What an interesting subject! To me for an intutive understanding of the BIBO/Lyap stability, you have all the ingrediants in the body of your question. Beyond this you have to go deeper into its mathematical iterpretations/equations.

BTW:

When I am teaching this subject, at the end of the session I put an RL circuit (R=1, L=1) in a paralle connection on the board. Then I asked them to think of its behaviour with regards to a unit step excitation... it is a passive circuit and expected to be stable ... then how we can interpret its behaviour as an BIBO system when the output state is the charge in C. Is it BIBO when the output of the intrest is the second state viz the current in the inductor...what if the current in the C is the output of the intrest?.... I want them to observe that for the internal stability we concern entire state variable but in I/O sense the input and output behaviour (a subset of the states / a linear combinations of the staes and the source....) maters...

What is the beaviour of the states when they are imposed with a disturbance that makes this Lyap stable (and not asymptotic)? If we exclude the source are all the conclsions still valid for the atonomous form?

Dear Dr. Nouri,

It is a nice simple example you identified. My comments:

1) Sorry, I do not understand why a capacitor is discussed when only an RL circuit is considered(?)

2) "Is it BIBO when the output of the intrest is the second state viz the current in the inductor": Here, the relation between the input voltage and the current is clearly integrative. Therefore, it cannot be BIBO nor asymptotically stable. I agree that when considering BIBO, we have to clearly define the output.

3) Please, reformulate the last paragraph - I do not unsertand your note. Lya. stability is not affected by any input disturbance. It is the case of system itself.

Thank you very much again!

Traditionally Lyapunov stability refers to asymptotic stability of non-linear dynamic systems. For linear systems, exponential stability is the normal concept used. An exponential stable system is one that will exponentially decay when given a finite time input, e.g., a finite pulse or "delta" function. However, given an input that is not finite duration, exponential stability does not guarantee exponential decay of the solution.

However, a BIBO system is one that defines a bounded linear operator on the space of bounded measurable functions. Since a step function is bounded in a BIBO system, the output will be bounded. A system is BIBO iff the impulse response is absolutely integrable.

This is similar to the concepts of the difference of exponential and ordinary dichotomies from dynamic systems.

https://en.wikipedia.org/wiki/Exponential_dichotomy BIBO, that is every bounded input has a bounded output,

Dear Dr. Truman Prevatt ,

Thank you for your response and a very interesting information about the dichotomy.

However, I still cannot find the answer to my question. Other issues you discussed are known for me very well. Btw, L. stability can be clearly and usefully defined also for a linear set of differential equations. It is, for instance, interesting that even a linear system can be L. stable but not assymtotic.

Thank you again!

Sorry Libor, I ment LC circuit.

Dear Behzad Nouri ,

Regarding the questions you identified - According to my opinion, I would explained them as follows:

There is a general problem that the system constitutes a parallel combination of purely derivative and integrative components here. By L. and BIBO stability definitions, both are neither BIBO nor asymptotically stable. Note that the integrator is L. stable (yet not asymptotically). Hence, the system itself cannot be L. (internally) stable.

1) Output is the capacitor charge: Since the output is proportional to the input (= input voltage), it is BIBO. It is exactly the case I introduced in my first question here.

2) Output is the inductor current: It is the pure integrator w.r.t the input voltage; therefore not BIBO one.

3) Output is the capacitor current: It is the purely derivative term w.r.t the input voltage; therefore not BIBO one as well.

Thank you very much for such an interesting example! It makes one thinking about it deeply. But, I may not be right... :)

Libor Pekař

Dear Libor,

I don’t know if you eventually found the answer to your question. I shall give my answer below.

The output y(t) of a system has two components: the free component y_L(t) generated by the initial conditions and the forced component y_F(t) generated by the input signal.

