What is the Laplace Transform of Rectifier Function?

Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if x(t) is a time domain function, then its Laplace transform is defined as −

$$\mathrm{L[x(t)]\:=\:X(s)\:=\:\int_{-\infty}^{\infty}\:x(t)e^{-st}\:dt}$$

Laplace Transform of Full-Wave Rectified Sine Wave Function

The full-wave rectified sine wave function is shown in Figure-1 and is given by,

$$\mathrm{x(t)\:=\: sin\: \omega t\:;\: \: for\: 0\:\lt\:t\:\lt\: \frac{\pi}{\omega}}$$

The full-wave rectified sine wave function $\mathrm{x(t)}$ is a periodic function with time period equal to (π⁄ω). Therefore, by using the periodicity property $\mathrm{[ie.,\: L[x(t)] \:=\: [1\:-\:e^{-sT}]^{-1} \int_{0}^{T}\:x(t)e^{-st}dt]}$ of Laplace transform, we get,

$$\mathrm{X(s)\:=\:L[x(t)]\:=\:\frac{1}{[1\:-\:e^{-(s\pi /\omega)}]}\int_{0}^{(\pi /\omega)}\:sin(\omega t)e^{-st}dt}$$

On solving the integration in RHS, we have,

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{1}{[1\:-\:e^{-(s\pi /\omega)}]}\left [\frac{e^{-st}(-s\:sin\omega t\:-\:\omega \: cos\:\omega t)}{s^{2}\:+\:\omega^{2}} \right ]_{0}^{(\pi /\omega)}}$$

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{1}{[1\:-\:e^{-(s\pi /\omega)}]}\left[\frac{(\omega e^{-(s\pi/\omega)}\:+\:\omega) }{s^{2}\:+\:\omega^{2}} \right ]\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}\left [\frac{1\:+\:e^{-(s\pi /\omega)}}{1\:-\:e^{-( s\pi /\omega)}} \right ]}$$

Now, substituting $\mathrm{1\:=\:[e^{(s\pi /2\omega)}/e^{(s\pi/2\omega)}]}$ in the RHS of the above expression, we get,

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}\left[\frac{e^{(s\pi /2\omega)}\:+\:e^{-(s\pi/2\omega )}}{e^{(s\pi/2\omega)}\:-\:e^{-(s\pi/2\omega)}} \right ]\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}\left[\frac{cosh(s\pi /2\omega)}{sinh( s\pi /2\omega)} \right ]}$$

$$\mathrm{\therefore\: X(s)\:=\:L[sin\: \omega t]\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}cot\:h\left(\frac{s\pi}{2\omega} \right)}$$

Laplace Transform of Half-Wave Rectified Sine Wave Function

The half-wave rectified sine wave function is shown in Figure-2 and is given by,

$$\mathrm{x(t)\:=\: {\begin{cases} sin \: \omega t\:;\:\:for\: 0\:\lt\:t\:\lt\: \frac{\pi}{\omega}\\\\ 0\:;\:\: for\: \frac{\pi}{\omega}\:\lt\:t\:\lt\:\frac{2\pi}{\omega} \ \end{cases} }}$$

The half-wave rectified sine wave function $\mathrm{\mathrm{x\left ( t \right )}}$ is a periodic function with time period equal to (2π⁄ω). Therefore, by using the periodicity property $\mathrm{\left[ie.,\: L[x(t)] \:=\: [1\:-\: e^{-sT}]^{-1}\int_{0}^{T}x(t)e^{-st}\:dt\right]}$ of Laplace transform, we get,

$$\mathrm{X(s)\:=\:L[x(t)]\:=\:\frac{1}{[1\:-\:e^{-(2\pi s/\omega)}]}\left [\int_{0}^{(\pi /\omega)}sin(\omega t)e^{-st}dt \:+\: 0 \right ]}$$

On solving the integration in RHS, we have,

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{1}{[1-e^{-( 2\pi s /\omega)}]}\left [ \frac{e^{-st}(-s\: sin\: \omega t\:-\:\omega \: cos\: \omega t )}{s^{2}\:+\:\omega^{2}} \right ]_{0}^{(\pi /\omega)}}$$

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{1}{[1\:-\:e^{-(2\pi s /\omega)}]}\left [\frac{(\omega e^{-( s\pi /\omega)} \:+\: \omega )}{s^{2}\:+\:\omega^{2}} \right ]\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}\left[ \frac{1\:+\:e^{-(s\pi /\omega)}} {1\:-\:e^{-(2\pi s /\omega)}} \right ]}$$

$$\mathrm{\Rightarrow\: X(s)\:=\:\frac{\omega}{s^{2}\:+\:\omega^{2}}\left\{\frac{1\:+\:e^{-( s\pi /\omega)}}{[1\:-\: e^{-(s \pi /\omega)}][1\:+\:e^{-(s \pi/\omega)}]} \right\}}$$

$$\mathrm{\therefore\: X(s)\:=\:L[sin\: \omega t]\:=\:\frac{\omega}{(s^{2}\:+\:\omega^{2})(1\:-\:e^{-s\pi /\omega})}}$$