Exercises — Laplace transform — definition, region of convergence

KYA karte hain: $f(t)=1$ ko definition mein plug karo. $F(s)={\int }_0^{\infty }{e}^{-st}\cdot 1dt={\left[\frac{-1}{s}{e}^{-st}\right]}_0^{\infty }.$ KYU upper limit ke liye ek condition chahiye: term $e^{-st}$ ki size $|e^{-st}|=e^{-\sigma t}$ hai jahan $\sigma =\mathrm{Re}(s)$. Ye $0$ tak sirf tabhi shrink karta hai jab $\sigma >0$ ho. $t=0$ par hume $e^0=1$ milta hai. $F(s)=0-\left(\frac{-1}{s}\right)=\boxed{\frac{1}{s}},\mathrm{Re}(s)>0.$

Pehle exponents combine karo (cleaner integral): $F(s)={\int }_0^{\infty }{e}^{-(s-a)t}dt={\left[\frac{-1}{s-a}{e}^{-(s-a)t}\right]}_0^{\infty }.$ Upper limit tab hi zero hota hai jab $\mathrm{Re}(s-a)>0$, yaani $\mathrm{Re}(s)>a$. $F(s)=\boxed{\frac{1}{s-a}},\mathrm{Re}(s)>a.$ Growth rate $a$, $F(s)$ ka pole ban jaata hai — $s$ ki woh value jahan denominator $s-a$ zero ho jaata hai aur $F$ blow up karta hai — aur ROC ka boundary line bhi. Ek hi number, do kaam.

Jhooth. Integrand $e^{-st}{e}^{t^2}=e^{t^2-st}$ hai. Koi bhi $s$ fix karo; $t\to \infty$ ke saath exponent $t^2-st\to +\infty$ kyunki $t^2$ eventually kisi bhi linear $st$ ko dominate kar leta hai. Toh integrand har $s$ ke liye grow without bound karta hai — integral har jagah diverge karta hai. $f$ exponential order ka nahi hai (koi bhi fixed $a$, $e^{t^2}$ ko bound nahi kar sakta), isliye koi transform nahi. Dekho Exponential Order and Growth Rates.

KYU integration by parts (IBP): integrand ek algebraic factor $t$ aur ek exponential $e^{-st}$ ka product hai. IBP awkward $t$ ko ek constant se trade kar deta hai. Lo $u=t,dv=e^{-st}dt\Rightarrow du=dt,v=-\frac{1}{s}{e}^{-st}.$ $F(s)={\left[-\frac{t}{s}{e}^{-st}\right]}_0^{\infty }+\frac{1}{s}{\int }_0^{\infty }{e}^{-st}dt.$ Boundary term $\mathrm{Re}(s)>0$ ke liye vanish ho jaata hai (exponential decay linear $t$ ko hara deta hai). Remaining integral $\mathcal{L}\{1\}=\frac{1}{s}$ hai. $F(s)=\frac{1}{s}\cdot \frac{1}{s}=\boxed{\frac{1}{s^2}},\mathrm{Re}(s)>0.$

Neeche ki picture IBP trade ko visually dikhati hai: tall factor $t$ (cyan) flat constant se swap ho jaata hai, jabki $e^{-st}$ (amber) $t$ ko crush kar deta hai toh boundary term $\infty$ par zero ho jaata hai.

KYU exponential form: hum pehle se $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ jaante hain. Cosine ko do exponentials mein split karna hume directly reuse karne deta hai IBP do baar karne ke bajaye. $F(s)=\frac{1}{2}\left(\frac{1}{s-i\omega }+\frac{1}{s+i\omega }\right)=\frac{1}{2}\cdot \frac{(s+i\omega )+(s-i\omega )}{s^2+{\omega }^2}=\frac{1}{2}\cdot \frac{2s}{s^2+{\omega }^2}.$ $F(s)=\boxed{\frac{s}{s^2+{\omega }^2}},\mathrm{Re}(s)>0.$ ROC reasoning: $a=\pm i\omega$ purely imaginary hain, toh dono ko $\mathrm{Re}(s)>0$ chahiye; cosine bounded hai (growth rate $0$).

Integral linear hai, toh term by term transform karo: $\mathcal{L}\{3e^{2t}\}=\frac{3}{s-2}(\mathrm{Re}(s)>2),\mathcal{L}\{5\}=\frac{5}{s}(\mathrm{Re}(s)>0).$ KYU sum ka ROC intersection hota hai: puri cheez converge hone ke liye dono integrals individually converge hone chahiye. Stricter condition jeetti hai: $\mathrm{Re}(s)>2$ implies $\mathrm{Re}(s)>0$, toh $F(s)=\boxed{\frac{3}{s-2}+\frac{5}{s}},\mathrm{Re}(s)>2.$

$F(s)={\int }_0^{\infty }t{e}^{at}{e}^{-st}dt={\int }_0^{\infty }t{e}^{-(s-a)t}dt.$ Ye exactly $\mathcal{L}\{t\}$ computation hai lekin $s$ ki jagah $(s-a)$ hai. Exercise 2.1 se, ${\int }_0^{\infty }t{e}^{-pt}dt=\frac{1}{p^2}$ for $\mathrm{Re}(p)>0$. $p=s-a$ set karo: $F(s)=\boxed{\frac{1}{(s-a{)}^2}},\mathrm{Re}(s)>a.$ YEH kya reveal karta hai: $f(t)$ ko $e^{at}$ se multiply karna $F(s)\to F(s-a)$ shift karta hai aur ROC boundary ko $0$ se $a$ par shift kar deta hai. Yahi first shifting theorem hai jo definition se nikal raha hai.

