Impulse response and transfer function (GNC connection): Question bank

Before the traps, three quick pictures build the vocabulary these questions lean on, so you never have to jump to the parent note.

Ground vocabulary you'll need here

Figure — Impulse response and transfer function (GNC connection)

True or false — justify

The impulse response

h (t)

can be nonzero for

t < 0

in a physical system.

False for any causal system — the output cannot precede the kick, so

h (t) = 0

for

t < 0

is built into the definition. A nonzero

h

before

t = 0

would mean the system responded before it was hit.

L {δ (t)} = 1

, so the delta "has no effect".

False — the transform equalling

1

means the delta injects all frequencies equally with unit weight; it is the maximally rich input, which is exactly why its response is a full fingerprint.

For an LTI system,

u * h = h * u

True — Convolution is commutative (see the overlap-area figure above): the substitution

τ^{'} = t - τ

maps one integral onto the other, so it does not matter which curve you hold still.

The transfer function

H (s)

fully describes the system's response for any initial conditions.

False —

H (s) = Y / U

is the zero-state relation only. Nonzero

y (0)

contributes a separate zero-input term (

s Y - y (0)

pieces) that

H

does not carry.

A pole at

s = 0

means the system is stable because nothing blows up exponentially.

False — it is only marginally stable. The double integrator gives

h (t) = t

, which grows without bound linearly; unbounded is unbounded, exponential or not.

If all poles have

ℜ (p) < 0

the impulse response decays to zero.

True — each pole contributes a term

\sim e^{pt}

, and

ℜ (p) < 0

forces the envelope

e^{ℜ (p) t} \to 0

, so the ringing dies out.

A system with complex poles must be unstable.

False — complex poles produce oscillation (

sin

cos

factors), but stability depends only on the real part. Poles at

- ζ ω_{n} \pm j ω_{d}

with

ζ > 0

oscillate and decay.

The convolution theorem

L {u * h} = U (s) H (s)

holds for any signals regardless of their region of convergence.

False — it holds only where the ROCs overlap. Both being causal gives right-half-plane ROCs that intersect; content before

t = 0

or non-overlapping ROCs breaks the clean product rule.

Spot the error

"Since

δ (t)

has zero width,

\int_{0}^{\infty} e^{- s t} δ (t) d t = 0

, so

L {δ} = 0

The error is treating zero width as zero area. The delta has unit area, and sifting gives

\int e^{- s t} δ (t) d t = e^{- s \cdot 0} = 1

, not

0

"For

y^{'} + 2 y = u

the pole is at

s = + 2

, so the system is unstable."

Sign error:

s Y + 2 Y = U

gives

H = 1/ (s + 2)

, whose denominator vanishes at

s = - 2

. The pole is at

- 2 < 0

, so it is stable.

"The impulse response of

\overset{y}{¨} + 2 ζ ω_{n} \overset{y}{˙} + ω_{n}^{2} y = u

e^{- ζ ω_{n} t} sin (ω_{d} t)

(with damped frequency

ω_{d} = ω_{n} 1 - ζ^{2}

)."

Missing the

1/ ω_{d}

factor. Inverting

1/ ((s + ζ ω_{n})^{2} + ω_{d}^{2})

uses the pair with numerator

ω_{d}

, so you must divide:

h = \frac{1}{ω _{d}} e^{- ζ ω_{n} t} sin (ω_{d} t)

"To convolve, integrate

\int_{- \infty}^{\infty} u (τ) h (t - τ) d τ

always."

For a causal system with input starting at

0

, the limits collapse to

0

t

. Using

\pm \infty

silently includes impossible future contributions (

h (t - τ) = 0

for

τ > t

and

u = 0

for

τ < 0

anyway).

"Because

H (s) = 1/ s^{2}

has DC gain

H (0)

, the spacecraft settles to a fixed angle."

H (0) = 1/0

is undefined — there is no finite DC gain. The double integrator never settles; a constant torque ramps the angle forever, which is why it needs feedback.

H (s)

is the Laplace transform of the input."

