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

4.6.33 · D3 · Maths › Ordinary Differential Equations › Impulse response and transfer function (GNC connection)

Shuru karne se pehle, notation ke teen chhote reminders, taaki koi symbol unexplained na rahe:

Recall Teen symbols jinka hum sahara lete hain

• $s$ ::: Laplace variable. Socho "ek knob jo measure karta hai ki cheezein kitni tezi se badhti hain ($\Re s>0$) ya decay hoti hain ($\Re s<0$)".
• $H(s)$ ::: impulse response $h(t)$ ka Laplace transform; zero initial conditions par $Y(s)/U(s)$ bhi hai.
• pole ::: $s$ ki woh value jahan $H(s)$ ka denominator zero ho jaata hai — system ke natural "notes". Dekho Stability and poles (left-half plane).

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Scenario matrix

Har LTI ODE problem in cells mein se kisi ek mein fit hoti hai. Neeche ke examples unke cells ke label se marked hain.

Cell	Kya cheez isse alag banati hai	Sign / limit jo matter karta hai	Example
C1 Real negative pole	$H=\frac{1}{s+a}$, $a>0$	$h$ decay karta hai → stable	Ex 1
C2 Real positive pole	$H=\frac{1}{s-a}$, $a>0$	$h$ badhta hai → unstable	Ex 2
C3 Pole at zero (integrator)	$H=\frac{1}{s}$	$h=$ const → marginal	Ex 3
C4 Repeated pole	$H=\frac{1}{s^2}$ ya $\frac{1}{(s+a{)}^2}$	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 add hota hai	trap: $H$ akela galat hai	Ex 8
C9 Word problem (GNC)	thruster → attitude	units & physical read	Ex 9
C10 Exam twist	numerator dynamics / degenerate	zero pole ko cancel karta hai	Ex 10

Upar ka figure map hai: baayein complex $s$-plane mein pole location, daayein matching impulse-response shape. Baar baar isko dekhte rehna — har example is map par ek dot move karne jaisa hai.

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Ex 1 — Real negative pole (C1)

Forecast: abhi andaza lagao — kya ringing mit jaayegi, blow up hogi, ya flat rahegi? Ek constant input kahan settle karega?

1. ODE ko zero state par transform karo. $y^{\prime }\to sY$, toh $sY+3Y=U$. Kyun? $\mathcal{L}\{y^{\prime }\}=sY-y(0)$ aur $y(0)=0$ (zero state), calculus ko algebra mein badal deta hai.
2. Divide karo. $(s+3)Y=U\Rightarrow H(s)=\frac{1}{s+3}$. Kyun? Definition se $H=Y/U$; single pole $s=-3$ par baithta hai.
3. Invert karo. Pair $\mathcal{L}\{e^{-at}\}=\frac{1}{s+a}$ ko $a=3$ ke saath use karo, milta hai $h(t)=e^{-3t}$ for $t\ge 0$. Kyun? Known table pair se match karna fastest legal inverse hai.
4. DC gain $=H(0)=\frac{1}{3}$. Kyun? $s=0$ set karna constant (zero-frequency) input ke response ko probe karta hai.

Verify: pole $-3<0$ → map (C1 dot) left half-plane mein baithta hai, $h$ decay karta hai. Ek step $u=1$ convolve karo: $y={\int }_0^t{e}^{-3(t-\tau )}d\tau =\frac{1}{3}(1-e^{-3t})\to \frac{1}{3}$ ✓ DC gain se match karta hai.

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Ex 2 — Real positive pole: instability (C2)

Forecast: Ex 1 se bas ek sign change hai. $h(t)$ ka kya hoga, predict karo.

1. Transform karo. $sY-Y=U\Rightarrow H(s)=\frac{1}{s-1}$. Kyun? Same algebra; ab pole $s=+1$ par hai.
2. Invert karo. ${\mathcal{L}}^{-1}\{\frac{1}{s-1}\}=e^{+t}$, $t\ge 0$. Kyun? Table pair $\frac{1}{s-a}\to e^{at}$, $a=+1$ ke saath.
3. Stability padho. $\Re (\text{pole})=+1>0$. Kyun? Positive real part wala koi bhi pole ek growing $e^{pt}$ inject karta hai — Characteristic polynomial and roots ka physical version.

