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The fundamental period of a signal is the smallest period over which the entire signal repeats. In other words, it's the smallest $T_0$ such that $f(t + T_0) = f(t)$ for all $t$.

The fundamental frequency is $\omega_0 = {1 \over T_0}$. For a signal that contains multiple frequencies, $\omega_1, \omega_2, \ldots, \omega_N$, the fundamental frequency $\omega_0$ is the highest frequency shared by all of the multiple frequencies. In other words, it's the greatest common factor of $\omega_1, \omega_2, \ldots, \omega_N$.

Note that the fundamental frequency is the highest frequency shared by the constituent frequencies. It's not the highest frequency that is present.

All frequencies present in the signal are an integer multiple of the fundamental frequency.

The fundamental wavenumber is $k_0 = 2 \pi \omega_0 = {2 \pi \over T_0}$.

Example

If $f(t)$ has a bounded domain and is, for example, defined on $[0,n-1]$, we can make it periodic with fundamental period $n$ by defining $f(t+n) = f(t)$. So the signal repeats every $n$ units and has a fundamental frequency of $1 \over n$.

Then all frequencies in the signal will be non-negative integer multiples of $1 \over n$: $${0 \over n}, {1 \over n}, {2 \over n}, {3 \over n}, \cdots$$

The complex Fourier transform is defined on both positive and negative wavenumbers (and, hence, frequencies), so we consider there to be "negative frequencies" present, too: $$\cdots, {-3 \over n}, {-2 \over n}, {-1 \over n}, {0 \over n}, {1 \over n}, {2 \over n}, {3 \over n}, \cdots$$

where the negative frequencies can be thought of as oscillating in the opposite direction. Recall that the positive and negative complex Fourier coefficients form conjugate pairs, each containing half of the energy, so that when added they yield all of the energy as a real value, with no imaginary value.