Visual walkthrough — Heaviside step function and Dirac delta function

4.6.31 · D2 · Maths › Ordinary Differential Equations › Heaviside step function and Dirac delta function

Prerequisites jo tumhare paas khule ho sakte hain: Laplace Transform — Definition and Existence, First and Second Shifting Theorems.

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Step 1 — Switch draw karo (the step function)

KYA. Hum Heaviside step $u(t-a)$ draw karte hain: ek graph jo $t=a$ se pehle har time ke liye floor pe chipka hua (height $0$) rehta hai, phir height $1$ par jump karta hai aur wahan hamesha ke liye ruk jaata hai.

KYUN. Is page par sab kuch change ke baare mein hai. Pehle hum "kitni tezi se change hota hai" ki baat kar sakein, pehle humein woh cheez dekhni chahiye jo change hoti hai. Step sabse simple possible change hai: kuch nahi, phir click, phir ek flat plateau.

PICTURE.

Figure ko term by term padhna:

• Horizontal axis $t$ hai — time. Left matlab pehle, right matlab baad mein.
• Vertical axis signal ki height hai (uski value).
• $t=a$ par red mark woh ek instant hai jahan poori kahani hoti hai: jump.
• $a$ ke left mein curve $0$ par baitha hai (switch off). Right mein $1$ par baitha hai (switch on).

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Step 2 — Jump differentiate karne ke liye bahut sharp hai, toh hum use widen karte hain

KYA. Hum instant jump ko ek ramp se replace karte hain jo ek chote se width $\epsilon$ (Greek letter "epsilon", bas ek tiny positive number ka naam) mein smoothly $0$ se $1$ tak chadh jaata hai. Is widened switch ko $u_{\epsilon }$ kehte hain.

KYUN. Derivative slope measure karta hai $=\frac{\text{rise}}{\text{run}}$. True step mein rise $1$ hai lekin run $0$ hai, aur $1/0$ koi number nahi hai. Hum ek vertical cliff ka slope measure nahi kar sakte. Ilaaj yeh hai: cliff ko ek tiny but nonzero run $\epsilon$ do. Ab slope ek honest fraction hai. Baad mein hum $\epsilon \to 0$ shrink karte hain aur dekhte hain slope kya karta hai.

PICTURE.

• Red ramp interval $[a,a+\epsilon ]$ mein upar chadha hai.
• Uski total rise $1$ hai (step ke bottom se top tak).
• Horizontal run $\epsilon$ hai (tiny width, double arrow se marked).

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Step 3 — Ramp ka slope measure karo

KYA. Ramp mein, $[a,a+\epsilon ]$ ke bahar har jagah graph flat hai, toh uska slope $0$ hai. Ramp par khud slope constant hai: $\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{1}{\epsilon }.$

KYUN. Yeh widened switch ki derivative hai. Hum literally $\frac{d}{dt}{u}_{\epsilon }(t-a)$ compute kar rahe hain. Jahan graph flat hai, rate of change kuch nahi hai. Jahan woh chadha hai, rate of change ek single tall value $1/\epsilon$ hai.

PICTURE.

• $\frac{1}{\epsilon }$ har term mein is tarah aata hai: numerator $1$ = total height jo chadhi; denominator $\epsilon$ = chadne mein laga time.
• Red block slope graph hai: $1/\epsilon$ height ka ek rectangle jo $[a,a+\epsilon ]$ ke upar baitha hai, aur baaki jagah zero.

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Step 4 — Notice karo ki box ka area hamesha 1 hota hai

KYA. Slope-box ek rectangle hai. Uska area height $\times$ width hai: $\text{area}=\frac{1}{\epsilon }\times \epsilon =1.$

KYUN. Yeh poore page ka pivot hai. Hum ramp ko kitna bhi thin banayein, uske slope-graph ke neeche area $1$ par locked hai. Thin box $\Rightarrow$ tall box, lekin product kabhi nahi badalta. Area = switch kitna total chadha = $1$. Woh "$1$" us impulse ki strength hai jo hum abhi banane wale hain.

