Question

The inverse Laplace transform of $F(s) = \frac{s^2 - 3s + 4}{s^3}$ is _______}

• A. $1 - 3t + 2t^2$
• B. $1 - 3t + 4t^2$
• C. $1 + 3t + 2t^2$
• D. $1 + 3t + 4t^2$

Hint:

Splitting rational functions into partial fractions or simpler components is almost always the easiest way to evaluate inverse Laplace transforms.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks for the inverse Laplace transform of a rational algebraic expression in the $s$-domain to find its corresponding time-domain function $f(t)$.

Step 2: Key Formula or Approach:
We use the standard inverse Laplace transform formulas:
\[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\} = \frac{t^{n-1}}{(n-1)!} \quad \text{for } n \ge 1 \] We also utilize the linearity property of the inverse Laplace transform.

Step 3: Detailed Explanation:

• Split the given fraction into simpler individual fractions:
\[ F(s) = \frac{s^2 - 3s + 4}{s^3} = \frac{s^2}{s^3} - \frac{3s}{s^3} + \frac{4}{s^3} \] \[ F(s) = \frac{1}{s} - \frac{3}{s^2} + \frac{4}{s^3} \]
• Apply the linearity property to find $f(t) = \mathcal{L}^{-1}\{F(s)\}$:
\[ f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - 3 \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} + 4 \mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} \]
• Calculate each inverse Laplace transform:

• $\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1$

• $\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t$

• $\mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{t^{3-1}}{(3-1)!} = \frac{t^2}{2} = 0.5 t^2$

• Combine the terms:
\[ f(t) = 1 - 3t + 4\left(\frac{t^2}{2}\right) \] \[ f(t) = 1 - 3t + 2t^2 \]

Step 4: Final Answer:
The inverse Laplace transform is $1 - 3t + 2t^2$, which corresponds to option (A).