Exercises — Heaviside step function and Dirac delta function

Symbol $u(t-a)$ ka matlab hai "jab tak $t<a$, 0; jab $t>a$ ho jaye, 1" — jaise $t=a$ par flip hone wala ek light switch. (Exactly $t=a$ par value unspecified chhodi jaati hai; yeh kabhi ek integral ko affect nahi karti.)

• $t=1$ par: kyunki $1<3$, hum switch se pehle hain, isliye $u(1-3)=u(-2)=0$.
• $t=3.5$ par: kyunki $3.5>3$, hum switch ke baad hain, isliye $u(3.5-3)=u(0.5)=1$.
• Difference $u(t-3)-u(t-7)$ ek boxcar hai: yeh $3<t<7$ ke liye $1-0=1$ hai aur baaki jagah $0$. Socho interval $(3,7)$ par rakha ek flat-topped box — Figure s01 mein teal box, jiska left edge $t=3$ par uthta hai aur right edge $t=7$ par girta hai.

Delta ka ek hi kaam hai — sifting: $\int f(t)\delta (t-a)dt=f(a)$. Yeh poore integrand ko ignore karta hai sivaaye single point $t=a=2$ ke. Yahan $f(t)=t^2+1$ hai, isliye answer hai $f(2)=2^2+1=5$. Yeh kaisa dikhta hai: Figure s01 mein plum spike exactly $t=2$ par khada hai; $f$ se multiply karke phir integrate karna sirf wahin curve ki height padh leta hai aur baki sab kuch phenk deta hai.

Dono seedhe defining integral $\mathcal{L}\{f\}={\int }_0^{\infty }{e}^{-st}f(t)dt$ se aate hain.

• Step: $\mathcal{L}\{u(t-4)\}={\int }_4^{\infty }{e}^{-st}dt=\frac{e^{-4s}}{s}$. (Lower limit $4$ hai kyunki $u(t-4)=0$ usse neeche hai; $1/s$ integral karne se aata hai, $e^{-4s}$ $4$ par shuru karne se aata hai.)
• Delta: $\mathcal{L}\{\delta (t-4)\}={\int }_0^{\infty }{e}^{-st}\delta (t-4)dt=e^{-4s}$. ($e^{-st}$ ko $t=4$ par sift karo; koi $1/s$ nahi kyunki integrate karne ke liye koi interval nahi hai — delta khud yeh karta hai.) Single difference — $1/s$ hai ya nahi — yahi "step vs impulse" ki pehchaan hai.

Yeh already shifted hai: jo cube ho raha hai woh $(t-2)$ hai, jo switch ke argument se match karta hai. Isliye directly delay stamp use karo.

• Identify karo: $f(t-2)=(t-2{)}^3\Rightarrow f(t)=t^3$, $a=2$ ke saath.
• $F(s)=\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$ (${\int }_0^{\infty }{e}^{-st}{t}^{3}dt$ se, standard $\mathcal{L}\{t^n\}=n!/s^{n+1}$).
• Delay stamp lagao: $\boxed{\mathcal{L}\{(t-2{)}^{3}u(t-2)\}=e^{-2s}\frac{6}{s^4}}$.

Switch $u(t-2)$ hai lekin multiplier $t$ hai, $(t-2)$ nahi. Delay stamp ki demand hai ki multiplier $(t-2)$ ke terms mein likha jaye. Isliye rewrite karo: $t=(t-2)+2.$ Tab $tu(t-2)=[(t-2)+2]u(t-2)=(t-2)u(t-2)+2u(t-2)$.

• Pehla piece: $f(t)=t\Rightarrow F(s)=1/s^2$, stamped: $e^{-2s}/s^2$.
• Doosra piece: $2u(t-2)\Rightarrow 2\cdot e^{-2s}/s$.
• Total: $\boxed{e^{-2s}\left(\frac{1}{s^2}+\frac{2}{s}\right)}$.

Sifting $f(a)$ pick karti hai jahan $f=\cos$, $a=\pi /3$ (jo $(0,\infty )$ ke andar hai, isliye count hoga). $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}.$ Answer: $1/2$.

