Heaviside step function and Dirac delta function

1. The Heaviside Step Function

WHY it matters. It lets us write piecewise inputs as one formula. Want a function that is $g(t)$ but only after time $a$ ? Multiply: $g(t)\,u(t-a)$ kills it for $t<a$ and reveals it for $t>a$ .

HOW to build windows. To turn something on at $a$ and off at $b$ : $u(t-a)-u(t-b)=\begin{cases}1 & a<t<b\\0&\text{otherwise}\end{cases}$ This "boxcar" is the workhorse of piecewise modelling.

Laplace transform of the step

Formula Second shifting theorem — derived

We want $\mathcal{L}\{f(t-a)\,u(t-a)\}$ . Compute: $\int_0^\infty e^{-st}f(t-a)u(t-a)\,dt=\int_a^\infty e^{-st}f(t-a)\,dt.$ Substitute $\tau=t-a$ , so $t=\tau+a$ , $dt=d\tau$ , limits $\tau:0\to\infty$ : $=\int_0^\infty e^{-s(\tau+a)}f(\tau)\,d\tau=e^{-as}\int_0^\infty e^{-s\tau}f(\tau)\,d\tau=e^{-as}F(s).$ $\boxed{\;\mathcal{L}\{f(t-a)\,u(t-a)\}=e^{-as}F(s)\;}$ The factor $e^{-as}$ is the delay stamp: shifting in time = multiplying by $e^{-as}$ in $s$ .

2. The Dirac Delta Function

HOW it arises as a limit. Define a narrow box of width $\varepsilon$ , height $1/\varepsilon$ (area $=1$ ), centred at $a$ : $\delta_\varepsilon(t-a)=\frac{1}{\varepsilon}\big[u(t-a)-u(t-a-\varepsilon)\big].$ As $\varepsilon\to 0^+$ , the box becomes infinitely tall and thin, area still $1$ — that is $\delta(t-a)$ .

Sifting property — derived from the box

Laplace transform of the delta

Figure — Heaviside step function and Dirac delta function

Worked example Solving an ODE with an impulse

Solve $y''+y=\delta(t-\pi)$ , $\;y(0)=0,\ y'(0)=0$ .

Take Laplace: $\;s^2Y - sy(0)-y'(0) + Y = e^{-\pi s}$ . Why? $\mathcal{L}\{y''\}=s^2Y-sy(0)-y'(0)$ ; RHS uses $\mathcal{L}\{\delta(t-\pi)\}=e^{-\pi s}$ .
With zero ICs: $(s^2+1)Y=e^{-\pi s}\Rightarrow Y=\dfrac{e^{-\pi s}}{s^2+1}$ .
Invert: $\mathcal{L}^{-1}\{1/(s^2+1)\}=\sin t$ , then apply the delay $e^{-\pi s}$ → shift & gate: $y(t)=\sin(t-\pi)\,u(t-\pi)=-\sin t\;\;u(t-\pi).$ Interpretation: Nothing happens until $t=\pi$ ; then the kick starts the oscillation. Physically: a hammer hits a still mass-spring at $t=\pi$ .

3. Common Mistakes (Steel-manned)

4. Active Recall

Recall Quick self-test (hide and answer)

State the sifting property and derive $\mathcal{L}\{\delta(t-a)\}$ .
Why is $\delta = \frac{d}{dt}u$ ?
Transform $(t-3)^3 u(t-3)$ .
What's missing if you write $\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)$ ?

Recall Feynman: explain to a 12-year-old

The step function is like a light switch: pitch dark, then click, fully bright — that "click" happens at one exact moment. The delta function is the click itself: a super-quick poke that delivers a fixed amount of "push" in no time at all, like flicking a marble. The neat trick: how fast the light brightness changes at the click is exactly the poke. So the poke (delta) is just the speed of the switch flipping (the slope of the step). In maths, when you ask "how much total push did the poke give?", the answer is always whatever the system felt right at that instant — the delta picks out that one value and ignores everything else.

Definition of

u(t-a)

0

for

t<a

1

for

t>a

; switches the value

1

on at

t=a

\mathcal{L}\{u(t-a)\}

\dfrac{e^{-as}}{s}

, for

s>0

Second shifting theorem

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

Sifting property of the delta

\int_{-\infty}^{\infty} f(t)\delta(t-a)\,dt = f(a)

\mathcal{L}\{\delta(t-a)\}

e^{-as}

(and

\mathcal{L}\{\delta(t)\}=1

Relation between step and delta

\delta(t-a)=\dfrac{d}{dt}u(t-a)

Boxcar (window) on

(a,b)

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

\mathcal{L}^{-1}\{e^{-as}F(s)\}

f(t-a)\,u(t-a)

(the gate is required!).

\mathcal{L}\{t\,u(t-2)\}

write

t=(t-2)+2

→

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

Why

\delta

is not a true function

It is a distribution defined only via integrals; it has no finite pointwise value at

a

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, do simple ideas hain. Heaviside step function $u(t-a)$ ek switch ki tarah hai: $t=a$ se pehle output zero (light off), $t=a$ ke baad output one (light on). Iska kaam hai piecewise inputs ko ek single formula mein likhna — jaise "yeh force sirf 2 second ke baad chalu hoti hai". Window banane ke liye $u(t-a)-u(t-b)$ use karo, jo sirf $a$ se $b$ ke beech 1 deta hai. Iska Laplace transform $e^{-as}/s$ hota hai, aur agar function delayed ho to second shifting theorem kehta hai $\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$ — yaani time mein delay = $s$ -domain mein $e^{-as}$ se multiply.

Dirac delta $\delta(t-a)$ ek instant ka jhatka (impulse) hai — hammer se ek zor ka tap. Yeh koi normal function nahi hai; isko hum sirf integral ke andar samajhte hain. Iska main rule sifting property hai: $\int f(t)\delta(t-a)\,dt=f(a)$ , matlab yeh poore function mein se sirf $t=a$ wali value utha leta hai, baaki sab ignore. Iska Laplace $e^{-as}$ hai (step se ek $1/s$ kam, kyunki delta step ka derivative hai).

Sabse important baat ek hi hai: delta step ka derivative hai, $\delta=\frac{d}{dt}u$ . Switch achanak on hota hai, aur us achanak change ka rate hi impulse hai. ODE solve karte waqt $\delta(t-\pi)$ ka matlab hai: $t=\pi$ tak kuch nahi, phir ek dum kick lag gaya. Ek common galti: inverse Laplace mein gate $u(t-a)$ lagana mat bhoolna, warna answer $t<a$ ke liye galat ho jayega. Aur $g(t)u(t-a)$ ko transform karne se pehle $g(t)$ ko $(t-a)$ ke terms mein likhna zaroori hai.