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,
\[ \boxed{\mathcal{L}\{f(t - T)\} = e^{-sT} F(s} \]

Answer 2

Rather than quoting the time-shifting property directly, it can be derived from the definition of the Laplace transform and then checked against each option.

By definition, \[ \mathcal{L}\{f(t-T)\} = \int_0^{\infty} f(t-T) e^{-st}\,dt \] Substituting \(\tau = t - T\), so \(t = \tau + T\) and \(dt = d\tau\), and shifting the limits accordingly (with \(f\) taken as zero for negative argument): \[ \mathcal{L}\{f(t-T)\} = \int_0^{\infty} f(\tau)e^{-s(\tau+T)}\,d\tau = e^{-sT}\int_0^{\infty} f(\tau)e^{-s\tau}\,d\tau = e^{-sT}F(s) \]

1. \(e^{sT}F(s)\): This would arise from a sign error in the substitution, effectively treating the shift as advancing \(f(t)\) in time rather than delaying it, which is not what \(f(t-T)\) represents.
2. \(e^{-sT}F(s)\): This matches exactly what the substitution above produces, with the factor \(e^{-sT}\) coming directly from splitting \(e^{-s(\tau+T)}\) into \(e^{-sT}e^{-s\tau}\).
3. \(\dfrac{F(s)}{1+e^{sT}}\): This form does not arise from the direct substitution and instead resembles an expression from periodic-function Laplace transforms, not a simple time shift.
4. None of the above: Ruled out since option (ii) is exactly what the derivation produces.

Therefore, the correct answer is \(e^{-sT}F(s)\).