I have been studying the book Digital Control of Dynamic Systems. In chapter 5 of this book the concept of the ideal sampler is introduced
and the Laplace transform of the sampled signal $r^*(t)$ is derived as $R^*(s) = \sum\limits_{k=-\infty}^{\infty}r(kT)\cdot e^{-skT}$. This derivation seemed to me to be clear but later in the book I have found following equation
$$ Y^*(s) = \left[G(s)\cdot M(s)^*\cdot\left(\frac{1-e^{-sT}}{s}\right)\right]^* = (1-e^{-sT})\cdot M^*(s)\cdot\left[\frac{G(s)}{s}\right]^* $$
That is my problem. I don't understand why the term $(1-e^{-sT})$ "passes" through the sampler without any change. The authors attempted to clarify that by following comment: "taking out the periodic parts, which are those in which $s$ appears only as $e^{-sT}$". Unfortunately this comment haven't helped me too much.
$(1-e^{-sT})F(s)$ is the transform of $g(t)=f(t)-f(t-T)$. The sampled signal $$ g^*(t)=\sum [f(kT)-f((k-1)T)]\delta(t-kT) $$ has again the form of a difference $f^*(t)-f^*(t-T)$. This only works because the shift is equal to the sampling period, else the values in the second term would be drawn from a different sample sequence. So in the end you get $$ [(1-e^{-sT})F(s)]^*=(1-e^{-sT})F^*(s). $$