Question

If F(s) denotes the Laplace transform of some function f(t), then the Laplace transform of e\(^{-at}\)f(t), where a is a real constant, is

• A. F(a \(-\) s)
• B. \(-F\)(s)
• C. F(s \(-\) a)
• D. F(s \(+\) a)

Hint:

First Shifting Theorem:
- Multiplying by \( e^{-at} \) in the time domain causes a shift of \( +a \) in the complex frequency domain:
\[ \mathcal{L}\{e^{-at} f(t)\} = F(s + a) \]
- Multiplying by \( e^{at} \) causes a shift of \( -a \):
\[ \mathcal{L}\{e^{at} f(t)\} = F(s - a) \]

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks for a standard theorem of Laplace transforms, specifically the first shifting (or frequency translation) theorem.

Step 2: Key Formula or Approach:
The definition of the Laplace transform of a function \( f(t) \) is:
\[ \mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt \]
We need to determine the Laplace transform of \( e^{-at}f(t) \).

Step 3: Detailed Explanation:

• Apply the Definition of Laplace Transform:
\[ \mathcal{L}\left\{e^{-at} f(t)\right\} = \int_0^{\infty} e^{-st} \left[ e^{-at} f(t) \right] dt \] - Group the exponential terms together:
\[ e^{-st} \cdot e^{-at} = e^{-(s+a)t} \] - Substitute this back into the integral:
\[ \mathcal{L}\left\{e^{-at} f(t)\right\} = \int_0^{\infty} e^{-(s+a)t} f(t) \, dt \]

• Compare with Standard Definition:
- Let a new variable \( s' = s + a \). The integral becomes:
\[ \int_0^{\infty} e^{-s't} f(t) \, dt = F(s') \] - Substituting back \( s' = s + a \):
\[ \mathcal{L}\left\{e^{-at} f(t)\right\} = F(s + a) \] - This is the First Shifting Theorem. If the exponential has a negative sign (\( e^{-at} \)), the shift is in the positive direction (\( s + a \)).


Step 4: Final Answer:
The Laplace transform of \( e^{-at}f(t) \) is \( F(s + a) \).
Therefore, the correct choice is option (D).