PICTURE.

• Shrinking $\epsilon$ ke liye teen boxes dikhaye gaye hain: wide-and-short, medium, thin-and-tall (sabse paatle ka rang red hai).
• Har ek apne $\frac{1}{\epsilon }\times \epsilon =1$ ke saath labelled hai: same area, alag shape.

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Step 5 — Box ko zero width tak squeeze karo

KYA. $\epsilon \to 0^{+}$ hone do (epsilon positive side se zero ki taraf shrink hota hai). Box infinitely thin aur infinitely tall ho jaata hai, phir bhi uska area abhi bhi exactly $1$ hai. Limiting object Dirac delta hai: $\delta (t-a)={\lim }_{\epsilon \to 0^{+}}{\delta }_{\epsilon }(t-a).$

KYUN. Yeh "delta, step ki slope hai" ka honest matlab hai. True step (Step 1) ramps (Step 2) ka limit hai; isliye uski slope ramp-slopes (Step 3) ka limit hai, aur woh limit yeh area $1$ ka infinite spike hai.

PICTURE.

• Red spike (ek arrow ke roop mein drawn, kyunki yeh kisi bhi page se zyada tall hai) $t=a$ par baitha hai.
• Uske saath label "area $=1$" yaad dilaata hai: height meaningless hai ($\infty$), sirf area count karta hai.

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Step 6 — Box ke through ek function feed karo (the sifting property)

KYA. Koi bhi smooth function $f(t)$ ko box se multiply karo aur integrate karo: ${\int }_{-\infty }^{\infty }f(t){\delta }_{\epsilon }(t-a)dt=\frac{1}{\epsilon }{\int }_a^{a+\epsilon }f(t)dt.$

KYUN. $[a,a+\epsilon ]$ ke bahar box $0$ hai, toh woh $f$ ka sab kuch us tiny interval ke upar wale sliver ke alawa delete kar deta hai. Right side exactly us sliver par $f$ ki average value hai ($f$ ka total area wahan, width $\epsilon$ se divided). Jab sliver single point $a$ par collapse karta hai, toh us par $f$ ka average sirf $f(a)$ ban jaata hai.

PICTURE.

• Black curve $f(t)$ hai.
• Red strip woh jagah hai jahan box rehta hai; sirf uske andar $f$ ka piece survive karta hai.
• Arrow strip ko $a$ par collapse hote dikhata hai, jahan $f$ single value $f(a)$ leta hai (red dot).

Limit ke term by term:

• $\frac{1}{\epsilon }{\int }_a^{a+\epsilon }fdt$ — "$\frac{1}{\epsilon }$" aur "width-$\epsilon$ integral" size mein cancel hote hain, ek average chhodke.
• Jab $\epsilon \to 0$, ek vanishing interval par $f$ ka average point par value hai: $f(a)$.

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Step 7 — Wohi trick Laplace transform deti hai

KYA. Special function $f(t)=e^{-st}$ choose karo (har Laplace transform ke andar ka ingredient) aur sift karo: $\mathcal{L}\{\delta (t-a)\}={\int }_0^{\infty }{e}^{-st}\delta (t-a)dt=e^{-as}.$

KYUN. Laplace transform kisi bhi cheez ka ${\int }_0^{\infty }{e}^{-st}(\cdot )dt$ hota hai. Jab "$\cdot$" delta hai, sifting sirf $t=a$ par $e^{-st}$ ki value read karta hai. Koi naya kaam nahi — Step 6 ki wohi picture ek specific curve ke saath.

PICTURE.

• Black curve $e^{-st}$ hai, $t$ badhne ke saath neeche slide karta hua.
• Red spike $a$ par wahan uski height pick karta hai: $e^{-as}$ (curve par red dot).

