Routh–Hurwitz Stability Check: For the denominator polynomial s^3 + s^2 + 2 s + 24, determine the qualitative nature of the network (stable, oscillatory, or unstable).

Introduction / Context:System stability for linear time-invariant (LTI) networks with rational transfer functions is determined solely by the denominator polynomial (the characteristic equation), provided there is no pole-zero cancellation in the right half-plane. The Routh–Hurwitz criterion gives a quick stability test without explicitly computing roots.

Given Data / Assumptions:

• Characteristic polynomial: s^3 + s^2 + 2 s + 24.
• We consider the denominator only (standard stability test scenario).
• No special pole-zero cancellations are indicated.

Concept / Approach:

For a cubic, construct the Routh array and inspect the signs of the first column. A sign change indicates roots in the right half-plane (RHP), implying instability. No sign changes imply stability; a row of zeros suggests marginal cases or symmetric root pairs.

Step-by-Step Solution:

Verification / Alternative check:

Because a sign change exists, the system cannot be stable. Numerical root solving would show one positive-real root; however, the Routh test suffices for a qualitative conclusion.

Why Other Options Are Wrong:

• stable: Requires all first-column entries positive—violated.
• oscillatory or marginally stable: Would correspond to purely imaginary roots or a zero row—none present.
• depends on numerator polynomial: Stability depends on poles (denominator), not zeros, assuming no cancellations.

Common Pitfalls:

• Arithmetic slip in computing the s^1 row element.
• Believing zeros can “stabilize” a system; zeros do not move poles.

Final Answer:

unstable