Questions: Transfer Function Poles and Zeros Interpretation

Question 1

A control system has two poles: one at s = −0.05 (close to the imaginary axis) and one at s = −50 (far into the left half plane). Which statement best describes the step response?

• A. Both poles contribute equally throughout the transient because both are in the stable left half plane
• B. The s = −0.05 pole dominates — it produces a sluggishly decaying mode with a time constant of 20 seconds, while the s = −50 mode vanishes in milliseconds
• C. The s = −50 pole dominates because poles with larger magnitude exert stronger influence on the response
• D. The system is marginally stable because one pole is very close to the imaginary axis

Answer: (B)

The time constant of a real pole at s = σ is τ = 1/|σ|. The s = −0.05 pole has τ = 20 seconds; the s = −50 pole has τ = 0.02 seconds — the latter is essentially gone in about 0.06 seconds. The dominant pole is the one closest to the imaginary axis, because it is the slowest mode and controls settling time. In multi-pole systems, poles far into the left half plane can often be ignored in analysis because they decay before the slow modes have barely moved. This is the concept of dominant pole approximation: keep the poles near the imaginary axis, ignore the rest.

Question 2

A transfer function has a zero in the right half plane at s = +3. Compared to a minimum-phase system with identical pole locations, this non-minimum-phase system is distinguished by:

• A. Instability — right half plane zeros cause the output to grow without bound
• B. An identical Bode magnitude plot but additional phase lag — the extra phase drop creates fundamental limits on achievable closed-loop bandwidth
• C. A faster step response, because the RHP zero adds derivative-like action to the numerator
• D. Sustained oscillation at the frequency equal to the imaginary part of the RHP zero

Answer: (B)

Zeros do not determine stability — only poles do. A RHP zero makes the system non-minimum phase: the Bode magnitude plot may appear normal, but the phase response decreases more than the minimum-phase case. This extra phase lag shrinks the phase margin as loop gain increases, limiting how aggressively feedback can be applied before the closed loop goes unstable. A visible symptom is that the step response initially moves in the wrong direction (undershoot) before recovering — characteristic of systems like liquid-level processes or unstable aircraft where actuator effects are initially counterintuitive.

Question 3

A pole at s = −2 + j8 produces a step response component that oscillates at 8 rad/s and decays with a time constant of 0.5 seconds.

• T. True
• F. False

Answer: True

The real part σ = −2 controls the decay rate: time constant τ = 1/|σ| = 1/2 = 0.5 seconds. The imaginary part ω = 8 rad/s is the oscillation frequency of the damped sinusoidal mode. The contribution from this pole (together with its conjugate at s = −2 − j8) is of the form e^(−2t)·cos(8t + φ): a sinusoid at 8 rad/s that decays to 1/e of its initial amplitude in 0.5 seconds. The pole location directly encodes both the decay rate (real part) and oscillation frequency (imaginary part) of its associated transient mode.

Question 4

A pole at s = −0.1 + j15 indicates a rapidly decaying oscillatory mode because the real part is negative and the overall pole magnitude is large.

• T. True
• F. False

Answer: False

The negative real part guarantees eventual stability (the mode decays), but 'rapidly decaying' is wrong. The time constant is τ = 1/|σ| = 1/0.1 = 10 seconds. The oscillation frequency is 15 rad/s, corresponding to a period of about 0.42 seconds — so the response completes roughly 24 oscillations per time constant before the amplitude diminishes significantly. This is classic lightly damped behavior: many slow-decaying oscillations. Rapidly decaying oscillations require large |σ| (pole deep in the LHP). The pole magnitude alone does not determine decay speed — only the real part matters for the time constant.

Question 5

How does the location of a pole in the s-plane determine both the stability and the qualitative shape of the transient response? Address the real part and imaginary part separately.

Model answer: The real part σ of a pole at s = σ + jω determines stability and decay speed. If σ < 0 (left half plane), the associated mode decays exponentially — the system is stable. If σ > 0 (right half plane), the mode grows without bound — unstable. The magnitude |σ| controls speed: large |σ| means fast decay (small time constant τ = 1/|σ|), while small |σ| near the imaginary axis means sluggish decay and poor damping. The imaginary part ω determines the oscillation frequency of the transient. A purely real pole (ω = 0) produces a pure exponential response; complex conjugate poles at σ ± jω produce a damped sinusoid oscillating at ω rad/s. Poles exactly on the imaginary axis (σ = 0) produce sustained, undamped oscillations.

The power of the pole-zero picture is that it makes all this information readable at a glance from a geometric diagram, without computing the full inverse Laplace transform. Controller design via root locus is essentially the art of moving poles to desired (σ, ω) positions: far enough left for fast decay, close enough to the real axis for adequate damping, with no poles crossing into the right half plane.