Heaviside step function and Dirac delta function: Worked examples

4.6.31 · D3 · Maths › Ordinary Differential Equations › Heaviside step function and Dirac delta function

Definition Is page par use ki gayi Laplace transform notation

Symbol $L {\cdot}$ hai Laplace transform — ek operator jo time $t$ ki function ko ek naye variable $s$ ki function mein badalta hai. Plain words mein, yeh ek machine hai: aap $f (t)$ feed karo, bahar aata hai $L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t .$

$s$ kya hai? Ek helper variable (socho "frequency-jaisa knob"). Hum ise real world mein kabhi measure nahi karte; yeh sirf transformed picture mein exist karta hai, aur integral converge hone ke liye yeh kaafi bada hona chahiye ( $s > 0$ yahan wali functions ke liye). Poori details Laplace Transform — Definition and Existence mein.
$F (s)$ kya hai? Bas output ka naam; capital letter lower-case input se match karta hai ( $f \to F$ , $y \to Y$ ).
Inverse $L^{- 1} {\cdot}$ machine ko ulta chalana hai: $F (s)$ diya, toh original $f (t)$ wapas milta hai. To $L^{- 1} {F (s)} = f (t)$ undo karta hai $L {f (t)} = F (s)$ ko, jaise $arctan$ undo karta hai $tan$ ko.
Ise use kyun karein? Yeh ek differential equation (mushkil: derivatives involved) ko $s$ mein ek algebra problem (aasaan: bas divide karo) mein convert karta hai, jise hum phir time mein wapas invert karte hain. Dekho Solving IVPs with Laplace Transforms.

Scenario matrix

Is topic ka har problem neeche diye gaye cells mein se ek (ya kuch ka mix) hai. Last column us worked example ka naam deta hai jo usse cover karta hai.

Cell	Scenario class	Tricky kya hai	Example
A	Un-shifted $g (t) u (t - a)$	$g$ pehle se $g (t - a)$ ke form mein nahi likha — pehle rewrite karna hoga	Ex 1
B	Multi-piece step assembly	kai boxcars, kai delay stamps	Ex 2
C	Delta inside an integral (sifting)	plain evaluation, koi ODE nahi	Ex 3
D	Degenerate $a = 0$	$u (t)$ aur $δ (t)$ bilkul shuruat mein	Ex 4
E	Delta at a shifted time jo ODE drive kare	output par gate + shift	Ex 5
F	Step forcing ek ODE mein, steady state dekho	$t \to \infty$ par limiting value	Ex 6
G	RHS par dono step aur delta	do responses superpose karo	Ex 7
H	Word problem (real impulse: hammer/kick)	physics ko $δ$ mein translate karo	Ex 8
I	Exam twist: $t = 0$ se pehle delta, ya window ke bahar delta evaluate karna	sift us jagah land karta hai jahan kuch measure nahi hota	Ex 9

Ab hum matrix ko upar se neeche sweep karte hain.

Ex 1 — Cell A: un-shifted $g (t) u (t - a)$

Worked example

$L {t u (t - 2)}$ transform karo Forecast: Pehle guess karo. Naive (galat) instinct hai $e^{- 2 s} \cdot \frac{1}{s ^{2}}$ . Kya $g (t) = t$ pehle se $g (t - 2)$ form mein hai? Nahi — theorem ko shifted argument chahiye. To ek baar jab hum yeh fix karein to hum expect karte hain ki ek extra term aayega. Kuch aisa predict karo jaise $e^{- 2 s} (\frac{1}{s ^{2}} + \frac{something}{s})$ .

Steps.

