Laplace Transform of Unit Impulse Function and Unit Step 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 $\mathrm{x\left(t\right)}$ is a time-domain function, then its Laplace transform is defined as −

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

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

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

Laplace Transform of Impulse Function

The impulse function is defined as,

$$\mathrm{\delta\left(t\right)\:=\:\begin{cases} 1\:; \text{ for } \: t \:=\: 0 \\\\ 0 \:;\: \text{ for } \: t\:=\: 0 \end{cases}}$$

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

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

$$\mathrm{\Rightarrow\: L\left[\delta\left(t\right)\right]\:=\:\left[e^{-st} \right]_{t=0}\:=\:1}$$

The region of convergence (ROC) of the Laplace transform of impulse function is the entire s-plane as shown in Figure-1. Hence, the Laplace transform of the impulse function along with its ROC is,

$$\mathrm{\delta\left(t\right)\overset{LT}{\leftrightarrow}1\::and\:\:ROC\:\to \:all\:s}$$

Laplace Transform of Step Function

The unit step function is defined as,

$$\mathrm{u\left(t\right)\:=\:\begin{cases} 1\: ;\:\text{ for }\: t\:\geq\:0 \\\\\ 0\: ;\:\text{ for }\: t\:\lt\: 0 \end{cases}}$$

Therefore, by the definition of the Laplace transform, we get,

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

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

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

The above integral converges when $\mathrm{Re\left(s \right ) \:\gt\:0}$, i.e., the ROC of Laplace transform of unit step function is $\mathrm{Re\left(s \right )\:\gt\:0}$ as shown in Figure-2. Thus, the Laplace transform of the step function along with its ROC is,

$$\mathrm{u\left(t\right)\overset{LT}{\leftrightarrow}\frac{1}{s}\:and\:ROC\:\to\: Re\left(s\right)\:\gt\:0}$$