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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 $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a time-domain function, then its Laplace transform is defined as −
$$\mathrm{\mathit{L}\mathrm{\left[ \mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-st}}\:\mathit{dt}\:\:\:\:\:\:...(1)}$$
Equation (1) gives the bilateral Laplace transform of the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,
$$\mathrm{\mathit{L}\mathrm{\left[ \mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-st}}\:\mathit{dt}\:\:\:\:\:\:...(2)}$$
Laplace Transform of Impulse Function
The impulse function is defined as,
$$\mathrm{\mathit{\delta}\mathrm{\left(\mathit{t}\right)}\mathrm{=}\begin{cases} 1& \text{ for } t= 0 \ 0 & \text{ for } t
eq 0 \end{cases}}$$
Thus, from the definition of Laplace transform, we have,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{s}\right)}\mathrm{=}\mathit{L\mathrm{\left[\mathit{\delta}\mathrm{\left(\mathit{t}\right)} \right]}}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{\delta}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-st}}\:\mathit{dt}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{\delta}\mathrm{\left(\mathit{t}\right)}\right]}}\:\mathrm{=}\:\mathrm{\left[\mathit{e^{-st}} \right]_{\mathit{t=\mathrm{0}}}}\:\mathrm{=}\:\mathrm{1}}$$
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,
$$\mathrm{\mathit{\delta}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathrm{1}\:\mathrm{and\:ROC\to all}\:\mathit{s}}$$
Laplace Transform of Step Function
The unit step function is defined as,
$$\mathrm{\mathit{u}\mathrm{\left(\mathit{t}\right)}\mathrm{=}\begin{cases} 1& \text{ for } t\geq 0 \ 0 & \text{ for } t< 0 \end{cases}}$$
Therefore, by the definition of the Laplace transform, we get,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathit{L\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{t}\right)} \right]}}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{u}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-st}}\:\mathit{dt}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{t}\right)} \right]}}\:\mathrm{=}\:\int_{\mathrm{0}}^{\infty}\mathit{e^{-st}}\:\mathit{dt}\:\mathrm{=}\:\mathrm{\left[\frac{\mathit{e^{-st}}}{-\mathit{s}}\right]^{\infty}_{\mathrm{0}}}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{t}\right)} \right]}}\:\mathrm{=}\:\mathrm{\left[\frac{\mathit{e^{-\infty}}-\mathit{e^{\mathrm{0}}}}{-\mathit{s}}\right]\:\mathrm{=}\:\frac{1}{\mathit{s}}}}$$
The above integral converges when $\mathit{Re}\mathrm{\left(\mathit{s} \right )}>\mathrm{0}$, i.e., the ROC of Laplace transform of unit step function is $\mathit{Re}\mathrm{\left(\mathit{s} \right )}>\mathrm{0}$ as shown in Figure-2. Thus, the Laplace transform of the step function along with its ROC is,
$$\mathrm{\mathit{u}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\frac{1}{\mathit{s}}\:\mathrm{and\:ROC\to Re\mathrm{\left(\mathit{s}\right)}}>\mathrm{0}}$$
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