The unit step function u(t) is defined as 0 for t < 0 and 1 for t ≥ 0. What is its Laplace transform?

The unit step function u(t) is defined as 0 for t < 0 and 1 for t ≥ 0. What is its Laplace transform?

• A. 1/s
• B. s
• C. 1/(s+1)
• D. 1/s^2

Answer: (A)

The key idea is how the Laplace transform handles a function that is zero for t < 0 and one for t ≥ 0. For such a unit step, you only integrate from 0 to infinity: the transform is ∫₀^∞ e^{-st} dt. Evaluating this integral gives 1/s, provided Re(s) > 0 so the integral converges. This is why the unit step’s Laplace transform is 1/s; a constant in time becomes 1/s in the s-domain. To see why other options don’t fit: 1/s^2 would come from multiplying the time-domain function by t (since L{t f(t)} = -d/ds F(s)); 1/(s+1) would come from a function like e^{-t} u(t); and s by itself doesn’t correspond to the transform of a unit-step function.

The key idea is how the Laplace transform handles a function that is zero for t < 0 and one for t ≥ 0. For such a unit step, you only integrate from 0 to infinity: the transform is ∫₀^∞ e^{-st} dt. Evaluating this integral gives 1/s, provided Re(s) > 0 so the integral converges. This is why the unit step’s Laplace transform is 1/s; a constant in time becomes 1/s in the s-domain.

To see why other options don’t fit: 1/s^2 would come from multiplying the time-domain function by t (since L{t f(t)} = -d/ds F(s)); 1/(s+1) would come from a function like e^{-t} u(t); and s by itself doesn’t correspond to the transform of a unit-step function.