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What is the difference between stable, asymptotically stable, marginally stable, and unstable systems?

I am familiar with the first term (stable) and last term (unstable) but I often get very confused in understanding the other two (asymptotically stable and marginally stable).

Which of these two systems is stable in BIBO sense?

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3 Answers 3

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Let's restrict ourselves to the practically relevant case of causal linear time-invariant (LTI) systems. Asymptotic stability refers to the internal behavior of a system, that's why it's also called internal stability. This is in contrast to external stability, as described by the bounded input / bounded output (BIBO) condition. External stability can be observed by looking at the inputs and outputs of a system.

For causal LTI systems described by differential equations with constant coefficients (or, in discrete time (DT), by difference equations with constant coefficients), a system is asymptotically stable if all its poles are in the left half-plane (DT: inside the unit circle). Asymptotic stability implies BIBO-stability.

A system is marginally stable if there are simple poles on the imaginary axis (DT: on the unit circle). A marginally stable system is BIBO-unstable.

A (causal) system is unstable if there is at least one pole in the right half-plane (DT: outside the unit circle), or if there are multiple roots on the imaginary axis (DT: on the unit circle). Such a system is clearly also BIBO-unstable. Note, however, that there are stable systems with poles in the right half-plane (DT: outside the unit circle), but they cannot be causal.

Asymptotic stability always implies BIBO stability, but not the other way around. There are (degenerate) cases of BIBO-stable systems that are asymptotically unstable. This is the case if the system is either uncontrollable or unobservable. Internally unstable systems can be externally stable. Hence, asymptotic stability is a stricter condition than BIBO-stability. However, in most practically relevant cases (i.e., controllable and observable systems), BIBO-stability and asymptotic stability can be treated as synonymous.

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  • $\begingroup$ "A system is unstable if there is at least one pole in the right half-plane (DT: outside the unit circle), or if there are multiple roots on the imaginary axis" multiple roots here mean roots that occur exactly at same point for example j2? or it can also means roots that are not exactly at same point,one at j2 and other at -j2? $\endgroup$
    – DSP_CS
    Sep 15, 2022 at 11:26
  • $\begingroup$ @engr: a multiple root means there's a factor $(s-a)^n$, $n>1$ in the denominator of the transfer function. $\endgroup$
    – Matt L.
    Sep 15, 2022 at 12:01
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marginally stable has poles on the unit circle. An example would be a true integrator, it has infinite gain at DC, is marginally stable, and BIBO unstable. An integrator will increase without bound due to a step input (making it BIBO unstable). However, if a finite duration transient is input to the system, the output will increase or decrease only for the duration of the transient. Once the transient is removed, the system will not continue to grow or decay - this type of behavior can be thought of as marginal stability. Usually people desire a stability property such that the system will return to a common steady state value after being perturbed by a transient input as opposed to the behavior of the marginal stability system (though not always - a good example is a proportional/integral control loop filter needs to have marginal stability to have memory in order for the loop to have zero error at steady state.)

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I will add to Matt's answer that when investigating stability one has to look at the unreduced transfer function. A reduced transfer function may yield inadequate information about the stability.

Example

The response of a linear system can be partitioned into a zero-state response and a zero-input response

$$y(t) = y_\text{zs}(t) + y_\text{zi}(t) \leftrightarrow$$ $$Y(s) = \underbrace{\frac{P(s)}{Q(s)}X(s)}_{\text{zero-state}} + \underbrace{\frac{I(s)}{Q(s)}}_{\text{zero-input}}. $$

For a system with differential equation $\ddot{y}(t)-y(t) = \dot{x}(t)-x(t) $ the response becomes

$$Y(s) = \frac{s-1}{(s+1)(s-1)}X(s) + \frac{I(s)}{(s+1)(s-1)} \Leftrightarrow $$ $$Y(s) = \frac{1}{s+1}X(s) + \frac{I(s)}{(s+1)(s-1)}$$

This system is BIBO-stable, because the zero-state response is bounded for any bounded input. However, the zero-input response is not (necessarily) stable because the pole in $s=1$ is not cancelled. Hence, the system is unstable because the initial conditions (which constitute $I(s)$) will cause the total response to diverge.

This example shows that the transfer function is inadequate to describe and LTI system in general, because it only reveals what happens at the external terminals (input and output) and not what happens internally (intial conditions).

The reason for this inadequacy is because the system is not observable, as demonstrated by this MATLAB code:

>> b = [1,-1];
>> a = [1,0,-1];
>> H = tf(b,a)

H =
 
   s - 1
  -------
  s^2 - 1

>> [A,B,C,D] = tf2ss(b,a);
>> rank(obsv(A,C))

ans =

     1

>> rank(ctrb(A,B))

ans =

     2
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