Heaviside step function and Dirac delta function: Question bank

4.6.31 · D5 · Maths › Ordinary Differential Equations › Heaviside step function and Dirac delta function

Set-up: pehle yeh pictures aur symbols samjho

Misconceptions dhundne se pehle, khud objects ko dekho. Is page par har cheez teen sketches se refer karti hai — figure s01 (step $u$ ), figure s02 (impulse $δ$ ), aur figure s03 (shrinking boxes). Jab bhi koi symbol unfamiliar lage, uss figure ko ek nazar dekho jahan se woh symbol aata hai.

Figure — Heaviside step function and Dirac delta function

Figure s01 yeh step draw karta hai: $0$ par flat, phir $t = a$ par $1$ par ek saaf jump (chhota open dot woh ek ambiguous point mark karta hai). Neeche har step-function item ke liye figure s01 ko dhyan mein rakho.

Figure s02 $δ (t - a)$ ko "area $1$ " ka ek arrow ki tarah draw karta hai jo bilkul $t = a$ par baitha hai — uski height symbolic hai, uski area hi real hai. Neeche har impulse item ke liye figure s02 ko dhyan mein rakho.

Figure s03 teen boxes $δ_{ε}$ draw karta hai shrinking $ε$ ke liye: har ek area $1$ maintain karta hai (shaded), aur arrow dikhata hai ki woh figure s02 ke ideal spike ki taraf ja rahe hain. Jab bhi koi delta item "area" ki baat kare, figure s03 ko dhyan mein rakho.

True or false — justify karo

Heaviside function $u (t - a)$ , $a$ se pehle $0$ hai aur baad mein $1$ hai; delta $δ (t - a)$ uski slope hai.

u (t - a)

ek genuine function hai jisme tum ek number plug kar ke output le sakte ho.

True — yeh har

t \neq = a

par defined hai (sirf ek point

t = a

ambiguous chhoda gaya hai, conventionally

\frac{1}{2}

), isliye yeh ek ordinary piecewise function hai,

δ

ke unlike.

δ (t - a)

ek genuine function hai jisme tum ek number plug kar ke output le sakte ho.

False — integral ke bahar iska koi usable pointwise value nahi hai ("

a

par

\infty

" koi number nahi hai); yeh ek distribution hai, yaani ek linear functional jo ek test function

φ

φ (a)

bhejta hai.

δ (t - a)

ke neeche area

1

ke barabar hai chahe

a

kahan bhi ho.

True — total impulse hamesha definition se

1

hota hai (figure s03 mein area-

1

boxes dekho);

a

shift karne se sirf kahan spike hai woh badalta hai, woh kitna carry karta hai woh kabhi nahi.

\int_{- \infty}^{\infty} δ (t - a) d t

aur

\int_{0}^{\infty} δ (t - a) d t

hamesha equal hote hain.

False — yeh sirf tabhi agree karte hain jab

a > 0

; agar

a < 0

toh spike

[0, \infty)

ke bahar baitha hai isliye doosra integral

0

hai jabki pehla

1

hai.

Boxcar

u (t - a) - u (t - b)

poore interval

[a, b]

par endpoints including

1

ke barabar hai.

False (harmlessly) — yeh open interval

(a, b)

par

1

hai aur dono switch points par undefined hai, lekin kyunki integrals single points ignore karte hain toh yeh kabhi Laplace transform ko affect nahi karta.

L {δ (t)} = 1

ka matlab hai ki delta mein "sab frequencies equally" hain.

True — saare

s

par

1

ka flat transform isliye hai ki

δ

ek system ko uniformly probe karta hai, iske response ko impulse response banata hai (dekho Impulse Response and Convolution).

Ek function ko

u (t - a)

se multiply karne se us function ki value

t > a

ke liye badal sakti hai.

False —

t > a

ke liye step

1

ke barabar hai, isliye

g (t) u (t - a) = g (t)

wahan; step sirf

t < a

ka hissa delete karta hai, visible part ko kabhi rescale nahi karta.

u (0)

ke liye choose ki gayi value Laplace transform

\frac{e ^{- a s}}{s}

ko badal deti hai.

