Properties — linearity, first - second shift theorems, scaling: Question bank

4.6.27 · D5 · Maths › Ordinary Differential Equations › Properties — linearity, first - second shift theorems, scali

True or false — justify

TRUE/FALSE:

L {f (t) + g (t)} = F (s) + G (s)

ke liye zaroori hai ki

f

aur

g

ki growth ka order same ho.

False. Linearity purely integral ke sum split karne se aati hai; iske liye sirf itna chahiye ki dono transforms alag-alag exist karein, koi bhi matching zaroori nahi.

TRUE/FALSE:

L {f (t) g (t)} = F (s) G (s)

False. Linearity sums ko handle karti hai, products ko nahi. Do time-functions ke product ka transform

s

mein ek convolution hota hai, product nahi — dekho Convolution Theorem.

TRUE/FALSE: First shift

L {e^{a t} f (t)} = F (s - a)

negative

a

ke liye bhi kaam karta hai.

True. Derivation mein kahin bhi

a > 0

assume nahi kiya gaya; jaise

e^{- t} f (t)

F (s + 1)

milta hai. Sign sirf decide karta hai ki transform

s

-axis par kis direction mein slide karega.

TRUE/FALSE: Second shift

L {f (t - a) u (t - a)} = e^{- a s} F (s)

a < 0

ke liye valid hai.

False. Derivation integral ko

t = a

par split karta hai aur iske liye

a \geq 0

chahiye taaki delay

[0, \infty)

ke andar ho. "Negative delay" matlab

t = 0

se pehle ki information maangna, jo transform ke domain se bahar hai.

TRUE/FALSE: Scaling

L {f (c t)} = \frac{1}{c} F (s / c)

kisi bhi real

c \neq = 0

ke liye hold karta hai.

False. Iske liye

c > 0

chahiye. Agar

c < 0

ho toh substitution

τ = c t

integration limits ko flip kar deta hai aur

f (c t)

negative time probe karta hai, derivation toot jaati hai.

TRUE/FALSE:

L {5} = \frac{5}{s}

sirf

L {1} = \frac{1}{s}

par linearity apply karna hai.

True. Constant

5 = 5 \cdot 1

hai, aur linearity

5

ko bahar le aati hai:

L {5} = 5 \cdot \frac{1}{s}

TRUE/FALSE: Kisi function ko

e^{a t}

se multiply karne se uske transform ki shape badal jaati hai.

False. Ye poore transform ko sideways slide karta hai (

s \to s - a

) bina form badlaaye; har

s

simply

s - a

ban jaata hai.

TRUE/FALSE: Time delay

F (s)

ki size (amplitude) badal deta hai.

False. Ye sirf

e^{- a s}

ka tag lagaata hai, jiski imaginary axis par magnitude ek hoti hai; underlying

F (s)

untouched rehta hai — "delay shape nahi badlata."

TRUE/FALSE: First aur second shift ek hi function par ek ke baad ek apply kiye ja sakte hain.

True. Ye alag-alag domains par kaam karte hain (ek

s

par, ek time delay ke zariye), to jaise

L {e^{2 t} f (t - 3) u (t - 3)}

ek legitimate combination hai — bas har rule carefully apply karo.

Why questions

WHY Laplace transform mein linearity bilkul kyu hold karti hai?

Kyunki

L

ek integral se bana hai, aur integration constants bahar nikalne aur sums split karne deta hai — ye do rules hi linearity hain.

WHY first shift, time mein

e^{a t}

se multiply karne ko

s

-shift mein kyun convert karta hai?

e^{- s t} e^{a t} = e^{- (s - a) t}

combine karne par integral bilkul

F

ki definition jaisa dikhta hai lekin

s

ki jagah

s - a

hota hai, to result

F (s - a)

hai.

WHY second shift ko unit step

u (t - a)

ki zaroorat kyun hai?

Iske bina integral

t = 0

par shuru hota hai, lekin substitution

τ = t - a

e^{- a s}

factor karta hai, tabhi cleanly kaam karti hai jab integrand

t = a

se neeche zero ho — exactly yahi

u (t - a)

guarantee karta hai.

WHY time ko

c

se scale karne par

s

-axis

1/ c

se stretch hoti hai?

Substitution

τ = c t

force karta hai ki

s

integral ke andar

s / c

ke roop mein aaye, to time squeeze karna (bada

c

) transform ko

s

ke along spread karta hai — Fourier jaisi reciprocal trade-off.

