Step 1: Interpreting the unit step function.
The unit step function \( u(t) \) represents a signal that suddenly turns ON at \( t = 0 \) and stays ON forever. So for all practical purposes after \( t = 0 \), the signal behaves like a constant signal of value 1.
Step 2: Laplace transform of a constant signal.
From basic Laplace transform properties, the transform of a constant signal is known:
\[ \mathcal{L}\{1\} = \frac{1}{s} \] Since \( u(t) = 1 \) for \( t \ge 0 \), it behaves exactly like a constant input in the Laplace domain.
Step 3: Verifying using area interpretation.
The Laplace transform can also be viewed as the weighted area under the curve:
\[ \mathcal{L}\{u(t)\} = \int_{0}^{\infty} e^{-st}\,dt \] The exponential term \( e^{-st} \) ensures convergence by gradually reducing the contribution of the signal as time increases.
Step 4: Final result.
Evaluating the integral gives:
\[ \mathcal{L}\{u(t)\} = \frac{1}{s} \] which is valid for \( s > 0 \).
Final Conclusion:
The Laplace transform of the unit step function is \( \dfrac{1}{s} \).