Let's say that I have the following continuous system:
$$G(s)= \frac{2}{1+s}$$
I could convert it to a discrete system using for example the Tustin approximation https://en.wikipedia.org/wiki/Bilinear_transform
So I replace s with:
$$s \rightarrow{} \frac{2(1-z^{-1})}{T_e(1+z^{-1})} $$
Hence I get the approx. discrete transfer function:
$$G(z)= \frac{2}{1+\frac{2(1-z^{-1})}{T_e(1+z^{-1})}}$$
Now my question is, how can I compute its frequency response ?
In the end, I would like to be able to compare the discrete approx. freqe. response with the freq. response of the continuous original transfer function.
Given
$$ \left|H(j\omega)\right| = \left|G(j\omega)\right|=\frac{2}{\sqrt{\omega^2+1}} $$
from $G'(z)$ we can obtain $G'(j\omega)$ by considering the transformation associated to the sampling phenomena
$$ z = e^{-j\omega T} $$
hence
$$ G'(z) = \frac{2 T (z+1)}{(T+2) z+T-2}\to G'(j\omega) = \frac{2 T \left(1+e^{i \omega T}\right)}{(T+2) e^{i \omega T}+T-2} $$
and
$$ \left| H'(j\omega)\right| = \sqrt{\frac{4 T^2 (\cos (\omega T)+1)}{\left(T^2-4\right) \cos (\omega T)+T^2+4}} $$
Attached the comparison between $|H'|$ (red) and $|H|$ (blue) for $T = 1$
and also $\angle{H}$,$\angle{H'}$