Talk:BIBO stability

Distinction between one-sided and two-sided transforms is important
In the definition and proofs in the first sections, two-sided transforms and convolutions are currently used. However, for proper linking to the frequency domain results (and conditions in terms of pole locations) one-sided transforms and convolutions are needed. Example: If a continuous-time transfer function with a single stable pole is considered, e.g. H(s)=1/(s+1), then the corresponding impulse response is h(t)=exp(-t)u(t), in which u(t) is the Heaviside step function. This implies that h(t)=0 for negative t, making the one-sided transforms more natural to work with. Using exp(-t) on all of the real axis is unwanted as it would violate the two-sided integral condition for BIBO stability. — Preceding unsigned comment added by 77.171.130.148 (talk) 13:39, 19 February 2023 (UTC)

Last section
In the last section, "Frequency domain condition," doesn't it leave out the condition that the number of zeros must be less than the number of poles? This is for both continuous and discrete time that I think this condition was left out.


 * I think you are right about continuous-time systems - for example, H(s)=s has no poles at all, but is not BIBO stable. In discrete-time, however, there is no such condition, for example H(z)=z is BIBO stable. 94.159.139.235 (talk) 23:59, 24 March 2012 (UTC)

Link to ISS
When the page is created, a link to input-to-state stability (ISS) should be added to the See also section. ISS is the nonlinear analog to BIBO stability. &mdash;TedPavlic (talk) 18:09, 30 January 2009 (UTC)

Definition of bounded signal
The definition of a bounded signal in the first section is maybe just confusing? Is it needed to define the input? If so it should be clear that the definition only is valid for the input signal. Since it is not valid for the output signal. Ex. if the output is h(t)=a, it is not integrable, hence it is not BIBO stable. Tibnor (talk) 10:04, 17 December 2010 (UTC)


 * The definition is valid for the output as well. If the output is y(t)=a, then it is a bounded output. 94.159.139.235 (talk) 00:02, 25 March 2012 (UTC)

Get confused
In "Frequency-domain condition for linear time invariant systems" section, the formula said
 * $$ = \int_{-\infty}^{\infty}{\left|h(t)\right| \left| e^{-j \omega t} \right| dt}$$


 * $$= \int_{-\infty}^{\infty}{\left|h(t) (1 \cdot e)^{-j \omega t} \right| dt}$$

According to the triangle inequality, is it need to be the following formula ?


 * $$ = \int_{-\infty}^{\infty}{\left|h(t)\right| \left| e^{-j \omega t} \right| dt}$$


 * $$\ge \int_{-\infty}^{\infty}{\left|h(t) (1 \cdot e)^{-j \omega t} \right| dt}$$

--Wolfch (talk) 15:46, 1 October 2013 (UTC)

condition for continuous-time signal
What is written here https://en.wikipedia.org/wiki/BIBO_stability#Continuous-time_signals is completely uncorrect, see https://fr.wikipedia.org/wiki/Stabilit%C3%A9_EBSB#Condition_dans_le_domaine_fr.C3.A9quentiel for a  correct derivation. 78.196.93.135 (talk) 14:08, 30 January 2017 (UTC)

Poles need not be strictly in the imaginary axis in order for BIBO stability(?)
In section “Frequency-domain condition for linear time-invariant systems”, subsection “Continuous-time signals”, at the end of the first paragraph it says “Therefore, all poles of the system must be in the strict left half of the s-plane for BIBO stability.” I think this is not true.

Consider a Laplace transform which consists exclusively of a simple pole at the origin in the frequency domain. Then the corresponding signal is a constant in the time domain. A constant signal is a bounded signal. Hence if the output is a constant (and so it has a simple pole at the origin of the s plane), the system should be considered BIBO stable, yet the article says otherwise. Why?

Perhaps the article refers to the poles of the transfer function, not to the poles of the Laplace transform of the output signal? --Alej27 (talk) 17:18, 18 January 2022 (UTC)