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

The Laplace transform of unit step function is equal to 1/s. As the unit step function is denoted by u(t), its Laplace is given by L{u(t)} = 1/s. Before we learn how to find it, let us recall the definition of the unit step function.

u(t) = 0 if t<0

u(t) = 1 if t≥0.

This means that a unit step function u(t) is a function of t (represents time), where u is 0 when time is negative and 1 when time is positive. The formula of the Laplace of the unit step function u(t) is as follows.

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

Laplace of u(t)

By the definition of Laplace transforms, the Laplace of a function f(t) is given by

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

So the Laplace of the unit step function u(t) is

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

Now by the above definition of u(t), we deduce that

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

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

= lim_t→∞ $\dfrac{e^{-st}}{-s}$ + e⁰/s

= 0 + 1/s as the limit lim_t→∞ e^-st = 0.

= 1/s.

So the Laplace transform of the unit step function u(t) is equal to 1/s, that is, L{u(t)} = 1/s, and this is obtained by using the definition of Laplace transforms.

More Laplace Transforms:

Laplace of u(t-a)

More generally, the Laplace transform of the shifted unit step function, denoted by u(t-a), is equal to e^-as/s, that is,

L{ u(t-a)} = e^-as/s.

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