What is the Laplace transform of u(t-1)?

The Laplace transform of u(t-1) is equal to e^-s/s, that is, L{u(t-1)} = e^-s/s. Note that u(t-1) is the shifted unit step function by 1 and it is defined as follows.

u(t-1) = 0 if t<1

u(t-1) = 1 if t≥1.

Let us now learn how to find the Laplace transform of u(t-1).

Laplace of u(t-1)

We will find the Laplace of u(t-1) by definition. The definition of Laplace transforms says that the Laplace of a function f(t) is given by the following integral formula

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

So the Laplace of u(t-1) is

L{u(t-1)} = $\int_0^\infty$ e^-st u(t-1) dt.

By the definition of u(t-1) given above, this integral will be equal to

L{u(t-1)}= $\int_1^\infty$ e^-st dt

= $\Big[ \dfrac{e^{-st}}{-s}\Big]_1^\infty$

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

= 0 + e^-s/s as we know that lim_t→∞ e^-st = 0.

= e^-s/s.

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

More Laplace Transforms:

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

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