Laplace Transform of Periodic Functions: Formula, Proof, Example

The Laplace transform formula of periodic functions is used to find the Laplace of a periodic function with period T, that is, f(t+T)=f(t). This formula says that the Laplace transform of f(t) is given by

L{f(t)} = $\dfrac{\int_0^T e^{-st} f(t) dt}{1-e^{-sT}}$.

Laplace Transform of a Periodic Function

Theorem: If f: [0, ∞)  ℝ is a periodic function with period T > 0, then its Laplace transform is given by L{f(t)} = $\dfrac{\int_0^T e^{-st} f(t) dt}{1-e^{-sT}}$.

Proof

By definition of Laplace transforms, we have that

L{f(t)} = $\int_0^\infty$ e-st f(t) dt …(∗)

As f(t) has period T, we have f(t+T) = f(t). Using this fact, from (∗) we deduce that

L{f(t)} = $\int_0^T$ e-st f(t) dt + $\int_T^{2T}$ e-st f(t) dt +$\int_{2T}^{3T}$ e-st f(t) dt + …

Let us compute $\int_{nT}^{(n+1)T}$ e-st f(t) dt.

Put t = u+nT.

tu
nT0
(n+1)TT

Therefore, $\int_{nT}^{(n+1)T}$ e-st f(t) dt

= $\int_{0}^{T}$ e-su-nsT f(u+nT) du

= e-nsT $\int_{0}^{T}$ e-su f(u) du as f has period T.

Combining steps 2 and 3, we obtain that

L{f(t)} = $\int_0^T$ e-st f(t) dt + e-sT$\int_0^{T}$ e-st f(t) dt e-2sT+$\int_{0}^{T}$ e-st f(t) dt + …

= (1 +e-sT +e-2sT + … + e-nsT +…) $\int_0^T$ e-st f(t) dt

= $\dfrac{1}{1-e^{-sT}}$ $\int_0^T$ e-st f(t) dt, as the series inside the bracket is an infinite geometric series with common ratio e-sT.

= $\dfrac{\int_0^T e^{-st} f(t) dt}{1-e^{-sT}}$.

So the Laplace transform of a periodic function f(t) with period T is equal to L{f(t)} = ( ∫0T e-st f(t)dt)/(1-e-sT), and this is proved by the definition of Laplace transforms. That is,

L{f(t)} = $\dfrac{\int_0^T e^{-st} f(t) dt}{1-e^{-sT}}$.

More Laplace Transfroms:

Laplace Transform: Definition, Table, Formulas, Properties
Change of Scale Property
First Shifting Property: Formula, Proof
Second Shifting Property: Formula, Proof
Laplace Transform of Derivatives
Laplace Transform of Integrals
Laplace transform of unit step function
Laplace of 1/t does NOT exist: Proof

Example

Find the Laplace transform of the function:

f(t) = sinat, 0< t < π/a

= 0, π/a< t < 2π/a

Note that f(t) is a periodic function with period 2π/a. Thus, by the above formula, its Laplace transform will be equal to

L{f(t)} = $\dfrac{1}{1-e^{-2\pi s/a}}$ ∫02π/a e-st f(t) dt

= $\dfrac{1}{1-e^{-2\pi s/a}}$ [ ∫0π/a e-st sinat dt + ∫π/a2π/a e-st ⋅ 0 dt ]

= $\dfrac{1}{1-e^{-2\pi s/a}}$ $\Big[\dfrac{e^{-st}(-s \sin at -a\cos at)}{s^2+a^2} \Big]_0^{\pi/a}$

= $\dfrac{a e^{-\pi s/a}+a}{(1-e^{-2\pi s/a})(s^2+a^2)}$.

= $\dfrac{a}{(1-e^{-\pi s/a})(s^2+a^2)}$.

FAQs

Q1: What is the Laplace transform formula of a periodic function?

Answer: If f(t) is a periodic function with a period T>0, then its Laplace transform formula is given by L{f(t)} = ( ∫0T e-st f(t)dt)/(1-e-sT).

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