Exercises — Properties — linearity, first - second shift theorems, scaling

KYA rule: linearity + standard block $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$. KYU: $5$ ek constant hai, isliye linearity usse transform ke bahar ride karne deti hai, bilkul untouched. $\mathcal{L}\{5e^{4t}\}=5\cdot \frac{1}{s-4}=\frac{5}{s-4}.$ ROC: $s>4$ ke liye valid hai (jahan defining integral converge hoti hai).

Block: $\mathcal{L}\{\cos 3t\}=\frac{s}{s^2+9}$ (yahan $b=3$, isliye $b^2=9$). $\mathcal{L}\{2\cos 3t\}=\frac{2s}{s^2+9}.$ ROC: $s>0$.

KYA: first shift, $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$, yahan $a=-2$. YEH rule KYU: hum dekh rahe hain $e^{(\text{const})t}$ ek ordinary function ko multiply kar raha hai — yahi exactly $s$-shift ka trigger hai. Start karo: $\mathcal{L}\{\sin 5t\}=\frac{5}{s^2+25}=F(s)$. Har jagah $s\to s-(-2)=s+2$ replace karo: $\mathcal{L}\{e^{-2t}\sin 5t\}=F(s+2)=\frac{5}{(s+2{)}^2+25}.$ ROC: shift convergence region ko apne saath drag karta hai, isliye $s>-2$.

Linearity dono terms ko split karti hai; har ek ek standard block use karta hai. $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4}$, aur $\mathcal{L}\{1\}=\frac{1}{s}$. $\mathcal{L}\{4t^3-7\}=4\cdot \frac{6}{s^4}-7\cdot \frac{1}{s}=\frac{24}{s^4}-\frac{7}{s}.$ ROC: $s>0$ (dono blocks ko chahiye).

KYA: second shift, $\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$, yahan $a=3$. KYU: argument already $(t-3)$ hai aur step $u(t-3)$ hai — matched form, isliye seedha plug in karo. Underlying $f(t)=t^2\Rightarrow F(s)=\frac{2}{s^3}$. $\mathcal{L}\{(t-3{)}^{2}u(t-3)\}=e^{-3s}\cdot \frac{2}{s^3}.$ ROC: $s>0$ ($e^{-3s}$ tag convergence nahi badalta).

KYA: scaling, $\mathcal{L}\{f(ct)\}=\frac{1}{c}F(s/c)$ for $c>0$. Yahan $f(t)=\cos t$, $c=4$. $1/c$ KYU aata hai: substitution $\tau =4t$ ek Jacobian $dt=d\tau /4$ carry karta hai, jo bahar ek factor ki tarah survive karta hai. $\mathcal{L}\{\cos 4t\}=\frac{1}{4}F\left(\frac{s}{4}\right)=\frac{1}{4}\cdot \frac{s/4}{(s/4{)}^2+1}.$ Upar aur neeche $16$ se multiply karo: $=\frac{1}{4}\cdot \frac{4s}{s^2+16}=\frac{s}{s^2+16}.$ ✔ (direct block se match karta hai). ROC: $s>0$.

$t^2$ ko $(t-3)$ ki powers mein rewrite karo taaki second shift apply ho sake: $t^2=((t-3)+3{)}^2=(t-3{)}^2+6(t-3)+9.$ Har term mein ab ek genuine $(t-3)$ argument hai (constant $9$ ek "$(t-3{)}^0$" piece ki tarah count hota hai, yaani $9\cdot 1$). Sabko $u(t-3)$ multiply karne ke saath, har block par $e^{-as}F(s)$, $a=3$, apply karo:

• $(t-3{)}^2\to \frac{2}{s^3}$
• $6(t-3)\to 6\cdot \frac{1}{s^2}$
• $9\to 9\cdot \frac{1}{s}$ $\mathcal{L}\{t^{2}u(t-3)\}=e^{-3s}\left(\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}\right).$ ROC: $s>0$.

