# Why is the condition: region of convergence of z-transform contains unit circle sufficient for BIBO stability of a discrete time system

I'm reading Signals and Systems by Oppenheim, and in the section 10.7.2 about stability there's a conclusion I don't understand:

For the impulse response h[n] of a discrete time system he summarizes:

BIBO stability $$\iff$$ h[n] absolutely integrable $$\Rightarrow$$ Fourier-Transform of h[n] exists $$\iff$$ Unit circle is within the ROC of the Z-Transform of h[n].

I didn't find a proof of it in the book or online, but he then concludes

ROC contains the unit circle $$\iff$$ BIBO stability.

As far as I know, absolute summability is sufficient for the Fourier-Transform to exist, but not necessary, so it can't be because of that. Where does the equivalence relationship come from?

Edit The part with the Fourier-Transform is a bit ambiguous. I think the equivalence with "ROC contains unit circle" only holds if the Fourier-Transform converges absolutely, in which case it is also equivalent with absolute summability, giving full equivalence for all statements.

If the ROC of $$H(z)$$ contains the unit-circle on the complex z-plane, then $$H(z)$$, evaluated at $$z= e^{j\omega}$$, must uniformly converge for all $$\omega$$

$$\left| H(z) \right|_{z=e^{j \omega}} = \left| \sum_{n=-\infty}^{\infty} h[n] e^{-j \omega n} \right| < \infty . \tag{1}$$

And the rest follows due to the fact that absolute summability of $$h[n]$$ is a necessary and sufficient condition for the uniform convergence of $$H(e^{j\omega})$$. And further, because it's also a necessary and sufficient condition for BIBO stability of LTI systems too. Hence, the condition: ROC contains unit-circle, is sufficient for the BIBO stability of the LTI system.

• I don't see why Eq. (1) implies absolute summability. I understand that it's a stronger condition than just normal convergence $$\left| \sum_{n = -\infty}^{\infty} h[n] \right| < \infty$$ because you can choose $\omega$ freely. So something like $-\frac{1}{n}u[n]$ wouldn't converge anymore for $\omega = \pi$. Commented Aug 2, 2023 at 8:29
• I believe this might be because convergence of a Laurent Series on $R_1 < z-z_0 < R_2$ also implies absolute convergence on this disk. So the ROC containing the unit circle$\Rightarrow$ absolute convergence on the unit disk $\Rightarrow$ absolute summability. Commented Aug 2, 2023 at 10:22
• In the comment above, I meant unit circle instead of unit disk in the second statement. Commented Aug 2, 2023 at 10:28
• @klingeron Eq-1 implies H(w) to exist, and that implies system is BIBO stable, and that implies h[n] is absolutely summable. If you have an issue to see this, then your actual problem is not about ROC vs stability, but about why (and how) absolute summabilty of h[n] is a nec & suf condition for BIBO stability of LTI systems. Specifically, you can't see why absolute summabiliy is a nec cond for BIBO stability. But that's a different question than your specific post. Commented Aug 2, 2023 at 12:02
• No I know why absolute summability is equivalent with BIBO stability. The unclear part was this part " Eq-1 implies H(w) to exist, and that implies system is BIBO stable". My conclusion now is that we already know from Analysis that convergence on a disk also implies absolute convergence, and therefore BIBO stability (if the ROC contains the unit circle). Commented Aug 2, 2023 at 12:35

In case someone has the same question, there are two answers:

1. The statement "Fourier-Transform of some function h exists" is ambiguous since there are different "levels" of convergence. Absolute summability of h[n] $$\sum_{n = -\infty}^{\infty}|h[n]| < \infty$$ is a sufficient condition, but there are also weaker conditions. If the Fourier-Transform of the impulse response h exists in this sense, then by definition h is absolutely summable which is equivalent with BIBO stability. One proof for that is in the book Signals and Systems by A. Oppenheim. Also, the equivalence of "Fourier-Transform exists" with "ROC contains the unit circle" only holds in this case. This results in equivalence of "ROC contains the unit circle" and "BIBO stability".

2. With the ROC of the z-transform, if somebody says H(z) exists on some region, usually only the inner part of the ROC is meant, not boundary points. On every point $$z$$ that is in the ROC and not on the boundary, $$H(z) = \sum_{n = -\infty}^{\infty}h[n]z^{-n}$$ converges absolutely (and uniformly): $$|\sum_{n = -\infty}^{\infty}h[n]z^{-n}| \le \sum_{n = -\infty}^{\infty}|h[n]z^{-n}| < \infty.$$ If the ROC contains the unit circle, this immediately implies absolute summability with $$z^{-n} = (e^{-j\omega})^n$$ and $$|e^{-j\omega n}| = 1$$: $$\sum_{n = -\infty}^{\infty}|h[n]e^{-j\omega n}| = \sum_{n = -\infty}^{\infty}|h[n]| < \infty.$$ Since absolute summability is equivalent with BIBO stability as above, this shows that "ROC contains unit circle" is equivalent with BIBO stability.