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

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 $$T$$.

• Assume $x(t+T)-x(t)=0$ for all $t$. Take derivatives of orders $4$ and $8$. You get a system $Av=0$ where $A$ is a matrix with constant coefficients and $v=(\sin(3t+3T)-\sin(t), \cos(2t/3+2T/3)-\cos(2t/3), \cos(\pi(t+T))-\cos(t))^T$. Since $A$ is non-singular you must have that $v=0$ for all $t$. Therefore $T$ would have to be a period of $\sin(3t),\cos(2t/3)$ and $\cos(\pi t)$ simultaneously. Commented Jan 16, 2018 at 1:25
• It can’t be periodic, since $\pi$ is irrational. Commented Jan 16, 2018 at 1:26

$x(t)$ 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.