Properties of Periodic Functions

Posted on 2024-05-02 Edited on 2025-06-17 In Mathematics Views: 30

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Periodic Functions

A periodic function $x (t)$ ¹ with fundamental period² $T_{0}$ has the property $f (x) = f (x + T_{0}) for all x .$

A periodic function takes the same values at intervals of $T_{0}$ .

The area under a periodic signal $x (t)$ over any interval of duration $T_{0}$ is the same; that is, for any real numbers $a$ and $b$ $\int_{a}^{a + T_{0}} f (x) d x = \int_{b}^{b + T_{0}} f (x) d x .$

This result follows from the fact that a periodic function takes the same values at intervals of $T_{0}$ . Hence, the values over any segment of duration $T_{0}$ are repeated in any other interval of the same duration. For convenience, the area under $x (t)$ over any interval of duration $T_{0}$ will be denoted by $\int_{T_{0}} f (x) d x$

The proof is simple.

First step: We need to prove that, for any contant $a$ , $\int_{a}^{a + t_{0}} f (x) d x = \int_{0}^{T_{0}} f (x) d x .$
Then, we can substitude $b$ into a. The result is: $\int_{b}^{b + T_{0}} f (x) d x = \int_{0}^{T_{0}} f (x) d x = \int_{a}^{a + T_{0}} f (x) d x .$ Q.E.D.

Now we prove step 1 (-->Source):

For simplicity, here we use period $T$ instead of minimal period $T_{0}$ . Let $a \in [0, T]$ and $N = ⌊ \frac{a}{T} ⌋$ , such that $0 \leq a - N T < T$ , we then do the variable change $u = x - N T, f$ being periodic then $f (u) = f (x)$ and $\int_{a}^{a + T} f (x) d x = \int_{a - N T}^{a - N T + T} f (u) d u$

Now, we know that $0 \leq a - N T \leq T \leq a - N T + T$ . So, $a - N T + T$ is greater than $T$ , thus we need to cut the "excedent", shift it again, and restick it to your main integral : $\int_{a - N T}^{a - N T + T} f (x) d x = \int_{a - N T}^{T} f (x) d x + \int_{T}^{a - N T + T} f (x) d x$

Do a variable change of $v = x - T$ again in the second integral, it's still invariant by $f$ being just a translation by a multiple of the period, and restick the two parts : $\int_{a - N T}^{T} f (x) d x + \int_{T}^{a - N T + T} f (x) d x = \int_{a - N T}^{T} f (x) d x + \int_{0}^{a - N T} f (v) d v = \int_{0}^{T} f (x) d x$

In signal processing we usually use the noataion $x (t)$ . You can use the more general form $f (x)$ .↩︎
For a periodic function, the smallest value of $T_{0}$ that satisfies this periodicity condition is called the fundamental period.↩︎

Periodic Functions

A periodic function takes the same values at intervals of T0.

A periodic function takes the same values at intervals of $T_{0}$ .