Periodic Functions

14 May 2023

Background

We say $f(t)$ is $T$-periodic if $f(t+T) = f(t) \quad \forall t \in \mathbb{R}$

An interesting fact is that if $T$ is irrational, then $f$ must be a constant function. Thus, you can restrict $T \in \mathbb{N}$ for all non-trivial cases. However, this will not affect our proof.

The Laplace Transform Proof

Given a $T$-periodic function $f(t)$

$$\begin{align} \mathcal{L}\{ f(t) \} &= \int_{0}^{\infty} f(t) e^{-st} \ dt \\[10pt] &= \int_{0}^{T} f(t) e^{-st} \ dt + \int_{T}^{\infty} f(t) e^{-st} \ dt \\[10pt] &\text{let } \tau = t - T \implies dt = d\tau \\[10pt] &= \int_{0}^{T} f(t) e^{-st} \ dt + \int_{0}^{\infty} f(\tau+T) e^{-s(\tau+T)} \ d\tau \\[10pt] &= \int_{0}^{T} f(t) e^{-st} \ dt + e^{-sT}\int_{0}^{\infty} f(\tau) e^{-s\tau} \ d\tau \\[10pt] &= \int_{0}^{T} f(t) e^{-st} \ dt + e^{-sT}\mathcal{L}\{ f(t) \} \end{align}$$

Now, we do a little algebra to isolate $\mathcal{L}{ f(t) }$, and we get

$$\mathcal{L}\{ f(t) \} = \frac{1}{1-e^{-sT}} \int_{0}^{T} f(t) e^{-st} \ dt$$