Laplace transforms of periodic functions

Differential equations and Laplace transforms: Laplace-transformations

Laplace transforms of periodic functions

Also for periodic functions there is a rule for the calculation of the Laplace transform.

Let $T$ be a positive number. A function $f$ on $\ivco{0}{\infty}$ is called periodic with period $T$ if, for all $t\ge0$ , we have $f(t+T) = f(t)$

The sine and cosine are periodic with period $2\pi$ .

A constant function is periodic with period $T$ for each positive value of $T$ .

Here is a formula for the Laplace transform of a periodic function:

If the Laplace transform of a periodic function $f$ with period $T\gt0$ exists, then it satisfies

$\laplace(f) (s) = \frac{1}{1-\ee^{-Ts}}\cdot \int_0^T\ee^{-st}\cdot f(t)\,\dd t$

The periodicity leads to a linear equation for the Laplace transform:

$\begin{array}{rcl} \laplace{(f)} (s) &=&\int_0^{\infty}f(t)\cdot \e^{-st}\,\dd t\\ &=& \int_0^T f(t)\cdot \e^{-st}\,\dd t+ \int_T^{\infty} f(t)\cdot \e^{-st}\,\dd t\\ &=&\int_0^T f(t)\cdot \e^{-st}\,\dd t+ \int_0^{\infty} f(t+T)\cdot \e^{-st-sT}\,\dd t\\&&\phantom{xxx}\color{blue}{\text{substitution }t = T+t}\\&=&\int_0^T f(t)\cdot \e^{-st}\,\dd t+ \int_0^{\infty} f(t)\cdot \e^{-st-sT}\,\dd t\\ &&\phantom{xxx}\color{blue}{\text{periodicity}}\\&=&\int_0^T f(t)\cdot \e^{-st}\,\dd t+\e^{-sT}\cdot \laplace{(f)}(s)\end{array}$

Solution of this linear equation in the unknown $\laplace{(f)}(s)$ gives

$\laplace{(f)}(s) = \frac{1}{1-\ee^{-Ts}}\cdot \int_0^T\ee^{-st}\cdot f(t)\,\dd t$

Compute the Laplace transform of the periodic function $f$ on $\ivco{0}{\infty}$ with period $8$ given by $f(t)=\begin{cases}4 &\text{if }\ 0\le t \lt 4\\ -4 &\text{if }\ 4\le t\lt 8\end{cases}$

Give your answer as a function of $s$ for $s\gt 0$ .

$\laplace{(f)}(s) =$ ${{4\cdot \euler^{4\cdot s}-4}\over{s\cdot \euler^{4\cdot s}+s}}$

According to the Laplace-transform of a periodic function we have
$\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle\frac{1}{1-\ee^{-8 s}}\cdot \int_0^{8}\ee^{-st}\cdot f(t)\,\dd t\\ &&\phantom{xx}\color{blue}{\text{The Laplace transform of a periodic function}}\\ &=&\displaystyle\frac{1}{1-\ee^{-8 s}}\cdot \left(\int_0^{4}4\ee^{-st}\,\dd t-4\int_{4}^{8}\ee^{-st}\,\dd t\right)\\ &&\phantom{xx}\color{blue}{\text{definition of }f}\\ &=&\displaystyle\frac{1}{1-\ee^{-8 s}}\cdot\left( \left[-{{4 \euler^ {- s t }}\over{s}}\right]_0^{4}+\left[{{4 \euler^ {- s t }}\over{s}}\right]_{4}^{8}\right)\\ &&\phantom{xx}\color{blue}{\text{antiderivative calculated}}\\ &=&\displaystyle\frac{1}{1-\ee^{-8 s}}\cdot\left( 4 \left({{1}\over{s}}-{{\euler^ {- 4 s }}\over{s}}\right)-4 \left({{\euler^ {- 4 s }}\over{s}}-{{\euler^ {- 8 s }}\over{s}}\right)\right)\\ &&\phantom{xx}\color{blue}{\text{antiderivatives evaluated at boundaries}}\\ &=&\displaystyle {{4\cdot \euler^{4\cdot s}-4}\over{s\cdot \euler^{4\cdot s}+s}}\\ &&\phantom{xx}\color{blue}{\text{simplified}}\\ \end{array}$

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