2
$\begingroup$

$G(z) = \frac{1-p}{z-p}$

If the value of p satisfies $ 0 \leq p < 1$ there are no oscillations in the transient response.

Question: Why is that $\uparrow$ true? I know roughly what a transient response is but how is the relation between the position of the pole and the nature of the transient response?

Edit: In another paper by the same authors:

Enforcing the stability of the controlled system means ensuring that the pole p is non-negative and less than 1.

Now I'm completely out, stability does not depend upon the position of the pole as long as they are inside the unit circle of the z-plane I have thought?

Paper 2:

http://www.iste.uni-stuttgart.de/fileadmin/user_upload/iste/zss/publications/supplementaryMaterial/2014-ICSE-Control-PaperReviewCopy01.pdf

Look at the sentence short above eq. (10) on page 5

Paper 1 is not available on the internet as fas as I know. It's Antonio Filieri et al. "Software Engineering Meets Control Theory"

$\endgroup$
  • $\begingroup$ Could you link to the paper? I think we're missing some context here. $\endgroup$ – Matt L. Jan 13 '18 at 17:17
  • $\begingroup$ Sure, look at my edit! $\endgroup$ – Alon Jan 13 '18 at 17:42
1
$\begingroup$

If you have a single real-valued pole, you get a term $k\,p^{(n-n_0)}u[n-n_0],n\ge 0$, with some constant $k$ and some delay $n_0\ge 0$ in the system's impulse response ($u[n]$ is the unit step). Clearly, if $|p|\ge 1$ the transient will never settle. If $-1<p<0$, the transient will alternate in sign (i.e., "oscillate"), like e.g.,

$$\left(-\frac12\right)^nu[n]=\left\{1,-\frac12,\frac14,-\frac18,\dots\right\}\tag{1}$$

So for a transient that decays (i.e., for the system to be stable) and that does not oscillate you need $0\le p<1$.

Note that if you only have one real-valued pole $p$, stability implies $$-1<p<1\tag{2}$$ If you exclude systems with oscillating impulse responses (i.e., with a negative pole), you're left with the requirement $0\le p<1$. But, as you've noticed, this requirement is not only dictated by stability (which only requires $(2)$), but also by the requirement of having no oscillations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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