The Laplace transform of unit step function is equal to 1/s. The unit step function, denoted by, u(t), is defined as follows:

$u(t) = \begin{cases} 0, & \text{if } t<0 \\ 1, & \text{if } t \geq 0. \end{cases}$

The Laplace transform of u(t), unit step function, is given by

L{u(t)} = $\dfrac{1}{s}$.

## Laplace transform of u(t)

The Laplace transform of f(t), by definition, is equal to

L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt.

∴ L{u(t)} = $\int_0^\infty$ u(t) e^{-st} dt

= $\int_0^\infty$ e^{-st} dt, by the above above definition of u(t).

= $\Big[ \dfrac{e^{-st}}{-s}\Big]_0^\infty$

= lim_{t→∞}$\Big[ \dfrac{e^{-st}}{-s}\Big] – \dfrac{e^{0}}{-s}$

= $0-\dfrac{1}{-s}$ as we know lim_{t→∞} e^{-st}=0.

= $\dfrac{1}{s}$.

So the Laplace of the unit step function, u(t), is equal to 1/s.

L{u(t)} = 1/s. |

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## FAQs

**Q1: What is the Laplace transform of unit step function u(t)?**

Answer: The Laplace transform of the unit step function is equal to 1/s.

**Q2: What is L{u(t)}?**

Answer: L{u(t)} = 1/s, provided that s>0.