Discussion
Let \(L\{f(t)\} = F(s)\) and \(u(t - a)\) be a unit step function. Then \(L\{f(t - a)\,u(t - a)\}\) is :
Answer: A
The transform is \(e^{-as}F(s)\).
This is the second shifting theorem. With the substitution \(\tau = t - a\),
\[ \begin{aligned} L\{f(t - a)u(t - a)\} &= \int_{a}^{\infty} e^{-st}f(t - a)\,dt \\ &= e^{-as}\int_{0}^{\infty} e^{-s\tau}f(\tau)\,d\tau \end{aligned} \]
which is \(e^{-as}F(s)\). The extra \(1/s\) in option B belongs to the transform of the step function alone, \(L\{u(t - a)\} = e^{-as}/s\), and a delay gives \(e^{-as}\), not \(e^{as}\).
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