Convolution theorem — proof, applications: Question bank

4.6.32 · D5 · Maths › Ordinary Differential Equations › Convolution theorem — proof, applications

Ek golden rule jo answering ke dauran dimaag mein rakhna hai:

Sahi hai ya galat — justify karo

Neeche har statement ya toh subtly sahi hai ya subtly galat. Reason do.

L^{- 1} {F (s) G (s)} = f (t) g (t)

, yani ordinary pointwise product.

Galat.

s

-world mein ek product,

t

-world mein convolution ban jaata hai, pointwise product nahi. Test karo:

L {1} \cdot L {1} = 1/ s^{2}

, lekin

1 \cdot 1 = 1

ka transform

1/ s \neq = 1/ s^{2}

hai; jabki

1 * 1 = t

ka transform

1/ s^{2}

hai. ✓

Convolution commutative hai, isliye

(f * g) (t) = (g * f) (t)

sabhi

t \geq 0

ke liye.

Sahi. Substitution

u = t - τ

se swap ho jaata hai ki kaun sa function

τ

carry karta hai aur kaun sa

t - τ

, aur limits wapas flip ho jaati hain, jo

g * f

deta hai. Tum hamesha easier-to-integrate factor ko "andar" rakh sakte ho.

L {f \cdot g} = F (s) \cdot G (s)

Galat. Theorem convolution

f * g

ke baare mein hai, pointwise product

f \cdot g

ke baare mein nahi. Generally

L {f \cdot g}

ka koi simple formula nahi hota — us direction mein ek complex-plane (Bromwich) convolution chahiye, jo bahut mushkil object hai.

Convolution integral ki upper limit ko

\infty

se replace kar sakte hain kyunki Laplace transforms

\infty

tak integrate karte hain.

Galat. Upper limit

t

hai kyunki causal functions ke liye

g (t - τ) = 0

jab

τ > t

ho jaata hai (argument

t - τ

negative ho jaata).

\infty

tak extend karna "future" se fictitious contributions add kar deta.

Convolution associative hai:

(f * g) * h = f * (g * h)

Sahi.

s

-world mein ye sirf

(F G) H = F (G H)

hai, functions ka ordinary multiplication, jo associative hota hai. Wapas transform karne par

t

-world mein bhi associativity free mein mil jaati hai.

Agar

f

aur

g

dono continuous hain, toh

f * g

bhi continuous hai.

Sahi. Convolution ek smoothing (averaging) operation hai: ye generally results ko inputs se smoother banata hai, kabhi rougher nahi, isliye continuity preserve hoti hai aur aksar improve bhi hoti hai.

δ * f = f

, jahan

δ

unit impulse hai.

Sahi. Impulse, convolution ka identity element hai — ek spike ko

f

par slide karna bas us instant par

f

ko read kar deta hai.

s

-world mein

L {δ} = 1

, aur

1 \cdot F (s) = F (s)

. ✓

1 * 1 = 1

Galat.

1 * 1 = \int_{0}^{t} 1 \cdot 1 d τ = t

, na ki

1

. Constant

1

convolution ka identity nahi hai — impulse

δ

hai. "Multiply by 1" aur "convolve with 1" ko confuse karna ek classic slip hai.

Convolution theorem require karta hai ki

f

aur

g

ka same Laplace transform ho.

Galat. Wo koi bhi do functions ho sakte hain jinke paas Laplace transform ho; theorem alag-alag

F (s)

aur

G (s)

ko stitches karta hai.

sin t * sin t

example mein bas ittifaq se do equal factors hain.

Agar

F (s) G (s)

ek rational function hai, toh convolution aur partial fractions dono same inverse denge.

Sahi. Ye dono same unique inverse Laplace transform tak pahunchne ke do raaste hain (jo continuous functions mein unique hota hai). Convolution aksar tab faster hota hai jab factors ke clean inverses hon; partial fractions tab jab wo naturally split na karein.

Error dhundo

Har line mein ek planted mistake hai. Use dhuncho aur correct version batao.