Verify: $h(t)=e^t\to \infty$ jab $t\to \infty$ — C2 dot right half-plane mein rehta hai. Sanity check: $t=1$ par, $h=e\approx 2.718>1$, already amplified. Unstable.

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

Forecast: jis system ka pole boundary par baitha ho, woh kya karta hai — decay, grow, ya hover?

1. Transform karo. $sY=U\Rightarrow H(s)=\frac{1}{s}$. Kyun? Ek derivative → $s$ ka ek factor; pole exactly $s=0$ par.
2. Invert karo. ${\mathcal{L}}^{-1}\{1/s\}=1$ (unit step) for $t\ge 0$. Kyun? $\mathcal{L}\{1\}=1/s$; impulse response ek constant hai jo kabhi zero par return nahi karta.
3. Classify karo. $\Re (\text{pole})=0$ → marginally stable: na badhta hai, na decay karta hai. Kyun? Left-half plane ki boundary; Dirac delta function input ek permanent offset chhod jaata hai.

Verify: $h(t)=1$ for all $t>0$ — bounded par non-decaying. Ek kick permanently level ko $1$ se shift kar deta hai, ${\int }_0^{\infty }\delta dt=1$ area se match karta hai.

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Ex 4 — Repeated pole: double integrator (C4)

Forecast: do integrators stacked — kya angle settle karega, oscillate karega, ya forever drift karega?

1. Transform karo. $s^2\Theta =U\Rightarrow H(s)=\frac{1}{s^2}$. Kyun? Do derivatives → $s^2$; $s=0$ par ek double pole.
2. Invert karo. ${\mathcal{L}}^{-1}\{1/s^2\}=t$, $t\ge 0$. Kyun? Pair $\mathcal{L}\{t\}=1/s^2$; repeated root ek constant ko ramp mein badal deta hai.
3. Classify karo. Axis par repeated pole → unbounded ramp → usable sense mein marginal bhi nahi. Kyun? Ek torque impulse angle ko linearly forever badhata rehta hai; craft kabhi rotate karna band nahi karta.

Verify: $h(2)=2$, $h(4)=4$ — strictly increasing without bound. Exactly isliye real attitude control ko Guidance Navigation and Control (GNC) ke andar Feedback control systems chahiye. (Ex 3 se compare karo: zero par ek pole → flat; zero par do poles → ramp.)

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Ex 5 — Complex pair: damped ring (C5)

Forecast: kya fingerprint clean decay hogi, ya decaying wiggle? Frequency ka andaza lagao.

1. Transform karo. $H(s)=\frac{1}{s^2+2s+5}$. Kyun? Har $y^{(k)}\to s^{k}Y$ zero state par.
2. Standard form se match karo $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$. Toh ${\omega }_n^2=5\Rightarrow {\omega }_n=\sqrt{5}$, aur $2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{1}{\sqrt{5}}\approx 0.447$. Kyun? Yeh form damping $\zeta$ aur natural frequency ${\omega }_n$ directly expose karta hai.
3. Poles. $s=-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}=-1\pm 2j$ (kyunki $0<\zeta <1$). Kyun? Under-damped → complex conjugates; real part $-1$ = decay rate, imaginary part $2$ = ring frequency ${\omega }_d$.
4. Invert karo. Square complete karo: denominator $=(s+1{)}^2+2^2$. Pair $\mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a{)}^2+b^2}$, $a=1,b=2$ ke saath, phir $b=2$ se divide karo: $h(t)=\frac{1}{2}{e}^{-t}\sin (2t),t\ge 0.$ Kyun? Hamara numerator $1$ hai, pair $b$ supply karta hai, toh hum $1/b$ se scale karte hain.

