Impulse response and transfer function (GNC connection): Exercises

Every symbol used below was built in the parent note, but let us restate the tools we lean on hardest so line one is readable by anyone:

Definition Region of convergence (ROC) — the edge case we must not skip

The Laplace integral $\int_{0}^{\infty} e^{- s t} f (t) d t$ only converges (gives a finite number) for some values of $s$ . For a signal that grows no faster than $e^{c t}$ , the integral converges when $ℜ (s) > c$ ; that set of $s$ is the region of convergence. For every causal impulse response here ( $h = 0$ for $t < 0$ ), the ROC is the half-plane to the right of the rightmost pole. Two consequences we rely on silently in every exercise:

The inversion $L^{- 1}$ back to a causal $h (t)$ is only valid when we take that right-side ROC; that is why we always write " $t \geq 0$ " and never a left-sided answer.
The convolution theorem $L {u * h} = U (s) H (s)$ and the pairs we quote are guaranteed only where the ROCs overlap — which, for causal inputs and causal stable systems, always includes $ℜ (s) > 0$ . For stable systems (all poles in $ℜ (s) < 0$ ) the ROC even contains the imaginary axis $ℜ (s) = 0$ , which is exactly why the DC gain $H (0)$ is a legitimate number to read off.

Level 1 — Recognition

Exercise 1.1

Read off the transfer function. For the ODE $3 y^{'} + 6 y = u$ , what is $H (s)$ ? Where is its pole?

Recall Solution 1.1

WHAT we do: take the Laplace transform of both sides, replacing $y^{'} \to s Y$ . WHY zero initial conditions: the exact rule is $L {y^{'}} = s Y (s) - y (0)$ (this comes from integration by parts on the Laplace integral). By assuming the system is at rest before $t = 0$ , i.e. $y (0) = 0$ , that $- y (0)$ term vanishes and the derivative maps cleanly to $s Y$ . This is essential: the transfer function describes only the zero-state (forced) response, so we must drop the initial-condition terms deliberately. $3 s Y + 6 Y = U \Rightarrow (3 s + 6) Y = U \Rightarrow H (s) = \frac{Y}{U} = \frac{1}{3 s + 6} .$ Pole: set the denominator to zero: $3 s + 6 = 0 \Rightarrow s = - 2$ . WHY it matters: the pole is at $s = - 2$ , real and negative, so the impulse response decays like $e^{- 2 t}$ — stable (and its ROC $ℜ (s) > - 2$ contains the imaginary axis).

Exercise 1.2

Identify the impulse response. Given $H (s) = \frac{1}{s + 5}$ , what is $h (t)$ ?

Recall Solution 1.2

WHAT we do: invert the Laplace transform. We recognise the standard pair $L {e^{- a t}} = \frac{1}{s + a}$ , here with $a = 5$ . $h (t) = L^{- 1} {\frac{1}{s + 5}} = e^{- 5 t}, t \geq 0.$ WHY this pair and not another: $H$ has a single real pole at $s = - 5$ , and a single real pole always inverts to a single decaying exponential $e^{pt}$ with $p$ the pole. No sine, no ramp — those come from complex or repeated poles. Choosing the causal (right-sided, $t \geq 0$ ) inverse fixes the ROC as $ℜ (s) > - 5$ .

Exercise 1.3

Recognise the delta. What is $L {δ (t)}$ , and hence what is $H (s)$ if the input is a unit impulse and the measured output transform is $Y (s) = \frac{2}{s ^{2} + 1}$ ?

Recall Solution 1.3

Step 1: $L {δ (t)} = \int_{0}^{\infty} e^{- s t} δ (t) d t = e^{0} = 1$ by the sifting property of the Dirac delta function. Step 2: since $Y = H \cdot U$ and $U (s) = L {δ} = 1$ , $H (s) = \frac{Y ( s )}{U ( s )} = \frac{Y ( s )}{1} = \frac{2}{s ^{2} + 1} .$ WHY: feeding a delta is exactly how you read off $H$ directly, because $U (s) = 1$ leaves $Y = H$ .

Level 2 — Application

Exercise 2.1

Solve for $h (t)$ from an ODE. Find the impulse response of $y^{'} + 4 y = u$ and confirm $h (0^{+}) = 1$ .

