I see two notions that describe the relationship between poles and system stability. But they are not the same from my understanding

  1. The system is BIBO stable if and only if all the poles are in the left half of the complex plane
  2. A LTI system with a rational system function H(z) is stable if and only if all of the poles of H(z) lie inside the unit circle.

Why these two notions are different? Is that in different conditions?


3 Answers 3


The two are both true, but they are for different cases. Case 1 is true for continuous-time systems, and the transform is the Laplace transform and the variable is the derivative operator, $s$. Case 2 is true for discrete-time systems, and the transform is the $z$-transform and the variable is the delay operator, $z$.

  • $\begingroup$ @Downvoter? Any reasoning behind it? Or just get out of bed on the wrong side? $\endgroup$
    – Peter K.
    Feb 16, 2015 at 16:43
  • $\begingroup$ Good answer! I just upvoted to offset the downvote! $\endgroup$ May 28, 2017 at 21:38

I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI systems described by rational transfer function is:

An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer function includes the $j\omega$-axis (for continuous-time systems), or the unit circle $|z|=1$ (for discrete-time systems).

Since for causal systems the ROC is to the right of the pole with the largest magnitude (or outside the circle with radius equal to the largest pole radius in the $z$-plane), the above statement coincides with the statements in the question. However, non-causal systems can have poles in the right half-plane (or outside the unit circle of the $z$-plane) and still be stable.

Take as an example the discrete-time transfer function

$$H(z)=\frac{z^2}{(z-a)(z-b)},\quad a<1<b\tag{1}$$

which has one pole inside the unit circle and one pole outside the unit circle. You can't say that the corresponding system is unstable, because it depends on which system you mean. The transfer function (1) has 3 possible ROCs, corresponding to 3 different systems:

  1. $|z|>b$
  2. $|z|<a$
  3. $a<|z|<b$

The first one corresponds to a causal system which is unstable, because one of its poles is outside the unit circle (the ROC does not include the unit circle). The second one corresponds to an anti-causal system which is also unstable, because it has one pole inside the unit circle (again, the ROC does not include the unit circle). Finally, the third ROC includes the unit circle, i.e. it corresponds to a stable system. It is a non-causal system with a two-sided impulse response which decays exponentially as the time index $n$ goes to $\pm \infty$.

  • 1
    $\begingroup$ Good point, Matt! $\endgroup$
    – Peter K.
    Feb 16, 2015 at 12:40

From your statements, "But they are not the same from my understanding", and "Why these two notions are different?", I think that it needs to be added that these two statements are related to each other.

For ease of understanding, consider the example of designing systems by ourselves (e.g., IIR filters).

Let us employ impulse invariance technique first. Using this technique, a causal + stable continuous-time prototype filter generates a causal + stable discrete-time filter. A pole at $s=p$ on the s-plane maps to a pole at $z=e^{pT}$ in the z-plane. So for a left-half pole in s-plane, $$p = \sigma + j\omega T $$ with $\sigma$ being negative implies $$|e^{pT}| = |e^{\sigma + j\omega T}| = e^{\sigma T} < 1$$ which depicts that this pole is inside the unit circle. So all poles on the left half of s-plane go to inside the unit circle.

A similar result can be proved for bilinear transformation $$s = \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}$$ which also maps all points in the left half of s-plane to points inside the unit circle in z-plane.


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