Delay stamp yaad karo: $L {f (t - a) u (t - a)} = e^{- a s} F (s)$ . Yeh step kyun? Yahi ek rule hai jo $u (t - a)$ factor handle karta hai — lekin isko demand hai ki multiplier $(t - a)$ ki function ho, $t$ ki nahi.
$t$ ko $(t - 2)$ se rewrite karo: $t = (t - 2) + 2$ . Yeh step kyun? Isse theorem ka zaroori " $(t - 2)$ " force hota hai; bacha hua $+ 2$ ek constant hai jise hum alag handle karte hain.
Split karo: $t u (t - 2) = (t - 2) u (t - 2) + 2 u (t - 2)$ . Yeh step kyun? Ab har piece genuinely $(t - 2)$ ki function hai times $u (t - 2)$ .
Har piece transform karo. $(t - 2) u (t - 2)$ ke liye: yahan $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ , deta hai $e^{- 2 s} \frac{1}{s ^{2}}$ . $2 u (t - 2)$ ke liye: yahan $f (t) = 2$ , $F (s) = \frac{2}{s}$ , deta hai $e^{- 2 s} \frac{2}{s}$ . Yeh step kyun? Stamp ko har shifted piece par alag apply karo.
Add karo: $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) .$

Verify: Forecast ka "extra term" exactly $\frac{2}{s}$ hai, jo rewrite ke $+ 2$ se match karta hai. Sanity check: agar hum step 2 bhool jaate to woh $\frac{2}{s}$ drop ho jaata — yahi woh classic mistake hai jo parent note mein steel-man ki gayi hai. ✓

Cell A ka Takeaway: jab bhi multiplier ek bare $g (t)$ ho, $t = (t - a) + a$ substitute karo aur theorem touch karne se pehle expand karo.

Ex 2 — Cell B: multi-piece input assemble karna

Worked example Piecewise

$f (t) = ⎩ ⎨ ⎧ 3 t 0 0 < t < 2 2 < t < 5 t > 5$ transform karo Forecast: Hum $f$ ko boxcars se build karenge aur predict karte hain ek plain $\frac{3}{s}$ plus kai $e^{- 2 s} (\dots)$ aur $e^{- 5 s} (\dots)$ stamps.

Steps.

Boxcar form scratch se banao. Ek boxcar $u (t - a) - u (t - b)$ equals $1$ on $(a, b)$ aur $0$ baaqi jagah (switch on at $a$ , off at $b$ ). Har piece isolate karo:
- Constant $3$ lives on $(0, 2)$ : likho $3 (u (t) - u (t - 2))$ .
- Ramp $t$ lives on $(2, 5)$ : likho $t (u (t - 2) - u (t - 5))$ .
- $t = 5$ ke baad function $0$ hai, to koi term zaroorat nahi. Add karke aur $u (t) = 1$ use karke $t > 0$ ke liye: $f = 3 - 3 u (t - 2) + t u (t - 2) - t u (t - 5)$ . Kyun? Har boxcar exactly ek interval reveal karta hai aur baaki chhupata hai, isliye unka sum piecewise graph reproduce karta hai.
$L {3} = \frac{3}{s}$ ; $L {- 3 u (t - 2)} = - 3 \cdot \frac{e ^{- 2 s}}{s}$ . Kyun? Direct step transform $L {u (t - a)} = \frac{e ^{- a s}}{s}$ .
$t u (t - 2)$ : Ex 1 ki trick reuse karo $a = 2$ ke saath: $= e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})$ . Kyun? Un-shifted multiplier.
$- t u (t - 5)$ : rewrite karo $t = (t - 5) + 5$ , deta hai $- e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s})$ . Kyun? Wahi fix, ab $a = 5$ .
Collect karo: $F (s) = \frac{3}{s} - \frac{3 e ^{- 2 s}}{s} + e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) - e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s}) .$ $e^{- 2 s}$ group simplify karo: $- \frac{3}{s} + \frac{2}{s} = - \frac{1}{s}$ , to $F (s) = \frac{3}{s} + e^{- 2 s} (\frac{1}{s ^{2}} - \frac{1}{s}) - e^{- 5 s} (\frac{1}{s ^{2}} + \frac{5}{s}) .$

Verify (value continuity check): $t \to 2^{-}$ par, $f = 3$ ; $t \to 2^{+}$ par, $f = t = 2$ . Neeche diya figure input dikhata hai; $t = 2$ par $3$ se $2$ tak ka genuine downward jump hai (function wahan discontinuous hai), aur humara formula ise reproduce karta hai kyunki $3 u$ term switch off hoti hai jabki $t u$ term switch on hoti hai. ✓

Figure — Ex 2 input graph. Blue segment constant $3$ hai $(0, 2)$ par; yellow line ramp $t$ hai $(2, 5)$ par; green segment $0$ hai $t = 5$ ke baad. Open circle $(2, 2)$ par aur filled circle $(2, 3)$ par genuine downward jump (red arrow) mark karte hain; doosra red arrow $t = 5$ par switch-off mark karta hai. Yahi woh graph hai jo humara boxcar sum reproduce karta hai.