(f * g) (t) = \int_{0}^{t} f (τ) g (τ) d τ

Dono factors

τ

carry kar rahe hain — ye ek pointwise product hai, convolution nahi. Ek factor ko flip aur slide hona chahiye:

g (t - τ)

, taaki arguments

t

mein add ho sakein.

(f * g) (t) = \int_{0}^{\infty} f (τ) g (t - τ) d τ

Upper limit galat hai.

[0, \infty)

par causal functions ke liye ye

\int_{0}^{t}

hona chahiye, kyunki

g (t - τ) = 0

jab

τ > t

ho.

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

se, hum

L^{- 1} {F (s) + G (s)} = (f * g) (t)

paate hain."

Theorem products

F (s) G (s)

ke baare mein hai, sums ke baare mein nahi. Sum ke liye, linearity se

L^{- 1} {F + G} = f (t) + g (t)

milta hai — ordinary sum, koi convolution nahi.

"Proof mein,

(σ, τ)

(t, τ)

mein change karne par ek Jacobian factor of

s

aa jaata hai."

Jacobian

1

hai,

s

nahi. Kyunki

σ = t - τ

aur

τ = τ

hai, transformation ek shear hai jiska determinant

1

hai; area preserve hota hai, isliye koi extra factor nahi aata.

"Kyunki convolution cheezein smear karta hai,

y (t) = t + \int_{0}^{t} y (τ) sin (t - τ) d τ

ko Laplace se solve nahi kar sakte."

Bilkul ulta — integral exactly

y * sin t

hai, isliye ye

Y \cdot \frac{1}{s ^{2} + 1}

mein Laplace ho jaata hai, aur integral equation ko

Y

ke liye simple algebra mein badal deta hai.

L^{- 1} {1/ (s^{2} + 1)^{2}} = sin t \cdot sin t = sin^{2} t

Galat operation — ye

sin t * sin t

hai, convolution, jo

\frac{1}{2} (sin t - t cos t)

ke barabar hai,

sin^{2} t

nahi.

t = σ + τ

ke baad naya integration region

0 \leq t \leq τ

hai har fixed

τ

ke liye."

Ulta hai. Constraints

σ = t - τ \geq 0

aur

τ \geq 0

0 \leq τ \leq t

milta hai jahan

t

0

\infty

tak range karta hai — inner variable

τ

t

se bounded hai, na ki ulta.

Why questions

Mechanism se jawab do, sirf restatement se nahi.

s

-world mein product kyun

t

-world mein integral ke corresponding hota hai?

Kyunki

F (s) G (s)

secretly quarter-plane par ek double integral hai, aur usse diagonals

σ + τ = t

ke hisaab se group karne par woh saari

(σ, τ)

pairs ikkatthe ho jaati hain jo kisi given

t

mein add hoti hain — woh "diagonal ke saath sum" precisely convolution integral hai.

Jab

F (s) G (s)

ko integrals ki tarah likhte hain toh hum deliberately alag dummy variables

σ

aur

τ

kyun use karte hain?

Taaki dono integrals independent axes par rahein aur ek honest double integral mein merge ho sakein. Same letter reuse karna do integration variables ko falsely bind kar deta.

Proof mein substitution

t = σ + τ

(say,

t = σ τ

ki jagah) kyun "sahi" move hai?

Hum standard Laplace form

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

ki taraf ja rahe hain. Kyunki exponent

e^{- s (σ + τ)}

hai,

t = σ + τ

choose karne se ye exactly

e^{- s t}

ban jaata hai — exponent khud bata deta hai ki naya time variable kya hona chahiye.

Convolution ek Volterra integral equation mein "help" kyun karta hai lekin ek pointwise-product integral nahi karta?

Ek Volterra kernel

\int_{0}^{t} y (τ) k (t - τ) d τ

ki tarah appear hota hai — arguments

t

mein add ho rahe hain, yani ek convolution — isliye ye

Y (s) K (s)

product mein Laplace ho jaata hai. Ek genuine pointwise product

\int y (τ) k (τ) d τ

ka aisa koi clean transform nahi hota.

