Impulse response and transfer function (GNC connection)

WHAT are we solving?

We study a Linear Time-Invariant (LTI) ODE: $a_n y^{(n)} + \dots + a_1 y' + a_0 y = u(t)$

• $y(t)$ = system output (e.g. attitude angle).
• $u(t)$ = input / forcing (e.g. thruster command).
• Linear ⇒ scaling & adding inputs scales & adds outputs.
• Time-invariant ⇒ kicking now vs. later gives the same shape, just shifted.

WHY does the impulse response give EVERYTHING? (Convolution)

WHAT we want: the output for an arbitrary input $u(t)$.

HOW we build it (derivation from scratch):

1. By sifting, write any input as a sum of shifted, scaled impulses: $u(t) = \int_{-\infty}^{\infty} u(\tau)\,\delta(t-\tau)\,d\tau$ Why this step? The delta picks out the value $u(\tau)$; integrating reassembles the whole signal from spikes.

2. Time-invariance: input $\delta(t-\tau)$ produces output $h(t-\tau)$.

3. Linearity: input $u(\tau)\,\delta(t-\tau)$ (scaled spike) produces $u(\tau)\,h(t-\tau)$; summing (integrating) over $\tau$: $\boxed{\,y(t) = \int_{-\infty}^{\infty} u(\tau)\,h(t-\tau)\,d\tau = (u * h)(t)\,}$

So $h$ alone determines all zero-state behaviour. That's the 80/20 punchline.

The Transfer Function $H(s)$

Convolutions are annoying. The Laplace transform turns them into multiplication.

Derivation of $H(s)$. Apply $\mathcal{L}$ to $L[y]=u$ with zero initial conditions. Each $y^{(k)}\to s^k Y(s)$: $(a_n s^n + \dots + a_1 s + a_0)\,Y(s) = U(s)$ $\boxed{\,H(s) := \frac{Y(s)}{U(s)} = \frac{1}{a_n s^n+\dots+a_0}\,}$

Worked Example 1 — First-order (a leaky tank / RC analog)

Solve $y' + 2y = u$, find $h(t)$ and $H(s)$.

• $H(s)$: $sY+2Y=U \Rightarrow H(s)=\dfrac{1}{s+2}$. Why? Replace $y'\to sY$, divide.
• $h(t)$: $\mathcal{L}^{-1}\{1/(s+2)\}=e^{-2t}$ for $t\ge0$. Why? Standard pair $\mathcal{L}\{e^{-at}\}=1/(s+a)$.
• Check by convolution: for a step input $u=1$, $y=\int_0^t e^{-2(t-\tau)}d\tau=\tfrac12(1-e^{-2t})$. Why this step? Convolve known $h$ with the input; settles to $1/2$ = DC gain $H(0)=1/2$. ✓ Pole at $s=-2<0 \Rightarrow$ stable.

Worked Example 2 — Spacecraft attitude (double integrator)

A thruster torque $u$ on inertia $J$: $J\ddot\theta = u$. Take $J=1$.

• $H(s)$: $s^2\Theta = U \Rightarrow H(s)=\dfrac{1}{s^2}$. Why? Two derivatives ⇒ $s^2$.
• $h(t)$: $\mathcal{L}^{-1}\{1/s^2\}=t$ for $t\ge0$. Why? Pair $\mathcal{L}\{t\}=1/s^2$. A single impulse of torque makes the angle ramp forever — repeated pole at $s=0$ ⇒ marginally unstable, so we add control. This is exactly why attitude needs feedback.

Worked Example 3 — Damped oscillator (a real GNC plant)

$\ddot y + 2\zeta\omega_n \dot y + \omega_n^2 y = u$.

• $H(s)=\dfrac{1}{s^2+2\zeta\omega_n s+\omega_n^2}$. Poles $s=-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}$.
• For $0<\zeta<1$ (underdamped): poles $=-\zeta\omega_n\pm j\omega_d$, $\omega_d=\omega_n\sqrt{1-\zeta^2}$.
• $h(t)=\dfrac{1}{\omega_d}e^{-\zeta\omega_n t}\sin(\omega_d t)$. Why? Inverse of $1/((s+\zeta\omega_n)^2+\omega_d^2)$ uses the pair $\mathcal{L}\{e^{-at}\sin bt\}=b/((s+a)^2+b^2)$, then divide by $\omega_d$. Decaying ring = the "fingerprint."

Active Recall

Connections

• 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)

Concept Map

Hinglish (regional understanding)