Time Periodicity Property of The Laplace Transform

If function $ f(t) $ is a periodic function such as shown in Fig. 1, it can be represented as the sum of time-shifted functions shown in Fig. 2. Thus,
$$\begin{aligned}f(t)=& f_{1}(t)+f_{2}(t)+f_{3}(t)+\cdots \\=& f_{1}(t)+f_{1}(t-T) u(t-T) \\&+f_{1}(t-2 T) u(t-2 T)+\cdots \end{aligned} \tag{1}$$
where $ f_{1}(t) $ is the same as the function $ f(t) $ gated over the interval $ 0 < $ $ t < T $, that is,
$$f_{1}(t)= \{ \begin{array}{ll}f(t), & 0 < t < T \\0, & \text { otherwise }\end{array}$$
A periodic function.
Fig. 1: A periodic function.
Fig. 2: Decomposition of the periodic function in Fig. 1.
We now transform each term in Eq. (1) and apply the time-shift property in Eq. (2).
$$\bbox[10px,border:1px solid grey]{\mathcal{L}[f(t-a) u(t-a)]=e^{-a s} F(s)} \tag{2}$$
We obtain
$$\begin{aligned} F(s)&=F_{1}(s)+F_{1}(s) e^{-T s}+F_{1}(s) e^{-2 T s}+F_{1}(s) e^{-3 T s}+\cdots \\ &=F_{1}(s)\left[1+e^{-T s}+e^{-2 T s}+e^{-3 T s}+\cdots\right] \end{aligned}$$
if $|x| < 1$. Hence,
$$ F(s)=\frac{F_{1}(s)}{1-e^{-T s}} \tag{3}$$
where $ F_{1}(s) $ is the Laplace transform of $ f_{1}(t) $; in other words, $ F_{1}(s) $ is the transform $ f(t) $ defined over its first period only. Equation (3) shows that the Laplace transform of a periodic function is the transform of the first period of the function divided by $ 1-e^{-T s} $.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024 realnfo.com
Privacy Policy