# Laplace transform of product of signal and impulse train

I'm reading 'Discrete Time Control Systems' book by Ogata and came across a few statements (specifically, (3-1) and (3-2)) which I have not been able to understand.

It is said that any continuous signal can be sampled and the output represented as $$y(t) = \sum_{n=- \infty}^{+\infty}x(nT)\delta(t-nT)$$

Now taking laplace transform \begin{align} Y(s) &= \sum_{n=- \infty}^{+\infty}x(nT)\mathscr{L}\{\delta(t-nT)\} \\ &= \sum_{n=- \infty}^{+\infty}x(nT)e^{-nTs} \\ \end{align}

Now I have a confusion:

Is the $\delta(t)$ function

1. the dirac delta function, so that $\mathscr{L}\{\delta(t-nT)\} = e^{-nTs}$ but then the signal representation makes no sense as there is infinite amplitude in the output signal at multiples of $nT$
2. or is it the unit impulse function (value $1$ at $t=0$ and value $0$ everywhere else) in which case how exactly has $Y(s)$ been evaluated?

since no one else seems to have said it, if the ideally-sampled $x(t)$ is defined as

$$x_\text{s}(t) \triangleq \sum_{n=-\infty}^{+\infty}x(nT)\delta(t-nT)$$

and we define discrete-time samples as $x[n] \triangleq x(nT)$, the Laplace transform of

\begin{align} X_\text{s}(s) &= \sum_{n=- \infty}^{+\infty}\mathscr{L}\{x(nT) \delta(t-nT)\} \\ &= \sum_{n=- \infty}^{+\infty}x[n] \mathscr{L}\{\delta(t-nT)\} \\ &= \sum_{n=- \infty}^{+\infty}x[n] e^{-nTs} \\ &= \sum_{n=- \infty}^{+\infty}x[n]z^{-n} \\ &= \mathcal{Z}\{x[n]\} \Bigg|_{z=e^{sT}} \\ \end{align}

or, if I abuse the notation a little and change the meaning of $X(\cdot)$, the Z-Transform of $x[n]$ is related to the Laplace Transform of

$$\mathcal{Z}\{x[n]\}\Bigg|_{z=e^{sT}} = X(z)\Bigg|_{z=e^{sT}} = \mathscr{L}\{x_\text{s}(t)\}$$

So the Z-Transform of a discrete-time signal is nothing other than the Laplace Transform of the ideally-sampled continuous-time corresponding.

$\delta(t)$ is indeed the Dirac delta impulse, which is not defined by its values (because it is no ordinary function), but by its properties under an integral.

The first expression in your question is a standard model for sampling a continuous-time signal, and since it's a mathematical model it does not represent actual physical sampling. You can replace the Dirac delta impulse by other impulse-like functions. If you use a rectangular impulse you get a zero-order hold. However, if you're not interested in any effects of non-ideal sampling, multiplication with a Dirac comb is a convenient and useful model for sampling.