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

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:


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"

  • $\begingroup$ Could you link to the paper? I think we're missing some context here. $\endgroup$
    – Matt L.
    Jan 13, 2018 at 17:17
  • $\begingroup$ Sure, look at my edit! $\endgroup$
    – Alon
    Jan 13, 2018 at 17:42

1 Answer 1


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.,


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.


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