Visual walkthrough — Properties — linearity, first - second shift theorems, scaling

4.6.27 · D2 · Maths › Ordinary Differential Equations › Properties — linearity, first - second shift theorems, scali

---

Step 0 — Yeh sab kab allowed hai? ($f$ aur $s$ ke prerequisites)

KYA. Kuch bhi machine mein daalne se pehle hume machine ke input rules jaanne chahiye — woh hypotheses jo integral ko finite number banate hain. $f$ par do conditions, aur $s$ par ek condition.

YEH DO KYUN. Fading curtain $e^{-st}$ tabhi $f$ ko tame kar sakta hai jab $f$ usse aage na nikal jaaye. Agar $|f(t)|\le M{e}^{{\sigma }_{0}t}$ aur $\mathrm{Re}(s)>{\sigma }_0$, to $|e^{-st}f(t)|\le M{e}^{-(\mathrm{Re}(s)-{\sigma }_0)t}$, jo decay karta hai — isliye area finite hai. Piecewise continuity guarantee karti hai ki pieces pehli jagah integrable hain.

AAGE KYA BADLEGA. ROC pocket mein rakh lo: Step 6 mein delay tag $e^{-as}$ ($a\ge 0$, $\sigma =\mathrm{Re}(s)>0$ ke saath) satisfy karta hai $|e^{-as}|=e^{-a\sigma }\le 1$, isliye yeh kabhi nahi badhata woh region jahan cheezein converge hoti hain — delayed signal ka ROC wahi $\sigma >{\sigma }_0$ hai jaise original ka. Kisi signal ko delay karna uski transform ki convergence ki jagah ko chhota nahi kar sakta; tag convergence ke liye harmless hai.

---

Step 1 — Hum kaunsi machine mein daal rahe hain? (integral, zero se)

KYA. Laplace transform ek machine hai. Tum ise time ka ek admissible function $f(t)$ (time-axis par bana ek wiggle) dete ho, aur yeh complex variable $s$ ka ek naya function $F(s)$ deta hai, valid $\mathrm{Re}(s)>{\sigma }_0$ ke liye:

$$\underset{\text{"transform of }f\text{"}}{\underbrace{\mathcal{L}\{f(t)\}}}=\underset{\text{output recipe}}{\underbrace{F(s)}}={\int }_0^{\infty }\underset{\text{fading weight}}{\underbrace{e^{-st}}}\underset{\text{your wiggle}}{\underbrace{f(t)}}dt.$$

PIECES KYUN. Integrand ko picture ke right-to-left padho:

• $f(t)$ woh signal hai jo tumhe care karta hai (admissible, Step 0 ke according).
• $e^{-st}$ ek fading curtain hai: real slice $\sigma =\mathrm{Re}(s)>0$ par yeh $t=0$ par $1$ se start karta hai aur $0$ ki taraf decay karta hai. Yeh early time ko zyada weight deta hai aur late time ko bahut kam.
• ${\int }_0^{\infty }\dots dt$ ka matlab hai "product ko saare future time par add karo." Har valid $s$ ke liye ek number niklega — un numbers ka collection hi $F(s)$ hai.

PICTURE. Neeche: blue mein wiggle $f(t)$, yellow mein fading curtain $e^{-st}$, aur unka product (pink, shaded) jiska total area ek akela value $F(s)$ hai.

---

Step 2 — "$a$ se delay" kaisa dikhta hai? (unit step build karo)

KYA. Hum ek signal ka transform chahte hain jo late start karta hai. "Time $a$ par start, pehle kuch nahi" describe karne ke liye hume ek switch chahiye. Pehle plain Heaviside unit step $u(t)$ define karo, phir use shift karo.

YEH KYUN CHAHIYE. Kisi bhi function ko $u(t-a)$ se multiply karna $t=a$ se pehle sab kuch erase kar deta hai aur baaki ko untouched chhod deta hai — yeh ek clean on-switch hai jo $t=a$ par flip hota hai.

Jo delayed signal hum transform karte hain woh hai $f(t-a)u(t-a)$:

• $f(t-a)$ $f$ jaise same shape hai, lekin right mein $a$ se slide hua (har feature $a$ seconds baad hota hai).
• $u(t-a)$ guarantee karta hai ki switch $t=a$ par flip hone tak yeh silent (zero) rahe.

