Questions: Characteristic Equation and Closed-Loop Stability

Question 1

A control engineer must stabilize an unstable robotic arm with open-loop transfer function G(s) = 1/(s − 1). She adds a proportional controller C(s) = K and closes the loop with unity feedback. What is the minimum gain K that achieves closed-loop stability?

• A. Any positive K will stabilize the system, since feedback always moves poles toward the left half-plane
• B. K > 1, because the characteristic equation s − 1 + K = 0 places the closed-loop pole at s = 1 − K, which is negative only when K > 1
• C. K < 1, because a small gain is needed to avoid exciting the unstable mode
• D. The system cannot be stabilized with a proportional controller because the open-loop pole at s = +1 is inherently unstable

Answer: (B)

The closed-loop characteristic equation is 1 + K·G(s) = 1 + K/(s−1) = 0, giving s − 1 + K = 0, so the closed-loop pole is at s = 1 − K. For left-half-plane stability: 1 − K < 0 requires K > 1. With K ≤ 1, the pole is at s ≥ 0 — unstable or marginally stable. This illustrates two key points: (1) the characteristic equation, not the open-loop poles, governs closed-loop stability, and (2) feedback CAN stabilize an unstable plant, but only with sufficient gain. Option A is the most tempting misconception — feedback does not automatically improve stability; it depends on design parameters.

Question 2

A unity-feedback system has plant G(s) = 2/(s² + 3s + 2) and proportional controller C(s) = K. An engineer wants to find all values of K for which the closed-loop system is stable. What is the correct first step?

• A. Find the poles of G(s) and check whether they are all in the left half-plane
• B. Compute the closed-loop transfer function T(s) and analyze the roots of its numerator
• C. Form the characteristic equation 1 + K·G(s) = 0 and analyze the roots of the resulting polynomial as a function of K
• D. Compute the open-loop frequency response and apply the Nyquist stability criterion

Answer: (C)

Closed-loop stability depends on the roots of the closed-loop denominator, which is 1 + K·G(s) — the characteristic equation. The poles of G(s) (option A) are the open-loop poles, which change when the loop is closed; checking them says nothing about closed-loop stability. The numerator of T(s) (option B) contains the closed-loop zeros, which affect response shape but not stability. Option D (Nyquist) is a valid stability method but is equivalent to analyzing the characteristic equation — and forming the characteristic polynomial directly is the more direct algebraic approach for this problem type.

Question 3

An open-loop unstable plant (with poles in the right half-plane) can be stabilized by closing a feedback loop with an appropriate controller, because the characteristic equation 1 + G(s)C(s) = 0 can have roots in different locations than the open-loop poles.

• T. True
• F. False

Answer: True

Feedback fundamentally alters the system's effective dynamics. The closed-loop poles — roots of 1 + G(s)C(s) = 0 — are generally different from the open-loop poles of G(s) alone. For example, G(s) = 1/(s−1) has an open-loop pole at s = +1 (unstable), but with K > 1, the closed-loop pole moves to s = 1−K < 0 (stable). Stabilization of inherently unstable plants (aircraft, inverted pendulums, chemical reactors) is one of the primary engineering reasons to use feedback control. The characteristic equation is the tool that shows exactly which controller parameters achieve this.

Question 4

Adding more gain to a stable feedback system generally makes it more robust, because a larger controller output provides stronger corrective action against disturbances.

• T. True
• F. False

Answer: False

High gain can destabilize a stable closed-loop system. As gain K increases, the characteristic equation's roots trace paths in the complex plane (the root locus). For many plants, this path crosses into the right half-plane at a critical gain K_c — beyond which the system oscillates with growing amplitude and is unstable. For example, G(s) = 1/(s(s+1)(s+2)) is stable for small K but becomes unstable beyond a critical gain found from Routh-Hurwitz analysis. 'More gain = more stability' is a common and dangerous misconception. The characteristic equation must be analyzed over the full range of design parameters, not just at a single operating point.

Question 5

Explain why the closed-loop stability of a feedback system cannot be determined by simply checking whether the open-loop plant poles are in the left half-plane.

Model answer: Closing a feedback loop creates a new effective system with different pole locations. The closed-loop poles are the roots of the characteristic equation 1 + G(s)C(s) = 0, not the poles of G(s) alone. An open-loop stable plant can become unstable when high gain or poor controller design pushes the characteristic equation's roots into the right half-plane — the closed-loop poles migrate as controller parameters change. Conversely, an open-loop unstable plant can be stabilized by feedback that moves the roots into the left half-plane. The open-loop poles are simply the roots of G(s)'s denominator; the characteristic equation creates a new denominator (1 + G(s)C(s)) whose roots govern the closed-loop natural modes and hence stability.

The intuitive confusion arises because students think of a 'stable plant' as intrinsically safe. But stability is a property of the closed-loop system, not the plant in isolation. Feedback fundamentally rewrites the system's characteristic equation — and the engineer's job is to ensure the new equation's roots are where desired. This is the entire motivation for techniques like root locus, Routh-Hurwitz, and Nyquist analysis: all are tools for analyzing the characteristic equation.