Question

Explain, with reasons, whether the LTIC systems described by the following equations are

(i) stable or unstable in the BIBO sense;
(ii) asymptotically stable, unstable, or marginally stable. Assume that the systems are controllable and observable.
a. (D² + 8D +12)y(t) = (D - 1)x(t)
b. D(D² + 3D + 2)y(t) = (D + 5)x(t)
c. D²(D² + 2)y(t) = x(1)
d. (D+1)(D2 - 6D + 5)y(t) = (3D + 1)x(t)

Answer 1

LTIC systems' stability is studied by examining the characteristic equations related to the systems. A LTIC system is BIBO stable if all poles have negative real parts, and it is asymptotically stable if the poles have strictly negative real parts. Systems with poles on the imaginary axis are marginally stable, while systems with any poles with positive real parts are unstable.

Step-by-step explanation:

The stability of Linear Time-Invariant Continuous (LTIC) systems can be assessed by examining their characteristic equations. Bounded Input Bounded Output (BIBO) stability requires that all the outputs of a system remain bounded for any bounded input. Asymptotic stability involves the system's response to perturbations from the equilibrium state over time.

In assessing the stability of the given LTIC systems, we focus on the characteristic polynomial of each system's differential equation, which is found by setting the input x(t) to zero. The roots of the characteristic polynomial (the system's poles) determine the stability as follows:

1. For a system to be BIBO stable, all poles must have negative real parts.
2. A system is asymptotically stable if all poles have strictly negative real parts.
3. A system with poles on the imaginary axis (excluding the origin) and no poles in the right half-plane is marginally stable.
4. A system with one or more poles with positive real parts, or with a pole at the origin and additional poles with non-negative real parts, is considered unstable.

For example:

• System (a) with characteristic equation D² + 8D + 12 = 0 has roots -2, -6, both with negative real parts, so it is both BIBO stable and asymptotically stable.
• System (b), characterized by D(D² + 3D + 2) = 0, has a pole at the origin and poles at -1 and -2, indicating a marginally stable system.
• For system (c), D²(D² + 2) = 0 has a double pole at the origin, making it unstable, as the repeated pole at the origin suggests an unbounded response over time.
• Lastly, system (d), with characteristic equation (D+1)(D² - 6D + 5) = 0, has poles at -1 (which is stable) and 1 and 5 (both unstable), resulting in an unstable system.

In conclusion, for system (a), we'd say it is stable and asymptotically stable. System (b) would be marginally stable due to the pole at the origin, whereas systems (c) and (d) are unstable.