False — Laplace ek integral hai, aur integrals ek single point ka contribution ignore karte hain, isliye

u (0) = 0, \frac{1}{2},

1

sab identical transform dete hain (ROC

ℜ (s) > 0

par valid).

Kisi signal ko

u (t - a)

se gate karna ya delay karna region of convergence ko shrink kar sakta hai.

False — delay stamp

e^{- a s}

(jab

a \geq 0

) ka modulus

e^{- a ℜ (s)} \leq 1

hai

ℜ (s) > 0

par, isliye convergence kabhi worse nahi hoti;

e^{- a s} F (s)

ka ROC wahi hai jo

F (s)

ka hai.

Spot the error

Koi likhta hai

L {t u (t - 2)} = e^{- 2 s} \frac{1}{s ^{2}}

Error — second shifting theorem (upar restate kiya gaya) ko shifted function

f (t - 2)

chahiye, lekin

t

shifted nahi hai. Pehle

t = (t - 2) + 2

rewrite karo, jo deta hai

e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})

Koi

e^{- π s} / (s^{2} + 1)

ko invert karke

y (t) = sin (t - π)

likhta hai.

Error — gate

u (t - π)

missing hai; correct inverse hai

sin (t - π) u (t - π)

, jo

t < π

ke liye

0

hai.

u

drop karne se answer

t < π

ke liye galat ho jaata hai.

Koi claim karta hai ki

δ (t - a)^{2}

"twice the impulse" deliver karta hai.

Error — ek distribution ko square karna undefined hai;

δ

ek linear functional hai aur sirf linear integrals ke andar rehta hai, isliye

δ^{2}

ka koi matlab hi nahi hai, doubled area toh bilkul bhi nahi.

Koi likhta hai

\int_{- \infty}^{\infty} f (t) δ (t - a) d t = f (t)

Error — sifting

f

ko spike location par evaluate karta hai, isliye result number

f (a)

hai, function

f (t)

nahi; integral variable

t

ko already erase kar chuka hai.

Koi kehta hai "

δ

u

ki derivative hai, isliye

δ

u

se smoother hai."

Error — yahan differentiate karna cheezein rougher banata hai:

u

mein ek jump hai, aur uski distributional derivative

δ

aur bhi wild hai (ek infinite spike), smoother nahi.

Koi

L {u (t - a)}

compute karta hai

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

integrate karke aur kuch bhi multiply nahi karta, aur

1/ s

paata hai.

Error — lower limit

a

honi chahiye kyunki

u (t - a) = 0

[0, a)

;

a

se integrate karne par correct answer milta hai

e^{- a s} / s

, ROC

ℜ (s) > 0

par, naa ki

1/ s

Koi "input

g (t)

starting at time

a

" ko sirf

g (t - a)

ki tarah model karta hai.

Error — bina gate

u (t - a)

ke shifted function

t < a

ke liye bhi nonzero hai; sahi "turn on at

a

" object hai

g (t - a) u (t - a)

Why questions

δ

ko ek formula jaise

δ (a) = \infty

ki jagah ek integral se kyun define karna padta hai?

Kyunki pointwise height "

\infty

" ko coherently add ya scale nahi kiya ja sakta, isliye koi ordinary function required behaviour reproduce nahi kar sakta. Iski jagah hum

δ

ko ek linear functional ki tarah define karte hain — ek machine jo har smooth test function

φ

ko number

φ (a)

map karti hai. Linearity plus sifting rule hi woh object hai; distributional framework (Distributions and Generalized Functions) ise full rigour deta hai.

Time mein function shift karna

s

-domain mein

e^{- a s}

se multiply karne mein kyun badal jaata hai?

Laplace integral mein

τ = t - a

substitute karne par har term se ek constant factor

e^{- a s}

alag ho jaata hai, isliye poora transform sirf delay factor

e^{- a s}

se stamp ho jaata hai; kyunki

∣ e^{- a s} ∣ \leq 1

ℜ (s) > 0

convergence untouched rehti hai (the First and Second Shifting Theorems).

