Impulse response and transfer function (GNC connection): Question bank

4.6.33 · D5 · Maths › Ordinary Differential Equations › Impulse response and transfer function (GNC connection)

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Ground vocabulary you'll need here

Figure — Impulse response and transfer function (GNC connection)

Definition Laplace variable

$s$ (yeh aata kahan se hai) Jab hum transform karte hain, hum $f (t)$ ko $e^{- s t}$ se multiply karke integrate karte hain: $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ . Yahan == $s = σ + j ω$ ek complex number hai==: iska real part $σ$ batata hai ki test-weight $e^{- σ t}$ kitni tezi se decay karta hai, aur iska imaginary part $ω$ ek oscillation frequency set karta hai. Toh $s$ ek "probe" hai jo poochta hai ki $f (t)$ ke andar kitna decaying-oscillating pattern hai. Ek pole woh value of $s$ hai jahan $H (s)$ blow up ho jata hai — yeh us natural pattern ko mark karta hai jisme system ring karta hai.

Integral sirf tab converge karta hai jab $σ$ kaafi bada ho; woh set of $s$ jahan yeh hota hai woh region of convergence (ROC) kehlata hai. Ek causal signal ke liye ROC ek right half-plane $σ > σ_{0}$ hota hai. Product rule $L {u * h} = U (s) H (s)$ tab hi valid hai jab dono ROCs overlap karein — wahi overlap multiplication ko legal banata hai.

True or false — justify karo

The impulse response

h (t)

kisi physical system mein

t < 0

ke liye nonzero ho sakti hai.

Kisi bhi causal system ke liye False hai — output kick se pehle nahi aa sakta, isliye

h (t) = 0

for

t < 0

definition mein built-in hai.

t = 0

se pehle nonzero

h

ka matlab hoga ki system hit hone se pehle respond kar gaya.

L {δ (t)} = 1

, toh delta ka "koi effect nahi hota".

False — transform ka

1

ke barabar hona iska matlab hai ki delta sabhi frequencies ko equally unit weight ke saath inject karta hai; yeh maximally rich input hai, aur isliye iska response ek poora fingerprint hota hai.

Ek LTI system ke liye,

u * h = h * u

True — Convolution commutative hai (upar overlap-area figure dekho): substitution

τ^{'} = t - τ

ek integral ko doosre pe map karta hai, toh koi fark nahi padta ki kaun sa curve still rakho.

Transfer function

H (s)

system ke response ko kisi bhi initial conditions ke liye poori tarah describe karta hai.

False —

H (s) = Y / U

sirf zero-state relation hai. Nonzero

y (0)

ek alag zero-input term contribute karta hai (

s Y - y (0)

pieces) jo

H

carry nahi karta.

s = 0

pe pole ka matlab hai system stable hai kyunki kuch exponentially blow up nahi hota.

False — yeh sirf marginally stable hai. Double integrator

h (t) = t

deta hai, jo linearly unbounded badhta hai; unbounded to unbounded hai, exponential ho ya na ho.

Agar sabhi poles ka

ℜ (p) < 0

ho toh impulse response zero pe decay ho jaata hai.

True — har pole ek

\sim e^{pt}

term contribute karta hai, aur

ℜ (p) < 0

envelope

e^{ℜ (p) t} \to 0

force karta hai, toh ringing khatam ho jaati hai.

Complex poles wale system ko unstable hona chahiye.

False — complex poles oscillation (

sin

cos

factors) produce karte hain, lekin stability sirf real part pe depend karti hai.

- ζ ω_{n} \pm j ω_{d}

pe poles jab

ζ > 0

ho toh oscillate bhi karte hain aur decay bhi karte hain.

Convolution theorem

L {u * h} = U (s) H (s)

kisi bhi signals ke liye unke region of convergence ki parwah kiye bina hold karta hai.

False — yeh sirf tab hold karta hai jab ROCs overlap karein. Dono causal hone se right-half-plane ROCs milte hain jo intersect karte hain;

t = 0

se pehle content ya non-overlapping ROCs clean product rule ko tod dete hain.

Error dhundho

"Kyunki

δ (t)

ki width zero hai,

\int_{0}^{\infty} e^{- s t} δ (t) d t = 0

, toh

L {δ} = 0

Error hai zero width ko zero area treat karna. Delta ki unit area hai, aur sifting deta hai

\int e^{- s t} δ (t) d t = e^{- s \cdot 0} = 1

0

nahi.

y^{'} + 2 y = u

ke liye pole

s = + 2

pe hai, toh system unstable hai."

Sign error:

s Y + 2 Y = U

deta hai

H = 1/ (s + 2)

, jiska denominator

s = - 2

pe zero hota hai. Pole

- 2 < 0

pe hai, toh yeh stable hai.

\overset{y}{¨} + 2 ζ ω_{n} \overset{y}{˙} + ω_{n}^{2} y = u

ki impulse response

e^{- ζ ω_{n} t} sin (ω_{d} t)

hai (damped frequency

ω_{d} = ω_{n} 1 - ζ^{2}

ke saath)."

1/ ω_{d}

factor missing hai.

1/ ((s + ζ ω_{n})^{2} + ω_{d}^{2})

ko invert karne mein woh pair use hota hai jisme numerator

ω_{d}

hota hai, toh divide karna padega:

h = \frac{1}{ω _{d}} e^{- ζ ω_{n} t} sin (ω_{d} t)

"Convolve karne ke liye, hamesha

\int_{- \infty}^{\infty} u (τ) h (t - τ) d τ

integrate karo."

