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:

block diagram

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?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.