PICTURE. Left: original $f(t)$. Middle: $f(t-a)$ — identical shape right mein $a$ se shift hua (green arrow slide dikhata hai). Right: on-switch $u(t-a)$ jo ise $a$ se pehle zero rakhta hai.

---

Step 3 — Delayed signal ko machine mein daalo

KYA. Step 1 ki definition likho, lekin is baar hamara delayed signal $f(t-a)u(t-a)$ plug in karo:

$$\mathcal{L}\{f(t-a)u(t-a)\}={\int }_0^{\infty }{e}^{-st}\underset{\text{slid shape}}{\underbrace{f(t-a)}}\underset{\text{on-switch}}{\underbrace{u(t-a)}}dt.$$

KYUN. Abhi koi clever move nahi — hum sirf definition apply kar rahe hain. Poori derivation yahi hai: is ek integral ko simplify karo jab tak hum ise pehchaan lein.

PICTURE. Fading curtain $e^{-st}$ (yellow) ab ek aise signal ko multiply kar raha hai jo $t=a$ tak flat-zero hai, phir wiggle karta hai. Shaded product sirf $a$ ke right mein exist karta hai.

---

Step 4 — Switch integral ka floor $t=a$ tak cut karta hai

KYA. $t<a$ ke liye, factor $u(t-a)=0$ hai, isliye poora integrand wahan zero hai. Woh parts area mein kuch contribute nahi karte. Isliye hum lower limit $0$ se $a$ tak raise kar sakte hain:

$${\int }_0^{\infty }{e}^{-st}f(t-a)u(t-a)dt={\int }_a^{\infty }{e}^{-st}f(t-a)dt.$$

Dekho $u(t-a)$ apna kaam kar ke gayab ho gaya — $t\ge a$ ke liye yeh sirf $1$ equals karta hai.

KYUN. Woh region cut karna jahan function exactly zero hai kisi integral ko kabhi nahi badal sakta. Yeh sabse important move hai: yeh ek step-function problem ko ordinary integral mein badal deta hai, sirf shifted lower limit ke saath. (Yahan exactly $a\ge 0$ use hota hai: new lower limit $a$ domain $[0,\infty )$ ke andar honi chahiye — Step 7b dekho kya toot ta hai agar $a<0$.)

PICTURE. Zero region ($a$ ke left mein) greyed out hai; live region ($a$ ke right mein) hi akela area matter karta hai. Lower limit visibly $0$ se $a$ tak jump karta hai.

---

Step 5 — Time-axis slide karo: substitution $\tau =t-a$

KYA. Integrand mein $f(t-a)$ hai, lekin $F$ define hua tha ${\int }_0^{\infty }{e}^{-s\tau }f(\tau )d\tau$ ke roop mein plain argument ke saath. Match karne ke liye, clock rename karo taaki shift gayab ho jaaye. Maano

$$\tau =t-a⟹t=\tau +a,dt=d\tau ,$$

aur limits move karti hain: jab $t=a$, $\tau =0$; jab $t\to \infty$, $\tau \to \infty$.

YEH TOOL (substitution) KYUN. Hum $u$-substitution use karte hain — chain rule ka reverse — specifically taaki delay $a$ ko variable mein absorb karein, taaki $f(t-a)$ ban jaaye $f(\tau )$. Yahi ek tarika hai integral ko literally $F$ ki definition jaisa banana ka.

Har piece transform hoti hai:

$${\int }_a^{\infty }\underset{e^{-s(\tau +a)}}{\underbrace{e^{-st}}}\underset{f(\tau )}{\underbrace{f(t-a)}}\underset{d\tau }{\underbrace{dt}}={\int }_0^{\infty }{e}^{-s(\tau +a)}f(\tau )d\tau .$$

PICTURE. Do axes stack karke: purana $t$-axis ($a$ par start) aur, seedha neeche, naya $\tau$-axis same wiggle ke saath lekin ab $0$ se starting. Poori picture sirf $a$ se left mein slide hui.

