Laplace transform of unit step function | L{u(t)}

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.

Related Topics:

Laplace transform of 0

Laplace transform of t

Laplace transform of t²

Laplace transform of 2

Laplace transform of e^t

Laplace transform of sint

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