Figure shift ko literal banata hai: $e^{at}$ ko $f$ se attach karna poori $s$-picture (pole aur ROC boundary dono saath mein) ko $a$ se right shift kar deta hai.

(i) $t^{10}$: koi bhi polynomial har positive exponential se peet jaati hai, yaani $t^{10}\le M{e}^{\epsilon t}$ har $\epsilon >0$ ke liye. Toh sabse chhota kaam karne wala rate $a=\boxed{0}$ hai. (ROC $\mathrm{Re}(s)>0$.)

(ii) $e^{5t}\sin (3t)$: sine $1$ se bounded hai, toh $|f|\le e^{5t}$. Envelope $e^{5t}$ ki tarah badhti hai, toh $a=\boxed{5}$.

(iii) $\cosh (4t)=\frac{1}{2}(e^{4t}+e^{-4t})$: dominant piece $e^{4t}$ hai, toh $a=\boxed{4}$. $e^{-4t}$ piece sirf convergence mein help karta hai.

Step 1 (modulus andar kheeencho): $|\int g|\le \int |g|$, aur $|e^{-st}|=e^{-\sigma t}$: $|F(s)|\le {\int }_0^{\infty }{e}^{-\sigma t}|f(t)|dt.$ Step 2 (exponential order use karo): upar ki definition se, $|f(t)|\le M{e}^{at}$, toh $|F(s)|\le M{\int }_0^{\infty }{e}^{-(\sigma -a)t}dt=\frac{M}{\sigma -a},\text{finite iff }\sigma >a.$ Step 3 (limit): jab $\sigma \to \infty$, bound $\frac{M}{\sigma -a}\to 0$, jo force karta hai $F(s)\to 0$. YEH kya prove karta hai: har genuine Laplace transform ko $\mathrm{Re}(s)\to \infty$ ke saath $0$ tak decay karna hoga. Toh $F(s)=1$ ya $F(s)=s$ jaisa "transform" kisi piecewise-continuous, exponential-order $f$ se kabhi nahi aa sakta.

Integrand $t=2$ ke baad zero hai, toh infinite integral finite hai: $F(s)={\int }_0^2{e}^{-st}dt={\left[\frac{-1}{s}{e}^{-st}\right]}_0^2=\frac{1}{s}\left(1-e^{-2s}\right).$ $F(s)=\boxed{\frac{1-e^{-2s}}{s}}.$ KYU ROC poori $s$-plane hai: kyunki interval finite hai, koi $t\to \infty$ tail nahi hai thamne ke liye — convergence ko kabhi $\mathrm{Re}(s)>0$ ki zaroorat nahi padi. Formula mein $s=0$ par pole dikhta hai, lekin woh pole removable hai: $1-e^{-2s}=2s-2s^2+\cdots$ expand karne se $F(s)=\frac{2s-2s^2+\cdots }{s}=2-2s+\cdots$ milta hai, toh ${\lim }_{s\to 0}F(s)=2$ (jo sirf pulse ka area hai, ${\int }_0^{2}1dt=2$). Kyunki apparent singularity cancel ho jaati hai, $F(s)$ har $s$ ke liye finite hai — ROC poori plane hai, koi point exclude nahi. Neeche ki picture dikhati hai kyun: ek signal jo switch off ho jaati hai woh kabhi infinite integral produce nahi kar sakti.

$\cos 2t=\frac{1}{2}(e^{2it}+e^{-2it})$ likho, phir $e^{-t}$ attach karo: $e^{-t}\cos 2t=\frac{1}{2}\left(e^{(-1+2i)t}+e^{(-1-2i)t}\right).$ Har term $e^{at}$ hai with $a=-1\pm 2i$, giving $\frac{1}{s-a}$: $F(s)=\frac{1}{2}\left(\frac{1}{s+1-2i}+\frac{1}{s+1+2i}\right)=\frac{1}{2}\cdot \frac{2(s+1)}{(s+1{)}^2+4}.$ $F(s)=\boxed{\frac{s+1}{(s+1{)}^2+4}}.$ ROC: dono poles ka $\mathrm{Re}(a)=-1$ hai, toh hume $\mathrm{Re}(s)>-1$ chahiye. $e^{-t}$ factor ne ROC boundary ko left mein $-1$ par dhakka diya (ye convergence mein help karta hai).

