## 2. Laplace transform of a periodic function

### Proof:

We may rewrite the integral from the Laplace Transform definition as follows

\displaystyle\mathscr{L}\{f(t)\}=\int_{0}^{\infty}e^{-st}f(t)dt=\int_{0}^{T}e^{-st}f(t)dt+\int_{T}^{\infty}e^{-st}f(t)dt

In the second integral, let:

t=u+T\Rightarrow dt=du

so, in the boundaries, for t=T, u=0 and u\to\infty as t\to\infty. Thus:

\displaystyle\int_{T}^{\infty}e^{-st}f(t)dt=\int_{0}^{\infty}e^{-s(u+T)}f(u+T)du

From the periodicity of f, we have

f(u+T)=f(u)

Therefore

\begin{aligned} \int_{T}^{\infty}e^{-st}f(t)dt &=\int_{0}^{\infty}e^{-su}e^{-sT}f(u)du \\ &=e^{-sT}\int_{0}^{\infty}e^{-su}f(u)du \\ &=e^{-sT}\mathscr{L}\{f(t)\} \end{aligned}

Hence, the initial expression becomes:

\displaystyle\mathscr{L}\{f(t)\}=\int_{0}^{T}e^{-st}f(t)dt+e^{-sT}\mathscr{L}\{f(t)\}

i.e.,

\begin{aligned} & \mathscr{L}\{f(t)\}-e^{-sT}\mathscr{L}\{f(t)\}=\int_{0}^{T}e^{-st}f(t)dt \\ & \mathscr{L}\{f(t)\}(1-e^{-sT})=\int_{0}^{T}e^{-st}f(t)dt \\ & \mathscr{L}\{f(t)\}=\frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-st}f(t)dt \end{aligned}

## Illustrative example

It is clear f has period T=a. Also, for 0\leq t<2a

f(t)=\left\{\begin{array}{ll} 1, & 0\leq t<a \\ 0, & a\leq t<2a \end{array}\right.

From the above theorem:

\displaystyle\mathscr{L}\{f(t)\}=\frac{1}{1-e^{-2as}}\int_{0}^{a}e^{-st}f(t)dt

Since

\begin{aligned} \int_{0}^{a}e^{-st}f(t)dt & =\int_{0}^{a}e^{-st}\cdot1dt+\int_{a}^{2a}e^{-st}\cdot0dt \\ & =\int_{0}^{a}e^{-st}dt \end{aligned}

Let u=-st\Rightarrow du=-sdt and as dt=-\dfrac{du}{s}, we get

\begin{aligned} \int_{0}^{a}e^{-st}dt & =\int_{0}^{-sa}e^{u}\left(-\frac{du}{s}\right) \\ & =-\frac{1}{s}\int_{0}^{-sa}e^{u}du \\ & =-\frac{1}{s}(\left.e^{u}\right|_{u=0}^{u=-sa}) \\ & =-\frac{1}{s}(e^{-sa}-1) \end{aligned}

Thus

\displaystyle\mathscr{L}\{f(t)\}=\frac{1}{1-e^{-2as}}\cdot\frac{1}{s}(1-e^{-sa})

We reached the end of this lesson about the Laplace transform of periodic functions. For a full list of Laplace transform properties, check this post!