Verify: ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=\sqrt{5}\cdot \sqrt{1-1/5}=\sqrt{5}\cdot \sqrt{4/5}=2$ ✓ imaginary part se match karta hai. Real part $-1<0$ → decaying ring (C5 dot left half-plane mein). Figure mein envelope $\pm \frac{1}{2}{e}^{-t}$ sine ko squeeze karte dikhti hai.

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

Forecast: damping bilkul remove ho jaaye toh, kya ring kabhi rukti hai?

1. Transform karo. $H(s)=\frac{1}{s^2+9}$. Kyun? Koi $\dot{y}$ term nahi → poles mein koi real part nahi.
2. Poles. $s^2=-9\Rightarrow s=\pm 3j$ — imaginary axis par. Kyun? $\zeta =0$; decay rate exactly zero hai.
3. Invert karo. Pair $\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$, $b=3$ ke saath; $3$ se divide karo: $h(t)=\frac{1}{3}\sin (3t),t\ge 0.$ Kyun? Numerator $1$, pair $b=3$ deta hai, $1/3$ se scale karo.
4. Classify karo. $\Re (\text{poles})=0$ → marginally stable — bounded par kabhi decay nahi karta.

Verify: amplitude $\frac{1}{3}$ kabhi nahi shrinkta; peaks har period $2\pi /3$ par repeat hote hain. Map par, C6 exactly axis par baitha hai, decaying C5 aur right-plane growing pole ke beech.

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

Forecast: ek decaying system mein growing input — kya output ramp track karega, usse lag karega, ya blow up karega?

1. Causal convolution set up karo. $y(t)={\int }_0^{t}u(\tau )h(t-\tau )d\tau ={\int }_0^t\tau e^{-3(t-\tau )}d\tau$. Kyun? Causal $h$ ⇒ limits $0$ to $t$; kabhi "future" $\tau >t$ par integrate mat karo.
2. Constant $e^{-3t}$ bahar nikalo: $y=e^{-3t}{\int }_0^t\tau e^{3\tau }d\tau$. Kyun? $e^{-3(t-\tau )}=e^{-3t}{e}^{3\tau }$; jo $\tau$ par depend karta hai use alag karo.
3. Integration by parts. ${\int }_0^t\tau e^{3\tau }d\tau ={\left[\frac{\tau e^{3\tau }}{3}\right]}_0^t-{\int }_0^t\frac{e^{3\tau }}{3}d\tau =\frac{t{e}^{3t}}{3}-\frac{e^{3t}-1}{9}$. Kyun? Standard $\int \tau e^{a\tau }$; parts polynomial factor ko hata deta hai.
4. Multiply back karo. $y(t)=\frac{t}{3}-\frac{1}{9}+\frac{1}{9}{e}^{-3t}$. Kyun? $e^{-3t}$ distribute karo; transient $\frac{1}{9}{e}^{-3t}$ khatam ho jaata hai, lagging ramp bacha rehta hai.

Verify: jab $t\to \infty$, $y\approx \frac{t}{3}-\frac{1}{9}$: output ramp ko slope $\frac{1}{3}$ (= Ex 1 ka DC gain) ke saath track karta hai, minus ek steady offset $\frac{1}{9}$. $t=0$ par: $y(0)=0-\frac{1}{9}+\frac{1}{9}=0$ ✓ (rest state). Laplace se cross-check: $Y=H(s)U(s)=\frac{1}{s+3}\cdot \frac{1}{s^2}$, aur partial fractions $\frac{1}{3s^2}-\frac{1}{9s}+\frac{1}{9(s+3)}$ reproduce karta hai ✓.

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

Forecast: $u=0$ hai lekin $y(0)\ne 0$, toh kya transfer function $H=\frac{1}{s+3}$ kuch bhi bata sakta hai?

