Worked examples — Impulse response and transfer function (GNC connection)

Before anything, three tiny reminders of notation, so no symbol is unearned:

Recall The three symbols we lean on

• $s$ ::: the Laplace variable. Think "a knob that measures how fast things grow ($\Re s>0$) or decay ($\Re s<0$)".
• $H(s)$ ::: the Laplace transform of the impulse response $h(t)$; also $Y(s)/U(s)$ at zero initial conditions.
• pole ::: a value of $s$ where the denominator of $H(s)$ is zero — the system's natural "notes". See Stability and poles (left-half plane).

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The scenario matrix

Every LTI ODE problem lives in one of these cells. The examples below are labelled by the cell(s) they hit.

Cell	What makes it distinct	Sign / limit that matters	Example
C1 Real negative pole	$H=\frac{1}{s+a}$, $a>0$	$h$ decays → stable	Ex 1
C2 Real positive pole	$H=\frac{1}{s-a}$, $a>0$	$h$ grows → unstable	Ex 2
C3 Pole at zero (integrator)	$H=\frac{1}{s}$	$h=$ const → marginal	Ex 3
C4 Repeated pole	$H=\frac{1}{s^2}$ or $\frac{1}{(s+a{)}^2}$	ramp / $t{e}^{-at}$	Ex 4
C5 Complex pair (underdamped)	$H=\frac{1}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$	decaying ring	Ex 5
C6 Purely imaginary pair	$H=\frac{1}{s^2+{\omega }^2}$	pure oscillation → marginal	Ex 6
C7 Non-impulse input via convolution	step / ramp input	steady-state limit	Ex 7
C8 Nonzero initial conditions	zero-input term added	trap: $H$ alone is wrong	Ex 8
C9 Word problem (GNC)	thruster → attitude	units & physical read	Ex 9
C10 Exam twist	numerator dynamics / degenerate	zero cancels pole	Ex 10

The figure above is the map: pole location in the complex $s$-plane on the left, the matching impulse-response shape on the right. Keep glancing back at it — every example is one dot moving on this map.

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Ex 1 — Real negative pole (C1)

Forecast: guess now — does the ringing die out, blow up, or stay flat? Where does a constant input settle?

1. Transform the ODE at zero state. $y^{\prime }\to sY$, so $sY+3Y=U$. Why this step? $\mathcal{L}\{y^{\prime }\}=sY-y(0)$ and $y(0)=0$ (zero state), turning calculus into algebra.
2. Divide out. $(s+3)Y=U\Rightarrow H(s)=\frac{1}{s+3}$. Why? $H=Y/U$ by definition; the single pole sits at $s=-3$.
3. Invert. Pair $\mathcal{L}\{e^{-at}\}=\frac{1}{s+a}$ with $a=3$ gives $h(t)=e^{-3t}$ for $t\ge 0$. Why? Matching to a known table pair is the fastest legal inverse.
4. DC gain $=H(0)=\frac{1}{3}$. Why? Setting $s=0$ probes the response to a constant (zero-frequency) input.

Verify: pole $-3<0$ → the map (C1 dot) sits in the left half-plane, $h$ decays. Convolve a step $u=1$: $y={\int }_0^t{e}^{-3(t-\tau )}d\tau =\frac{1}{3}(1-e^{-3t})\to \frac{1}{3}$ ✓ matches the DC gain.

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Ex 2 — Real positive pole: instability (C2)

Forecast: the only sign change from Ex 1 is a minus. Predict the fate of $h(t)$.

1. Transform. $sY-Y=U\Rightarrow H(s)=\frac{1}{s-1}$. Why? Same algebra; now the pole is at $s=+1$.
2. Invert. ${\mathcal{L}}^{-1}\{\frac{1}{s-1}\}=e^{+t}$, $t\ge 0$. Why? Table pair $\frac{1}{s-a}\to e^{at}$ with $a=+1$.
3. Read stability. $\Re (\text{pole})=+1>0$. Why? Any pole with positive real part injects a growing $e^{pt}$ — Characteristic polynomial and roots made physical.

Verify: $h(t)=e^t\to \infty$ as $t\to \infty$ — the C2 dot lives in the right half-plane. Sanity: at $t=1$, $h=e\approx 2.718>1$, already amplified. Unstable.

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Ex 3 — Pole exactly at zero: the pure integrator (C3)

Forecast: what does a system whose pole sits on the boundary do — decay, grow, or hover?

1. Transform. $sY=U\Rightarrow H(s)=\frac{1}{s}$. Why? One derivative → one factor of $s$; pole exactly at $s=0$.
2. Invert. ${\mathcal{L}}^{-1}\{1/s\}=1$ (the unit step) for $t\ge 0$. Why? $\mathcal{L}\{1\}=1/s$; the impulse response is a constant that never returns to zero.
3. Classify. $\Re (\text{pole})=0$ → marginally stable: doesn't grow, doesn't decay. Why? Boundary of the left-half plane; the Dirac delta function input leaves a permanent offset.