---

Step 6 — Exponential split karo aur delay ko bahar nikalo (constant-factor rule)

KYA. Exponent law $e^{-s(\tau +a)}=e^{-s\tau }\cdot e^{-as}$ use karo. Factor $e^{-as}$ mein koi $\tau$ nahi hai, isliye integrals ke constant-factor rule se, $\int g(\tau )[\text{const}]d\tau =[\text{const}]\int g(\tau )d\tau$, yeh bahar nikal aata hai:

$${\int }_0^{\infty }{e}^{-s\tau }{e}^{-as}f(\tau )d\tau \stackrel{\text{const. out}}{=}{e}^{-as}\underset{=F(s)\text{(Step 1 definition!)}}{\underbrace{{\int }_0^{\infty }{e}^{-s\tau }f(\tau )d\tau }}=e^{-as}F(s).$$

KYUN (explicitly). Equality ${\int }_0^{\infty }cg(\tau )d\tau =c{\int }_0^{\infty }g(\tau )d\tau$ constant $c=e^{-as}$ ke liye integral ki linearity hai scalar multiple ke liye applied. Kyunki $e^{-as}$ integration variable $\tau$ ke saath vary nahi karta, yeh legitimate constant $c$ hai aur seedha bahar aata hai — koi approximation nahi, ek exact algebraic move. Pehle splitting ne time-part $e^{-s\tau }$ (jo $F$ rebuild karta hai) ko pure delay-part $e^{-as}$ (the "main late aaya" stamp) se alag kiya.

PICTURE. Integral $F(s)$ ki definition mein collapse ho jaata hai (Step 1 se same blue-and-yellow area), ek yellow tag $e^{-as}$ front par laga hua.

---

Step 7 — Edge case $a=0$: koi delay nahi, koi tag nahi

KYA. $a=0$ set karo. Tab $u(t-0)=u(t)=1$ sabhi $t\ge 0$ ke liye, aur $f(t-0)=f(t)$. Formula plain transform mein reduce hona chahiye:

$$\mathcal{L}\{f(t)\cdot 1\}=e^{-0\cdot s}F(s)=1\cdot F(s)=F(s).✓$$

YEH KYUN MATTER KARTA HAI. Ek accha formula apna trivial input survive karna chahiye. Zero delay ⇒ tag $e^{-as}=e^0=1$ ⇒ kuch nahi badlta. Consistent hai.

PICTURE. Delay slider $a=0$ par drag kiya hua: "shifted" curve exactly original ke upar baith jaata hai, tag $1$ read karta hai.

---

Step 7b — Edge case $a<0$: theorem ise kyun forbid karta hai

KYA. Maano koi negative shift $a<0$ try karta hai (ek advance, yani signal $t=0$ se pehle start hota hai). Step 4 dekho: humne lower limit $0$ ko $a$ se replace kiya tha. Agar $a<0$, to $a$ integration domain $[0,\infty )$ ke bahar pad jaata hai — Laplace integral ke paas $t<0$ ke baare mein koi information nahi hai, kyunki yeh $0$ se start hota hai. "Floor ko $a$ tak raise karo" move "floor ko $0$ se neeche le jao" ban jaata hai, jo definition nahi kar sakti.

YEH KYUN FAIL HOTA HAI, concretely. $a<0$ ke liye switch $u(t-a)=1$ sabhi $t\ge 0$ ke liye (yeh past mein fire ho chuka hai), isliye $u(t-a)$ kuch nahi karta — lekin $f(t-a)$ ab negative arguments par $f$ ki values mangta hai $t-a$ ke liye jo range karte hain $t=0\Rightarrow$ argument $-a>0$ mein... lekin saath hi advanced signal ka early part (true start aur $t=0$ ke beech ka part) lower limit $0$ se throw away ho jaata hai. Clean substitution toot jaati hai; ek single tag nahi nikal sakta. Isliye theorem sirf $a\ge 0$ ke liye state ki gayi hai.

PICTURE. Delay slider $a<0$ par drag kiya hua: shape $t=0$ wall se left mein push ho jaati hai; shaded early lobe $t<0$ mein cross kar jaata hai aur integral ke hard floor se $0$ par chop off ho jaata hai.

