Transfer function — Laplace domain, poles and zeros: Question bank

Before you start, every symbol you'll meet below is defined once, right here, so nothing is used before it's earned.

And the geometry vocabulary, anchored to a picture:

$s = σ + j ω$ is a complex frequency: real part $σ$ = growth/decay rate, imaginary part $ω$ = oscillation frequency.
"LHP" = left half of the complex plane ( $σ < 0$ ); "RHP" = right half ( $σ > 0$ ); the vertical line $σ = 0$ is the imaginary axis.

The map above is the whole game: where a pole sits in this plane tells you how its mode moves in time. Every trap below is really a question about reading this map.

True or false — justify

A transfer function fully describes the system's response to any input.

False — only for zero initial conditions. The transfer function is the machine's fingerprint; a non-zero start state adds extra

f (0)

terms the transfer function deliberately throws away.

Two systems with identical poles always behave identically.

False — same poles means the same modes

e^{pt}

, but different zeros (and gain) reweight those modes, so the transient shapes and overshoot can differ.

Moving a pole further into the left half-plane always makes the whole system faster.

Not necessarily — it speeds up that mode (its time constant

τ = 1/∣ σ ∣

shrinks), but overall speed is set by the dominant (slowest, rightmost) pole. Dragging a far-left non-dominant pole even further left barely changes the response, because the sluggish rightmost mode still governs the tail. You only get faster overall when you move the dominant pole left.

If a system has more zeros than poles, that's fine and common.

False in practice — a physical (causal) system needs the denominator degree

\geq

numerator degree (proper); more zeros than poles implies pure differentiation, which amplifies noise without bound and isn't realisable.

A pole exactly on the imaginary axis (

s = j ω

) gives a stable response.

False — it's marginally stable:

e^{j ω t}

neither grows nor decays, so the system oscillates forever without settling. Not "stable" and not "unstable" — the knife's edge.

A real pole can produce oscillation.

False — oscillation needs a non-zero imaginary part (

ω \neq = 0

). A purely real pole

σ

gives a pure exponential

e^{σ t}

, monotone rise or decay, no ringing.

Cancelling a pole and a zero at the same location removes that mode entirely.

Dangerous-true — algebraically they cancel, but if the pole was unstable, the hidden mode still exists in the real hardware and can blow up; pole-zero cancellation on paper can mask a real instability.

The slowest pole dominates the long-term response.

True — the pole closest to the imaginary axis (smallest

∣ σ ∣

) has the longest time constant, so its

e^{σ t}

term lingers longest and dictates the tail of the response. This "dominant pole" idea is exactly why the previous trap needed care.

Adding a zero can never cause overshoot.

False — a zero adds a derivative-like term to the response; a zero close to the imaginary axis can inject a fast kick that overshoots the final value even in an otherwise well-damped system.

Spot the error

L {f^{'} (t)} = s F (s)

, always."

The full rule is

s F (s) - f (0)

(from integration by parts on the defining integral). Dropping

f (0)

is only legal because transfer functions assume zero initial conditions — you must state that, not assume it silently.

"This system has a pole in the RHP but a zero also in the RHP, so they balance and it's stable."

Only poles decide stability. A RHP pole means an

e^{+ σ t}

term that grows without bound; no zero can cancel that growth in the exponent — the zero only reweights coefficients.

G (s) = \frac{5}{s ^{2} - s - 6}

has all negative coefficients... wait, it has a

- 6

, but the leading coefficient is positive so it's stable."

Sign of the leading coefficient tells you nothing. Factor it:

(s - 3) (s + 2)

, giving a pole at

s = + 3

(RHP) → unstable. A missing or negative coefficient in

D (s)

is a red flag (see Stability - Routh-Hurwitz criterion).

"The zero at

s = - 3

\frac{s + 3}{( s + 1 ) ( s + 2 )}

sits left of both poles, so it slows the system down."

Zeros don't set decay rates — poles do. The zero at

s = - 3

shapes how the response gets going (its early curvature), not how fast it settles.

s

is just a dummy variable, so its units don't matter."

s

has units of

1/ s

(inverse seconds) because it stands in for

\frac{d}{d t}

. Its real part is a rate and imaginary part a frequency — that's why plotting poles on a plane with real/imaginary axes is physically meaningful.

"The transfer function tells me the output for input

u (t) = sin (2 t)

starting from a moving vehicle."

Not directly — if the vehicle already has non-zero velocity/position, those initial conditions add response terms the transfer function omits. It gives the forced response from rest only.

Why questions

Why does a pole become

e^{pt}

in the time domain?

Because of a three-step chain (see figure s02): (1) partial fractions rewrite

G (s)

as a sum

\sum_{k} \frac{A _{k}}{s - p _{k}}

; (2) each simple term

\frac{1}{s - p}

is, by the defining integral, exactly the transform of

e^{pt}

(check:

\int_{0}^{\infty} e^{pt} e^{- s t} d t = \frac{1}{s - p}

); (3) so inverting term-by-term gives

\sum_{k} A_{k} e^{p_{k} t}

. The pole location

p

picks the exponent, the coefficient

A_{k}

picks the weight.