Haath se convolution compute karte waqt tum dono factors mein se kisi ko bhi

t - τ

argument kyun de sakte ho?

Commutativity (

f * g = g * f

) se tum roles freely flip kar sakte ho, isliye jo function zyada mushkil ho usse

g (t - τ)

par park karo ya jo easier ho — jo bhi integral simpler ho — answer change nahi hoga.

Kisi function ko unit step

u (t)

ke saath convolve karne par uska running integral

\int_{0}^{t} f (τ) d τ

kyun milta hai?

Kyunki

u (t - τ) = 1

sabhi

τ \leq t

ke liye aur uske baad

0

hai, isliye

f * u = \int_{0}^{t} f (τ) \cdot 1 d τ

— step ek "ab tak sab kuch rakho" mask ki tarah kaam karta hai, jo integration hai.

s

-world mein ye

F (s) \cdot \frac{1}{s}

hai, jo known integration rule hai.

Edge cases

Boundary aur degenerate inputs — woh scenarios jinhe formula ko survive karna chahiye.

Kisi bhi bounded

f, g

ke liye

(f * g) (0)

kya hai?

Zero. Integral

\int_{0}^{0} (\dots) d τ

ek empty interval mein collapse ho jaata hai, isliye har convolution

0

par start hota hai — kisi bhi answer par ek useful sanity check.

f * 0

(zero function ke saath convolution) kya hai?

Zero function. Integrand identically zero hai, isliye integral sabhi

t

ke liye

0

hai.

s

-world mein ye

F (s) \cdot 0 = 0

hai. ✓

Kya

c

(ek constant) convolution ka identity hai, yaani kya kisi

c

ke liye

c * f = f

hota hai?

Koi bhi constant kaam nahi karta —

c * f = c \int_{0}^{t} f (τ) d τ

hai jo

f

ka integral hai,

f

khud nahi. Identity impulse

δ

hai, jo ordinary function hi nahi hai.

Agar

f

e^{2 t}

ki tarah grow karta hai aur

g

e^{3 t}

ki tarah, toh kya

f * g

exist karta hai?

Haan, har finite

t

ke liye integral ki tarah (ek finite interval

[0, t]

hamesha ek continuous function ko integrate karta hai). Transform

F (s) G (s)

tab sirf

Re (s) > 5

ke liye exist karta hai, dono convergence ke abscissas ka sum.

Ek spiky function ko ek wide smooth function ke saath convolve karne par spike ka kya hota hai?

Spike smear (blur) ho jaata hai smooth function ki width par — convolution ek weighted local average hai, isliye sharp features soften ho jaati hain. Ye exactly parent note waali "glowing sand grain" picture hai.

Agar

g (t - τ)

ek delayed step hai jo sirf

t = a

par on hota hai (yaani

g (t) = u (t - a)

), toh effective lower limit kaise change hoti hai?

Contributions sirf tab start hoti hain jab

t - τ \geq a

ho, yaani

τ \leq t - a

, isliye effective integral

\int_{0}^{t - a} f (τ) d τ

ban jaata hai aur

t < a

ke liye exactly

0

hota hai — delay simply saara output postpone kar deta hai.

s

-world mein ye shift

e^{- a s} / s

hai.

Recall Ek-line summary jo saath le jaao

Is page ka har trap ek hi do ideas ka disguise hai: (1) product in $s$ ↔ convolution (product nahi) in $t$ , aur (2) integrand ke arguments $t$ mein add hone chahiye, integral $0$ se $t$ tak run karta hai kyunki functions causal hote hain. Kisi bhi convolution claim ko in donon ke against check karo, aur tum bahut kam dhokhaa khaoge.

Connections

Parent: full proof & worked examples
Laplace Transform — definition and properties
Inverse Laplace Transform — partial fractions
Solving Linear ODEs with Laplace Transforms
Volterra Integral Equations
Fourier Transform Convolution Theorem
Transfer Functions and Impulse Response