The forced component is the convolution between the causal impulse response h_C(t) (the inverse Laplace of the transfer function H(s)) and the input signal u(t) and has two sub-components: the transitory component y_T(t) related to the system’s structure (influenced by the signs of the poles) and the permanent component y_P(t) related to the input signal (influenced by the input’s poles).

If the linear system is exponential stable, then the free- and transitory components of the output response vanish as the time increase sufficiently, i.e. y_L(t) -> 0 and y_T(t) -> 0 as t-> infinity. Thus, the only component which remains is y_P(t) and moreover, for exponentially stabile systems the permanent component is of the same shape as the input signal as t-> infinity.

BIBO-stability reflects the input-output stability and it is evaluated on the transfer function, H(s). On the state space model one evaluates Lyapunov, asymptotic, exponential and other kind of stability properties.

Consequently:

- If the linear system is internally Lyapunov exponential stable, then for bounded inputs the output responses are bounded, i.e. THE SYSTEM IS BIBO-STABLE;

- If the linear system is internally unstable, but it is also non-minimal such that all the uncontrollable or the unobservable eigenvalues are those with positive sign (within the open right complex half -plane, inducing instability), then THE SYSTEM IS BIBO-STABLE. Explanation: taking into account that the transfer function obtained from the state-space model is always the irreducible, the unstable roots of the characteristic polynomial (poles of H(s)) will be simplified by the similar zeros of the nominator of H(s); consequently, H(s) will have only negative poles which means that the transitory component vanish as time tends to infinity and for bounded inputs the permanent component of the output response will be also bounded as determined by the poles of the input (the Laplace transform of u(t)).

- Conversely, if the transfer function has all its poles with negative real part, i.e. the system is BIBO stable, we cannot know if it is also Lyapunov exponentially stable (the characteristic polynomial can have uncontrollable or unobservable eigenvalues with positive real parts). From a stable transfer function one obtains a minimal state realization which is thus exponentially stable, but this will not reflect the internal state of the system which can be evaluated only on the system mathematical model of differential equations – i.e. the equations which result directly from the mathematical modeling process.

- A transfer function with all its poles in the open left complex half-plane has the impulse response absolutely integrable, i.e. it is L1-stable and vice-versa.

Thank you very much, Daniela!

This is an interesting topic. Instead of trying to give an answer, I would like to add another dimension to this problem. Usually a Lyapunov function is a representation of the abstract energy of the system, which could be easily defined for many systems, especially LTI, with no delays. However, failing to prove asymptotic stability might be because of the wrong choice of the Lyapunov function, especially when dealing with complex nonlinear systems. Proving BIBO stability is a much easier task, I think :)

A nice related topics could be found here:

https://www.researchgate.net/post/What_is_the_relationship_between_passivity_and_stability

https://www.researchgate.net/post/Relation_between_BIBO_criterion_location_of_poles_Polar_plot_Bode_plot_and_Nyquist_plot

Thanks, Ashraf, especially for the links.

Bounded Input and Bounded Output (BIBO) stability is a definition for a stable system based on what we can observe physically (or by simulation), that is, if the system is stable in the sense of BIBO, then if we change the input by a certain finite amount, the output also will change by a finite amount.

Upon mathematical analysis, e.g., in Laplace Domain, a BIBO system must have all of its poles strictly on the left-hand side (LHS) of the complex plane. In other words, such a system will be stable in the sense of BIBO definition. The system may show oscillations with decaying amplitude over time, if some of its poles have complex parts and negative real parts, or strictly asymptotic decay if all poles are negative reals. Indeed, this definition is a general one based on what we can observe physically (within certain ranges) or by simulation.

Thank you, @Jobrun Nandong.

Dear Libor Pekař ,

For a comprehensive thought on the subject, please, see the classical textbook by Zadeh and Desoer (Chapter 7, p.369):

https://archive.org/details/LinearSystemTheoryTheStateSpaceApproachZadehDesoer/page/n387/mode/2up

Wishes,

J. Mário

Thank you, José Mário, I will go through the text.