Signals and Systems – 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 $\mathrm{\mathit{x\left ( t \right )}}$ is a time domain function, then its Laplace transform is defined as −

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

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,

$$\mathrm{\mathit{x\left ( t \right )=\mathrm{sin}\: \omega t;\; \; \mathrm{for\: 0}< \mathit{t}< \frac{\pi }{\omega }}}$$

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

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

On solving the integration in RHS, we have,

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

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

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

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

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

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,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=} \left\{\begin{matrix} \mathrm{sin}\: \omega t;\;\; \mathrm{for\: 0}< t< \frac{\pi }{\omega }\ \mathrm{0;\;\; for\:} \frac{\pi }{\omega } < t<\frac{\mathrm{2}\pi }{\omega } \ \end{matrix}\right. }}$$

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

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

On solving the integration in RHS, we have,

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

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

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

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

Manish Kumar Saini

Updated on: 05-Jan-2022

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