Understanding the Inverse Laplace Transform Shifting Property

The question asks us to find the inverse Laplace transform of a function of the form $f(s-a)$, given that the inverse Laplace transform of $f(s)$ is $f(t)$. This involves a fundamental property of the Laplace transform known as the First Shifting Property or s-shifting property. We are given that: $$ \mathcal{L}^{-1}[f(s)] = f(t) $$ Now we need to find $ \mathcal{L}^{-1}[f(s - a)] $.

Applying the First Shifting Property of Inverse Laplace Transform

The First Shifting Property for inverse Laplace transforms states that if $ \mathcal{L}^{-1}[F(s)] = f(t) $, then for any constant 'a', the inverse Laplace transform of $F(s-a)$ is given by $ e^{at}f(t) $. Mathematically, the property is: $$ \text{If } \mathcal{L}^{-1}[F(s)] = f(t), \text{ then } \mathcal{L}^{-1}[F(s - a)] = e^{at}f(t) $$ In our specific problem, $F(s)$ is given as $f(s)$. So, if $ \mathcal{L}^{-1}[f(s)] = f(t) $, applying the First Shifting Property directly gives us: $$ \mathcal{L}^{-1}[f(s - a)] = e^{at}f(t) $$ Since we know from the problem statement that $ f(t) = \mathcal{L}^{-1}[f(s)] $, we can substitute this back into the result: $$ \mathcal{L}^{-1}[f(s - a)] = e^{at} \mathcal{L}^{-1}[f(s)] $$ This matches one of the given options. Let's look at the options provided:

• $e^{at} \mathcal{L}^{-1}[f(s)]$
• $e^{-at} \mathcal{L}^{-1}[f(s)]$
• $ \mathcal{L}^{-1}[f(s)]$
• $ \mathcal{L}^{-1}[f\’(s)]$

Comparing our derived result $ \mathcal{L}^{-1}[f(s - a)] = e^{at} \mathcal{L}^{-1}[f(s)] $ with the options, we see that the first option matches perfectly. Therefore, using the First Shifting Property of the inverse Laplace transform, the inverse Laplace transform of $f(s-a)$ is $e^{at}$ times the inverse Laplace transform of $f(s)$.