Questions: Pole-Zero Plots and Stability Analysis

Question 1

A system has poles at s = −2 and s = 1 + j3. Is the system stable? Explain.

Model answer: No. The pole at s = 1 + j3 has a positive real part (σ = +1), placing it in the right half-plane. This contributes a growing mode e^t·cos(3t + φ), which grows without bound. A single RHP pole is sufficient to make the system unstable regardless of where other poles are.

Stability requires ALL poles in the open left half-plane. The pole at s = −2 contributes a decaying mode (stable), but the pole at s = 1 + j3 contributes an exponentially growing oscillation. The system's output will diverge for any nonzero initial condition or input.

Question 2

Two systems have complex pole pairs: System A at −0.5 ± j10, System B at −5 ± j10. Compare their transient responses.

Model answer: Both oscillate at approximately 10 rad/s, but System A decays much more slowly (time constant 1/0.5 = 2 s) with low damping ratio ζ ≈ 0.05, showing many cycles before settling. System B decays rapidly (time constant 1/5 = 0.2 s) with ζ ≈ 0.45, settling in less than one oscillation cycle.

The imaginary part of the pole determines the oscillation frequency (~ω_d ≈ Im(s) for lightly damped systems). The real part magnitude |σ| is the decay rate — larger |σ| means faster decay. System A's poles are close to the imaginary axis (lightly damped resonance); System B's poles are far left (heavily damped, fast settling). In a Bode plot, System A would show a sharp resonance peak near 10 rad/s; System B would show a broad, low peak.