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Laplace Transform of Periodic Functions 📂Odinary Differential Equations

Laplace Transform of Periodic Functions

Formulas

Let ff be a periodic function with period TT. Then f(t+T)=f(t)f(t+T)=f(t) and the Laplace transform of f(t)f(t) is as follows.

L{f(t)}=∫0∞e−stf(t)dt=∫0Te−stf(t)dt1−e−st \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.

∫0∞e−stf(t)dt=∫0Te−stf(t)dt+∫T2Te−stf(t)dt+∫2T3Te−stf(t)dt+⋯ \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′+Tt=t^{\prime}+T. Then,

∫T2Te−stf(t)dt=∫0Te−s(t′+T)f(t′+T)dt′ \int_{T}^{2T} e^{-st}f(t)dt=\int_{0}^T e^{-s(t^{\prime}+T)}f(t^{\prime}+T)dt^{\prime}

Since ff is a periodic function with a period of TT, f(t′+T)=f(t′)f(t^{\prime}+T)=f(t^{\prime}) and if we pull out the constant, it is organized as below.

e−sT∫0Te−st′f(t′)dt′=e−sT∫0Te−stf(t)dt 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′+2T,t=t′+3T,⋯t=t^{\prime}+2T, t=t^{\prime}+3T, \cdots. Then, the Laplace transform of f(t)f(t) can be represented as follows.

∫0∞e−stf(t)dt=∫0Te−stf(t)dt+e−sT∫0Te−stf(t)dt+e−2sT∫0Te−stf(t)dt+⋯=(1+e−sT+e−2sT+⋯ )∫0Te−stf(t)dt \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*}

Geometric Series

When ∣r∣<1|r|<1,

∑n=1∞arn−1=a1−r \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.

11−e−sT \frac{1}{1-e^{-sT}}

Therefore, the following is obtained.

∫0∞e−stf(t)dt=∫0Te−stf(t)dt1−e−sT \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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See Also