Laplace Transform of Periodic Functions
Formulas
Let $f$ be a periodic function with period $T$. Then $f(t+T)=f(t)$ and the Laplace transform of $f(t)$ is as follows.
$$ \mathcal{L} \left\{ f(t) \right\} = \int_{0}^\infty e^{-st}f(t)dt = \frac{\displaystyle \int_{0}^T e^{-st}f(t)dt}{1-e^{-st}} $$
Derivation
From the definition of Laplace transform, split the integral like this.
$$ \int_{0}^\infty e^{-st}f(t)dt = \int_{0}^T e^{-st}f(t)dt + \int_{T}^{2T} e^{-st}f(t)dt + \int_{2T}^{3T}e^{-st}f(t)dt + \cdots $$
At this point, to make the integration range of the second term the same as the first term, let’s substitute with $t=t^{\prime}+T$. Then,
$$ \int_{T}^{2T} e^{-st}f(t)dt=\int_{0}^T e^{-s(t^{\prime}+T)}f(t^{\prime}+T)dt^{\prime} $$
Since $f$ is a periodic function with a period of $T$, $f(t^{\prime}+T)=f(t^{\prime})$ and if we pull out the constant, it is organized as below.
$$ e^{-sT}\int_{0}^T e^{-st^{\prime}}f(t^{\prime})dt^{\prime}=e^{-sT}\int_{0}^T e^{-st}f(t)dt $$
The third term and those following can similarly be reorganized by substituting like $t=t^{\prime}+2T, t=t^{\prime}+3T, \cdots$. Then, the Laplace transform of $f(t)$ can be represented as follows.
$$ \begin{align*} & \int_{0}^\infty e^{-st}f(t)dt \\ &= \int_{0}^T e^{-st}f(t)dt + e^{-sT}\int_{0}^T e^{-st}f(t)dt + e^{-2sT}\int_{0}^T e^{-st}f(t)dt + \cdots \\ &= \left( 1+e^{-sT} + e^{-2sT} + \cdots \right) \int_{0}^T e^{-st}f(t)dt \end{align*} $$
When $|r|<1$,
$$ \sum_{n=1}^{\infty} a r^{n-1} = { a \over {1-r}} $$
If we represent the terms grouped at the front using the geometric series formula, it is as follows.
$$ \frac{1}{1-e^{-sT}} $$
Therefore, the following is obtained.
$$ \int_{0}^\infty e^{-st}f(t)dt = \dfrac{\displaystyle \int_{0}^T e^{-st}f(t)dt}{1-e^{-sT}} $$
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