Boxcars se build karo (ek boxcar $u(t-a)-u(t-b)$ sirf $(a,b)$ par 1 hota hai):

• $2$ on $(0,1)$: $2(u(t)-u(t-1))$.
• $0$ on $(1,3)$: kuch contribute nahi karta.
• $t$ for $t>3$: $tu(t-3)$ (ek half-line jo $3$ par switch on hoti hai). Isliye $f(t)=2-2u(t-1)+tu(t-3)$. ($t=1,3$ par values aage ke liye irrelevant hain — single points integral ko affect nahi karte.)

Transform. Pehle base fact yaad karo: $\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}dt=\frac{1}{s}$; linearity se $\mathcal{L}\{2\}=\frac{2}{s}$. $u(t-1)$ ke liye step formula deta hai $e^{-s}/s$. $tu(t-3)$ ke liye, $t=(t-3)+3$ split karo (taaki multiplier $(t-3)$ ki function ban jaye, jaise delay stamp ki requirement hai): $tu(t-3)=(t-3)u(t-3)+3u(t-3)\Rightarrow e^{-3s}\left(\frac{1}{s^2}+\frac{3}{s}\right).$ Total: $F(s)=\frac{2}{s}-\frac{2e^{-s}}{s}+e^{-3s}\left(\frac{1}{s^2}+\frac{3}{s}\right).$

Spike $t=5$ par hai. Ek finite integral isko tabhi "catch" karta hai jab $5$ integration ke interval ke andar ho.

• $I_1={\int }_0^2{e}^t\delta (t-5)dt$: $5$ par spike $[0,2]$ ke bahar hai, isliye integral $0$ hai.
• $I_2={\int }_0^6{e}^t\delta (t-5)dt$: ab $5\in [0,6]$ hai, isliye sifting deta hai ${\mathbf{e}}^5$. Moral: delta tabhi contribute karta hai jab uski spike location limits ke andar ho — yeh baat students tab bhool jaate hain jab limits $-\infty$ se $\infty$ ki jagah finite hon.

Delay strip karo: maan lo $F(s)=\frac{1}{s^2+4}$. Kyunki $\mathcal{L}\{\sin (2t)\}=\frac{2}{s^2+4}$ hai, hame milta hai $F(s)=\frac{1}{2}\cdot \frac{2}{s^2+4}$, isliye $f(t)=\frac{1}{2}\sin (2t)$. Factor $e^{-\pi s}$ ka matlab hai "$f$ ko $\pi$ se right shift karo AUR isse $u(t-\pi )$ se gate karo": ${\mathcal{L}}^{-1}\left\{\frac{e^{-\pi s}}{s^2+4}\right\}=\frac{1}{2}\sin (2(t-\pi ))u(t-\pi ).$ Ab sine ki periodicity use karke phase simplify karo (period $2\pi$): $\sin (2t-2\pi )=\sin (2t)$ kyunki angle se ek full period $2\pi$ ghatane par sine unchanged rehta hai. Isliye yeh $\frac{1}{2}\sin (2t)u(t-\pi )$ ke barabar hai. Case se likha hua (boundary value $t=\pi$ par limit hai, isliye $0$): $y(t)=\{\begin{array}{ll}0& 0\le t<\pi \\ \frac{1}{2}\sin (2t)& t>\pi .\end{array}$

Dono sides transform karo. $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)$ aur $\mathcal{L}\{\delta (t-\pi )\}=e^{-\pi s}$ use karke: $s^{2}Y+4Y=e^{-\pi s}\Rightarrow (s^2+4)Y=e^{-\pi s}\Rightarrow Y=\frac{e^{-\pi s}}{s^2+4}.$ Invert karo. Yeh bilkul Exercise 3.3 wala expression hai: $y(t)=\frac{1}{2}\sin (2(t-\pi ))u(t-\pi )=\frac{1}{2}\sin (2t)u(t-\pi ).$ Cases mein ($t=\pi$ par boundary value limit hai, isliye $y(\pi )=0$): $y(t)=\{\begin{array}{ll}0& 0\le t<\pi \\ \frac{1}{2}\sin (2t)& t>\pi .\end{array}$ Motion: mass $t=\pi$ tak bilkul still baitha rehta hai; $t=\pi$ par hammer-kick ek steady oscillation launch karta hai jiska amplitude $1/2$ aur angular frequency $2$ hai. Figure s02 mein plum kick aur orange oscillation dekho.

