Laplace Transform of Ramp Function and Parabolic Function

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

$$\mathrm{L[x(t)]\:=\:X(s)\:=\:\int_{-\infty}^{\infty}\:x(t)e^{-st}\:dt\:\: \dotso\: (1)}$$

Equation (1) gives the bilateral Laplace transform of the function $\mathrm{x(t)}$ . But for the causal signals, the unilateral Laplace transform is applied, which is defined as,

$$\mathrm{L\left [x(t) \right]\:=\:X(s)\:=\:\int_{0}^{\infty}\:x\left( t \right)e^{-st}\:dt\:\:\dotso\:(2)}$$

Laplace Transform of Ramp Function

The ramp function is defined as,

$$\mathrm{x\left ( t \right )\:=\:t\: u\left ( t \right )}$$

Thus, from the definition of Laplace transform, we have,

$$\mathrm{L\left [ x(t) \right ]\:=\:L\left [ t\: u\left( t \right ) \right]\:=\:\int_{0}^{\infty}t\:u(t)e^{-st}\:dt}$$

$$\mathrm{\Rightarrow\: L\left [ t\: u\left ( t \right ) \right ]\:=\:\int_{0}^{\infty }\:t\: e^{-st}\: dt}$$

$$\mathrm{\Rightarrow\:L\left[t\:u(t)\right]\:=\:\left[\frac{t\:e^{-st}}{-s}\right]_{0}^{\infty}\:-\:\int_{0}^{\infty}(1) \frac{e^{-st}}{-s}\:dt }$$

$$\mathrm{\Rightarrow\: L\left[ t\: u(t) \right]\:=\:0\:-\:\left[ \frac{e^{-st}}{s^{2}} \right]_{0}^{\infty}\:=\:\left (0 \:-\:\frac{1}{s^{2}}\right )\:=\:\frac{1}{s^{2}}}$$

The region of convergence (ROC) of the Laplace transform of the ramp function $\mathrm{[tu(t)]}$ is Re(s) > 0 as shown in Figure-1. Hence, the Laplace transform of the ramp function along with its ROC is,

$$\mathrm{t\:u(t)\overset{LT}{\leftrightarrow}\frac{1}{s^{2}}\:\:and\:\:ROC\:\:\rightarrow\:\: Re(s)\:\gt\:0}$$

Laplace Transform of Parabolic Function

The parabolic function is defined as,

$$\mathrm{x\left ( t \right )\:=\:t^{2}\:u\left ( t \right )}$$

Now, from the definition of the Laplace transform, we have,

$$\mathrm{L\left[ x(t) \right]\:=\:L\left [ t^{2}u(t) \right ]\:=\:\int_{0}^{\infty}\:t^{2}\:u(t)e^{-st}\: dt} $$

$$\mathrm{\Rightarrow\: L\left [ t^{2}u\left ( t \right ) \right ]\:=\:\int_{0}^{\infty}\:t^{2}\: e^{-st}\: dt\:=\:\left [ \frac{t^{2}e^{-st}}{-s} \right ]_{0}^{\infty}\:-\:\int_{0}^{\infty}\left (2t \right )\frac{e^{-st}}{-s}dt}$$

$$\mathrm{\Rightarrow\: L\left [t^{2}u(t)\right]\:=\:0\:+\:\frac{2}{s}\int_{0}^{\infty}\:t\: e^{-st}\: dt}$$

$$\mathrm{\Rightarrow\: L\left [t^{2}u\left ( t \right ) \right ]\:=\:\frac{2}{s}\left\{\left [ \frac{te^{-st}}{-s} \right ]_{0}^{\infty}\:-\:\int_{0}^{\infty }\:\left (1 \right )\frac{e^{-st}}{-s} \: dt\right\}} $$

$$\mathrm{\Rightarrow\: L\left [ t^{2}u\left ( t \right ) \right ]\:=\:\frac{2}{s}\left\{0\:-\:\left [ \frac{e^{-st}} {s^{2}} \right ]_{0}^{\infty } \right\}\:=\:\frac{2}{s^{3}}\left [ e^{-st} \right ]_{0}^{\infty }}$$

$$\mathrm{\therefore\: L\left [ t^{2}u\left ( t \right ) \right ]\:=\:\frac{2}{s^{3}}}$$

The ROC of Laplace transform of the parabolic function $\mathrm{\left [ t^{2}u\left ( t \right ) \right ]}$ is also Re(s) > 0, which is shown in Figure-1. Therefore, the Laplace transform of the parabolic function along with its ROC is,

$$\mathrm{t^{2}u\left ( t \right )\overset{LT}{\leftrightarrow}\frac{2}{s^{3}}\:\:and\:\:ROC\:\to\: Re(s)\:\gt\:0}$$