=============================================
 Pairs of the (Unilateral) Laplace Transform
=============================================


.. list-table:: **Laplace Transform Properties**
   :header-rows: 1
   :class: tablefullwidth

   * - 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
     - $e^{-at}\cos(\omega t) u(t)$
     - $\frac{s+a}{(s+a)^2+\omega^2}$
     - $Re(s)>0$



.. rubric:: Unit Pulse

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

.. math::
   X(s) = \int_{0}^{\infty} \delta(t) e^{-st} dt = 1


.. rubric:: Unit Step Function

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

.. math::
   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 $\Re(s)>0$.


.. rubric:: Exponential Decay

Consider the exponential decay function

.. math::
   \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}

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.


.. rubric:: Sine Function

.. math::
   \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}


..
   rubric:: Exponentially Decaying Cosine