Textbooks normally give the following 'derivation' (or justification, if you prefer) of the z-Transform from the Laplace Transform. Let $x(t)$ be a signal defined on $t\geq 0$, and write a discretized version of $x(t)$ as: $$ x_d(t)=\sum\limits_{n=0}^{\infty} x(n\Delta t)\delta(t-n\Delta t) $$ where $\delta()$ is the Dirac delta function. Take the Laplace Transform of $x_d(t)$: $$ \mathcal{L}\{x_d(t)\}= \sum\limits_{n=0}^{\infty} x(n\Delta t)\mathcal{L}\{\delta(t-nT)\}=\sum\limits_{n=0}^{\infty} x(n\Delta t) e^{-sn\Delta t} =^{z=e^{s\Delta t}} \sum\limits_{n=0}^{\infty} x(n\Delta t) z^{-n} = X(z) $$ I wish demonstrate to someone who is already familiar with the $z$ Transform that the Laplace Transform is its continuous analogue, so I wish to do the above process in reverse. I am trying the following approach:
Consider a discrete-time signal $x[n]$ as a sampled version of a continuous signal: $x[n]=x(n\Delta t)$ and let $z=e^{s\Delta t}$: $$ X(z)=\sum\limits_{n=0}^{\infty}x(n\Delta t) e^{-sn\Delta t} $$
What I'm going for is to argue that as $\Delta t \rightarrow 0$, this sum becomes an integral and therefore the Laplace Transform. But there is a 'missing' $\Delta t$ inside the sum for it to match the definition of a Riemann integral.
I would be grateful if someone could tell me what I am missing.
Notes:
- This video from MIT ignores the problem altogether and just writes an integral $dt$.
- This attempt from Stanford does something weird which seems wrong to me.
The whole problem lies in the fact that the discretized version of the $x$ is not what is claimed unless $\Delta t=1$. In fact, the right discrete version should be $$ x_d(t)=\sum_{n=0}^\infty x(n\Delta t)\color{red}{\Delta t}\,\delta(t-n\Delta t) $$ which converges in the sense of distributions to $x(t)$, and not the proposed one. Indeed, if $\varphi$ is a test function then $$ \langle x_d,\varphi\rangle=\sum_{n=0}^\infty x(n\Delta t)\,\varphi(n\Delta t)\,\Delta t \mathop{\longrightarrow}_{\Delta t\to0}\int_0^\infty x(t)\varphi(t)dt $$ while clearly this does not hold when we consider the version of discretized signal proposed in the OP's question.
Now, with the new formula of the $x_d$ every thing works.