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 \[f(t+T) = f(t)\]
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
\[\laplace(f) (s) = \frac{1}{1-\ee^{-Ts}}\cdot \int_0^T\ee^{-st}\cdot f(t)\,\dd t\]
#\laplace{(f)}(s) = # #{{\euler^{3\cdot s}-1}\over{s\cdot \euler^{3\cdot s}+s}}#
According to the Laplace-transform of a periodic function we have
\[\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot \int_0^{6}\ee^{-st}\cdot f(t)\,\dd t\\
&&\phantom{xx}\color{blue}{\text{The Laplace transform of a periodic function}}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot \left(\int_0^{3}\ee^{-st}\,\dd t-\int_{3}^{6}\ee^{-st}\,\dd t\right)\\
&&\phantom{xx}\color{blue}{\text{definition of }f}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot\left( \left[-{{\euler^ {- s t }}\over{s}}\right]_0^{3}+\left[{{\euler^ {- s t }}\over{s}}\right]_{3}^{6}\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivative calculated}}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot\left( {{1}\over{s}}-{{\euler^ {- 3 s }}\over{s}}+{{\euler^ {- 6 s }}\over{s}}-{{\euler^ {- 3 s }}\over{s}}\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivatives evaluated at boundaries}}\\
&=&\displaystyle {{\euler^{3\cdot s}-1}\over{s\cdot \euler^{3\cdot s}+s}}\\
&&\phantom{xx}\color{blue}{\text{simplified}}\\
\end{array}\]
According to the Laplace-transform of a periodic function we have
\[\begin{array}{rcl}\laplace{(f)}(s) &=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot \int_0^{6}\ee^{-st}\cdot f(t)\,\dd t\\
&&\phantom{xx}\color{blue}{\text{The Laplace transform of a periodic function}}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot \left(\int_0^{3}\ee^{-st}\,\dd t-\int_{3}^{6}\ee^{-st}\,\dd t\right)\\
&&\phantom{xx}\color{blue}{\text{definition of }f}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot\left( \left[-{{\euler^ {- s t }}\over{s}}\right]_0^{3}+\left[{{\euler^ {- s t }}\over{s}}\right]_{3}^{6}\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivative calculated}}\\
&=&\displaystyle\frac{1}{1-\ee^{-6 s}}\cdot\left( {{1}\over{s}}-{{\euler^ {- 3 s }}\over{s}}+{{\euler^ {- 6 s }}\over{s}}-{{\euler^ {- 3 s }}\over{s}}\right)\\
&&\phantom{xx}\color{blue}{\text{antiderivatives evaluated at boundaries}}\\
&=&\displaystyle {{\euler^{3\cdot s}-1}\over{s\cdot \euler^{3\cdot s}+s}}\\
&&\phantom{xx}\color{blue}{\text{simplified}}\\
\end{array}\]
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