Step 1 — square hatao (KYU): hamare paas ${\cos }^2$ ke liye koi table entry nahi hai, isliye identity ${\cos }^{2}t=\frac{1}{2}(1+\cos 2t)$ use karo taaki yeh unhi cheezoon mein badal jaaye jo hum jaante hain. $\mathcal{L}\{{\cos }^{2}t\}=\frac{1}{2}\mathcal{L}\{1\}+\frac{1}{2}\mathcal{L}\{\cos 2t\}=\frac{1}{2}\cdot \frac{1}{s}+\frac{1}{2}\cdot \frac{s}{s^2+4}.$ Isse $F(s)=\frac{1}{2s}+\frac{s}{2(s^2+4)}$ kaho (ROC $s>0$). Step 2 — first shift apply karo $a=3$ ke saath ($s\to s-3$): $\mathcal{L}\{e^{3t}{\cos }^{2}t\}=\frac{1}{2(s-3)}+\frac{s-3}{2((s-3{)}^2+4)}.$ ROC: $s>3$ (shift region ko $3$ se right mein move karta hai).

(a) First shift: $f(t)=t\Rightarrow F(s)=\frac{1}{s^2}$. $a=2$ ke saath, $s\to s-2$: $\mathcal{L}\{t{e}^{2t}\}=\frac{1}{(s-2{)}^2}.$ (b) Direct table entry $\mathcal{L}\{t^n{e}^{at}\}=\frac{n!}{(s-a{)}^{n+1}}$ with $n=1,a=2$: $\frac{1!}{(s-2{)}^2}=\frac{1}{(s-2{)}^2}$. ✔ Same. ROC: $s>2$.

Yahan do rules stack hote hain. $e^{-t}(t-1{)}^{3}u(t-1)$ ko $e^{-t}\cdot h(t)$ treat karo jahan $h(t)=(t-1{)}^{3}u(t-1)$ hai. Step 1 — $h$ transform karo (second shift, $a=1$, base $f=t^3$, $F=\frac{6}{s^4}$): $H(s)=\mathcal{L}\{h(t)\}=e^{-s}\cdot \frac{6}{s^4}.$ Step 2 — $e^{-t}$ ke liye first shift apply karo (toh $a=-1$, yaani $H$ mein $s\to s+1$): $\mathcal{L}\{e^{-t}h(t)\}=H(s+1)=e^{-(s+1)}\cdot \frac{6}{(s+1{)}^4}=\frac{6e^{-1}{e}^{-s}}{(s+1{)}^4}.$ Dhyan do ki $e^{-1}$ (ek constant) tab nikalta hai jab $s\to s+1$ $e^{-s}$ factor ko hit karta hai — miss karna aasaan hai. ROC: $s>-1$.

Yeh second shift (delayed $u(t-\frac{\pi }{6})$ ek matched $(t-\frac{\pi }{6})$ argument ke saath) aur first shift ($e^{2t}$ multiplier) stack karta hai. $a=\frac{\pi }{6}$ consistently poore time use karo. Step 1 — $e^{2t}$ peel karo. Poori cheez $e^{2t}\cdot h(t)$ likho jahan $h(t)=\sin 3\left(t-\frac{\pi }{6}\right)u\left(t-\frac{\pi }{6}\right).$ Step 2 — $h$ transform karo (second shift, $a=\frac{\pi }{6}$). Argument already $(t-\frac{\pi }{6})$ hai, isliye koi rewriting nahi chahiye. Base $f(t)=\sin 3t$, $F(s)=\frac{3}{s^2+9}$: $H(s)=e^{-\pi s/6}\cdot \frac{3}{s^2+9}.$ Step 3 — $e^{2t}$ ke liye first shift apply karo ($a=2$, isliye $H$ mein har jagah $s\to s-2$): $\mathcal{L}\{e^{2t}h(t)\}=H(s-2)=e^{-\pi (s-2)/6}\cdot \frac{3}{(s-2{)}^2+9}=\frac{3e^{\pi /3}{e}^{-\pi s/6}}{(s-2{)}^2+9}.$ Constant $e^{\pi /3}$ ka dhyan rakho jo tab nikalta hai jab $s\to s-2$ $e^{-\pi s/6}$ factor ko hit karta hai (kyunki $e^{-\pi (s-2)/6}=e^{\pi /3}{e}^{-\pi s/6}$). ROC: $s>2$.

