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

Directly integrating the definition, \(\mathcal{L}\{f(t-T)\} = \int_0^\infty f(t-T)e^{-st}\,dt\). Substituting \(\tau=t-T\) turns this into \(e^{-sT}\displaystyle\int_0^\infty f(\tau)e^{-s\tau}\,d\tau = e^{-sT}F(s)\).

1. \(e^{sT}F(s)\): Wrong sign on the exponent; this would correspond to shifting the function earlier in time, not later.
2. \(e^{-sT}F(s)\): Matches the substitution exactly.
3. \(F(s)/(1+e^{sT})\): Not produced by a simple time-shift substitution.
4. None of the above: Ruled out.

Answer 2

The result can also be confirmed using the standard transform pair for a delayed unit step multiplying a delayed function, which is a cornerstone of Laplace transform tables. The rule states that delaying a time function by \(T\) is equivalent, in the s-domain, to multiplying the original transform by \(e^{-sT}\), because the exponential kernel \(e^{-st}\) itself picks up exactly this multiplicative factor whenever the time origin of the function is pushed forward by \(T\).

1. \(e^{sT}F(s)\): A positive exponent would correspond to the transform of a function shifted backward in time (advanced), which grows without bound for causal signals and is not the standard delay property.
2. \(e^{-sT}F(s)\): This is precisely the standard table entry for the time-delay property, since every point of \(f(t)\) is pushed later by \(T\), and later times are weighted by smaller values of \(e^{-st}\), captured exactly by the extra factor \(e^{-sT}\).
3. \(F(s)/(1+e^{sT})\): This form belongs to transforms of periodic waveforms built from repeated copies of a pulse, not to a single time-shifted function.
4. None of the above: Not applicable since the standard delay property matches option (ii) exactly.

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