1. IC term ke saath transform karo. $\mathcal{L}\{y^{\prime }\}=sY-y(0)=sY-4$. Toh $sY-4+3Y=0$. Kyun? Nonzero $y(0)$ woh $-y(0)$ piece rakhta hai jise hum zero-state cases mein drop kar dete the.
2. Solve karo. $(s+3)Y=4\Rightarrow Y=\frac{4}{s+3}$. Kyun? IC ek "internal source" ki tarah act karta hai — external input $U$ nahi.
3. Invert karo. $y(t)=4e^{-3t}$. Kyun? Same pole, lekin amplitude $y(0)$ se aata hai, $U$ se nahi.
4. Lesson. Yahan $U(s)=0$, toh $H\cdot U=0$ — transfer function zero predict karta hai, jabki asli output $4e^{-3t}$ hai. Kyun? $H$ sirf zero-state (forced) response describe karta hai; yeh zero-input response hai, ek alag additive term.

Verify: $y(0)=4e^0=4$ ✓ aur $y^{\prime }(0)+3y(0)=-12+12=0$ ✓ ODE satisfy karta hai. IC-wale driven problem ka total response = zero-state ($u\ast h$) plus yeh zero-input term.

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Ex 9 — Word problem: thruster se attitude tak (C9)

Forecast: ek clean torque pulse ke baad, kya craft ek fixed naye angle par settle karega, ya drift karta rahega?

1. Transform karo. $H(s)=\frac{1}{s^2+0.4s}=\frac{1}{s(s+0.4)}$. Kyun? Poles $s=0$ par (integrator) aur $s=-0.4$ par (damped) — ek rate jo decay karta hai plus ek position jo hold karti hai.
2. Partial fractions. $\frac{1}{s(s+0.4)}=\frac{2.5}{s}-\frac{2.5}{s+0.4}$. Kyun? Ek table-ready constant term aur ek decaying term mein split karo; $\frac{1}{0.4}=2.5$.
3. Invert karo. $\theta (t)=h(t)=2.5(1-e^{-0.4t})$, $t\ge 0$. Kyun? $\frac{1}{s}\to 1$, $\frac{1}{s+0.4}\to e^{-0.4t}$.
4. Final angle padho. ${\lim }_{t\to \infty }\theta =2.5$ rad. Kyun? Decaying rate term vanish ho jaata hai; integrator ka pole-at-zero ek permanent offset lock kar leta hai — physically, wheel ki spin absorb ho gayi.

Verify: $\theta (0)=2.5(1-1)=0$ ✓ (rest angle se shuru). Angular rate $\dot{\theta }(t)=e^{-0.4t}\to 0$ ✓ (craft turning band ho jaata hai). 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: ek zero jo pole cancel karta hai (C10)

Forecast: kya $s=-2$ par pole actually impulse response mein contribute karta hai, ya yeh ek decoy hai?

1. Common factor dhundho. Numerator $s+2$, denominator ke $(s+2)$ ko cancel karta hai, $s\ne -2$ ke liye. Kyun? ek zero (numerator root) wahin jahan pole hai cancel ho jaata hai — mode "unobservable/uncontrollable" hai; yeh $h$ mein kabhi appear nahi karta.
2. Reduced transfer function. $H(s)=\frac{1}{s+5}$. Kyun? Cancellation ke baad sirf $s=-5$ par genuine pole bachta hai.
3. Invert karo. $h(t)=e^{-5t}$, $t\ge 0$. Kyun? $a=5$ ke saath standard pair.
4. Stability padho. Sirf pole $-5<0$ → stable; phantom $-2$ ne kabhi matter nahi kiya. Kyun? Cancelled poles ki Characteristic polynomial and roots expansion mein zero coefficient hoti hai.

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 agree karta hai chahe pehle cancel karo ya baad mein.

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Recall Matrix ka one-line summary

$\Re (\text{pole})$ ka sign fate decide karta hai — negative decay karta hai, zero holds/ramps karta hai (multiplicity!), positive blow up karta hai; complex pairs ring karte hain; aur $H$ kabhi initial conditions ya cancelled modes nahi dekhta.

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)