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)} = ${\displaystyle \frac{{\int }_0^T{e}^{-st}f(t)dt}{1-e^{-sT}}}$.

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)} = ${\displaystyle \frac{{\int }_0^T{e}^{-st}f(t)dt}{1-e^{-sT}}}$.

Proof

By definition of Laplace transforms, we have that

L{f(t)} = ${\int }_0^{\mathrm{\infty }}$ e^-st f(t) dt …(∗)

Step 1:

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 + …

Step 2:

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

Put t = u+nT.

t	u
nT	0
(n+1)T	T

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.

Step 3:

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

= ${\displaystyle \frac{1}{1-e^{-sT}}}$ ${\int }_0^T$ e^-st f(t) dt, as the series inside the bracket is an infinite geometric series with common ratio e^-sT.

= ${\displaystyle \frac{{\int }_0^T{e}^{-st}f(t)dt}{1-e^{-sT}}}$.

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

L{f(t)} = ${\displaystyle \frac{{\int }_0^T{e}^{-st}f(t)dt}{1-e^{-sT}}}$.

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

Answer:

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)} = ${\displaystyle \frac{1}{1-e^{-2\pi s/a}}}$ ∫₀^2π/a e^-st f(t) dt

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

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

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

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

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