Relation between continuous time transfer function and sampled approximation

Question

Suppose I have some continuous time system and associated transfer function: $$ y(n)=x(n)+x(n-1)$$ $$ H(j \omega) = 1+e^{j \omega (-T_s)} $$

Now suppose I create a discrete-time approximation of this system like this:

The digital system contained within the converter blocks has the transfer function:

$$ H(z) = 1 + z^{-1} $$

I know that the actual transfer function from $V_{in}(t)$ to $V_{out}(t)$ is something else entirely, but is there any meaning in saying that if $H(z)$ is "evaluated" for $z=e^{j \omega T_s}$, then $H(z)=H(j \omega)$? What is the significance of this observation, if any?

Answer 1

$H(jω)$, which is DTFT of the transfer function $h(n)$, is basically a special case of the Z transform $H(z)$, where $z=e^jω$ i.e. $|z|=1$.

In this special case, the unit circle ($|z|=1$) should be included in the region of convergence (ROC) of $H(z)$. In other words, the system (which is LTI in your example) should be BIBO stable.

The system in your example is BIBO stable since it has one pole in $z=0$, therefore its ROC ($z≠0$) includes the unit circle. Hence, we can say that $H(z)=H(jw)$ only when $|z|=1$ i.e. on the unit circle.