how to Find the period of complex exponential function?

Question

How we can find period of this sequence?

$x[n]= e^{jn2π/3}$

Is it equal to $T=2π/(2π/3) $ or not?

I mean relation $T=2π/\omega$ will be valid in this case or not?

Answer 1

$e^{jn2\pi / 3} = cos(2n\pi/3) + i\cdot sin(2 n \pi/3)$ From there, it is easy to see that the period is $\frac{2 \pi}{2\pi/3} = 3$

Answer 2

Yes it is, as the complex exponent have the properties: $$ e^{z_1+z_2}=e^{z_1}e^{z_2},\text{ and }e^{2\pi i}=1. $$

Consider now the function: $$ f(x)=e^{i2\pi\frac xT}, $$ where $T\in\mathbb R$.

Can you find the relation between $f(x)$ and $f(x+T)$? Can you link the result to your particular function?

PS. I have used $i$ for $\sqrt{-1}$.

Answer 3

The sequence $x[n]=e^{jn2\pi/3}$ is actually a discrete complex exponential sequence. One thing to keep in mind is that the values of $n$ will always be an integer and never a fraction. To find the period of a discrete sequence we can use the fact that, $$x[n+N]=x[n]$$ where $N \gt 0$, and the $N$ is the least positive integer for which the sequence repeats itself. That is to say, since $x[n]$ is periodic with period $N$ it repeats itself after $N$ samples. And it is also periodic with the period $2N$,$3N$,$4N$, and so on so we choose the smallest possible positive integer $N$. So: $$e^{j(n+N)2\pi/3}=e^{jn2\pi/3}$$ Solving you will get,$$e^{jN2\pi/3}=1$$ $$\cos(\frac{2N\pi}{3})+ j\sin(\frac{2N\pi}{3})=1+j0$$

$$\cos(\frac{2N\pi}{3}) =1=\cos 0,\cos2\pi,\cos4\pi...$$ Or, $$\sin(\frac{2N\pi}{3}) =0=\sin 0,\sin2\pi,\sin4\pi...$$ Both are same. We will choose the value $2\pi$ because the period should be positive and least. So, $$\frac{2N\pi}{3}=2\pi$$ giving you $$N=3$$ So the period is $N$, now you can also see why the formula $$T=\frac{2\pi}{\omega}$$ actually works.Graphically this sequence can be plotted as a continuous sinusoid sampled at 1 second, Hope this helps...