# Discrete signal testing for periodicity

How would one go about determining if the following discrete time signal x[n] is periodic, and if it is, determine its fundamental period?

I understand that the period for the second exponential term is 6, but apart from that I am unable to further my calculation.

The answer to the question above is:

but I fail to understand how the equation highlighted in yellow comes to be. Any help with this matter would be appreciated.

A discrete signal is periodic only if it's normalized frequency can be expressed as a rational number. In your case: $$$$e^{-2}e^{j(n\frac{\pi}{3} - \frac{\pi}{4})}$$$$ so $$$$\begin{array}{rcl} 2 \pi f & = & \frac{\pi}{3} \\ f & = & \frac{1}{6} \end{array}$$$$ Since $$1/6$$ is rational, it's periodic. Now, you reduce $$f$$ to its canonical form and apply the definition of $$f$$: $$$$\begin{array}{rcl} f & = & \frac{k}{N} \\ \frac{1}{6} & = & \frac{k}{N} \\ N & = & 6 \end{array}$$$$
The signal has a fundamental period of $$N=6$$.
Solve the same, however, for $$$$e^{-2}e^{j(n\frac{1}{3} - \frac{\pi}{4})}$$$$ you'll get $$$$\begin{array}{rcl} 2 \pi f & = & \frac{1}{3} \\ f & = & \frac{1}{\pi 6} \end{array}$$$$ which cannot be expressed as a rational number, hence not periodic.
it is more clear if you utilize the Euler's formula \begin{align} x[n+N] &= e^{-2}e^{j(\frac{\pi}{3}(n+N)-\frac{\pi}{4})} \\ &= e^{-2}\Big[ \cos(\frac{\pi}{3}n+\frac{\pi}{3}N-\frac{\pi}{4}) -j \sin(\frac{\pi}{3}n+\frac{\pi}{3}N-\frac{\pi}{4}) \Big] \end{align} Using trigonometric identities ( I will leave it as an exercise to you), $$x[n]=x[n+N]$$ holds if N=6 which is the smallest value that makes the equality to hold. Indeed, the signal is periodic with a fundamental period $$N_o=6$$.