WHY scaling formula mein extra

1/ c

aata hai lekin shift formulas mein nahi?

Sirf scaling substitution differential badalta hai (

d t = d τ / c

); shift derivations ya toh exponentials combine karte hain ya limits shift karte hain bina

d t

rescale kiye, to koi Jacobian factor nahi aata.

WHY scratch se re-integrate karne ki jagah

L {e^{a t} f (t)} = F (s - a)

prefer kiya jaata hai?

Ye pehle se pata transform reuse karta hai aur sirf

s \to s - a

substitute karta hai, fresh integral se bachta hai — yahi table of properties banane ka poora point hai.

WHY hum

L {δ (t - a)}

ko second shift ke saath

f = 1

leke nahi likh sakte?

Dirac delta koi ordinary delayed function

f (t - a) u (t - a)

nahi hai; ye ek distribution hai aur directly transform hoti hai

L {δ (t - a)} = e^{- a s}

— dekho Unit Step and Dirac Delta Functions.

Edge cases

EDGE CASE:

L {f (t - a) u (t - a)}

kya hoga jab

a = 0

Step

u (t) = 1

ban jaata hai

t \geq 0

ke liye aur factor

e^{0} = 1

hota hai, to ye plain

L {f (t)} = F (s)

ban jaata hai — "no delay" case, general rule ke consistent.

EDGE CASE: First shift kya deta hai jab

a = 0

e^{0 \cdot t} f (t) = f (t)

aur

F (s - 0) = F (s)

, to kuch nahi badlata — identity case, ek sanity check ki rule gracefully degrade karta hai.

EDGE CASE: Scaling

c = 1

par kya deta hai?

\frac{1}{1} F (s /1) = F (s)

: koi squeeze nahi, koi change nahi — phir identity, formula ki boundary par consistency confirm karta hai.

EDGE CASE: Scaling ko

c \to 0^{+}

par apply karna (bahut slow time).

\frac{1}{c} F (s / c)

blow up karta hai:

1/ c

prefactor diverge karta hai aur

F (s / c)

s = 0

ke paas evaluate hota hai, reflecting karta hai ki infinitely-slowed signals transform ki convergence kho dete hain. Formula har fixed

c > 0

ke liye valid rehta hai lekin limit mein degenerate ho jaata hai.

EDGE CASE:

L {f (t) u (t - a)}

jab

f

f (t - a)

ke roop mein nahi likha gaya — kya phir bhi second shift use kar sakte hain?

Directly nahi. Pehle

f (t)

(t - a)

ki powers mein re-express karna hoga (jaise Taylor-expand ya algebraically shift karo), har delayed piece transform karo, phir har ek par

e^{- a s}

apply karo. Raw form ka koi shortcut nahi hai.

EDGE CASE: Kya linearity se

∣ f (t) ∣

F (s)

se transform kar sakte hain?

Nahi. Absolute value ek nonlinear operation hai, to

L {∣ f ∣}

F (s)

se linearity ke zariye koi relation nahi —

∣ f ∣

ko ek naye function ki tarah scratch se transform karna hoga.

EDGE CASE: Agar

F (s)

sirf

Re (s) > 0

ke liye exist karta hai, to

L {e^{2 t} f (t)} = F (s - 2)

kahaan converge karega?

Re (s - 2) > 0

ke liye, yaani

Re (s) > 2

: convergence ka region

2

se right shift ho jaata hai transform ke saath. Shift sirf formula nahi, convergence strip bhi move karta hai.

Recall Har trap ki ek-line summary

Linearity: sums split hote hain, products nahi. First shift: $s \to s - a$ har jagah substitute karo, $a$ ka koi bhi sign ho. Second shift: $u (t - a)$ chahiye, $a \geq 0$ chahiye, argument $(t - a)$ ke roop mein likhna chahiye, aur $e^{- a s}$ stamp hoga. Scaling: sirf $c > 0$ , aur $1/ c$ kabhi mat chhodna.

Connections

4.6.27 Properties — linearity, first - second shift theorems, scaling (Hinglish)
Laplace Transform — Definition and Existence
Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
Inverse Laplace Transform and Partial Fractions
Unit Step and Dirac Delta Functions
Convolution Theorem

Question bank — Properties — linearity, first - second shift theorems, scaling

True or false — justify

Spot the error

Why questions

Edge cases

Connections