Figure — Heaviside step function and Dirac delta function

Ex 3 — Cell C: pure sifting, koi ODE nahi

Worked example

$\int_{- \infty}^{\infty} cos (t) δ (t - \frac{π}{3}) d t$ aur $\int_{0}^{\infty} e^{- 2 t} δ (t - 1) d t$ evaluate karo. Forecast: Sifting rule kehta hai $t = a$ par delta bas integrand ke baaki hisse ki value $t = a$ par read off kar leta hai. To predict karo $cos (π /3) = \frac{1}{2}$ aur $e^{- 2}$ .

Steps.

Sifting: $\int_{- \infty}^{\infty} f (t) δ (t - a) d t = f (a)$ . Kyun? Yahi $δ$ ki poori definition hai — yeh ek value pick out karta hai.
Pehla integral: $f (t) = cos t$ , $a = \frac{π}{3}$ $\Rightarrow$ answer $= cos \frac{π}{3} = \frac{1}{2}$ . Answer ek plain number kyun hai? Delta poore integral ko ek evaluation tak collapse kar deta hai.
Doosra integral: $f (t) = e^{- 2 t}$ , $a = 1$ . Check karo ki spike $t = 1$ range $[0, \infty)$ ke andar hai — hai. $\Rightarrow$ answer $= e^{- 2}$ . Range check kyun? Agar $a$ limits ke bahar hota to integral $0$ hota (Ex 9 cover karta hai woh).

Verify: $cos (π /3) = 0.5$ exactly; $e^{- 2} \approx 0.1353$ . Dono numerically check hain. ✓

Ex 4 — Cell D: degenerate $a = 0$

Worked example

$L {u (t)}$ aur $L {δ (t)}$ nikalo, aur solve karo $y^{'} + 3 y = δ (t)$ , $y (0^{-}) = 0$ . Forecast: $a = 0$ ke saath delay stamp $e^{- a s}$ ban jaata hai $e^{0} = 1$ . To expect karo sabse simple possible transforms: step ke liye $\frac{1}{s}$ , delta ke liye $1$ . ODE phir bas ek decaying exponential hogi jo $t = 0$ par kick se start hogi.

Steps.

$L {u (t)} = \frac{e ^{- 0 \cdot s}}{s} = \frac{1}{s}$ — $L {1}$ ke identical $t > 0$ ke liye. Kyun? $u (t) = 1$ saare $t > 0$ ke liye; point $t = 0$ ki zero width hai aur integral change nahi kar sakta.
$L {δ (t)} = e^{- 0 \cdot s} = 1$ . Itna sundar kyun hai? Origin par delta woh input hai jiska transform flat $1$ hai — "saari frequencies equally," to output system ka raw impulse response hai.
ODE: $s Y - y (0^{-}) + 3 Y = 1 \Rightarrow (s + 3) Y = 1 \Rightarrow Y = \frac{1}{s + 3}$ . $y (0^{-}) = 0$ kyun? Hum initial condition kick se ठीक पहले lete hain, to delta ka punch response mein dikhta hai, IC mein nahi.
Invert karo: $y (t) = L^{- 1} {\frac{1}{s + 3}} = e^{- 3 t} u (t)$ . $u (t)$ kyun? $t = 0$ se pehle kuch nahi; kick $t = 0^{+}$ par response mein sudden jump of $1$ create karta hai.