Ek causal system ke liye jisme input

0

pe start hota hai, limits

0

t

tak collapse ho jaati hain.

\pm \infty

use karna silently impossible future contributions include karta hai (

h (t - τ) = 0

for

τ > t

aur

u = 0

for

τ < 0

waise bhi, lekin phir bhi).

"Kyunki

H (s) = 1/ s^{2}

ka DC gain

H (0)

hai, spacecraft ek fixed angle pe settle ho jaata hai."

H (0) = 1/0

undefined hai — koi finite DC gain nahi hai. Double integrator kabhi settle nahi hota; ek constant torque angle ko forever ramp karta hai, isliye ise feedback chahiye.

H (s)

input ka Laplace transform hai."

Nahi —

H (s) = L {h (t)}

, impulse response ka transform hai, jo

U (s) = 1

set karke milta hai (delta ka transform). Input ka transform

U (s)

hai.

Why questions

h (t)

akela jaankar har input ke liye output kyun determine hota hai?

Kyunki koi bhi input shifted, scaled impulses ka sum hai (sifting), aur linearity + time-invariance har impulse ko

h

ki shifted, scaled copy pe bhejti hai; superpose karne se

y = u * h

milta hai.

Hum seedha convolve karne ki bajaye Laplace transform kyun use karte hain?

Convolution ek integral hai jise tum har input ke liye recompute karte ho; Laplace ise plain multiplication

Y = H U

mein badal deta hai, aur derivatives ko

s

ki powers mein, ODE ko algebra mein convert karta hai.

Complex variable

s

poles ko ek nazar mein stability reveal kyun karne deta hai?

Kyunki

s = σ + j ω

decay rate aur frequency ko package karta hai, ek pole location seedha kehta hai "system frequency

ω

pe ring karta hai jabki uska envelope rate

σ

pe grow/decay karta hai" —

σ

ka sign stability hai.

Poles characteristic polynomial ke roots ke saath same kyun hote hain?

Transfer function denominator hi

a_{n} s^{n} + \dots + a_{0}

hai, aur ise zero karna bilkul wahi characteristic equation hai jo homogeneous ODE ki hai. Same polynomial, same roots.

Ek controller ko poles ko left-half plane mein kyun push karna chahiye?

Poles

e^{pt}

terms govern karte hain; sirf

ℜ (p) < 0

unhe decay karta hai. GNC mein, right-half-plane pole ka matlab hai spacecraft ka attitude error badhta hai aur woh tumble karta hai.

Time-invariance humein delta ki shifted response ko

h (t - τ)

kyun likhne deti hai?

Time-invariance kehti hai ki

τ

time pe kick karna

0

ki jagah same shape produce karta hai bas shifted; toh

δ (t - τ)

ka response

h

evaluated at

t - τ

hai, koi reshaping nahi.

Transfer function specifically GNC design ke liye useful kyun hai?

Yeh poore plant ko ek rational function mein compress karta hai jiske pole locations instantly stability aur response speed reveal karte hain, engineers ko feedback se unhe reshape karne dete hain thruster fire karne se pehle.

Edge cases

h (t)

kya hoga agar input impulse apply ho lekin system ka nonzero

y (0)

ho?

h (t)

sirf zero initial conditions pe defined hai; nonzero

y (0)

ke saath total output

h * u

plus ek alag zero-input response hai, toh

h

khud unchanged hai lekin poora answer nahi hai.

h (t) = \frac{1}{ω _{d}} e^{- ζ ω_{n} t} sin (ω_{d} t)

ka kya hoga jab

ζ \to 1^{-}

ω_{d} = ω_{n} 1 - ζ^{2} \to 0

, aur ratio

\frac{s i n ( ω _{d} t )}{ω _{d}} \to t

, jo critically-damped limit

h (t) = t e^{- ω_{n} t}

deta hai — ek smooth, oscillation-free bridge.

Ek repeated pole (jaise

s = 0

do baar) simple pole ke comparison mein response mein kya add karta hai?

Ek polynomial factor: double pole ek

t e^{pt}

term contribute karta hai (isliye

1/ s^{2}

ke liye

h = t

), toh repeated poles exponential ko

t

ki powers se multiply karte hain.

Kya system

y^{'} + 2 y = u

tab bhi stable hai agar hum impulse ki jagah step input add karein?

Haan — stability poles ki property hai, input ki nahi. Step response

\frac{1}{2} (1 - e^{- 2 t})

finite DC gain

H (0) = 1/2

pe settle hoti hai precisely isliye kyunki pole

- 2

pe rehta hai.

Jab ek pole exactly

s = 0

pe ho toh DC gain (unit step ke liye steady output) kya hai?

Undefined/infinite —

H (0)

diverge karta hai, matlab ek constant input unbounded output drive karta hai (double integrator ki ramp), toh koi steady value exist nahi karti.

0 < ζ < 1

ke liye, poles ke baare mein kaun sa geometric fact decay aur oscillation dono guarantee karta hai?

Poles

- ζ ω_{n} \pm j ω_{d}

left half-plane mein hote hain (negative real part

\Rightarrow

decay) lekin real axis se door (nonzero imaginary part

\Rightarrow

oscillation), jaise upar pole picture mein sketch kiya gaya hai.

Ek causal impulse response ka ROC kya hai, aur yahan yeh kyun matter karta hai?

Ek right half-plane

σ > σ_{0}

jahan

σ_{0}

rightmost pole hai; yeh matter karta hai kyunki

U (s) H (s) = L {u * h}

sirf tab valid hai jab input aur system ke ROCs overlap karein.

Connections

Parent: Impulse response & 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)