Verify: $h(t)=1$ for all $t>0$ — bounded but non-decaying. One kick permanently shifts the level by $1$, matching ${\int }_0^{\infty }\delta dt=1$ area.

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Ex 4 — Repeated pole: the double integrator (C4)

Forecast: two integrators stacked — does the angle settle, oscillate, or drift forever?

1. Transform. $s^2\Theta =U\Rightarrow H(s)=\frac{1}{s^2}$. Why? Two derivatives → $s^2$; a double pole at $s=0$.
2. Invert. ${\mathcal{L}}^{-1}\{1/s^2\}=t$, $t\ge 0$. Why? Pair $\mathcal{L}\{t\}=1/s^2$; the repeated root turns a constant into a ramp.
3. Classify. Repeated pole on the axis → unbounded ramp → not even marginal in the usable sense. Why? A single torque impulse makes the angle grow linearly forever; the craft never stops rotating.

Verify: $h(2)=2$, $h(4)=4$ — strictly increasing without bound. This is exactly why real attitude control needs Feedback control systems within Guidance Navigation and Control (GNC). (Compare Ex 3: one pole at zero → flat; two poles at zero → ramp.)

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Ex 5 — Complex pair: the damped ring (C5)

Forecast: will the fingerprint be a clean decay, or a decaying wiggle? Guess the frequency.

1. Transform. $H(s)=\frac{1}{s^2+2s+5}$. Why? Each $y^{(k)}\to s^{k}Y$ at zero state.
2. Match the standard form $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$. So ${\omega }_n^2=5\Rightarrow {\omega }_n=\sqrt{5}$, and $2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{1}{\sqrt{5}}\approx 0.447$. Why? This form exposes damping $\zeta$ and natural frequency ${\omega }_n$ directly.
3. Poles. $s=-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}=-1\pm 2j$ (since $0<\zeta <1$). Why? Under-damped → complex conjugates; real part $-1$ = decay rate, imaginary part $2$ = ring frequency ${\omega }_d$.
4. Invert. Complete the square: denominator $=(s+1{)}^2+2^2$. Pair $\mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a{)}^2+b^2}$ with $a=1,b=2$, then divide by $b=2$: $h(t)=\frac{1}{2}{e}^{-t}\sin (2t),t\ge 0.$ Why? Our numerator is $1$, the pair supplies $b$, so we scale by $1/b$.

Verify: ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=\sqrt{5}\cdot \sqrt{1-1/5}=\sqrt{5}\cdot \sqrt{4/5}=2$ ✓ matches the imaginary part. Real part $-1<0$ → decaying ring (C5 dot in left half-plane). The figure shows the envelope $\pm \frac{1}{2}{e}^{-t}$ squeezing the sine.

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Ex 6 — Purely imaginary pair: undamped oscillation (C6)

Forecast: with damping removed entirely, does the ring ever stop?

1. Transform. $H(s)=\frac{1}{s^2+9}$. Why? No $\dot{y}$ term → no real part in the poles.
2. Poles. $s^2=-9\Rightarrow s=\pm 3j$ — on the imaginary axis. Why? $\zeta =0$; the decay rate is exactly zero.
3. Invert. Pair $\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$ with $b=3$; divide by $3$: $h(t)=\frac{1}{3}\sin (3t),t\ge 0.$ Why? Numerator $1$, pair gives $b=3$, scale by $1/3$.
4. Classify. $\Re (\text{poles})=0$ → marginally stable — bounded but never decays.

Verify: amplitude $\frac{1}{3}$ never shrinks; peaks recur every period $2\pi /3$. On the map, C6 sits exactly on the axis between the decaying C5 and a growing right-plane pole.

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Ex 7 — Non-impulse input by convolution (C7)

Forecast: a growing input into a decaying system — does the output track the ramp, lag it, or blow up?

1. Set up causal convolution. $y(t)={\int }_0^{t}u(\tau )h(t-\tau )d\tau ={\int }_0^t\tau e^{-3(t-\tau )}d\tau$. Why? Causal $h$ ⇒ limits $0$ to $t$; never integrate over "future" $\tau >t$.
2. Pull out the constant $e^{-3t}$: $y=e^{-3t}{\int }_0^t\tau e^{3\tau }d\tau$. Why? $e^{-3(t-\tau )}=e^{-3t}{e}^{3\tau }$; separate what depends on $\tau$.
3. Integrate by parts. ${\int }_0^t\tau e^{3\tau }d\tau ={\left[\frac{\tau e^{3\tau }}{3}\right]}_0^t-{\int }_0^t\frac{e^{3\tau }}{3}d\tau =\frac{t{e}^{3t}}{3}-\frac{e^{3t}-1}{9}$. Why? Standard $\int \tau e^{a\tau }$; parts turns the polynomial factor away.
4. Multiply back. $y(t)=\frac{t}{3}-\frac{1}{9}+\frac{1}{9}{e}^{-3t}$. Why? Distribute $e^{-3t}$; the transient $\frac{1}{9}{e}^{-3t}$ dies, leaving a lagging ramp.