Verify: Kick ke immediately baad, $y (0^{+}) = 1$ — ek unit impulse se caused unit jump, exactly wahi jo strength $1$ ka impulse first-order system ko karna chahiye. $t \to \infty$ par, $y \to 0$ : system kick bhool jaata hai. ✓

Ex 5 — Cell E: shifted time par delta ek ODE drive karta hai

Worked example Solve karo

$y^{''} + 4 y = 3 δ (t - 2)$ , $y (0) = 0, y^{'} (0) = 0$ . Forecast: Ek still oscillator (natural frequency $ω = 2$ , kyunki $y^{''} + 4 y = 0$ , $cos 2 t, sin 2 t$ ki tarah oscillate karta hai) ko $t = 2$ par strength $3$ ka kick milta hai. Predict karo: $t = 2$ tak bilkul flat, phir angular frequency $2$ ki sine, kick se scale ki gayi, aur $u (t - 2)$ se gate ki gayi.

Steps.

Transform karo: $s^{2} Y + 4 Y = 3 e^{- 2 s}$ . $e^{- 2 s}$ kyun? $L {δ (t - 2)} = e^{- 2 s}$ ; zero ICs $sy (0), y^{'} (0)$ terms ko kill kar dete hain.
Algebraically solve karo: $Y = \frac{3 e ^{- 2 s}}{s ^{2} + 4}$ . Kyun? Bas divide karo.
Recognize karo $L^{- 1} {\frac{1}{s ^{2} + 4}} = \frac{1}{2} sin 2 t$ . $\frac{1}{2}$ kyun? Kyunki $L {sin ω t} = \frac{ω}{s ^{2} + ω ^{2}}$ with $ω = 2$ , to $\frac{1}{s ^{2} + 4}$ isolate karne ke liye $ω = 2$ se divide karte hain.
Delay stamp apply karo (shift aur gate): $y (t) = 3 \cdot \frac{1}{2} sin (2 (t - 2)) u (t - 2) = \frac{3}{2} sin (2 (t - 2)) u (t - 2) .$ Yeh step kyun? $e^{- 2 s}$ factor ko invert karne par (Ex 1 mein delay stamp dekho) shifted-and-gated version $f (t - 2) u (t - 2)$ milna chahiye; gate bhoolna galti se $y$ ko kick se pehle nonzero bana dega.

Verify (position ki continuity): $t = 2^{+}$ par, $sin (2 \cdot 0) = 0$ , to $y (2^{+}) = 0$ — position continuous hai, jaisa second-order kick ke liye hona chahiye ( $y^{''}$ mein delta velocity jump karta hai, position nahi). Velocity jump: $y^{'} (2^{+}) = \frac{3}{2} \cdot 2 cos 0 = 3$ , impulse strength $3$ se match karta hai. ✓

Figure — Ex 5 kicked oscillator. Green segment kick se pehle $y = 0$ dikhata hai; $t = 2$ par red arrow impulse $3 δ (t - 2)$ hai; blue curve resulting sine $\frac{3}{2} sin (2 (t - 2))$ hai jo sirf $t > 2$ ke liye switch on hoti hai. Note karo ki curve height $0$ se start hoti hai (position continuous) lekin nonzero slope ke saath (velocity kicked).

Ex 6 — Cell F: step forcing aur steady-state limit

Worked example Solve karo

$y^{'} + 2 y = 4 u (t - 1)$ , $y (0) = 0$ . Forecast: Parent ke step example jaisi hi shape lekin $4$ se scale ki gayi. $y^{'} + 2 y = 4$ ka steady state hai $y = 2$ ( $y^{'} = 0$ set karo). Predict karo: $t = 1$ tak $0$ , phir $2$ ki taraf rise.

Steps.

Transform karo: $s Y + 2 Y = \frac{4 e ^{- s}}{s} \Rightarrow Y = \frac{4 e ^{- s}}{s ( s + 2 )}$ . Kyun? $L {u (t - 1)} = \frac{e ^{- s}}{s}$ .
Partial fractions: $\frac{1}{s ( s + 2 )} = \frac{1}{2} (\frac{1}{s} - \frac{1}{s + 2})$ . Kyun? Standard inverses $\frac{1}{s} \to 1$ , $\frac{1}{s + 2} \to e^{- 2 t}$ hit karne ke liye.
To $\frac{4}{s ( s + 2 )} = 2 (\frac{1}{s} - \frac{1}{s + 2}) \to 2 (1 - e^{- 2 t})$ . Factor $2$ kyun? $4$ times partial fractions ka $\frac{1}{2}$ .
Delay apply karo: $y (t) = 2 (1 - e^{- 2 (t - 1)}) u (t - 1) .$ Kyun? Factor $e^{- s}$ $1$ se shift aur gate $u (t - 1)$ ke roop mein invert hota hai.

