Answer to Question #197448 in Differential Equations for M Kamran

Question

Find the Laplace Transform of half-wave and full-wave

rectified sine wave given in the following figures.

Take w = 2

Answer 1

$f(t)=sin\omega t$ ,

For full wave-

$L[sin\omega t]=\dfrac{1}{1-e^{-\frac{2\pi i}{\omega}}} \int_0^{2\pi} e^{-st} sin\omega t dt$

        $=\dfrac{1}{1-e^{-\frac{2\pi i}{\omega}}} (\dfrac{e^{-st}}{s^2+\omega^2}[-s sin\omega t -\omega cos\omega t])_0^{\frac{\pi}{\omega}}$

 

        $=\dfrac{1}{1-e^{-\frac{2\pi i}{\omega}}} (\dfrac{e^{-\frac{s\pi}{\omega}} \omega+\omega}{s^2+\omega^2})$

       $=\dfrac{\omega}{(1-e^{-\frac{\pi i}{\omega}} )(s^2+\omega^2)}$

For half wave-

$L[sin\omega t]=\dfrac{1}{1-e^{-\frac{2\pi i}{\omega}}} \int_0^{\pi} e^{-st} sin\omega t dt$

        $=\dfrac{1}{1-e^{-\frac{2\pi i}{\omega}}} (\dfrac{e^{-st}}{s^2+\omega^2}[-s sin\omega t -\omega cos\omega t])_0^{\frac{\pi}{2\omega}}$

 

        $=\dfrac{1}{1-e^{-\frac{\pi i}{\omega}}} (\dfrac{e^{-\frac{s\pi}{2\omega}} \omega+\omega}{s^2+\omega^2})$

       $=\dfrac{2\omega}{(1-e^{-\frac{\pi i}{\omega}} )(s^2+\omega^2)}$