Transform karo. $\mathcal{L}\{y^{\prime }\}=sY-y(0)=sY$, aur $\mathcal{L}\{u(t-2)\}=e^{-2s}/s$: $sY+3Y=\frac{e^{-2s}}{s}\Rightarrow Y=\frac{e^{-2s}}{s(s+3)}.$ Partial fractions un-delayed core par: $\frac{1}{s(s+3)}=\frac{1}{3}\left(\frac{1}{s}-\frac{1}{s+3}\right)$, jiska inverse hai $\frac{1}{3}(1-e^{-3t})$. Delay stamp apply karo (shift by 2 + gate): $y(t)=\frac{1}{3}(1-e^{-3(t-2)})u(t-2).$ Cases mein ($t=2$ par boundary limit ki tarah padhi jaati hai, $y(2)=0$ deta hai): $y(t)=\{\begin{array}{ll}0& 0\le t<2\\ \frac{1}{3}(1-e^{-3(t-2)})& t>2.\end{array}$ Jab $t\to \infty$, $e^{-3(t-2)}\to 0$, isliye $y\to 1/3$ — $y^{\prime }+3y=1$ ka steady state.

Har piece transform karo.

• $\mathcal{L}\{u(t)\}=1/s$, $\mathcal{L}\{u(t-2\pi )\}=e^{-2\pi s}/s$, $\mathcal{L}\{\delta (t-\pi )\}=e^{-\pi s}$.
• Zero ICs ke saath LHS: $(s^2+1)Y$. Isliye $(s^2+1)Y=\frac{1}{s}-\frac{e^{-2\pi s}}{s}+e^{-\pi s}.$ $Y=\frac{1}{s(s^2+1)}-\frac{e^{-2\pi s}}{s(s^2+1)}+\frac{e^{-\pi s}}{s^2+1}.$

Har term invert karo. Pehle, $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$, jiska inverse $1-\cos t$ hai. Ise $g(t)=1-\cos t$ kaho. Saath hi ${\mathcal{L}}^{-1}\{1/(s^2+1)\}=\sin t$.

• Term 1: $g(t)=1-\cos t$ (koi delay nahi).
• Term 2: $-g(t-2\pi )u(t-2\pi )=-(1-\cos (t-2\pi ))u(t-2\pi )$. Ab $\cos (t-2\pi )=\cos t$ kyunki cosine ka period $2\pi$ hai (ek full turn ghatane par woh unchanged rehta hai), isliye yeh $-(1-\cos t)u(t-2\pi )$ hai.
• Term 3: $\sin (t-\pi )u(t-\pi )$. Ab $\sin (t-\pi )=-\sin t$ — sine ko half period $\pi$ shift karne se uska sign flip ho jaata hai (ek odd-symmetry / half-period reflection). Isliye yeh term $-\sin tu(t-\pi )$ hai.

Assemble karo: $y(t)=(1-\cos t)-(1-\cos t)u(t-2\pi )-\sin tu(t-\pi ).$

Cases mein (har interval par switches evaluate karo; $t=\pi ,2\pi$ par boundary values limits hain, aur hum neeche continuity verify karte hain):

Step transform ek 1/s carry karta hai; delta transform ek bare exponential hai. Delay stamp ke liye multiplier ko (t − a) ki function ki tarah likha jaana chahiye. $e^{-as}F(s)$ invert karte waqt hamesha gate u(t − a) rakho. Ek finite integral $\delta (t-a)$ ko tabhi catch karta hai jab $a$ limits ke andar ho. Ek delta jo 2nd-order ODE force karta hai, velocity jump karta hai, position nahi.