Recall Solution 2.1

Step 1 (to Laplace world): using $L {y^{'}} = s Y - y (0)$ with $y (0) = 0$ , $s Y + 4 Y = U \Rightarrow H (s) = \frac{1}{s + 4}$ . Step 2 (back to time): $h (t) = L^{- 1} {1/ (s + 4)} = e^{- 4 t}$ for $t \geq 0$ (ROC $ℜ (s) > - 4$ ). Step 3 (check $h (0^{+})$ ): $h (0^{+}) = e^{0} = 1$ . ✓ WHY the check works: plug $t = 0$ directly into $h (t) = e^{- 4 t}$ : $e^{- 4 \cdot 0} = 1$ . More generally, a first-order system $a_{1} y^{'} + a_{0} y = u$ has $H (s) = \frac{1}{a _{1} s + a _{0}}$ , whose impulse response starts at $h (0^{+}) = \frac{1}{a _{1}}$ — here $a_{1} = 1$ (the coefficient of $y^{'}$ ), so $h (0^{+}) = 1$ .

Exercise 2.2

Step response by convolution. For $y^{'} + 4 y = u$ with a unit step input $u (t) = 1$ (for $t \geq 0$ ), compute $y (t)$ using $y = u * h$ .

Recall Solution 2.2

WHAT we do: convolve the impulse response $h (t) = e^{- 4 t}$ with the input, using causal limits $0$ to $t$ (see Convolution). The causal limits are legitimate exactly because both signals live in the right-sided ROC. $y (t) = \int_{0}^{t} u (τ) h (t - τ) d τ = \int_{0}^{t} 1 \cdot e^{- 4 (t - τ)} d τ .$ Step: substitute $v = t - τ$ , $d v = - d τ$ ; as $τ : 0 \to t$ , $v : t \to 0$ : $y (t) = \int_{0}^{t} e^{- 4 v} d v = [\frac{- e ^{- 4 v}}{4}]_{0}^{t} = \frac{1 - e ^{- 4 t}}{4} .$ Sanity check: as $t \to \infty$ , $y \to \frac{1}{4} = H (0) = \frac{1}{0 + 4}$ , the DC gain. ✓

Exercise 2.3

Two derivatives. Find $H (s)$ and $h (t)$ for $\overset{y}{¨} + 3 \overset{y}{˙} + 2 y = u$ .

Recall Solution 2.3

Step 1: $s^{2} Y + 3 s Y + 2 Y = U \Rightarrow H (s) = \frac{1}{s ^{2} + 3 s + 2} = \frac{1}{( s + 1 ) ( s + 2 )}$ . Step 2 (partial fractions): find $A, B$ with $\frac{1}{( s + 1 ) ( s + 2 )} = \frac{A}{s + 1} + \frac{B}{s + 2}$ . Cover-up: $A = \frac{1}{( - 1 + 2 )} = 1$ , $B = \frac{1}{( - 2 + 1 )} = - 1$ . WHY partial fractions: each simple pole inverts to one exponential, but only after we split the fraction into single-pole pieces. Step 3 (invert): $h (t) = e^{- t} - e^{- 2 t}, t \geq 0.$

Level 3 — Analysis

Exercise 3.1

Classify stability from poles. For each $H (s)$ decide stable / marginal / unstable: (a) $\frac{1}{s ^{2} + 2 s + 5}$ , (b) $\frac{1}{s ^{2} + 4}$ , (c) $\frac{1}{s ^{2} - s - 2}$ .

Recall Solution 3.1

A system is stable iff every pole has $ℜ (p) < 0$ (strictly left of the imaginary axis). The figure plots all three cases as $\times$ marks on the complex plane so you can see the verdict: green crosses (a) sit in the shaded left half-plane, yellow crosses (b) sit exactly on the vertical imaginary axis, and the red cross (c) at $s = + 2$ pokes into the forbidden right half-plane.

(a) poles $s = \frac{- 2 \pm 4 - 20}{2} = - 1 \pm 2 j$ . Real part $- 1 < 0$ → stable (damped oscillation). (b) poles $s = \pm 2 j$ . Real part $= 0$ → marginal (pure oscillation that never decays). (c) $s^{2} - s - 2 = (s - 2) (s + 1)$ , poles $s = 2$ and $s = - 1$ . One pole has $ℜ = + 2 > 0$ → unstable (blows up like $e^{2 t}$ ).

