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.

Periodic function 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:

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 #10# given by \[f(t)=\begin{cases}9 &\text{if }\ 0\le t \lt 5\\ -9 &\text{if }\ 5\le t\lt 10\end{cases}\]

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

#\laplace{(f)}(s) = # #{{9\cdot \euler^{5\cdot s}-9}\over{s\cdot \euler^{5\cdot s}+s}}#

According to the Laplace-transform of a periodic function we have
\[\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle\frac{1}{1-\ee^{-10 s}}\cdot \int_0^{10}\ee^{-st}\cdot f(t)\,\dd t\\
&&\phantom{xx}\color{blue}{\text{The Laplace transform of a periodic function}}\\
&=&\displaystyle\frac{1}{1-\ee^{-10 s}}\cdot \left(\int_0^{5}9\ee^{-st}\,\dd t-9\int_{5}^{10}\ee^{-st}\,\dd t\right)\\
&&\phantom{xx}\color{blue}{\text{definition of }f}\\
&=&\displaystyle\frac{1}{1-\ee^{-10 s}}\cdot\left( \left[-{{9 \euler^ {- t s }}\over{s}}\right]_0^{5}+\left[{{9 \euler^ {- t s }}\over{s}}\right]_{5}^{10}\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivative calculated}}\\
&=&\displaystyle\frac{1}{1-\ee^{-10 s}}\cdot\left( 9 \left({{1}\over{s}}-{{\euler^ {- 5 s }}\over{s}}\right)-9 \left({{\euler^ {- 5 s }}\over{s}}-{{\euler^ {- 10 s }}\over{s}}\right)\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivatives evaluated at boundaries}}\\
&=&\displaystyle {{9\cdot \euler^{5\cdot s}-9}\over{s\cdot \euler^{5\cdot s}+s}}\\
&&\phantom{xx}\color{blue}{\text{simplified}}\\
\end{array}\]

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