Verify: as $t\to \infty$, $y\approx \frac{t}{3}-\frac{1}{9}$: output tracks the ramp with slope $\frac{1}{3}$ (= DC gain from Ex 1) minus a steady offset $\frac{1}{9}$. At $t=0$: $y(0)=0-\frac{1}{9}+\frac{1}{9}=0$ ✓ (rest state). Cross-check via Laplace: $Y=H(s)U(s)=\frac{1}{s+3}\cdot \frac{1}{s^2}$, and partial fractions reproduce $\frac{1}{3s^2}-\frac{1}{9s}+\frac{1}{9(s+3)}$ ✓.

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Ex 8 — Nonzero initial conditions: the $H(s)$ trap (C8)

Forecast: with $u=0$ but $y(0)\ne 0$, does the transfer function $H=\frac{1}{s+3}$ tell you anything?

1. Transform WITH the IC term. $\mathcal{L}\{y^{\prime }\}=sY-y(0)=sY-4$. So $sY-4+3Y=0$. Why? Nonzero $y(0)$ keeps the $-y(0)$ piece we dropped in the zero-state cases.
2. Solve. $(s+3)Y=4\Rightarrow Y=\frac{4}{s+3}$. Why? The IC acts like an "internal source" — not an external input $U$.
3. Invert. $y(t)=4e^{-3t}$. Why? Same pole, but the amplitude comes from $y(0)$, not from $U$.
4. The lesson. Here $U(s)=0$, so $H\cdot U=0$ — the transfer function predicts zero, yet the true output is $4e^{-3t}$. Why? $H$ describes only the zero-state (forced) response; this is the zero-input response, a separate additive term.

Verify: $y(0)=4e^0=4$ ✓ and $y^{\prime }(0)+3y(0)=-12+12=0$ ✓ satisfies the ODE. Total response of a driven-with-IC problem = zero-state ($u\ast h$) plus this zero-input term.

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Ex 9 — Word problem: thruster to attitude (C9)

Forecast: after one clean torque pulse, does the craft settle to a fixed new angle, or keep drifting?

1. Transform. $H(s)=\frac{1}{s^2+0.4s}=\frac{1}{s(s+0.4)}$. Why? Poles at $s=0$ (integrator) and $s=-0.4$ (damped) — a rate that decays plus a position that holds.
2. Partial fractions. $\frac{1}{s(s+0.4)}=\frac{2.5}{s}-\frac{2.5}{s+0.4}$. Why? Split into a table-ready constant term and a decaying term; $\frac{1}{0.4}=2.5$.
3. Invert. $\theta (t)=h(t)=2.5(1-e^{-0.4t})$, $t\ge 0$. Why? $\frac{1}{s}\to 1$, $\frac{1}{s+0.4}\to e^{-0.4t}$.
4. Read final angle. ${\lim }_{t\to \infty }\theta =2.5$ rad. Why? The decaying rate term vanishes; the integrator's pole-at-zero locks in a permanent offset — physically, the wheel's spin was absorbed.

Verify: $\theta (0)=2.5(1-1)=0$ ✓ (starts at rest angle). Angular rate $\dot{\theta }(t)=e^{-0.4t}\to 0$ ✓ (craft stops turning). Final Value Theorem: $\lim sH(s)\cdot 1={\lim }_{s\to 0}\frac{1}{s+0.4}=2.5$ ✓. Units: torque-impulse/(rate coeff) → rad, consistent.

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Ex 10 — Exam twist: a zero that cancels a pole (C10)

Forecast: does the pole at $s=-2$ actually contribute to the impulse response, or is it a decoy?

1. Spot the common factor. Numerator $s+2$ cancels the $(s+2)$ in the denominator, for $s\ne -2$. Why? A zero (numerator root) at the same place as a pole cancels — the mode is "unobservable/uncontrollable"; it never appears in $h$.
2. Reduced transfer function. $H(s)=\frac{1}{s+5}$. Why? After cancellation only the genuine pole at $s=-5$ survives.
3. Invert. $h(t)=e^{-5t}$, $t\ge 0$. Why? Standard pair with $a=5$.
4. Stability read. Only pole is $-5<0$ → stable; the phantom $-2$ never mattered. Why? Cancelled poles carry zero coefficient in the Characteristic polynomial and roots expansion.

Verify: $h(0)=1$, $h(t)\to 0$ ✓. Cross-check: ${\int }_0^{\infty }hdt=\frac{1}{5}=H(0)=\frac{0+2}{(2)(5)}=\frac{2}{10}=\frac{1}{5}$ ✓ — DC gain agrees whether or not you cancel first.

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Recall One-line summary of the matrix

Sign of $\Re (\text{pole})$ decides fate — negative decays, zero holds/ramps (multiplicity!), positive blows up; complex pairs ring; and $H$ never sees initial conditions or cancelled modes.

Connections

• 4.6.33 Impulse response and transfer function (GNC connection) (Hinglish) (parent)
• 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)