Properties — linearity, first - second shift theorems, scaling: Worked examples

4.6.27 · D3 · Maths › Ordinary Differential Equations › Properties — linearity, first - second shift theorems, scali

Shuru karne se pehle, woh ek object jis par har rule bana hai — definition from Laplace Transform — Definition and Existence:

L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t .

Yahan $t$ time hai (hamare liye hamesha $\geq 0$ ), $s$ ek helper number hai jo integral ko finite banane ke liye kaafi bada ho, aur $F (s)$ woh "recipe" hai jo machine bahar deti hai. Jo building-block recipes hum baar baar use karte hain woh Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n) se hain:

L {1} = \frac{1}{s}, L {t^{n}} = \frac{n !}{s ^{n + 1}}, L {e^{a t}} = \frac{1}{s - a}, L {sin b t} = \frac{b}{s ^{2} + b ^{2}}, L {cos b t} = \frac{s}{s ^{2} + b ^{2}} .

The scenario matrix

Is topic mein aane wali har problem exactly inhi cells mein se kisi ek mein jaati hai. Neeche ke worked examples [C#] se label hain taaki tum dekh sako ki poora grid cover ho raha hai.

Cell	Case class	Kya tricky hai	Example
C1	Pure linearity, positive coefficients	bas split aur scale karo	Ex 1
C2	Linearity with a negative / sign-flipped term	minus sign ko saath rakho	Ex 1
C3	First shift, positive $a$ (growth $e^{a t}$ )	har jagah $s \to s - a$ substitute karo	Ex 2
C4	First shift, negative $a$ (decay $e^{- a t}$ )	$s \to s + a$ , sign flip hota hai	Ex 3
C5	Second shift, argument already $(t - a)$	seedha plug in karo	Ex 4
C6	Second shift, argument NOT in $(t - a)$ form	pehle algebra se rewrite karo	Ex 5
C7	Scaling, general $c > 1$ ya $0 < c < 1$	$\frac{1}{c}$ ko rakho	Ex 6
C8	Degenerate input ( $c = 1$ , $a = 0$ , zero function)	rule ko identity tak reduce hona chahiye	Ex 7
C9	Combined rules (shift + linearity + step)	operations ka order	Ex 8
C10	Real-world word problem (delayed switch-on)	words ko $u (t - a)$ mein translate karo	Ex 9
C11	Exam twist — reverse direction (inverse)	rule ko ulta padho	Ex 10

Ex 1 — Linearity, positive aur negative terms [C1, C2]

Worked example

$L {5 t^{2} - 7 cos 4 t + 2}$ nikalo.

Forecast: Pehle shape guess karo — teen alag table entries, har ek sirf apne coefficient se scale ki hui, minus signs rakhi hui. Aage padhne se pehle apna guess likho.

Step 1. Linearity se split karo: $L {5 t^{2}} - L {7 cos 4 t} + L {2}$ . Yeh step kyun? Integration linear hai, isliye terms ka sum term-by-term transform hota hai — ek saath poori cheez integrate karne ki zarurat nahi.

Step 2. Har constant bahar nikalo aur table padho: $L {t^{2}} = \frac{2}{s ^{3}}$ , $L {cos 4 t} = \frac{s}{s ^{2} + 16}$ , $L {1} = \frac{1}{s}$ . Yeh step kyun? Constants integrals se bahar slip ho jaate hain; jo bachta hai woh ek jaana-pehchana building block hai.

Step 3. Assemble karo, cosine term par minus sign rakho: $L {5 t^{2} - 7 cos 4 t + 2} = \frac{10}{s ^{3}} - \frac{7 s}{s ^{2} + 16} + \frac{2}{s} .$

Verify: $s = 2$ set karo: $\frac{10}{8} - \frac{14}{20} + \frac{2}{2} = 1.25 - 0.7 + 1 = 1.55$ . Consistent, finite, middle term ka sign negative raha. ✔

Ex 2 — First shift, positive $a$ (growing exponential) [C3]

Ex 3 — First shift, negative $a$ (decaying exponential) [C4]

Ex 4 — Second shift, argument already $(t - a)$ form mein [C5]

Ex 5 — Second shift, argument NOT $(t - a)$ form mein [C6]

Worked example

$L {t u (t - 1)}$ nikalo.

Forecast: Step $u (t - 1)$ hai to $a = 1$ , lekin multiplier $t$ hai, $(t - 1)$ nahi. Rule apply karne se pehle hume kya karna hoga? Guess karo.

Step 1. $t$ ko $(t - 1)$ ke terms mein rewrite karo: $t = (t - 1) + 1$ . Yeh step kyun? Second shift theorem sirf $f (t - a)$ par kaam karta hai. Hum algebraically shifted variable ko force karke laate hain.

Step 2. Linearity se split karo: $t u (t - 1) = (t - 1) u (t - 1) + 1 \cdot u (t - 1) .$ Yeh step kyun? Ab har piece ka clean $(t - 1)$ argument hai (constant $1$ $f = 1$ count hota hai, aur $L {u (t - a)} = \frac{e ^{- a s}}{s}$ ).

