Turn aperiodic function to periodic

Question

I got a quick question regarding aperiodic functions. Let's suppose I have an aperiodic function $$ f(x) = \left\{ \begin{array}{l l} \exp(-t) & \quad -2\leq t \leq2,\\ 0 & \quad \text{anywhere else,} \end{array} \right. $$ and I want this to become periodic with period $\ T=4 $. What are the steps involved in such a conversion? I know that $\ f(x+nT) = f(x) $ must hold in resulting function but I am really stuck at the moment. Thank you in advance!

Answer 1

You could represent the function as a Fourier series:

$$g(x) = \sum_{n=-\infty}^{\infty} c_n \, e^{i n \pi x/2}$$

where

$$\begin{align}c_n &= \frac1{4} \int_{-2}^2 dt\, e^{-t} \, e^{-i n \pi t/2}\\ &= \frac1{4} \int_{-2}^2 dt \, e^{-(1 + i n \pi/2) t}\\ &= \frac{(-1)^n}{4(1 + i n \pi/2)}\left ( e^2-\frac1{e^2}\right ) \end{align}$$

Answer 2

You don't really "convert" it to a periodic function. You produce a periodic function that agrees with $f$ on a certain interval. Perhaps that interval should be $[-2,2)$. So if $g$ is the periodic function, you want $g(t + 4 n) = f(t) = \exp(-t)$ for $-2 \le t < 2$ and integer $n$. You could take $g(x) = f(h(x))$ where $h$ is a "sawtooth" function, $h(x) = t$ where $-2 \le t < 2$ and $(x-t)/4$ is an integer. Except at the discontinuities, you could take $h(x) = \dfrac{\pi}{4} \arctan(\tan(\pi x/4))$.