Question

The Laplace transform of the function \(f(t) = 3t\) is

• A. \(\frac{3}{s^3}\)
• B. \(\frac{3}{s^4}\)
• C. \(\frac{3}{s^2}\)
• D. \(\frac{3}{s}\)

Hint:

Standard transforms to memorize: - \(\mathcal{L}\{\text{Step: } 1\} = \frac{1}{s}\) - \(\mathcal{L}\{\text{Ramp: } t\} = \frac{1}{s^2}\) - \(\mathcal{L}\{\text{Parabola: } t^2\} = \frac{2}{s^3}\)

Answer 1

The Correct Option is C

Concept: The Laplace transform converts time-domain differential models into easily solvable \(s\)-domain algebraic models. The standard integral definition of the Laplace transform for a continuous function \(f(t)\) is: \[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \]

Step 1: Utilizing the standard identity formula.
We use the general identity for polynomial time functions: \[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \end{cases} \] For a linear ramp function, \(n = 1\). Substituting this integer into our standard formula yields: \[ \mathcal{L}\{t^1\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2} \]

Step 2: Applying the linearity property.
The linearity property of Laplace transforms states that constants can be factored outside the operator: \[ \mathcal{L}\{c \cdot f(t)\} = c \cdot \mathcal{L}\{f(t)\} \] Given \(f(t) = 3t\), we factor out the constant scalar 3: \[ \mathcal{L}\{3t\} = 3 \cdot \mathcal{L}\{t\} = 3 \cdot \left(\frac{1}{s^2}\right) = \frac{3}{s^2} \] This calculation confirms option (C) as the correct choice.