δ (t)

ka response itna special kyun hai ki uska apna naam hai?

Kyunki

L {δ} = 1

, isliye

Y (s) = H (s) \cdot 1

raw system transfer function hai, aur uska inverse — impulse response — kisi bhi input ke saath convolve karke har doosra response deta hai (dekho Impulse Response and Convolution).

y^{''} + y = δ (t - π)

mein ek impulse

δ (t - π)

ek ongoing oscillation produce karta hai, sirf ek blip nahi — kyun?

Kick ek instant tak rehta hai lekin state change kar deta hai (yeh system ko velocity de deta hai); ek baar mass chalti hai, spring use hamesha ke liye oscillate karata rehta hai.

Box

δ_{ε}

ki area

1

kyun maintain ki jaati hai jaise

ε \to 0

, fixed height ki jagah?

Kyunki "total impulse = 1" woh physically conserved quantity hai; height ko bilkul blow up karna padhega jaise width shrink hoti hai taaki area (diya gaya punch) constant rahe — bilkul wahi area-

1

boxes figure s03 mein drawn hain.

Derivative link

δ = \frac{d}{d t} u

believable kyun lagta hai jabki

u

a

par koi ordinary derivative nahi hai?

Averaged box

δ_{ε} = \frac{1}{ε} [u (t - a) - u (t - a - ε)]

literally

u

ke rise ka average slope hai width

ε

par;

ε \to 0

lene par yeh (distributional) derivative ban jaata hai —

1

ka ek jump "ek instant ke liye infinite rate" hai.

Edge cases

\int_{- \infty}^{\infty} f (t) δ (t - a) d t

kya hai jab

f

bilkul

a

par discontinuous ho?

Yeh ambiguous hai — sifting ko

f

a

par continuous hona chahiye; agar

f

wahan jump kare toh "value

f (a)

" well-defined nahi hai, isliye integral ka koi clean answer nahi hai (ek symmetric-limit reading jump ka midpoint return karegi).

L {δ (t - a)} = e^{- a s}

limit

a \to 0^{+}

mein kya hota hai?

Yeh

e^{0} = 1

ki taraf tend karta hai, jo

L {δ (t)} = 1

se match karta hai; integration domain ke bilkul edge par spike abhi bhi poori tarah capture hoti hai (transform saare

s

ke liye valid, ROC poora plane).

Kya

δ (t - a)

meaningful hai

a < 0

ke liye ek one-sided Laplace transform ke andar?

Koi usable effect nahi — spike

t = a < 0

par hai,

[0, \infty)

ke bahar, isliye

L {δ (t - a)} = 0

; impulse "system ke sunna shuru karne se pehle" ho chuka tha.

Boxcar

u (t - a) - u (t - b)

kya banta hai agar

a = b

Yeh har jagah

0

par collapse ho jaata hai (zero-width window koi area carry nahi karta), jo identical steps subtract karne ke consistent hai.

y^{'} + 2 y = u (t - 1), y (0) = 0

ke liye, switch instant

t = 1

par

y

kya hai?

Effectively

0

— solution

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

t = 1

par value

0

hai aur tabhi

\frac{1}{2}

ki taraf rise karta hai; switch flip hone ke instant par kuch bhi accumulate nahi hua hai.

\int_{a}^{\infty} δ (t - a) d t

kya hai — kya integral ko bilkul spike par shuru karna use count karta hai?

Yeh genuinely ambiguous hai. Symmetric-limit convention sharp spike ko symmetric boxes

δ_{ε}

se replace karta hai jo

a

par centred hain aur

ε \to 0

leta hai; tab area ka exactly aadha

t \geq a

par hota hai, jo value

\frac{1}{2}

deta hai (wahi reason jisse

u (0) = \frac{1}{2}

). Hum prefer karte hain ki

a > 0

strictly

[0, \infty)

ke andar ho taaki koi half-spike kabhi boundary par naa baithe.