---

Step 8 — Classic trap: argument pehle se hi $(t-a)$ padhna chahiye

KYA. Theorem ko shape $(t-a)$ ke roop mein likhi chahiye — shifted argument. Agar tum $u(t-a)$ ke saath ek function dekhte ho jiska argument plain $t$ hai, to pehle use rewrite karna padega.

Example. $t^{2}u(t-2)$ transform karo. Argument $t^2$ hai, $(t-2{)}^2$ nahi, isliye abhi rule nahi pakad sakte. $a=2$ ke around rewrite karo:

$$t^2=((t-2)+2{)}^2=(t-2{)}^2+4(t-2)+4.$$

Ab har piece ka argument $(t-2)$ hai, aur hum $e^{-2s}F(s)$ term-by-term apply karte hain $\mathcal{L}\{t^2\}=\frac{2}{s^3},\mathcal{L}\{t\}=\frac{1}{s^2},\mathcal{L}\{1\}=\frac{1}{s}$ ke saath:

$$\mathcal{L}\{t^{2}u(t-2)\}=e^{-2s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right).$$

KYUN. Naive answer $e^{-2s}\cdot \frac{2}{s^3}$ silently assume karta hai ki shape $(t-2{)}^2$ thi — ek alag signal. Rewrite argument ko legal form mein force karta hai: sirf woh term jo delayed variable $(t-2)$ mein likhi hai $e^{-2s}$ tag inherit kar sakti hai, isliye pehle expand karo aur phir har shifted piece ko alag transform karo.

PICTURE. $t=2$ par switch ke baad do alag curves: galat reading $(t-2{)}^2$ versus actual $t^2$ — visibly same nahi, yahi wajah hai ki rewrite mandatory hai.

---

Ek picture mein summary

Sab kuch compressed: $f$ admissible aur $\mathrm{Re}(s)>{\sigma }_0$ ke saath, time world mein $a\ge 0$ ki delay shape ko right mein slide karti hai aur $u(t-a)$ se on karti hai; substitution $\tau =t-a$ use plain shape par slide wapas karta hai; constant-factor rule ek akela tag $e^{-as}$ s world mein free karta hai. Shape $F(s)$ aur uska ROC preserved, tag attached.

Recall Feynman retelling (plain words mein)

Ek aisi machine imagine karo jo time-wiggle ko ek compact recipe $F(s)$ mein squish karti hai, ise fading yellow curtain se multiply karke aur area add karke. Machine sirf reasonable wiggles accept karti hai — jinmein sirf finite jumps hain aur jo kisi exponential se zyada tez nahi badhte — aur tab bhi sirf $s$ ke knob settings ke liye jo kaafi right mein hain (real part $\sigma$ kaafi bada) taaki area finite ho. Ab same wiggle lo lekin ise $a$ seconds late start karo ($a$ negative nahi), $u(t-a)$ light-switch ke saath jo tab tak off rehta hai. Jab tum yeh late signal daalo, switch ka matlab hai ki sari early "kuch nahi" kuch contribute nahi karti — isliye tum $t=a$ par area count karna shuru kar sakte ho. Phir apna clock $a$ se wapas slide karo (ise $\tau$ kaho); wiggle bilkul original jaisi lagti hai, isliye uski recipe phir se $F(s)$ hai. Clock slide karne se sirf ek constant $e^{-as}$ bachta hai — aur kyunki yeh constant hai yeh seedha integral se bahar nikal aata hai aur ek sticker ke roop mein bahar khada ho jaata hai jiska matlab hai "main $a$ seconds late aaya." Time mein delay = $s$ mein sticker $e^{-as}$, aur kyunki sticker kabhi nahi badhta, achhe knob settings ka region unchanged rehta hai. Agar delay zero hai to sticker sirf $1$ hai. Agar koi negative delay try kare, trick toot jaati hai — signal ka ek hissa $t=0$ se pehle pad jaata aur chop ho jaata. Aur dhyan raho: rule tabhi kaam karta hai jab tumhara wiggle genuinely shifted shape $f(t-a)$ ke roop mein likha ho — agar plain $t$ likha hai, pehle expand karo.

---

Connections

• Laplace Transform — Definition and Existence
• Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
• Unit Step and Dirac Delta Functions
• Solving ODEs with Laplace Transforms
• Inverse Laplace Transform and Partial Fractions
• Convolution Theorem