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

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## 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 …(∗)

**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

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

More Laplace Transfroms:

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

= $\dfrac{1}{1-e^{-2\pi s/a}}$ [ ∫_{0}^{π/a} e^{-st} sinat dt + ∫_{π/a}^{2π/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)} = ( ∫_{0}^{T} e^{-st} f(t)dt)/(1-e^{-sT}).