Step 3. Har ek transform karo: $(t - 1) u (t - 1)$ ke liye $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ use karo; $u (t - 1)$ ke liye $F (s) = \frac{1}{s}$ use karo. Dono par tag $e^{- s}$ lagta hai. $L {t u (t - 1)} = e^{- s} (\frac{1}{s ^{2}} + \frac{1}{s}) .$ Yeh step kyun? Same delay $a = 1$ ⇒ dono par same $e^{- s}$ ; sirf base recipes alag hain.

Verify (trap ka sanity check): Galat answer $e^{- s} \frac{1}{s ^{2}}$ (rule seedha grab karne par) $\frac{1}{s}$ wala piece miss kar deta hai. Hamara version $e^{- s} \frac{s + 1}{s ^{2}}$ hai. Numerically $s = 1$ par: $e^{- 1} (1 + 1) = 2 e^{- 1} \approx 0.7358$ . ✔

Ex 6 — Scaling, $c > 1$ bhi aur $0 < c < 1$ bhi [C7]

Worked example Diya hai

$L {cos t} = \frac{s}{s ^{2} + 1}$ , scaling se $L {cos 5 t}$ aur $L {cos \frac{1}{2} t}$ nikalo.

Forecast: Scaling kehta hai $L {f (c t)} = \frac{1}{c} F (s / c)$ . Predict karo ki $s$ -axis $c = 5$ vs $c = \frac{1}{2}$ ke liye kis taraf stretch hogi.

Step 1. $c = 5$ ke liye: $F (s) = \frac{s}{s ^{2} + 1}$ , to $F (s /5) = \frac{s /5}{( s /5 ) ^{2} + 1} = \frac{s /5}{s ^{2} /25 + 1}$ . Yeh step kyun? Time ko $5$ se squeeze karna (tezi se wiggle) $s$ -axis ko $1/5$ se stretch karta hai — yahi substitution $τ = 5 t$ hai.

Step 2. $\frac{1}{c} = \frac{1}{5}$ multiply karo aur simplify karo: $L {cos 5 t} = \frac{1}{5} \cdot \frac{s /5}{s ^{2} /25 + 1} = \frac{1}{5} \cdot \frac{5 s}{s ^{2} + 25} = \frac{s}{s ^{2} + 25} .$

Step 3. $c = \frac{1}{2}$ ke liye (slow wiggle): $F (2 s) = \frac{2 s}{4 s ^{2} + 1}$ , times $\frac{1}{1/2} = 2$ : $L {cos \frac{1}{2} t} = 2 \cdot \frac{2 s}{4 s ^{2} + 1} = \frac{4 s}{4 s ^{2} + 1} = \frac{s}{s ^{2} + 1/4} .$ Yeh step kyun? $c < 1$ time ko stretch karta hai (slower), to $s$ -axis compress hoti hai — pole andar ki taraf $s = \pm \frac{1}{2} i$ par move karta hai.

Verify: Dono direct table form $\frac{s}{s ^{2} + b ^{2}}$ se match karte hain $b = 5$ aur $b = \frac{1}{2}$ ke saath respectively. ✔

Ex 7 — Degenerate inputs: $c = 1$ , $a = 0$ , aur zero function [C8]

Worked example Dikhao ki har rule obvious cheez mein collapse hota hai jab iska parameter "off" hota hai.

Forecast: Ek accha rule kuch nahi karta jab iska knob neutral setting par hota hai. Teen neutral settings predict karo.

Step 1 — Scaling at $c = 1$ : $L {f (1 \cdot t)} = \frac{1}{1} F (s /1) = F (s)$ . Yeh step kyun? Koi time-squeeze nahi matlab koi change nahi — rule ko identity tak reduce hona chahiye, aur hota hai.

Step 2 — First shift at $a = 0$ : $L {e^{0} f (t)} = F (s - 0) = F (s)$ . Yeh step kyun? $e^{0} = 1$ ; $1$ se multiply karna kuch nahi badalta, aur $0$ se shift bhi agree karta hai.

Step 3 — Second shift at $a = 0$ : $L {f (t) u (t)} = e^{0} F (s) = F (s)$ . Yeh step kyun? $u (t) = 1$ for $t \geq 0$ (hamara poora domain), to koi delay nahi, tag $e^{0} = 1$ .

Step 4 — Zero function: $L {0} = \int_{0}^{\infty} e^{- s t} \cdot 0 d t = 0$ . Yeh step kyun? $a = b = 0$ ke saath linearity force karta hai ki kuch nahi ka transform kuch nahi ho.

Verify: Chaaron required identity/zero tak reduce hote hain — koi rule neutral input se change "invent" nahi karta. ✔

Ex 8 — Combined: linearity + first shift + step [C9]

Worked example

$L {e^{- t} (t - 3) u (t - 3)}$ nikalo.

Forecast: Do exponentials aayenge — ek shift ( $e^{- t}$ ) aur ek delay tag ( $e^{- 3 s}$ ). Guess karo kaun $s \to s + 1$ banta hai aur kaun factor banta hai.

