A system's transfer function has poles at s = −1 ± 2j and a zero at s = +3. What is the stability of this system, and which feature determines it?
AUnstable — the zero in the right half-plane causes the system to diverge
BStable — only the pole locations determine stability, and both poles are in the left half-plane
CMarginally stable — the imaginary parts of the poles cause undamped oscillation
DUnstable — complex poles always lead to instability when paired with a right-half-plane zero
Stability is determined entirely by pole locations. The poles at s = −1 ± 2j have negative real parts (σ = −1), so each contributes a decaying sinusoidal term e^(−t)·cos(2t + φ) to the natural response — the system is stable. The zero at s = +3 is in the right half-plane (non-minimum-phase), which creates undershoot in the step response, but zeros do not cause instability. Mixing up poles and zeros on stability is the most common error in this topic.
Question 2 Multiple Choice
A closed-loop system has a pole at s = 2 + j. Describe the expected behavior of its step response.
AA decaying oscillation settling to the step value, because the imaginary part dominates
BA pure sinusoid at 1 rad/s that neither grows nor decays, because the real and imaginary parts balance
CAn exponentially growing oscillation — the positive real part causes divergence regardless of the imaginary part
DA step response identical to a first-order system — the imaginary part only affects frequency response
A pole at s = 2 + j contributes a term proportional to e^(2t)·cos(t + φ) to the response. The real part (+2) drives exponential growth; the imaginary part (1) modulates the oscillation frequency. Because the real part is positive, the envelope grows without bound — the system is unstable. No matter how small the imaginary part, a positive real part always causes divergence.
Question 3 True / False
A system with most poles in the left half-plane but a zero in the right half-plane (non-minimum-phase zero) is unstable.
TTrue
FFalse
Answer: False
False. Stability is determined solely by pole locations. A system with all poles in the left half-plane is stable regardless of where its zeros are. A right-half-plane (non-minimum-phase) zero causes the step response to initially move in the wrong direction (undershoot) before settling, and it limits how tightly you can close a feedback loop, but it does not cause instability on its own. Confusing zeros with poles on stability questions is a very common error.
Question 4 True / False
A system with a pair of purely imaginary poles (e.g., at s = ±3j) is called marginally stable because its natural response neither grows nor decays.
TTrue
FFalse
Answer: True
True. Purely imaginary poles at s = ±jω₀ contribute a pure sinusoid cos(ω₀t + φ) to the natural response — amplitude constant, neither increasing nor decreasing. The system oscillates indefinitely at frequency ω₀. This is called marginal stability: technically BIBO (bounded-input, bounded-output) stability fails because a sinusoidal input at ω₀ produces a response that grows without bound (resonance), but zero-input response is bounded. Most control systems aim for poles strictly in the left half-plane to guarantee convergence.
Question 5 Short Answer
What is the relationship between the location of a pole in the s-plane and the time-domain behavior it contributes to the system's natural response?
Think about your answer, then reveal below.
Model answer: Each pole p contributes a term of the form e^(pt) to the natural response. The real part of p (σ) determines growth or decay: if σ < 0, the term decays exponentially (stable contribution); if σ > 0, the term grows exponentially (unstable); if σ = 0, the term is a constant or pure sinusoid (marginally stable). The imaginary part (jω) determines oscillation frequency — complex conjugate pole pairs contribute damped sinusoids e^(σt)·cos(ωt + φ). The real part is the key: left half-plane means stability, right half-plane means instability.
This pole-to-time-domain correspondence is exact (via inverse Laplace transform partial fractions) and makes the s-plane a complete map of system dynamics. Engineers can read off decay rate (|σ|), oscillation frequency (ω), and damping ratio (ζ = −σ/|p|) directly from the pole location — no need to solve differential equations. This geometric intuition is why root locus and pole placement are such powerful design methods.