Figure — Impulse response and transfer function (GNC connection)

Reading the figure: the horizontal axis is $ℜ (s)$ , the vertical axis $ℑ (s)$ ; the light-green band is the stable region $ℜ (s) < 0$ . A case is stable only if every cross of its colour lands inside that band.

Exercise 3.2

Analyse an underdamped ring. For $\overset{y}{¨} + 2 \overset{y}{˙} + 5 y = u$ find $ζ$ , $ω_{n}$ , $ω_{d}$ , and $h (t)$ .

Recall Solution 3.2

WHAT we do: match to the standard form $\overset{y}{¨} + 2 ζ ω_{n} \overset{y}{˙} + ω_{n}^{2} y = u$ . Compare coefficients: $ω_{n}^{2} = 5 \Rightarrow ω_{n} = 5$ ; $2 ζ ω_{n} = 2 \Rightarrow ζ = \frac{1}{ω _{n}} = \frac{1}{5} \approx 0.447$ . Since $0 < ζ < 1$ it is underdamped. Damped frequency: $ω_{d} = ω_{n} 1 - ζ^{2} = 5 \cdot 1 - \frac{1}{5} = 5 \cdot \frac{4}{5} = 2.$ Impulse response (parent's underdamped pair): $h (t) = \frac{1}{ω _{d}} e^{- ζ ω_{n} t} sin (ω_{d} t) = \frac{1}{2} e^{- t} sin (2 t), t \geq 0.$ WHY $e^{- t}$ : $ζ ω_{n} = \frac{1}{5} \cdot 5 = 1$ , so the decay rate is exactly $1$ — matching the real part of the poles $- 1 \pm 2 j$ from 3.1(a). The figure shows this fingerprint: the blue curve is $h (t)$ oscillating at $ω_{d} = 2$ , hugged by the yellow dashed envelope $\pm \frac{1}{2} e^{- t}$ whose decay rate $1$ is the pole's real part. The ring dies out, so the system is stable — you can literally read stability off the shrinking envelope.

Reading the figure: horizontal axis is time $t$ , vertical axis the impulse response $h (t)$ ; the dashed envelope's shrink rate equals $ζ ω_{n} = 1$ , and the wiggle period equals $2 π / ω_{d} = π$ .

Exercise 3.3

Effect of a pole crossing. In $\overset{y}{¨} + 2 b \overset{y}{˙} + y = u$ , for which real $b$ is the system unstable? What happens at $b = 0$ ?

Recall Solution 3.3

Poles: $s = \frac{- 2 b \pm 4 b ^{2} - 4}{2} = - b \pm b^{2} - 1$ . Real part governs stability.

If $b > 0$ : for $∣ b ∣ < 1$ poles are $- b \pm j 1 - b^{2}$ , real part $- b < 0$ → stable. For $b \geq 1$ , both roots real and negative (their product is $+ 1$ , sum $- 2 b < 0$ ) → stable.
If $b = 0$ : poles $\pm j$ , real part $0$ → marginal (undamped sinusoid, no decay).
If $b < 0$ : real part becomes $- b > 0$ → unstable. Conclusion: unstable for $b < 0$ ; marginal at $b = 0$ ; stable for all $b > 0$ . The damping term's sign is the whole story.

Level 4 — Synthesis

Exercise 4.1

Design a pole location. The plant $\overset{y}{˙} = u$ (a pure integrator, $H_{p} (s) = 1/ s$ ) is only marginal. We wrap proportional feedback $u = - k y$ (so $\overset{y}{˙} = - k y$ ). Choose $k$ so the closed-loop pole sits at $s = - 3$ . Then give the closed-loop impulse response.