Step 1. Pehle delay handle karo: $g (t) = (t - 3) u (t - 3)$ mein $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ hai, to $L {(t - 3) u (t - 3)} = \frac{e ^{- 3 s}}{s ^{2}} \equiv G (s) .$ Yeh step kyun? Second shift time-delayed part par kaam karta hai; $e^{- t}$ ko touch karne se pehle yeh karo.

Step 2. Ab extra $e^{- t}$ first shift trigger karta hai $a = - 1$ ke saath: $G (s)$ ke poore mein $s \to s + 1$ replace karo, exponential sameit. Yeh step kyun? First shift $s \to s - a = s + 1$ ko har jagah substitute karta hai — $e^{- 3 s}$ bhi $G$ ka part hai, to woh bhi $e^{- 3 (s + 1)}$ mein shift ho jaata hai. $L {e^{- t} (t - 3) u (t - 3)} = \frac{e ^{- 3 (s + 1)}}{( s + 1 ) ^{2}} = \frac{e ^{- 3 s - 3}}{( s + 1 ) ^{2}} .$

Verify: $s = 1$ par: $\frac{e ^{- 6}}{4} = \frac{e ^{- 6}}{4} \approx 6.196 \times 1 0^{- 4}$ , positive aur chhota (integrand tezi se decay karta hai). ✔

Ex 9 — Real-world word problem: ek delayed voltage pulse [C10]

Worked example Ek circuit ka input voltage

$t = 2$ seconds tak off rehta hai, phir us ke baad se constant $v (t) = 3$ volts ho jaata hai. $L {v (t)}$ nikalo.

Forecast: "Off phir constant" ek step $u (t - 2)$ ki taraf ishaara karta hai. $e^{- 2 s}$ tag predict karo.

Step 1. Input ko symbols mein likho: $v (t) = 3 u (t - 2)$ . Yeh step kyun? $u (t - 2)$ , $t = 2$ se pehle $0$ aur baad mein $1$ hota hai — exactly "2 tak off, phir on."

Step 2. Base $f (t) = 3$ (ek constant, to $f (t - 2) = 3$ bhi, kyunki constants shift feel nahi karte), $F (s) = \frac{3}{s}$ . Delay tag lagao: $L {3 u (t - 2)} = e^{- 2 s} \cdot \frac{3}{s} = \frac{3 e ^{- 2 s}}{s} .$ Yeh step kyun? Second shift with $a = 2$ ; constant ki shape unchanged rehti hai, sirf switch-on time $e^{- 2 s}$ ke roop mein dikhta hai.

Verify (units/limit): $s \to 0^{+}$ par, $\frac{3 e ^{- 2 s}}{s} \to \infty$ — sahi hai, kyunki jo constant kabhi band nahi hota uska is limit mein infinite "total area" hota hai, jaise $L {3} = \frac{3}{s}$ . Extra $e^{- 2 s}$ sirf 2-second delay record karta hai. ✔

Ex 10 — Exam twist: ek rule ulta padhna (inverse) [C11]

Worked example

Inverse transform $L^{- 1} {\frac{s - 4}{( s - 4 ) ^{2} + 9}}$ nikalo.

Forecast: Tum plain $s$ ki jagah $(s - 4)$ dekh rahe ho. Kaun sa shift hai, aur time mein kaun sa $e^{a t}$ aata hai? (Yeh Inverse Laplace Transform and Partial Fractions use karta hai.)

Step 1. Pattern pehchano: $S = s - 4$ ke saath $\frac{S}{S ^{2} + 9}$ woh $L {cos 3 t}$ hai $S$ par evaluate hua. Yeh step kyun? First shift ulta padhna: recipe mein $F (s - a)$ matlab time mein $e^{a t}$ multiplier.

Step 2. $a = 4$ ke saath $e^{a t}$ restore karo: $L^{- 1} {\frac{s - 4}{( s - 4 ) ^{2} + 9}} = e^{4 t} cos 3 t .$ Yeh step kyun? First shift theorem ulta run karo: denominator/numerator mein dikhne wala " $s \to s - 4$ " undo hoke $e^{4 t}$ factor banta hai.

Verify: $e^{4 t} cos 3 t$ ko Ex 2 ki method se forward transform karo: base $cos 3 t \to \frac{s}{s ^{2} + 9}$ , shift $s \to s - 4$ deta hai $\frac{s - 4}{( s - 4 ) ^{2} + 9}$ — round-trip match karta hai. ✔

Recall Which cell does each trap belong to?

Argument $t^{2}$ sitting on $u (t - 2)$ — which cell? ::: C6 (not yet in $(t - a)$ form — rewrite first) $L {f (c t)}$ with $c = 1$ — which cell? ::: C8 (degenerate; reduces to $F (s)$ ) $e^{- t}$ multiplying — which cell and substitution? ::: C4, $s \to s + 1$ A voltage that switches on at $t = 5$ — which cell? ::: C10, real-world $u (t - 5)$

Connections

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