Answer to Question #197461 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

Laplace transform of half wave rectified sine wave is given by -

f(t)= { $sin(\omega t)$ , 0 < t < $\dfrac{\pi}{\omega}$

{ 0 , $\dfrac{\pi}{\omega}$ < t < $\dfrac{2{\pi}}{\omega}$

we know that the time period of half wave rectifier is given by $T=\dfrac{2{\pi}}{w}$

If we know the time period of given wave function , then laplace transfrom can be calculated by using formula which is given by

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}\int_{0}^{T}e^{-st}f(t)dt$

$L[{f(t)}]=\dfrac{1}{1-e^{(-s\dfrac {2{\pi}}{\omega})}}\int_{0}^{T}e^{-st}f(t)dt$

Now , we will split this integeral in the domain which is given above ,

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}[\int_{0}^{\dfrac {{\pi}}{\omega}}e^{-sT}sin({\omega}t)dt$ +$\int_{\dfrac{\pi}{\omega}}^{\dfrac {2{\pi}}{\omega}}e^{-sT}0 \ dt$ ].....1)

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}[\int_{0}^{\dfrac{\pi}{\omega}}e^{-sT}sin({\omega}t)dt]$

We know integeration of $\int{e^{at}sin(bt)}$ =$\dfrac{e^{at}}{a^{2}+b^{2}}[asin(bt)-bcos(bt)]$

now applying the above result in , in the integeral $\int_{0}^{\dfrac{\pi}{\omega}}e^{-sT}sin(\omega t)dt$

$=\dfrac{e^{-st}}{s^{2}+({\omega })^{2}}[-ssin{\omega}t-{\omega}cos{\omega}t]_{0}^{\dfrac{\pi}{\omega}}$

$=$ now putting the limits , in above integeral , we conclude ,

$=\dfrac{e^{-s\dfrac{\pi}{\omega}}}{s^{2}+({\omega })^{2}}[{-ssin{\pi}-{\omega}cos{\pi}}]-\dfrac{e^{0}}{s^{2}+({\omega })^{2}}[{0-{\omega}cos0}]$

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

now putting these integeration result in equation....... 1 ) we get -

$\dfrac{1}{1-e^{\dfrac{-2s{\pi}}{\omega}}}$ [$\dfrac{\omega}{s^{2}+{\omega}^{2}}(1+e^{\dfrac{s{\pi}}{\omega}})$ ]

This is the laplace transfrom of half wave rectifier .

Laplace transform of full wave rectifier sin equation is -

f(t)= { $sin(\omega t)$ , 0 < t < $\dfrac{\pi}{\omega}$

The period of function is $\dfrac{\pi}{\omega}$

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}\int_{0}^{T}e^{-st}f(t)dt$

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}\int_{0}^{T}e^{-st}sin{\omega}tdt$

$L[{f(t)}]=\dfrac{1}{1-e^{(-sT)}}[\int_{0}^{\dfrac {{\pi}}{\omega}}e^{-sT}sin({\omega}t)dt]$

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