For AR HP filter

  1. zeros, to the right of the imaginary axis
  2. poles outside the unit-circle
  3. zeros on the real axis
  4. poles, to the left of the imaginary axis

Apparently, the right answer is (4). Why?

We did learn in class some properties of the zeros and poles, and I probably need to relate it to the fact that it's a AR HP filter.

I'd be glad for your help.


I know that AR is an all-poles filter so it must be (1) or (4). What is the meaning of the position of the pole with relation to the imaginary axis?

Is it because the large $\omega$'s (frequencies as angles of the unit circle) are at the left side of the imaginary axis and so we get an HP filter?


1 Answer 1


An AR process is defined in a way that it is inherently causal. It depends on past and present values, but not on the future. Thus, its ROC must be:

  • In the continuous case, $s$ such that $|s|>\sigma$ for some $\sigma\in\mathbb{R}$.
  • In the discrete case, $z$ such that $|z|>r$ for some $r\in\mathbb{R}$.

For an AR filter, all the zeros are at the origin. This could make the 3rd option you wrote correct, as the origin is indeed on the real line. The first one is definitely incorrect.

Regarding the poles, it depends on the coefficient of the process, $a$. If $|a|<1$, the filter will be stable and all the poles will be on the left half plane of the $s$-plane, or inside the unit circle in the $z$-domain. If $|a|>1$, the contrary will happen.

So the answer would be the 4th, if the filter is required to be stable (which makes sense). If the filter is unstable, then the 2nd one would be the one to chose.


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.