Why do we insist on zero initial conditions when defining a transfer function?

So the answer depends only on the system, not on one particular starting state. The transfer function is meant to be a reusable fingerprint of the machine.

Why is

e^{- s t}

the right kernel for the Laplace integral?

It is the eigenfunction of differentiation (

\frac{d}{d t} e^{- s t} = - s e^{- s t}

), which is exactly what converts derivatives into multiplication by

s

— turning calculus into algebra.

Why do zeros shape the transient but not stability?

Zeros never appear in the exponents

e^{p_{k} t}

— those come from poles. Zeros only set the coefficients

A_{k}

multiplying each mode, so they change the shape/weighting, not whether anything grows.

Why can a right-half-plane zero cause "wrong-way" initial motion?

A RHP zero adds a response term with the opposite sign early on, so the output first moves away from its final value before correcting — the non-minimum-phase aircraft that sinks when you pull up.

Why prefer the transfer function over solving the ODE each time?

It converts a differential equation into a polynomial ratio you read off once; you then predict any input's response by algebra and by inspecting pole locations, without re-integrating.

Edge cases

Each edge case below is a special place on the complex-plane map from figure s01 — the pole-response gallery (figure s03) shows what these look like in time.

What does a pole at the origin,

s = 0

, mean physically?

It contributes

e^{0 \cdot t} = 1

, a constant — this is an integrator. The output holds/accumulates rather than decaying; marginally stable, and it kills steady-state error in feedback loops (see Block diagrams and feedback loops).

What happens with a repeated pole, e.g.

\frac{1}{( s + 1 ) ^{2}}

Partial fractions produce a

t e^{- t}

term alongside a plain exponential. The extra factor of

t

makes the response rise then decay (it doesn't jump instantly), but the exponential envelope still decides stability — still LHP, still settling.

What if the denominator is a constant (no

s

), like

G (s) = \frac{s + 1}{4}

There are no finite poles, so no dynamic modes. Written out,

Y (s) = \frac{1}{4} (s + 1) U (s)

, which in time is

y (t) = \frac{1}{4} (\overset{u}{˙} (t) + u (t))

— a scaled copy of the input plus its derivative. So a step input gives an instantaneous spike (the

\overset{u}{˙}

term) atop a scaled step: memoryless, and the bare

s

makes it improper/non-physical on its own.

Complex poles must come in what arrangement, and why?

In conjugate pairs

σ \pm j ω

, because the system's coefficients are real. The two

e^{(σ \pm j ω) t}

terms combine into a real

e^{σ t} cos (ω t + ϕ)

— a decaying (or growing) oscillation with no leftover imaginary part.

What does

ζ = 0

do to the second-order poles

s = - ζ ω_{n} \pm j ω_{n} 1 - ζ^{2}

The real part vanishes, leaving poles at

s = \pm j ω_{n}

on the imaginary axis — undamped, ringing forever at

ω_{n}

. Marginally stable, the boundary between decay and growth (see Second-order systems - damping ratio and natural frequency).

What does

ζ > 1

(overdamping) do to those poles?

The square root

1 - ζ^{2}

becomes imaginary, so

j ω_{n} 1 - ζ^{2}

turns into a real

\pm ω_{n} ζ^{2} - 1

— giving two real negative poles. No oscillation at all, just two decaying exponentials; the system crawls to its target without overshoot.

What happens as a stable pole is dragged toward the imaginary axis?

Its time constant

τ = 1/∣ σ ∣

grows without bound; the mode decays ever more slowly, and at

σ = 0

it stops decaying entirely — the onset of marginal stability.

The whole page in two rules

Figure s03 is your visual cheat-sheet: the left column is a pole plotted on the plane, the right column is the time-response it produces. Read it left-to-right and the two rules below become obvious — you can literally see stability living in the poles.

Recall Self-test: did you catch the pattern?

Every trap above reduces to two rules. Name them. Rule 1 ::: Poles decide stability (only they sit in the exponents $e^{pt}$ ; LHP = decaying = stable, RHP = growing = unstable, imaginary axis = marginal). Rule 2 ::: Zeros (and gain) only shape the transient (they reweight coefficients, never create or remove growth).

Connections

Parent topic — the full build-up of $G (s)$ , poles and zeros
Laplace transform — the $L {\cdot}$ operator and kernel defined above, in depth
Stability - Routh-Hurwitz criterion — decides pole side (LHP vs RHP) without factoring $D (s)$
Root locus — a plane-plot showing how the poles migrate as a gain knob turns, so you can see stability being won or lost in real time
Second-order systems - damping ratio and natural frequency — where $ζ$ and $ω_{n}$ place the poles
Block diagrams and feedback loops — how transfer functions multiply along signal paths