Verify (limiting value): $t \to \infty$ par, $e^{- 2 (t - 1)} \to 0$ , to $y \to 2$ — forecast steady state se match karta hai. $t = 1^{+}$ par: $y = 2 (1 - 1) = 0$ , continuous. ✓

Figure — Ex 6 step-forced rise. Green segment $t = 1$ par switch se pehle $y = 0$ hai (red arrow); yellow curve exponential rise hai; dashed blue line steady value $2$ mark karta hai jisski taraf curve approach karta hai lekin kabhi overshoot nahi karta.

Ex 7 — Cell G: step AUR delta saath mein

Worked example Solve karo

$y^{'} + y = u (t - 1) + δ (t - 2)$ , $y (0) = 0$ . Forecast: Do events. Pehle $t = 1$ par ek switch on hota hai ( $y$ ko $1$ ki taraf drive karta hai), phir $t = 2$ par ek kick sudden jump add karta hai. Linearity se total answer do individual responses ka sum hai. Predict karo $t = 1$ se rising curve, phir $t = 2$ par discontinuous upward jump.

Steps.

Transform karo: $s Y + Y = \frac{e ^{- s}}{s} + e^{- 2 s} \Rightarrow Y = \frac{e ^{- s}}{s ( s + 1 )} + \frac{e ^{- 2 s}}{s + 1}$ . Unhe add kyun karein? RHS ek sum hai, aur Laplace linear hai.
Step part: $\frac{1}{s ( s + 1 )} = \frac{1}{s} - \frac{1}{s + 1} \to 1 - e^{- t}$ ; delay $\Rightarrow (1 - e^{- (t - 1)}) u (t - 1)$ . Kyun? Standard partial fractions + delay stamp.
Delta part: $\frac{1}{s + 1} \to e^{- t}$ ; delay $\Rightarrow e^{- (t - 2)} u (t - 2)$ . Split karne ki zaroorat kyun nahi? Delta ka transform bas $e^{- 2 s}$ hai, jo ek clean $\frac{1}{s + 1}$ chhod deta hai.
Sum karo: $y (t) = (1 - e^{- (t - 1)}) u (t - 1) + e^{- (t - 2)} u (t - 2) .$ Simply add kyun karein? Kyunki equation linear hai: inputs ke sum ke liye response individual responses ka sum hai, to hum step-response aur impulse-response superpose karte hain.

Verify (kick par jump): $t = 2$ se just pehle: $y (2^{-}) = 1 - e^{- 1} \approx 0.632$ . Just baad: delta $e^{- (2 - 2)} = e^{0} = 1$ add karta hai, to $y (2^{+}) \approx 1.632$ . First-order system exactly impulse strength ( $= 1$ ) se jump karta hai — confirmed. ✓ (Dekho Impulse Response and Convolution for kyun har kick $e^{- t}$ ki ek copy shift karta hai.)

Ex 8 — Cell H: ek real-world hammer blow

Worked example Ek mass–spring jo rest mein hai, satisfy karta hai

$y^{''} + 9 y = 0$ (position $y$ , natural angular frequency $3$ ). $t = \frac{π}{6}$ par ek hammer strength $6$ ka impulse deliver karta hai. Motion nikalo. Forecast: Model hai $y^{''} + 9 y = 6 δ (t - \frac{π}{6})$ , $y (0) = y^{'} (0) = 0$ . Strike tak kuch nahi halta; phir angular frequency $3$ ki sine. Amplitude kick ke saath scale honi chahiye.

Steps.

