Periodic signal

A signal $g(t)$ is periodic if there is some positive $T$ such that $g(t+T)=g(t)$ for every $t$. The smallest such $T$ is the fundamental period $T_0$, and the corresponding fundamental frequency is $f_0=1/T_0$.

A single sinusoid $A\cos (2\pi f_{0}t+\theta )$ has fundamental period $1/f_0$.

The period of a sum of sinusoids

A sum of two sinusoids is periodic if and only if the ratio of their individual periods is rational. When it is periodic, the fundamental period of the sum is the least common multiple of the individual periods.

Take $g(t)=10\sin (12\pi t)+4\cos (18\pi t)$. The first sinusoid has period $T_1=2\pi /(12\pi )=1/6\text{s}$; the second has $T_2=2\pi /(18\pi )=1/9\text{s}$. Both periods are rational, so the sum is periodic. Find the smallest $T_0$ such that $T_0/T_1$ and $T_0/T_2$ are both integers: $T_0=1/3$ works ($1/3÷1/6=2$, $1/3÷1/9=3$), and it’s the smallest. So $T_0=1/3\text{s}$ and $f_0=3\text{Hz}$.

Now $g(t)=10\sin (12\pi t)+4\cos (18t)$ — note the missing $\pi$ in the second term. First period $1/6$; second period $2\pi /18=\pi /9$, which is irrational. The ratio of the two periods is irrational, so no finite $T$ works, and the sum is aperiodic.

Sanity check: if your sum involves a sinusoid with $\pi$ in the angular frequency and another without, the sum is almost certainly aperiodic.

Edge cases

A constant signal $g(t)=1$ is technically periodic with any period, so it has no fundamental period in the strict sense. The complex exponential $e^{-j60\pi t}$ is periodic with period $2\pi /(60\pi )=1/30$. A signal like $3+t^2$ has no period — it grows without bound.

A non-oscillating signal that grows or decays cannot be periodic. The real exponential is aperiodic.

Why periodicity matters

Periodic signals have Fourier series — discrete spectra concentrated at integer multiples of the fundamental frequency. Aperiodic signals have continuous spectra given by the Fourier transform. The dividing line at periodicity determines which tool you reach for.