Need a justification for the following discretization process. For a transfer function $G(s)$, its discretization by taking inverse Laplace transform of $\frac{G(s)}{s}$ and z-transforming its result, then multiplying it by $1-z^{-1}$ to simulate the zero order hold. Specifically why $\frac{G(s)}{s}$, and why not $G(s)$, is inverse Laplace transformed?
This process is used in the post: discretize a function using $z$-transform
On
The transfer function of a zero-order-hold is $$G_{h0}(s)=L\left(s(t)-s(t-T)\right)=\frac{1-e^{-Ts}}{s}$$ with $s(t)$ the unit step function. Imagine now an input signal $u(t)$ with z-transform $U(z)$ that passes through a sample and hold unit (ideal sampler+ZOH) followed by a system with transfer function $G(s)$. The z-transform of the output $y(t)$ is then $$Y(z)=Z[G_{h0}(s)G(s)]U(z)$$ where $$Z[G_{h0}(s)G(s)]=Z\left[\frac{1-e^{-Ts}}{s}G(s)\right]=(1-z^{-1})Z\left[\frac{G(s)}{s}\right]$$
Off the top of my head, isn't it because a discrete-time controller generates and input to a plant that piecewise-constant during with the sample periods?