Lets compare continuous time and discrete time representations of shifted delta functions:

\begin{align*}
\delta(n) &\ztp 1 & \delta(t) &\ltp 1 \\
\delta(n-N) &\ztp z^{-N}  & \delta(t-T) &\ltp e^{-sT}\\
\end{align*}

We can see that the $Z$-transform will give us a rational function in $z$ when we shift, whereas the Laplace Transform will not produce a rational function in $s$. A rational function in $s$ would be polynomial in $s$ over another polynomial in $s$:

$$
\frac{P(s)}{Q(s)} = \frac{ b_m s^m + b_{m-1} s^{m-1} + \cdots + b_1 s + b_0 }{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}
$$

The shifting property of the Laplace Transform,
$$x(t-T) \ltp \hat{X}(s) e^{-sT}$$

tells us that we produce a transform which is not rational. Recall that in discrete time, a rational $Z$-transform can be produce by linear combinations of $\alpha^n u(n)$, $\alpha^n u(-n)$, $n^k \alpha^n u(n)$, $n^k \alpha^n u(-n)$, where $k \in \N$, and shifted versions thereof. Note that the differentiation property for Laplace Transforms, 

$$ \frac{d}{dt} x(t) \ltp s\hat{X}(s) $$

tells us that we can produce a rational transform in $s$. In continuous time, a rational Laplace transform can be produce by linear combinations of $e^{-\alpha t} u(t)$, $e^{-\alpha t}u(-t)$, $t^k e^{-\alpha t}u(t)$, $t^k, e^{-\alpha t}u(-t)$, where $k \in \N$, and differentiated versions thereof with respect to $t$. The only difference between CT and DT when producing rational transforms is we have shifted versions in DT and differentiated versions in CT.

\begin{example}

Consider two causal signals, one in discrete time, and one in continuous time: $y(n) = \alpha y(n-1) + x(n)$ and $\dot{y}(t) = -\alpha y(t) + x(t)$. One is a first order difference equation, the other a first order differential equation. 

\begin{align*}
y(n) &= \alpha y(n-1) + x(n) & \dot{y}(t) &= -\alpha y(t) + x(t) \\
\hat{Y}(z) &= \alpha z^{-1}\hat{Y}(z) + \hat{X}(z) & s\hat{Y}(s) &= -\alpha \hat{Y}(s) + \hat{X}(s) \\
\hat{H}(z) &= \frac{1}{1 - \alpha z^{-1}} & \hat{H}(s) &= \frac{1}{s + \alpha} \\
h(n) &= \alpha^n u(n) & h(t) &= e^{-\alpha t}u(t) \\
\end{align*}

\end{example

\begin{example}

Consider a first order RC-circuit. The voltage across its capacitor is governed by the following equation: $y(t) = \frac{1}{C} \int_{-\infty}^t g(\tau) d\tau$, where $g(t)$ is the current, and $C$ is the capacitance.

% \img{images/circuits/combo/combo.ps}


Ohm's law tells us that $I = \frac{V}{R}$, where $V$ is the voltage drop across the circuit, and $R$ is the resistance. In our case, $I = g(t)$, and $V = x(t) - y(t)$. This gives us $g(t) = \frac{x(t) - y(t)}{R}$. Now we can solve for $y(t)$:

\begin{align*}
y(t) &= \frac{1}{C} \int_{-\infty}^t g(\tau) d\tau \\
y(t) &= \frac{1}{C} \int_{-\infty}^t \frac{x(\tau) - y(\tau)}{R} d\tau \\
y(t) &= \frac{1}{RC} \int_{-\infty}^t x(\tau)d\tau  -  \frac{1}{C} \int_{-\infty}^t y(\tau) d\tau \\
\frac{d}{dt} y(t) &= \frac{d}{dt} \frac{1}{RC} \int_{-\infty}^t x(\tau)d\tau  -  \frac{d}{dt} \frac{1}{RC} \int_{-\infty}^t y(\tau) d\tau \\
\end{align*}

Now we have an equation we can solve using Laplace Transforms:

\begin{align*}
\dot{y}(t) &= \frac{1}{RC}x(t) - \frac{1}{RC}y(t) \\
s\hat{Y}(s) &= \frac{1}{RC} \hat{X}(s) - \frac{1}{RC} \hat{Y}(s) \\
\hat{Y}(s) \lr{ s + \frac{1}{RC}} &= \frac{1}{RC} \hat{X}(s)  \\
\hat{H}(s) &= \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} \\
\end{align*}

So we find that the system function is $\hat{H}(t) = \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} $ with $RoC(h) = \{ s \st \Real{s} \gt -\frac{1}{RC} \}$.

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


We know the time domain representation (assuming causal):

\begin{align*}
\hat{H}(s) &= \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} \\
h(t) &= \frac{1}{RC} e^{-\frac{1}{RC}t}u(t) \\
\end{align*}

Here the $RoC$ includes the $i\omega$-axis, thus, 

$$
H(\omega) = \left. \hat{H}(s) \right|_{s = i\omega}
$$

We can find the magnitude response to determine what type of filter this circuit is:

$$
\abs{H(\omega)} = \frac{\frac{1}{RC}}{\abs{i\omega + \frac{1}{RC}}}
$$

% \img{images/roc/laplace/circuit/mag.ps}


At $i\omega = 0$, $\abs{H(\omega)}$ is at a maximum. As $\omega$ approaches $\pm \infty$, the magnitude is 0. This first order circuit is thus a low-pass filter.

% \img{images/graphs/ctlowpass/graph.ps}

\end{example}

Rational Laplace Transforms