# Relation between continuous time transfer function and sampled approximation

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?

$$H(jω)$$, 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.