Hammer ko $t = \frac{π}{6}$ par strength $6$ ke delta se model karo. Delta kyun? Ek hammer near-zero time par act karta hai lekin fixed total impulse deliver karta hai — exactly parent note ka "tap ko zero tak shrink karo" idea.
Transform karo: $s^{2} Y + 9 Y = 6 e^{- π s /6} \Rightarrow Y = \frac{6 e ^{- π s /6}}{s ^{2} + 9}$ . Kyun? Zero ICs, $L {δ} = e^{- a s}$ .
$L^{- 1} {\frac{1}{s ^{2} + 9}} = \frac{1}{3} sin 3 t$ . $\frac{1}{3}$ kyun? $L {sin 3 t} = \frac{3}{s ^{2} + 9}$ , $ω = 3$ se divide karo.
Shift & gate: $y (t) = 6 \cdot \frac{1}{3} sin (3 (t - \frac{π}{6})) u (t - \frac{π}{6}) = 2 sin (3 t - \frac{π}{2}) u (t - \frac{π}{6}) = - 2 cos (3 t) u (t - \frac{π}{6}) .$ Yeh step kyun? $e^{- π s /6}$ invert karne par $\frac{π}{6}$ se shift aur gate force hota hai; phir identity $sin (3 t - \frac{π}{2}) = - cos 3 t$ ise tidier form mein rewrite karta hai.

Verify (units & velocity jump): Position continuous: $t = \frac{π}{6}^{+}$ par, $y = - 2 cos \frac{π}{2} = 0$ . Velocity jump $y^{'} (\frac{π}{6}^{+}) = 2 \cdot 3 sin \frac{π}{2} = 6$ — impulse strength ko mass ( $= 1$ ) se divide karne ke barabar. Amplitude $2 = impulse / ω = 6/3$ , as expected. ✓

Ex 9 — Cell I: exam twist, spike window ke bahar

Worked example

$\int_{0}^{\infty} t^{2} δ (t + 1) d t$ aur $\int_{0}^{5} e^{t} δ (t - 7) d t$ evaluate karo. Forecast: Dono mein spike integration range ke bahar hai. Delta tab contribute karta hai jab uski spike limits ke andar ho; warna integrand har jagah zero hai jahan matter karta hai. Predict karo dono integrals $= 0$ .

Steps.

Pehla: $δ (t + 1) = δ (t - (- 1))$ spike karta hai $t = - 1$ par. Yeh matter kyun karta hai? Range $t = 0$ se start hoti hai; $- 1$ par spike kabhi include nahi hoti, to $\int_{0}^{\infty} (\dots) = 0$ . (Contrast: $\int_{- \infty}^{\infty}$ par yeh $(- 1)^{2} = 1$ deta.)
Doosra: $t = 7$ par spike, lekin upper limit $5$ hai. Kyunki $7 > 5$ , spike exclude hai, to integral $= 0$ . Dono limits kyun check karein? Sift tabhi fire karta hai jab $a$ $[lower, upper]$ mein ho; yahan hum confirm karte hain ki spike na lower limit se neeche hai na upper limit se upar, conclusion karne se pehle.

Verify: Dono integrals $0$ return karte hain kyunki $a \in /$ integration interval. Yahi woh trap hai jo zyaadatar exams set karte hain — sifting apply karne se pehle hamesha spike locate karo. ✓

Recall

Recall Kaun sa cell, kaun sa fix? (chhupaao aur jawaab do)

$g (t) u (t - a)$ ke liye jab $g$ shifted nahi hai, pehle $g (t)$ ko rewrite karo as ::: $g ((t - a) + a)$ , phir expand karo. First-order $y^{'} + cy$ par $t = a$ par strength $k$ ka delta $y$ ko $t = a$ par jump karata hai exactly ::: $k$ se. Second-order system ke RHS par delta ::: velocity jump karta hai (position nahi — position continuous rehti hai). $\int_{0}^{\infty} f (t) δ (t - a) d t$ equals $f (a)$ tabhi jab ::: $a \geq 0$ (spike $[0, \infty)$ ke andar hai); warna yeh $0$ hai. Delay stamp $e^{- a s} F (s)$ invert hota hai ::: $f (t - a) u (t - a)$ mein — gate $u (t - a)$ kabhi mat bhulo. $L^{- 1} {F (s)}$ ka matlab hai ::: Laplace machine ulta chalao — transform $F (s)$ se time function $f (t)$ recover karo.