4.1.3. Pairs of the (Unilateral) Laplace Transform

Table 4.3 Laplace Transform Properties

Function	Time Domain	\(s\)-Domain	ROC
Unit pulse	\(\delta(t)\)	\(1\)	\(\mathbb C\)
Delayed pulse	\(\delta(t-\tau)\)	\(e^{-\tau s}\)	\(Re(s)>0\)
Unit step	\(u(t)\)	\(\frac{1}{s}\)	\(Re(s)>0\)
Ramp	\(t u(t)\)	\(\frac{1}{s^2}\)	\(Re(s)>0\)
Exponential decay	\(e^{-\alpha t} u(t)\)	\(\frac{1}{s+\alpha}\)	\(Re(s)>-\alpha\)
Sine	\(\sin(\omega t) u(t)\)	\(\frac{\omega}{s^2+\omega^2}\)	\(Re(s)>0\)
Cosine	\(\cos(\omega t) u(t)\)	\(\frac{s}{s^2+\omega^2}\)	\(Re(s)>0\)
Exponentially Decaying Sine	\(e^{-at}\sin(\omega t) u(t)\)	\(\frac{\omega}{(s+a)^2+\omega^2}\)	\(Re(s)>0\)
Exponentially Decaying Cosine	\(e^{-at}\cos(\omega t) u(t)\)	\(\frac{s+a}{(s+a)^2+\omega^2}\)	\(Re(s)>0\)

Unit Pulse

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

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

Unit Step Function

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

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

Note that for the limit for \(t\rightarrow\infty\) to exist we must have that \(\Re(s)>0\).

Exponential Decay

Consider the exponential decay function

\[\begin{split}\LT\left\{ e^{-at} u(t) \right\} &= \int_0^\infty e^{-at}e^{-st}dt\\ &= \int_0^\infty e^{-(s+a)t}dt\\ &= \frac{-1}{s+a}\left[ e^{-(s+a)t} \right]_0^\infty\\ &= \frac{-1}{s+a}\left( 0 - 1\right)\\ &= \frac{1}{s+a}\end{split}\]

The limit of \(e^{-(s+a)t}\) for \(t\rightarrow\infty\) only converges to zero in case \(\Re(s+a)>0\), i.e. \(\Re(s)>-\Re(a)\). Note that we have not assumed that \(a\in\setR\). This allows us to use this result to tackle the sin and cosine functions.

Sine Function

\[\begin{split}\LT\left\{ \sin(\w t) u(t) \right\} &= \LT\left\{ \frac{e^{j\w t}-e^{-j\w t}}{2j} \right\}\\ &= \frac{1}{2j}\left( \LT\left\{e^{j\w t}\right\} - \LT\left\{e^{-j\w t}\right\}\right)\\ &= \frac{1}{2j}\left( \frac{1}{s+j\w} - \frac{1}{s-j\w} \right)\\ &= \frac{1}{2j}\left( \frac{2j\w}{s^2+\w^2} \right)\\ &= \frac{\w}{s^2+\w^2}\end{split}\]