Question

The Laplace transform of the continuous-time signal $x(t)=e^{-3t}\,u(t-5)$ is ____, where $u(t)$ is the unit step.

• A. $\displaystyle \frac{e^{-5s}}{s+3},\ \text{Real}\{s\}>-3$
• B. $\displaystyle \frac{e^{-5(s-3)}}{s-3},\ \text{Real}\{s\}>3$
• C. $\displaystyle \frac{e^{-5(s+3)}}{s+3},\ \text{Real}\{s\}>-3$
• D. $\displaystyle \frac{e^{-5(s-3)}}{s+3},\ \text{Real}\{s\}>-3$

Hint:

If the signal is {gated} by $u(t-a)$ (not shifted), write the Laplace integral from $t=a$ and change variables: it pulls out a factor $e^{-a(s+\alpha)}$ when $f(t)=e^{-\alpha t}$.

Answer 1

The Correct Option is C

Step 1: Use step–gating property
$\displaystyle \mathcal{L}\{f(t)\,u(t-a)\}=e^{-as}\int_{0}^{\infty} f(t+a)e^{-st}\,dt$.
Step 2: Substitute $f(t)=e^{-3t}$, $a=5$
$f(t+5)=e^{-3(t+5)}=e^{-15}e^{-3t}$. Hence \[ X(s)=e^{-5s}\,e^{-15}\int_{0}^{\infty} e^{-(s+3)t}\,dt = e^{-5(s+3)} . \frac{1}{s+3}, \text{Re}\{s\}>-3. \] Final Answer: $\displaystyle \frac{e^{-5(s+3)}}{s+3}$ with ROC $\text{Re}\{s\}>-3$ (Option C).