Do tools: $e^{-t}$ ke liye first shift, aur $t\times (\cdots )$ multiplier jo transform ka $-\frac{d}{ds}$ hai (from Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n): $\mathcal{L}\{tg(t)\}=-G^{\prime }(s)$). Step 1 — first shift $\cos t$ par $e^{-t}$ ke liye ($a=-1$, $s\to s+1$): $\mathcal{L}\{e^{-t}\cos t\}=\frac{s+1}{(s+1{)}^2+1}=:G(s)$. Step 2 — $t$ se multiply karo ⇒ $-G^{\prime }(s)$. $D=(s+1{)}^2+1=s^2+2s+2$ lo, numerator $N=s+1$. $G^{\prime }(s)=\frac{N^{\prime }D-N{D}^{\prime }}{D^2}=\frac{(1)(s^2+2s+2)-(s+1)(2s+2)}{(s^2+2s+2{)}^2}.$ Numerator $=s^2+2s+2-(2s^2+4s+2)=-s^2-2s=-(s^2+2s)$. $-G^{\prime }(s)=\frac{s^2+2s}{(s^2+2s+2{)}^2}.$ Toh $\mathcal{L}\{t{e}^{-t}\cos t\}=\frac{s^2+2s}{((s+1{)}^2+1{)}^2}$. ROC: $s>-1$.

Step 1 — ODE transform karo ($\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)$ use karo from Solving ODEs with Laplace Transforms, aur right side par second shift $\mathcal{L}\{u(t-1)\}=\frac{e^{-s}}{s}$): $sY-0+2Y=\frac{e^{-s}}{s}\Rightarrow (s+2)Y=\frac{e^{-s}}{s}\Rightarrow Y(s)=\frac{e^{-s}}{s(s+2)}.$ Step 2 — partial fractions $\frac{1}{s(s+2)}=\frac{1/2}{s}-\frac{1/2}{s+2}$ par (table-ready pieces mein split karo; dekho Inverse Laplace Transform and Partial Fractions): $Y(s)=e^{-s}\left(\frac{1/2}{s}-\frac{1/2}{s+2}\right).$ Step 3 — second shift se ulta invert karo. $e^{-s}$ ka matlab "1 se delay karo aur $u(t-1)$ se switch on karo." Bracket $\frac{1}{2}-\frac{1}{2}{e}^{-2t}$ mein invert hota hai, toh use $t-1$ par evaluate karo: $y(t)=\left[\frac{1}{2}-\frac{1}{2}{e}^{-2(t-1)}\right]u(t-1)=\frac{1}{2}(1-e^{-2(t-1)})u(t-1).$ Sanity — sab cases: $t<1$ ke liye, $u=0$ isliye $y=0$ (abhi kuch switch on nahi hua). $t=1$ par, $y=\frac{1}{2}(1-1)=0$ (continuous start). $t\to \infty$ pe, $y\to \frac{1}{2}$ (steady state jahan $y^{\prime }\to 0$ deta hai $2y=1$). ✔

Step 1 — base transform: $f(t)=t{e}^{-t}\Rightarrow F(s)=\frac{1}{(s+1{)}^2}$ ($\mathcal{L}\{t\}=\frac{1}{s^2}$ par first shift). Step 2 — scaling, $c=2$: $\mathcal{L}\{f(2t)\}=\frac{1}{2}F(s/2)=\frac{1}{2}\cdot \frac{1}{(s/2+1{)}^2}$. Simplify karo: ${\left(\frac{s}{2}+1\right)}^2=\frac{(s+2{)}^2}{4}$, toh $\mathcal{L}\{f(2t)\}=\frac{1}{2}\cdot \frac{4}{(s+2{)}^2}=\frac{2}{(s+2{)}^2}.$ Directly verify karo: $f(2t)=2t{e}^{-2t}$, aur $\mathcal{L}\{2t{e}^{-2t}\}=2\cdot \frac{1}{(s+2{)}^2}=\frac{2}{(s+2{)}^2}.$ ✔ ROC: $s>-2$.

Step 1 — $e^{-2s}$ strip karo (second shift, $a=2$): baaki jo invert hoga usse $g(t)$ kaho; poora answer $g(t-2)u(t-2)$ hoga. Step 2 — $\frac{s-1}{(s-1{)}^2+9}$ invert karo (first shift, $s\to s-1$ matlab $e^t$ se multiply karo): un-shifted $\frac{s}{s^2+9}$ $\cos 3t$ mein invert hota hai, toh $-1$ shift ke saath: $g(t)=e^t\cos 3t$. Step 3 — combine karo: ${\mathcal{L}}^{-1}\{\cdots \}=e^{t-2}\cos (3(t-2))u(t-2).$ Cases: $t<2$ ke liye yeh $0$ hai; $t\ge 2$ ke liye yeh ek growing cosine hai jo $t=2$ par "start" hota hai. ✔