Recall Solution 4.1

WHAT feedback does: substitute $u = - k y$ into $\overset{y}{˙} = u$ : $\overset{y}{˙} + k y = 0 \Rightarrow closed-loop pole at s = - k .$ (For the forced response, the closed-loop transfer function from a reference $r$ is $\frac{H _{p}}{1 + k H _{p}} = \frac{1/ s}{1 + k / s} = \frac{1}{s + k}$ .) See Feedback control systems. Choose the pole: we want $s = - 3$ , so set $k = 3$ . Closed-loop impulse response: $H_{c l} (s) = \frac{1}{s + 3} \Rightarrow h_{c l} (t) = e^{- 3 t}$ , $t \geq 0$ . WHY this is the punchline: an open-loop marginal system (pole at $0$ ) is stabilised purely by choosing feedback gain $k > 0$ ; larger $k$ pushes the pole further left (faster decay) — exactly the GNC idea from the parent's double-integrator remark.

Exercise 4.2

Stabilise the double integrator. A spacecraft's attitude angle — call it $θ (t)$ , the angle by which the vehicle is rotated from its target orientation — obeys $\ddot{θ} = u$ (torque $u$ on unit inertia; $H_{p} = 1/ s^{2}$ , poles both at $s = 0$ — see Guidance Navigation and Control (GNC)). Apply PD control $u = - k_{p} θ - k_{d} \dot{θ}$ . Find the closed-loop characteristic polynomial and choose $k_{p}, k_{d}$ for poles at $s = - 2 \pm 2 j$ .

Recall Solution 4.2

Here $θ$ is the output we want to steer to zero, $\dot{θ}$ its rate (angular velocity), and $u$ the commanded torque. Step 1 (close the loop): $\ddot{θ} = u = - k_{p} θ - k_{d} \dot{θ}$ gives $\ddot{θ} + k_{d} \dot{θ} + k_{p} θ = 0.$ Characteristic polynomial: $s^{2} + k_{d} s + k_{p} = 0$ . Step 2 (target poles): we want roots $- 2 \pm 2 j$ . Build the polynomial with those roots: $(s + 2 - 2 j) (s + 2 + 2 j) = (s + 2)^{2} + 2^{2} = s^{2} + 4 s + 8.$ Step 3 (match coefficients): $k_{d} = 4$ , $k_{p} = 8$ . WHY PD and not P alone: a pure $k_{p}$ gives $s^{2} + k_{p} = 0$ , poles $\pm j k_{p}$ — still marginal (undamped oscillation). The derivative term $k_{d} \dot{θ}$ supplies damping (the $s^{1}$ coefficient), pulling the poles off the imaginary axis into the stable left half-plane. Position feedback alone cannot damp; you need rate feedback too.

Level 5 — Mastery

Exercise 5.1

Full pipeline: ODE → $H$ → $h$ → step response → stability verdict. For $\overset{y}{¨} + 2 \overset{y}{˙} + 2 y = u,$ (a) find $H (s)$ and its poles, (b) find $h (t)$ , (c) find the step response $y_{step} (t)$ (input $u = 1$ , $t \geq 0$ ) by Laplace, (d) give the final value and the stability verdict.

Recall Solution 5.1

(a) Transfer function & poles. $H (s) = \frac{1}{s ^{2} + 2 s + 2} = \frac{1}{( s + 1 ) ^{2} + 1} .$ Poles: $s = \frac{- 2 \pm 4 - 8}{2} = - 1 \pm j$ . Both have $ℜ = - 1 < 0$ . Here $ζ ω_{n} = 1$ and $ω_{d} = 1$ (from $(s + 1)^{2} + 1$ ), so $ω_{n} = 2$ , $ζ = 1/ 2 \approx 0.707$ .

(b) Impulse response. Match the pair $L {e^{- a t} sin b t} = \frac{b}{( s + a ) ^{2} + b ^{2}}$ with $a = 1, b = 1$ : $H (s) = \frac{1}{( s + 1 ) ^{2} + 1} = \frac{1}{1} \cdot \frac{1}{( s + 1 ) ^{2} + 1} \Rightarrow h (t) = e^{- t} sin t, t \geq 0.$ (Since $b = 1$ , no division needed.)

(c) Step response by Laplace. With $U (s) = L {1} = 1/ s$ : $Y (s) = H (s) U (s) = \frac{1}{s ( s ^{2} + 2 s + 2 )} .$ Partial fractions $\frac{1}{s ( s ^{2} + 2 s + 2 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 2 s + 2}$ . Multiply out: $1 = A (s^{2} + 2 s + 2) + (B s + C) s$ .

