Problems calculating Z-transform I am trying to solve a class exercise in which I am given the following, in Laplace domain:

$$G(s)=\dfrac{e^{-Ts}}{s+3}$$
$$H(s)=\dfrac{1}{s}$$

And I need to calculate $\dfrac{C(z)}{R(z)}$, which is defined as:
$$\dfrac{C(z)}{R(z)}=\dfrac{G(z)}{1+H(z)G(z)}$$

In order to do so, I have to transform $G(s)$ and $H(s)$ into the $Z$ domain and operate to compute $\dfrac{C(z)}{R(z)}$. 

I have tried to compute $G(z)$ and $H(z)$, and I think I have failed in calculating $G(z)$, because I am not getting the right result.

I obtained that $H(z)=\dfrac{1}{1-z^{-1}}$ (that I think is correct).

Therefore, my question is how to obtain $G(z)$, I know that $\boxed{e^{sT}=z}$, but I don't know how to relate it with it, because initially I obtained: $G(z)=\dfrac{z^{-1}}{1-e^{-3}z^{-1}}$ but apparently is not correct. What am I doing wrong?

Note: in the statement of the exercise I am told to consider the period of one second (T=1s).



Solution: $$\dfrac{C(z)}{R(z)}=\dfrac{(1-z^{-1})z^{-1}}{1-(1+e^{-3})z^{-1}+(\dfrac{2}{3}e^{-3}+\dfrac{1}{3})z^{-2}}$$
We have
$$G(z)=\dfrac{z^{-1}}{1-e^{-3}z^{-1}}$$
and
$$H(z)=\dfrac{1}{1-z^{-1}}$$
and we want to calculate
$$\dfrac{C(z)}{R(z)}=\dfrac{G(z)}{1+H(z)G(z)}$$

So let's do that:
$$
\require{cancel}
\dfrac{\dfrac{z^{-1}}{1-e^{-3}z^{-1}}}{1+\dfrac{1}{1-z^{-1}}\dfrac{z^{-1}}{1-e^{-3}z^{-1}}} = \frac{z^{-1}(1-z^{-1})}{(1-z^{-1})(1-e^{-3}z^{-1}) + z^{-1}}\\
= \frac{z^{-1}(1-z^{-1})}{1 - z^{-1} - e^{-3}z^{-1} + e^{-3}z^{-2} + z^{-1}}
= \frac{z^{-1}(1-z^{-1})}{1 \cancel{- z^{-1}} - e^{-3}z^{-1} + e^{-3}z^{-2} \cancel{+ z^{-1}}}\\
= \frac{z^{-1}(1-z^{-1})}{1 - e^{-3}z^{-1} + e^{-3}z^{-2} }
$$
so it looks like either the solution you have is wrong, or that $G(z)$ is not right.



So, let's see what could be wrong. Either
$$
z \not = e^{sT}\tag{1}
$$
or
$$
s \not = 1 - z^{-1}\tag{2}
$$
or
$$
G(z) \not = \dfrac{1}{1-e^{-3}z^{-1}} \cdot {\cal T}_{s \rightarrow z}(e^{-s})\tag{3}
$$
where ${\cal T}_{s \rightarrow z}$ is the $s$ to $z$ conversion.

Comparing (1) and (2), it looks like (1) is assuming the starred transform and (2) is assuming the impulse invariance method --- which can't both be right.