos\left({2\over3}t\right) + \cos(\pi t)$ and I need to examine whether the function is periodic and, if so, find its period.

What I have done:

$ \begin{align} x(t) & = \sin(3t) + \cos\left({2\over3}t\right) + \cos(\pi t)\\ & = \sin\left(2\pi {{3}\over{2\pi }}t\right) + \cos\left(2\pi{{1}\over{3\pi}}t\right) + \cos\left(2\pi{{1}\over{2}}t\right)\Rightarrow T_1 = {{2π}\over{3}},\ T_2 = 3π,\ T_3 = 2. \end{align} $

Where I'm stuck:

$ \begin{align} T = aT_1 = bT_2 = cT_3 = {{{2πa}\over{3}}} = 3πb = 2c \end{align}. $

Is $x(t)$ actually periodic? Since it all its components are, it should, but I'm not sure how to find the actual value of $T$.

Answer 1

$x(t)$ is not periodic, despite being comprised by three periodic, sine waves, because there are no 3 integers that fulfil the following equation $ T = aT_1 = bT_2 = cT_3$:

$ T = aT_1 = bT_2 = cT_3 = {{{2πa}\over{3}}} = 3πb = 2c. \\ {{{2πa}\over{3}}} = 3πb\Leftrightarrow 2πa = 9πb\Leftrightarrow {a\over{b}} = {9\over2}\Rightarrow (a, b) = (9, 2). \\ 3πb = 2c\Leftrightarrow c = 3π\Rightarrow (a, b, c) = (9, 2, 3π). $

Based on the value $c = 3π$, it is apparent that $x(t)$ is not periodic, because $π$ is an irrational number and therefore a Least Common Multiple of $9$, $2$ and $3π$ is not defined.