For the same reasons we discussed Z-transforms as a way to manipulate functions where the Fourier Transform could not (in discrete time), we now discuss the Laplace Transform for continuous time.

\begin{nicebox}
The \emph{Laplace Transform} of a function $h(t)$ is defined as 
\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-st} dt \quad s \in \C \\
\end{align*}
where $ s = \sigma + i\omega$. The associated Region of Convergence is
\begin{align*}
RoC(h) &\bydef \{ s = \sigma + i\omega \in \C \st \int_{-\infty}^{\infty} \abs{h(t)e^{-st}} dt \lt \infty \} \\
\end{align*}
\end{nicebox}


We can now handle systems that don't have Fourier Transforms. Because $s = \sigma + i\omega$, we can ``veer off'' the $i\omega$ axis:

\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-st} dt \\
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-(\sigma + i\omega)t} dt \\
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-\sigma t} e^{-i\omega t} dt \\
\hat{H}(s) &= \ftrans{h(t) e^{-\sigma t}} \\
\end{align*}

If $h(t)$ is growing exponentially, we can tame that growth with $e^{-\sigma t}$. In the definition of the Laplace transform, we see that $h(t)e^{-st}$ must be absolutely integrable. Does the $\omega$ is $s = \sigma + i\omega$ matter?

\begin{align*}
&\int_{-\infty}^{\infty} \abs{h(t)e^{-st}} dt \\
&\int_{-\infty}^{\infty} \abs{h(t)e^{-\sigma t}}\abs{e^{-i\omega t}} dt \\
&\int_{-\infty}^{\infty} \abs{h(t)e^{-\sigma t}} dt \\
\end{align*}

Clearly only $\sigma$ matters for absolute integrability since $e^{-i\omega t} = 1$. So what determines the $RoC$ is $\sigma = \Real{s}$.

\begin{claim}
Right-sided signals have right-sided $RoC$s.
\end{claim}

\begin{proof}

Let $x$ be right-sided such that $x(t) = 0 $ $ \forall t \lt T$. Then
\begin{align*}
\hat{X}(s) &= \int_{T}^{\infty}e^{-st} \\
\hat{X}(s) &= \int_{T}^{0}e^{-st} + \int_{0}^{\infty}e^{-st} \\
\end{align*}

We only care about the integral with limits from 0 to $\infty$, since the other integral is finite and thus converges.

\begin{align*}
&\mbox{Suppose } \sigma_0 \in RoC(x) \\
&\Rightarrow \int_0^\infty \abs{x(t) e^{-\sigma_0 t} } dt \lt \infty \\
\end{align*}

Let $\sigma_1 \gt \sigma_0$ be arbitrary. Then we have the following relations:

\begin{align*}
\sigma_1 &\gt \sigma_0 \\
-\sigma_1 &\lt -\sigma_0 \\
-\sigma_1t &\lt -\sigma_0t \quad (\mbox{for } t \in (0,\infty))\\
e^{-\sigma_1t} &\lt e^{-\sigma_0t} \\
\end{align*}

Therefore $e^{-\sigma_1t}$ decays faster than $e^{-\sigma_0t}$. Now we use this to show how this affects absolute integrability:

\begin{align*}
e^{-\sigma_1t} &\lt e^{-\sigma_0t} \\
\abs{x(t)e^{-\sigma_1t}} &\lt \abs{x(t)e^{-\sigma_0t}} \\
\int_0^\infty \abs{x(t)e^{-\sigma_1t}} dt &\lt \int_0^\infty \abs{x(t)e^{-\sigma_0t}} dt \lt \infty \\
\int_0^\infty \abs{x(t)e^{-\sigma_1t}} dt &\lt \infty \\
\end{align*}

Therefore if $\sigma_1 \gt \sigma_0$ and $\sigma_0 \in RoC(x)$, then $\sigma_1 \in RoC(x)$.

\end{proof}

Below is the $RoC$ of a right-sided signal. In this case the $RoC$ includes the $i\omega$-axis, or $\Imag{s}$-axis:

% \imgsize{0.75}{images/roc/laplace/causal.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-3,-1)(3,3)
    
\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{\psframe(-1,-2)(2,2)}
\psline[linestyle=dashed](-1,-2.2)(-1,2.2)
\psline[linestyle=dashed](-0.5,-2.2)(-0.5,2.2)

\rput(-0.5,-2.4){$\sigma_1$}
\rput(-1,-2.4){$\sigma_0$}

% x-axis
\rput(2.5,0.3){ $\Real{s}$ }
\psline{->}(-2.5,0)(2.5,0)

% y-axis
\rput(0.6,2.5){ $\Imag{s}$ }
\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}

Here is an $RoC$ for an left-sided system. In this case the $RoC$ includes the $i\omega$-axis:

% \imgsize{0.75}{images/roc/laplace/anticausal.ps}

\begin{center}
\begin{pspicture}(-3,-1)(3,3)
    
