Question

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{f(t - T)\}$ is equal to,

• A. $e^{sT} F(s)$
• B. $e^{-sT} F(s)$
• C. $\dfrac{F(s)}{1 + e^{sT}}$
• D. none of the above

Hint:

A shift in time domain by $T$ seconds introduces a multiplying factor of $e^{-sT}$ in the Laplace domain.

Answer 1

The Correct Option is B

Step 1: Recall the time-shifting property of Laplace Transform.
If
\[ \mathcal{L}\{f(t)\} = F(s) \]
then the Laplace transform of a delayed function is given by
\[ \mathcal{L}\{f(t - T) u(t - T)\} = e^{-sT} F(s) \]
Step 2: Apply the property to the given expression.
The expression $f(t - T)$ represents a time delay of $T$ seconds.
Hence, its Laplace transform is
\[ e^{-sT} F(s) \]
Step 3: Conclusion.
Therefore,
\[ \boxed{\mathcal{L}\{f(t - T)\} = e^{-sT} F(s} \]