$G(s)=\frac{1}{2}\left(\frac{1}{s-3}-\frac{1}{s+3}\right)=\frac{1}{2}\cdot \frac{(s+3)-(s-3)}{s^2-9}=\frac{1}{2}\cdot \frac{6}{s^2-9}=\boxed{\frac{3}{s^2-9}}.$ ROC reasoning (key insight): term $e^{3t}$ ko $\mathrm{Re}(s)>3$ chahiye; term $-e^{-3t}$ ko $\mathrm{Re}(s)>-3$ chahiye. Dono hold hone chahiye, toh intersection stricter wala deta hai, $\mathrm{Re}(s)>3$. Sabse tez badhne wala piece hamesha boundary set karta hai — isliye $+3$, $-3$ se jeetta hai.

$I_n={\int }_0^{\infty }{t}^n{e}^{-st}dt$ lo. IBP use karo with $u=t^n,dv=e^{-st}dt$: $I_n={\left[-\frac{t^n}{s}{e}^{-st}\right]}_0^{\infty }+\frac{n}{s}{\int }_0^{\infty }{t}^{n-1}{e}^{-st}dt.$ Boundary term $\mathrm{Re}(s)>0$ ke liye vanish ho jaata hai (exponential kisi bhi power ko hara deta hai). Toh $I_n=\frac{n}{s}{I}_{n-1}.$ Recurrence unwind karo $I_0=\mathcal{L}\{1\}=\frac{1}{s}$ se: $I_n=\frac{n}{s}\cdot \frac{n-1}{s}\cdots \frac{1}{s}\cdot \frac{1}{s}=\frac{n!}{s^{n+1}}.$ $\boxed{\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}},\mathrm{Re}(s)>0.$ Check $n=1$: $\frac{1!}{s^2}=\frac{1}{s^2}$ ✓, Exercise 2.1 se match karta hai.

Known result se shuru karo with $a=i\omega$: $\mathcal{L}\{e^{i\omega t}\}=\frac{1}{s-i\omega },\mathrm{Re}(s)>\mathrm{Re}(i\omega )=0.$ Rationalise karo top aur bottom ko $s+i\omega$ se multiply karke: $\frac{1}{s-i\omega }=\frac{s+i\omega }{s^2+{\omega }^2}=\underset{\text{real}}{\underbrace{\frac{s}{s^2+{\omega }^2}}}+i\underset{\text{imag}}{\underbrace{\frac{\omega }{s^2+{\omega }^2}}}.$ Lekin $e^{i\omega t}=\cos \omega t+i\sin \omega t$, toh linearity se real part $\mathcal{L}\{\cos \omega t\}$ hai aur imaginary part $\mathcal{L}\{\sin \omega t\}$ hai: $\boxed{\mathcal{L}\{\cos \omega t\}=\frac{s}{s^2+{\omega }^2}},\boxed{\mathcal{L}\{\sin \omega t\}=\frac{\omega }{s^2+{\omega }^2}},\mathrm{Re}(s)>0.$ KYU ek calculation se do answers milte hain: ek single complex identity ke real aur imaginary parts match karna do real integrals se do baar zyada efficient hai — yahi trick Fourier Transform mein bhi kaam aati hai.

Do convergence checks (KYU dono matter karte hain):

1. $t=0$ ke paas: $t^{-1/2}$ unbounded hai, lekin integral ${\int }_0{t}^{-1/2}dt$ phir bhi converge karta hai kyunki exponent $-\frac{1}{2}$ greater than $-1$ hai (yaad karo ${\int }_0{t}^{-p}dt$ exactly tabhi converge karta hai jab $p<1$). Toh singularity integrable hai — theek hai.
2. $t=\infty$ ke paas: factor $e^{-st}$ tail ko tab tak thamta hai jab tak $\mathrm{Re}(s)>0$ ho.

Computation step by step. $u=st$ substitute karo (toh $t=u/s$ aur $dt=du/s$), $\mathrm{Re}(s)>0$ ke liye valid: $F(s)={\int }_0^{\infty }{t}^{-1/2}{e}^{-st}dt={\int }_0^{\infty }{\left(\frac{u}{s}\right)}^{-1/2}{e}^{-u}\frac{du}{s}.$ $s$ ke constant powers bahar nikalo. Note karo ${\left(\frac{u}{s}\right)}^{-1/2}=u^{-1/2}{s}^{1/2}$, aur $dt$ se aur ek $\frac{1}{s}$ hai, toh combined $s$-factor $s^{1/2}\cdot s^{-1}=s^{-1/2}$ hai: $F(s)=s^{-1/2}{\int }_0^{\infty }{u}^{-1/2}{e}^{-u}du.$ Gamma function pehchano. Definition se $\Gamma (z)={\int }_0^{\infty }{u}^{z-1}{e}^{-u}du$; yahan $z-1=-\frac{1}{2}$ toh $z=\frac{1}{2}$, giving ${\int }_0^{\infty }{u}^{-1/2}{e}^{-u}du=\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$. Isliye $F(s)=s^{-1/2}\cdot \sqrt{\pi }=\frac{\sqrt{\pi }}{\sqrt{s}}=\boxed{\sqrt{\frac{\pi }{s}}},\mathrm{Re}(s)>0.$