\newcommand{\Real}[1]{{\Re \mit{e}\left\{{#1}\right\}}}
\newcommand{\Imag}[1]{{\Im \mit{m}\left\{{#1}\right\}}}

\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
  \psframe(-2,-2)(1,2)
}

\psline[linestyle=dashed](1,-2.2)(1,2.2)
\rput(1,-2.4){$\sigma_0$}

% x-axis
\rput(2.5,0.3){ $\Real{s}$ }
\psline{->}(-2.5,0)(2.5,0)

% y-axis
\rput(0.6,2.5){ $\Imag{s}$ }
\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}


Here is an example $RoC$ for a two-sided system:

% \imgsize{0.75}{images/roc/laplace/leftright.ps}

\begin{center}
\begin{pspicture}(-3,-1)(3,3)
    
\newcommand{\Real}[1]{{\Re \mit{e}\left\{{#1}\right\}}}
\newcommand{\Imag}[1]{{\Im \mit{m}\left\{{#1}\right\}}}

\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
  \psframe(-1,-2)(1,2)
}

\psline[linestyle=dashed](-1,-2.2)(-1,2.2)
\psline[linestyle=dashed](1,-2.2)(1,2.2)
\rput(1,-2.4){$\sigma_1$}
\rput(-1,-2.4){$\sigma_0$}

% x-axis
\rput(2.5,0.3){ $\Real{s}$ }
\psline{->}(-2.5,0)(2.5,0)

% y-axis
\rput(0.6,2.5){ $\Imag{s}$ }
\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}


\begin{example}

Find the Laplace Transform for the function $h(t) = e^{-2t} u(t)$.
% \img{images/graphs/expun/graph.ps}

\begin{center}
\begin{pspicture}(-3,-1)(3,3)
    
\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
  \fileplot[linewidth=0pt]{fillgraph.dat}
}
\fileplot[linewidth=1.5pt]{graph.dat}
% function name
\rput(1.7,1.30815){$h(t) = e^{-2t} u(t)$}

% x-axis
\psline{->}(-1.4, 0)(2.4, 0)
\rput(2.4,-0.3){$t$}

% begin x-axis labels:
\psline(-1, -0.1)(-1, 0.1)
\psline(-0.5, -0.1)(-0.5, 0.1)
\psline(0.5, -0.1)(0.5, 0.1)
\psline(1, -0.1)(1, 0.1)
\psline(1.5, -0.1)(1.5, 0.1)
\psline(2, -0.1)(2, 0.1)
% end x-axis labels:

% y-axis
\psline{->}(0,-0.4)(0,1.4)
% begin x-axis labels:
\psline(-0.1, 1)(0.1, 1)
% end y-axis labels:

\end{pspicture}
\end{center}

\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} e^{-2t} u(t) e^{-st} dt \\
\hat{H}(s) &= \int_{0}^{\infty} e^{-2t}e^{-st} dt \\
\hat{H}(s) &= \int_{0}^{\infty} e^{-(2+s)t} dt \\
\hat{H}(s) &= \left. \frac{e^{-(2+s)t}}{-(2+s)} \right|_{0}^{\infty} \\
\hat{H}(s) &= \lim_{t \to \infty} \frac{e^{-(2+s)t}}{-(2+s)} - \frac{1}{-(2+s)} \\
\hat{H}(s) &= \frac{1}{(2+s)} \\
\end{align*}

In order for absolute integrability, $\Real{ -(2+s) } \lt 0$. Otherwise the limit in the above formulae would not have converged to zero as $t$ approaches $\infty$. In other words,
\begin{align*}
\Real{-(2+s)} \lt 0 \\
-2 - \sigma \lt 0 \\
-2 \lt \sigma \\
\end{align*}

So $RoC(h) : \{ s \st \Real{s} \gt -2 \}$:

% \img{images/roc/laplace/example/causal.ps}

\begin{center}
\begin{pspicture}(-3,-1)(3,3)

\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
  \psframe(-1,-2)(2,2)
}

\psline[linestyle=dashed](-1,-2)(-1,2)

\rput(-1,2.2){(-2)}
\rput(-1.75,1){$RoC(h)$}

% x-axis
\rput(2.5,0.3){ $\Real{s}$ }
\psline{->}(-2.5,0)(2.5,0)

% y-axis
\rput(0.6,2.5){ $\Imag{s}$ }
\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}

Notice that the Fourier Transform for this function is related to the Laplace Transform because the $RoC$ includes the $i\omega$-axis:

\begin{align*}
\ftrans{h(t)} &= \frac{1}{2 + i\omega} \\
\ltrans{h(t)} &= \frac{1}{2 + s} \\
\ftrans{h(t)} &= \left. \ltrans{h(t)} \right|_{s = i\omega} \\
\end{align*}

\end{example}

The Laplace Transform