Question

Match List - I with List - II

List - I (Property)	List - II (Expression)
A. Change of Scale Property	III. \(L\{f(at)\}=\frac{1}{a}F\left(\frac{s}{a}\right)\)
B. Final Value Theorem	IV. \(\lim_{s\to 0}[sF(s)] = \lim_{t\to \infty} f(t)\)
C. Heaviside's Shifting Theorem	II. \(L\{g(t)\} = e^{-as}F(s)\)
D. First Shifting Theorem	I. \(L[e^{at}f(t)] = F(s-a)\)

• A. A-III, B-II, C-I, D-IV
• B. A-III, B-IV, C-I, D-II
• C. A-III, B-IV, C-II, D-I
• D. A-I, B-II, C-III, D-IV

Hint:

First Shifting: shift in s (frequency). Second Shifting: shift in t (time).

Answer 1

The Correct Option is C

Concept: Laplace Transform properties allow us to find the transforms of complex functions by manipulating simpler, known transforms.

Step 1: Match Scale and Value theorems.
- Change of Scale Property (A): If $L\{f(t)\} = F(s)$, then $L\{f(at)\} = \frac{1}{a}F(\frac{s}{a})$. Match: A-III. - Final Value Theorem (B): Relates the steady-state value of a function in the time domain to its behavior in the s-domain as $s \to 0$. Match: B-IV.

Step 2: Match Shifting theorems.
- Heaviside's Shifting Theorem (C): Also known as the Second Shifting Theorem, it deals with functions shifted in time (using the unit step function $u(t-a)$). Match: C-II. - First Shifting Theorem (D): States that multiplying by an exponential in the time domain results in a shift in the s-domain: $L\{e^{at}f(t)\} = F(s-a)$. Match: D-I.

Step 3: Final Mapping.
The matches are A-III, B-IV, C-II, D-I. This corresponds to Option (3).