Impulse response and transfer function (GNC connection)

4.6.33 · Maths › Ordinary Differential Equations

WHAT — hum kya solve kar rahe hain?

Hum ek Linear Time-Invariant (LTI) ODE padhte hain: $a_n{y}^{(n)}+\cdots +a_1{y}^{\prime }+a_{0}y=u(t)$

• $y(t)$ = system ka output (jaise attitude angle).
• $u(t)$ = input / forcing (jaise thruster command).
• Linear ⇒ inputs ko scale aur add karo toh outputs bhi scale aur add hote hain.
• Time-invariant ⇒ abhi kick karo ya baad mein — same shape milega, bas shift hoga.

WHY — impulse response SAB KUCH kyun deta hai? (Convolution)

WHAT chahiye: kisi bhi arbitrary input $u(t)$ ke liye output.

HOW banate hain (derivation scratch se):

1. Sifting se, koi bhi input shifted, scaled impulses ki sum ke roop mein likho: $u(t)={\int }_{-\infty }^{\infty }u(\tau )\delta (t-\tau )d\tau$ Yeh step kyun? Delta $u(\tau )$ ki value pick karta hai; integrate karne se poora signal spikes se reassemble ho jaata hai.

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

3. Linearity: input $u(\tau )\delta (t-\tau )$ (scaled spike) produce karta hai $u(\tau )h(t-\tau )$; $\tau$ ke upar sum (integrate) karne par: $\boxed{y(t)={\int }_{-\infty }^{\infty }u(\tau )h(t-\tau )d\tau =(u\ast h)(t)}$

Toh $h$ akela zero-state behaviour poora determine karta hai. Yahi 80/20 punchline hai.

Transfer Function $H(s)$

Convolutions tedious hote hain. Laplace transform unhe multiplication mein badal deta hai.

$H(s)$ ki derivation. $L[y]=u$ par $\mathcal{L}$ apply karo zero initial conditions ke saath. Har $y^{(k)}\to s^{k}Y(s)$: $(a_n{s}^n+\cdots +a_{1}s+a_0)Y(s)=U(s)$ $\boxed{H(s):=\frac{Y(s)}{U(s)}=\frac{1}{a_n{s}^n+\cdots +a_0}}$

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

Solve $y^{\prime }+2y=u$, $h(t)$ aur $H(s)$ nikalo.

• $H(s)$: $sY+2Y=U\Rightarrow H(s)=\frac{1}{s+2}$. Kyun? $y^{\prime }\to sY$ replace karo, divide karo.
• $h(t)$: ${\mathcal{L}}^{-1}\{1/(s+2)\}=e^{-2t}$ for $t\ge 0$. Kyun? Standard pair $\mathcal{L}\{e^{-at}\}=1/(s+a)$.
• Convolution se check: step input $u=1$ ke liye, $y={\int }_0^t{e}^{-2(t-\tau )}d\tau =\frac{1}{2}(1-e^{-2t})$. Yeh step kyun? Jaane huye $h$ ko input ke saath convolve karo; $1/2$ par settle hota hai = DC gain $H(0)=1/2$. ✓ Pole at $s=-2<0\Rightarrow$ stable.

Worked Example 2 — Spacecraft attitude (double integrator)

Inertia $J$ par thruster torque $u$: $J\ddot{\theta }=u$. $J=1$ lo.

• $H(s)$: $s^2\Theta =U\Rightarrow H(s)=\frac{1}{s^2}$. Kyun? Do derivatives ⇒ $s^2$.
• $h(t)$: ${\mathcal{L}}^{-1}\{1/s^2\}=t$ for $t\ge 0$. Kyun? Pair $\mathcal{L}\{t\}=1/s^2$. Torque ka ek impulse angle ko hamesha ke liye ramp karwa deta hai — $s=0$ par repeated pole ⇒ marginally unstable, isliye hum control add karte hain. Yahi exact reason hai ki attitude ko feedback chahiye.

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

$\ddot{y}+2\zeta {\omega }_n\dot{y}+{\omega }_n^{2}y=u$.

• $H(s)=\frac{1}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$. Poles $s=-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}$.
• $0<\zeta <1$ ke liye (underdamped): poles $=-\zeta {\omega }_n\pm j{\omega }_d$, ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$.
• $h(t)=\frac{1}{{\omega }_d}{e}^{-\zeta {\omega }_{n}t}\sin ({\omega }_{d}t)$. Kyun? $1/((s+\zeta {\omega }_n{)}^2+{\omega }_d^2)$ ka inverse pair $\mathcal{L}\{e^{-at}\sin bt\}=b/((s+a{)}^2+b^2)$ use karta hai, phir ${\omega }_d$ se divide karo. Decaying ring = woh "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)

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