4.1.3. The pulse, step and exponential function

The Laplace transform of the pulse function \(\delta(t)\) is:

\[X(s) = \int_{0}^{\infty} \delta(t) e^{-st} dt = 1\]

The Laplace transform of the step function \(u(t)\) is:

\[X(s) = \int_{0}^{\infty} u(t) e^{-st} dt = \int_{0}^{\infty} e^{-st}dt = \left[ -\frac{1}{s} e^{-st}\right]_0^{\infty} =-\frac{1}{s} \left(0-1\right) = \frac{1}{s}\]

Consider the exponential function

\[x(t) = A e^{a t}\]

this function has Laplace transform:

\[X(s) = \frac{A}{s+a}\]

Summarizing we have:

\[\begin{split}\delta(t) &\ltarrow 1 \\ u(t) &\ltarrow \frac{1}{s} \\ A e^{at} &\ltarrow \frac{A}{s+a}\end{split}\]

Observe that the pulse is the derivative of the step function and the Laplace transform of the pulse is \(s\) times the Laplace transform of the step function. This relation between the Lt s of a function and its derivative is valid in general.