No —

H (s) = L {h (t)}

, the transform of the impulse response, obtained by setting

U (s) = 1

(the delta's transform). The input's transform is

U (s)

Why questions

Why does knowing

h (t)

alone determine the output for every input?

Because any input is a sum of shifted, scaled impulses (sifting), and linearity + time-invariance send each impulse to a shifted, scaled copy of

h

; superposing gives

y = u * h

Why do we bother with the Laplace transform instead of just convolving directly?

Convolution is an integral you must recompute for every input; Laplace turns it into plain multiplication

Y = H U

, and turns derivatives into powers of

s

, converting the ODE into algebra.

Why does the complex variable

s

let poles reveal stability at a glance?

Because

s = σ + j ω

packages a decay rate and a frequency, a pole location directly says "the system rings at frequency

ω

while its envelope grows/decays at rate

σ

" — the sign of

σ

is stability.

Why are the poles the same as the roots of the characteristic polynomial?

The transfer function denominator is

a_{n} s^{n} + \dots + a_{0}

, and setting it to zero is exactly the characteristic equation of the homogeneous ODE. Same polynomial, same roots.

Why must a controller push poles into the left-half plane?

Poles govern the

e^{pt}

terms; only

ℜ (p) < 0

makes them decay. In GNC, a right-half-plane pole means the spacecraft's attitude error grows and it tumbles.

Why does time-invariance let us write the delta's shifted response as

h (t - τ)

Time-invariance says kicking at time

τ

instead of

0

produces the same shape just shifted; so the response to

δ (t - τ)

h

evaluated at

t - τ

, no reshaping.

Why is the transfer function useful for GNC design specifically?

It compresses a whole plant into a rational function whose pole locations instantly reveal stability and speed of response, letting engineers reshape them with feedback before ever firing a thruster.

Edge cases

What is

h (t)

if the input impulse is applied but the system has nonzero

y (0)

h (t)

is defined only at zero initial conditions; with nonzero

y (0)

the total output is

h * u

plus a separate zero-input response, so

h

itself is unchanged but is not the whole answer.

What happens to

h (t) = \frac{1}{ω _{d}} e^{- ζ ω_{n} t} sin (ω_{d} t)

ζ \to 1^{-}

ω_{d} = ω_{n} 1 - ζ^{2} \to 0

, and the ratio

\frac{s i n ( ω _{d} t )}{ω _{d}} \to t

, giving the critically-damped limit

h (t) = t e^{- ω_{n} t}

— a smooth, oscillation-free bridge.

What does a repeated pole (like

s = 0

twice) add to the response compared to a simple pole?

A polynomial factor: a double pole contributes a

t e^{pt}

term (hence

h = t

for

1/ s^{2}

), so repeated poles multiply the exponential by powers of

t

Is the system

y^{'} + 2 y = u

still stable if we add a step input instead of an impulse?

Yes — stability is a property of the poles, not the input. The step response

\frac{1}{2} (1 - e^{- 2 t})

settles to the finite DC gain

H (0) = 1/2

precisely because the pole stays at

- 2

What is the DC gain (steady output for a unit step) when a pole sits exactly at

s = 0

Undefined/infinite —

H (0)

diverges, meaning a constant input drives an unbounded output (the ramp of the double integrator), so no steady value exists.

For

0 < ζ < 1

, what geometric fact about the poles guarantees decay yet oscillation?

The poles

- ζ ω_{n} \pm j ω_{d}

lie in the left half-plane (negative real part

\Rightarrow

decay) but off the real axis (nonzero imaginary part

\Rightarrow

oscillation), as sketched in the pole picture above.

What is the ROC of a causal impulse response, and why does it matter here?

A right half-plane

σ > σ_{0}

where

σ_{0}

is the rightmost pole; it matters because

U (s) H (s) = L {u * h}

is only valid where the input's and system's ROCs overlap.

Connections

Parent: Impulse response & transfer function
Dirac delta function
Laplace transform
Convolution
Characteristic polynomial and roots
Linear constant-coefficient ODEs
Stability and poles (left-half plane)
Feedback control systems
Guidance Navigation and Control (GNC)