$e^{-as}$ padhna: "$-a$" batata hai kitni der intezaar kiya (delay), "$s$" transform variable hai. Saath mein $e^{-as}$ delay stamp hai — wahi stamp jo First and Second Shifting Theorems mein appear karta hai.

Recall Special case check karo

$a=0$ set karo. Tab $\mathcal{L}\{\delta (t)\}$ ::: $e^0=1$ — delta woh input hai jiski transform exactly $1$ hai, isliye uska response system ka impulse response hota hai.

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Step 8 — Edge case: agar $a=0$ ho ya kick screen se bahar lande?

KYA. Do degenerate situations:

1. $a=0$ — spike Laplace window $[0,\infty )$ ke bilkul start par baitha hai.
2. $a<0$ — spike $t=0$ se pehle land karta, window ke bahar.

KYUN. Laplace integral $t=0$ par start hota hai. Agar poora spike $0$ par ya uske baad live kare toh capture ho jaata hai; agar strictly $0$ se pehle live kare toh $[0,\infty )$ par integral use kabhi nahi dekhta, aur transform $0$ hota hai. Yeh boundary case hai jis par ek solver ko kabhi trip nahi karna chahiye.

PICTURE.

• Left: spike exactly $t=0$ par (red) — convention se half-caught, transform $=1$ (hum $a\ge 0$ lete hain).
• Right: spike $a<0$ par (red, greyed window) — integration window $0$ par start hota hai, toh kuch collect nahi hota, transform $=0$.

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Ek-picture summary

Ek image, poori chain:

• Top row: step $u(t-a)$ ko width $\epsilon$ ke ramp mein soften kiya (red), phir wapas cliff par sharpen kiya.
• Bottom row: uski slope — height $1/\epsilon$ aur area $1$ ka ek box — red spike $\delta (t-a)$ mein squeeze ho gaya.
• Rows ke beech vertical arrow operation "$\frac{d}{dt}$" hai jo neeche jaata hai, aur "$\int dt$" upar jaata hai. Switch differentiate karo, kick milti hai; kick integrate karo, switch wapas ban jaata hai.

Recall Feynman retelling (poora walkthrough plain words mein explain karo)

Ek light switch imagine karo. Off, phir click, phir on — dark se bright tak ka woh jump step function hai, aur yeh ek exact moment $t=a$ par hota hai. Ab, "kitni tezi se roshan hua?" ek fair sawaal hai, lekin ek real switch ke liye honest jawab hai: bahut tezi se, ek tiny sliver of time $\epsilon$ mein. Us sliver ke dauran brightness $0$ se $1$ tak chadhi, toh brightening ki speed $1$ divided by woh tiny time hai — ek bada number, $1/\epsilon$. Woh speed draw karo aur tumhe ek thin tall box milta hai. Yahan magic hai: box $\frac{1}{\epsilon }$ tall aur $\epsilon$ wide hai, toh uska area hamesha $1$ hai, click kitna bhi jaldi ho. Switch ko aur tezi se flip karo ($\epsilon \to 0$) aur box aur taller aur thinner hota jaata hai lekin uska area $1$ par chipka rehta hai. Limit mein yeh area $1$ ka ek infinitely tall, infinitely thin needle hai — delta function, "click" khud. Kyunki yeh needle hai, uski height poochna silly hai (woh infinite hai). Iske bajaye tum poochte ho woh kya karta hai: ise kisi bhi curve $f$ ke neeche slide karo aur yeh exactly ek value, $f(a)$, grab karta hai, baaki sab ignore karta hai — yahi sifting hai. Ise Laplace ingredient $e^{-st}$ ke neeche slide karo aur yeh $e^{-as}$, delay stamp, grab karta hai. Aur woh ek sentence jo sab kuch bandhta hai: kick switch flipping ki speed hai — delta, step ki derivative hai.

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Yeh bhi dekho: Solving IVPs with Laplace Transforms, Piecewise and Periodic Forcing Functions, Distributions and Generalized Functions.