$s = 0$ : $1 = 2 A \Rightarrow A = \frac{1}{2}$ .
$s^{2}$ coeff: $0 = A + B \Rightarrow B = - \frac{1}{2}$ .
$s^{1}$ coeff: $0 = 2 A + C \Rightarrow C = - 1$ . So $Y (s) = \frac{1/2}{s} - \frac{\frac{1}{2} s + 1}{s ^{2} + 2 s + 2}$ . Rewrite the second term over $(s + 1)^{2} + 1$ : $\frac{1}{2} s + 1 = \frac{1}{2} (s + 1) + \frac{1}{2}$ , so $Y (s) = \frac{1/2}{s} - \frac{1}{2} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} - \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + 1} .$ Invert term by term (using $L {e^{- t} cos t} = \frac{s + 1}{( s + 1 ) ^{2} + 1}$ and $L {e^{- t} sin t} = \frac{1}{( s + 1 ) ^{2} + 1}$ ): $y_{step} (t) = \frac{1}{2} - \frac{1}{2} e^{- t} cos t - \frac{1}{2} e^{- t} sin t, t \geq 0.$

(d) Final value & verdict. As $t \to \infty$ , the $e^{- t}$ terms vanish: $y_{step} (\infty) = \frac{1}{2} = H (0) = \frac{1}{0 + 0 + 2} . ✓$ Poles at $- 1 \pm j$ have negative real part → stable, an underdamped decay to the DC gain $1/2$ . (Stability places the imaginary axis inside the ROC, so $H (0)$ is a valid number to read.)

Exercise 5.2

Design + verify. A GNC channel has plant $H_{p} (s) = \frac{1}{s ^{2}}$ , whose output is again the attitude angle $θ (t)$ (rotation from target). Using PD control $u = - k_{p} θ - k_{d} \dot{θ}$ , choose gains so the closed loop has $ω_{n} = 3$ and $ζ = 1$ (critically damped, no overshoot). Then state the repeated pole and the impulse response.

Recall Solution 5.2

Closed-loop char. poly: from Ex 4.2, $s^{2} + k_{d} s + k_{p}$ . Match to $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ :

$ω_{n}^{2} = 9 \Rightarrow k_{p} = 9$ .
$2 ζ ω_{n} = 2 (1) (3) = 6 \Rightarrow k_{d} = 6$ . Poles: $s^{2} + 6 s + 9 = (s + 3)^{2} = 0$ → repeated pole $s = - 3$ (critically damped). Impulse response: $H_{c l} (s) = \frac{1}{( s + 3 ) ^{2}}$ , and using the pair $L {t e^{- a t}} = \frac{1}{( s + a ) ^{2}}$ with $a = 3$ : $h_{c l} (t) = t e^{- 3 t}, t \geq 0.$ WHY critically damped: $ζ = 1$ is the boundary between overshoot ( $ζ < 1$ , ringing) and sluggishness ( $ζ > 1$ ); the response rises to target as fast as possible without oscillating — ideal for a settling manoeuvre. The repeated real pole produces the $t e^{- 3 t}$ shape: rises, peaks, then decays.

Recall Quick self-check ledger (answers only)

1.1 $H = \frac{1}{3 s + 6}$ , pole $- 2$ ::: 1.2 $h = e^{- 5 t}$ ::: 1.3 $H = \frac{2}{s ^{2} + 1}$ 2.1 $h = e^{- 4 t}$ ::: 2.2 $y = \frac{1 - e ^{- 4 t}}{4}$ ::: 2.3 $h = e^{- t} - e^{- 2 t}$ 3.1 stable / marginal / unstable ::: 3.2 $ζ = 1/ 5, ω_{n} = 5, ω_{d} = 2, h = \frac{1}{2} e^{- t} sin 2 t$ ::: 3.3 unstable $b < 0$ , marginal $b = 0$ 4.1 $k = 3, h = e^{- 3 t}$ ::: 4.2 $k_{d} = 4, k_{p} = 8$ 5.1 $h = e^{- t} sin t$ , step $\to \frac{1}{2}$ , stable ::: 5.2 $k_{p} = 9, k_{d} = 6$ , pole $- 3$ (double), $h = t e^{- 3 t}$

Connections

Parent: Impulse Response and 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)